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C O L L E G E A L G E B R A

• For First Year and

• Pre-Degree Student

B Y

T. G. KULKARNI , M. A., Professor of Mathematics Jai Hind College and Basantsing Institute of Science, Bombay

and

M. K. KELKAR, M.SC., B.T. , Professor of Mathematics and Head of the Department of Mathematics and Statistics, Ismail Yusuf College, Bombay

1973

Published by : Smt. SlNDHU KULKARNI, B. a , T D., Umasadan, 4th Lane, 102, Hindu Colony. Dadar, Bombay 14

All rights including those of translation and reproduction as well as those of preparing and publishing a key giving the solutions of the examples in this book are reserved by the Authors.

First Edition Second Edition Third Edition Fourth Edition Fifth Edition Revised Sixth Edition Revised Seventh Edition Revised Eighth Edition Revised Nineth Edition

; 1958 1959 1961 1962 1964 .1965 1968 1970 1973

Printed by :

R. S. Gupte, Aryabhushan Press, 915/1, Shivajinagar, Poona 4.

REVISED SYLLABUS IN ALGEBRA

for

F. Y. Arts and Science, Bombay University

1. Elements of set theory : sets, subsets, empty sets, union and intersection of sets, complementation, Venn diagrams.

2. Number systems : Natural numbers, integers, rational numbers, real numbers, complex numbers.

The real number system : correspondence of real numbers with points on a straight line ; concept of order and existence of real number between two given numbers to be introduced informally. Approximation of irrational numbers by rational numbers.

Surds: Rational operations with binomial quadratic surds ; conjugate surds and rationalising factors; Theorem : If a + yjb = c + \[d, then a '= c and b = d under prescribed conditions. Properties of real numbers with reference to closure for elementary operations, commutativity, associativity and distributivity. The complex number system : Correspondence between complex numbers and points in a co-ordinate plane referred to rectangular axes; conjugate complex numbers ; rational operations with com-plex numbers ; reduction of given complex expressions to the form a + ib (a, b real ) .

3. Theory of Quadratic Equations with real coefficients: Solution of quadratic equations ; nature of roots ; relations between roots and coefficients; generalisation to cubic and quartic equations (s ta tement on ly) . Simple symmetric functions of roots of a quadratic. Equations with roots related in a simple way to roots of a given quadratic equation.

iii

Exponents and Logarithms:

Definition of a" for a > 0 and m rational, Theorems: m n m + n , t \ m rn i m /• "l <

a Xa =a ,{ab) —a b , (a ) =a , m, n being rational numbers (proofs for positive integral exponents only) . Informal discussion of a" when x is irrational. Definition of log,, x, a > 0, a 1 and x > 0. Theorems on logarithms of product, quotient, power and change of base.

Permutations and Combinations : Linear permutations with distinct objects. Combinations (case of repetitions excluded ).

Theorems: M+V _V 4- V • "r — "r

Relation between nPr and "C,.

Simple Illustrations for the use of Mathematical Induction : Formulae for

2 a + ( r - l ) d , 2 ^ / ;

r=t r—0

n n ft 2 r, 2 r2, 2 r3. r= 1 r=l r=l

The Binomial Theorem for a positive integral exponent ( Exclude determination of greatest coefficient, greatest term, properties of coefficients).

iv

Preface to the Revised Eighth Edition, 1 9 7 3

This book, first published in 1958, is today running nto its tkineth edition. All along we have been very keen to incrase the utility of this book to the students. In view of the Itest decisions taken by all the Indian Universities to introduce :)me essential mathematical concepts to the First Year student: we had introduced an elementary treatment of the theory of ses in the very first chapter. The chapter on the number system had been thoroughly revised and made more logical to suit the £eds of the present tendencies in developing this topic. The ch pter on Method of Induction now has been taken earlier so as to :take the use of this method wherever possible. Determinants, th>ugh not included in the syllabus of the University of Bombay,'nave been considered in the last chapter to increase the utility cf the book to a genera] reader of the First Year standard. Tt is sujgest -ed that the first two chapters on Set Theory and the Nunber System need not be studied in detail at the first reading. Thf new ideas introduced in these chapters can be better understood f the book is read " backwards and fo rwards" as suggested by G. Chrystal a famous author of a book on Algebra.

Some Test Papers have been added at the end ii the Appendix to give the students some practice for the examiiation under the new course. We are thankful to the Authorities of Bombay University for allowing us to include their papers it our book. The copy right of these papers vests with the University of Bombay.

Our thanks are due to Shri. S. A. Bhide and the staff of Aryabhushan Press for helping us in all respects in the printing of this book.

Any suggestions for the further improvement of the book will be gratefully received.

T . G . K U L K A R N I ,

M . K . KELKAR.

v

PREFACE TO THE FIRST EDITION

The Bombay University, from this academic year, has revised he syllabus in mathematics for the First Year students. The

Poona University and some other Universities have started the Pre-degree classes. The present book has been written so as to meet the needs of the new syllabus in Algebra prescribed for the First Year and Pre-degree Courses.

The Authors have kept in view the standard of the present students passing the S. S. C. Examination with Elementary Mathematics in their regional languages. Special effort has been made to present the subject matter in simple language without sacrificing the mathematical rigour. Necessary important results from School Algebra are given in the Appendix for immediate reference. Ample illustrative problems have been worked out to illustrate every new principle and formula. Few examples have been given immediately at the end of every important article to enable the students to use and remember the result of the article. Every effort has been made to prepare carefully the gradation and selection of examples for every chapter. Attempt has been made t o complete the subject matter in every chapter instead of restricting ourselves completely to the exact syllabus of any particular University. Teachers may omit some articles to suit their needs. The arrangements of the various chapters is flexible and permits of any sequence desired by the teacher. We hope that this book will enable the students to understand the subject matter and also will create interest in the subject.

Our sincere thanks are due to the Manager Shri V. A. Patwardhan, and the staff of the Aryabhushan Press for the care they have taken throughout the printing of the book.

T . G . KULKARNI ,

M . K . KELKAR.

vi

CONTENTS

Chapter Page

1 Sets .... t

2 Real Numbers ..... 43

3 Complex Numbers 79

4 Indices ( Exponents ) ... 92

5 Logarithms ••• 115

6 Surds — 141

7 Quadratic Equations -- 177

8 Method of Induction . . . 213

9 Progressions -•• 222

10 Summation of Series '- ... 265

11 Permutations and Combinations ... 280

12 Binomial Theorem --• 307

13 Determinants ••• 325

Appendix

1 Logarithmic and A.nti-logarithmic Tables ... 355

2 Important Formulae and Results ... 359

3 Test Papers ••• 369

Bombay University Papers

vii

Symbols and Abbreviations used in this book

A — \ a, b, c, ••• A is a set of elements a, b, c, ... A = \ x j x satisfies a given property P

A is a set of elements x such that x satisfies a given property P.

s. t. such that iff if and only if 0 the empty set A' complement a set A e belongs to £ does not belong to g there exists V for all c subset C proper subset

=> implies o implies and is implied by U union f) intersection N set of positive integers JV set of negative integers J set of integers Q set of rational numbers Qi set of irrational numbers R set of real numbers C set of co7nplex numbers

A discriminant of a quadratic equation

2 summation n ! or ]_» 1 • 2 • • 3 • • • n "p number of permutations of n

distinct objects taken r at a time

*Cr number of combinations of n distinct objects taken r at a time.

Chapter 1

Sets

1. The concept of a set. 2. Notation. 3. Representation of a set 4. Finite. Infinite and Empty sets. 5 Subsets, Equal and Equivalent sets. 6. Universal set, complement of a set. Exercise 1 (a). 7. Union of two sets. 8. Intersection of two sets. 9 Extension to three sets. 10 Operations onse t s . 11. Illustrative Examples. Exercise 1 ( b ) .

1. The Concept of a Set. The idea of a set forms the basis of all the higher branches

of Mathematics. We, therefore, begin with a brief discussion of sets.

A set is a well defined collection of objects. Consider the following collections :

( i ) The colleges affiliated to the Bombay University.

( ii ) The books on Algebra in your college library.

( i i i ) The points on a given straight line. In ( i ), we have a collection of colleges affiliated to

the Bombay University. The colleges not affiliated to the Bombay University are not included in this collection. The objects in this collection are " colleges" and they are all given by the rule that they are affiliated to the Bombay University.

In ( ii ) , we have a collection of books on Algebra in your college library. The objects in this collection are " b o o k s " . These objects are given by the rule that " they are written on Algebra and that they belong to your library ". The books which are not in your library or those of the books in your library which are not written on Algebra are not included in this collection.

Similarly in ( iii ) , we have a collection of points lying on a given line. The objects in this collection are the points

2 : COLLEGE ALGEBRA

The objects are given by the property that they all lie on a given straight line. The points not lying on the given straight line are not included in this collection.

Thus, when we talk about a set we have in mind

( i ) the collection or the set,

( ii ) the objects in the set,

( i i i ) the rule or the property which enables us to decide whether a particular object belongs to the given set or not.

Thus, a collection of objects, defined by a given rule, forms a set.

Note : Just as in the study of plane geometry, points and lines are undefined terms, similarly we consider the term set as undefined ; but we agree intuitively that a set is a collection of objects described or identified in such a way that there is no doubt as to whether a particular object does or does not belong to the set.

2. Notation. Sets are usually denoted by capital letters such as A, B, C,

P,Q,R, X,Y,Z.

The objects in a set are called " members " or " elements " of the set and they are usually represented by small letters such as a, b, c , .„ , x, y, z.

While referring to an object * of a given set A we say that " x is in A " or " x belongs to A " or that " x is a member of A " and write this symbolically as " x e A". The symbol e ( t h e Greek letter epsilon ) stands in the context of sets for " lies in ", " belongs to ", or " is a member o f " . Similarly, when we want to say that " the object x does not belong to the set A " or " the object x is not in A ", or that " x is not a member of A", we write it symbolically as " x (£ A ".

Thus, the symbol ; stands for "be longs t o " or " i s i n " and the symbol £ stands for " does not belong t o " or " is not in ".

SETS : 3

Represention of a Set.

We will consider the following three forms in which sets may be represented.

I. Tabular Form. In this form we enumerate or list all the elements. The names of the individual elements arc separated by commas and are written within a set of braces ; for example the set A of even numbers between 1 and 11 may be written as

A = ) 2 , 4 , 6 , 8, 10 This may be read as " A is a set of elements 2, 4, 6, 8

and 10".

Tabular form of describing a set is known as the Roster Method.

II. The Rule Method. In this method we indicate a set by enclosing in brace brackets a general element and describe it in words by a property or the rule satisfied by it. Thus the set A of even numbers between 1 and 11 may be indicated as :

A = ) x | x is an even integer and 1 < x < 11 ( and may be read as:

"A is a set of all elements x such that x is an even integer and x is greater than one and less than eleven. "

The symbol " \ " which looks like a small vertical line, in the context of set notation, will stand for " such that '*.

The Tabular Method of indicating a set is particularly useful •when the number of elements in the set is small; for example the Set B of all the vowels of the English alphabets may be indicated as

B = } a, e, i, o, u (.

Sometimes all objects in a set may not possess any common property and in this case it becomes convenient to describe the set by listing all its elements. Thus we may have a set C of a table, a bottle, a taxi, a pen and the letter k. This may be expressed in the Tabular form as

C = \ a table, a bottle, a taxi, a pen, the letter k (. However, if the elements in a set are large or if they can

not all be listed then we have to describe a set only by the

4 I COLLEGB ALGEBRA.

Rule Method. For example a set D of positive even integers may be indicated as

D = } x — In ( n is a positive integer $.

We may, however, write this set as D = J 2, 4, 6, 8, 10, f, the dots in the braces

indicate that all the elements can not be listed and there are infinite number of elements in this set.

The rule method of describing a set is also known as set-builder notation.

Remark 1. If we make a list to show the elements of a set, the order of listing is not important. For example, the sets \ a, b,c,d\ and ) c,b, a, d\ are considered as identical. Similarly the sets } 1, 2, 3, 4, 5 | and j 5, 3, 4, 2, 1 j are the same.

Remark 2. When we indicate the elements in a set, it is conventional not to count the same member more than once. For example, in a set { 0, 1, 2, 1, 0, ] there are only three elements namely 0, 1 and 2.

III. Diagrammatic Representation of a Set. Many of the ideas about sets which we will be considering

in this book can be made more clear by means of diagrams called Venn diagrams (sometimes also called as Venn-Euler diagrams. )

v- '1

Set A, element x, e A. Fig. 1

Tn these diagrams a set, say, A is represented by some closed curve such as a square, an ellipse, a circle or a rectan-gle. The elements of the set A are represented by points in the closed figure; this enables us to show clearly the idea o f

SETS : 5

<; belonging to a set" . The Figure 1 shows how the element * indicated by a point belongs to the set A.

4. Finite, Infinite and Empty Sets. If we start counting the elements of a set, one of two

possibilities may arise ; ( i ) a definite number is reached beyond which no more number is needed because each element has already been counted;

( i i ) a definite number can not be reached to exhaust the counting of all the elements of a set.

Definition. If the elements of a set can be counted by a finite number, then that set is called a finite set; otherwise it is called an infinite set.

The set of positive integers less than 100 may be indicated as ) 1, 2, 3, ... 99 ( and is a finite set ; whereas the set of positive integers may be indicated as \ 1, 2, 3, 4, ... $ and is an infinite set.

It should be noted that a set containing a very large number of members is not an infinite se t : for example the set of ail Indians in the world is a finite set. The difference between a finite set and an infinite set is that the process of counting the number of elements in the finite set has an end; while in the infinite set the process of counting the elements is an endless one-For example, the set

E — } x | x is an even positive integer | (i . e. E = \ 2, 4, 6, 8 , . . . ( ) is an infinite set; whereas the

set F = \ x | x is an even positive integer < 14 {

( i . e. F = \ 2, 4, 6, 8 ,10, 12 O is a finite set.

Definition. A set with one element is called a singleton or a unit set.

For example, a set of real numbers possessing neither a positive nor a negative sign is a singleton set and has zero as its element.

Note : It is important to note that ) a [ & a, since \ a | denotes a set having an element a, whereas ' a' denotes only the element.

6 : COLLEGE ALGEBRA.

Empty Set. Definition. A set which has no elements in it is called an empty or the null or the void set. The symbol which is used to denote the empty set is 0. The symbol (p is S c a n d i n a v i a n letter and not the Greek letter phi. Whenever this symbol appears, in the context of sets, read it as ' the empty set " or " the null set" . Here the rule or the property describing a given set is such that no element can be included in the set. It is obvious that the set A of all married bachelors in the world is an empty set. Similarly the sets B, C, D given below are empty sets.

B = ) x i s a college in Delhi ; x is affiliated to Bombay University j ;

C = \ « | « is a negative number satisfying the equation : x2 - 3x + 2 = 0 { ;

D = $ x | x is a perfect square of an integer; 65 < x < 80

We know that there is no college in Delhi affiliated to the Bombay University. The roots of the equation x2 - 3x + 2 = 0 are 1 and 2 and hence this equation has no negative root. Similarly there is no number lying between 65 and 80 which is a perfect square of an integer. Thus, there are no elements which can be counted in the sets B, C, and D and hence they are null sets. In general, we may say that laying down an impossible condition for the formation of a set immediately produces an empty set.

Example. State with reasons whether the sets given below are empty, finite or infinite:—

(1 ) A = \ x I x = a3; a is an integer 5 • ( 2 ) A = j 1, 2, 3, 4, 500 \ . (3 ) A is the set of integers x, where x is a perfect square of an integer

and 26 < * < 35. (4 ) A is the set of all houses in the city of Bombay (5 ) A = $ * 1 * is a positive integer which is a multiple of 3 ,, An?. ( 1 ) Infinite. (2 J Finite. ( 3 ) Empty. ( 4 ) Finite

( 5 ) Infinite.

5. Subsets, Equal and Equivalent Sets. Subset. Definition. Set A is a subset of set B if and only

if every element of A is an element of B.

SETS : If

We write this relationships as ACB an d read it as "A is a subset of B " or " A is contained in B " or " A is included in B ". Some times the same statement is also written as B Oi A. and is read as " B is a superset of A " or " B contains A " or " B includes A". Observe that ACB means that every element of A is an element of B. This puts no restrictions on B other than that the set B includes the set A.

A^B As B Fig. 2 ( i ) Fig. 2 ( i i )

We can represent by Venn diagrams the relation i c B . We draw closed figures for sets A and B; but since every element of A is also in B, the closed figure for A should be completely in or coincident with the closed figure for B. This is shown in the figures 2 ( i ) and ( i i ) . Figure 2 ( i ) shows that every set is a subset of itself.

For example

( i ) If A = p , 2, 3 ) and B = ) 1, 2, 3, 4 then every element of A viz. 1, 2, 3 is also the element in B and hence ACB.

( ii ) If A = } I, 2 ( and B = J 1, 2 then A CB. ( i i i) If A is a set of schools in Bombay and B is a set of

schools in India then it is clear that every school in Bombay is also a school in India, Bombay being itself in India. Hence ACB.

Thus i C B i f f x e A^> xe B. [ We shall use the abbreviation " i f f " to denote if and

only if and the symbol for " implies ". We shall also use symbol o for "implies and is implied b y " . Thus " P = > Q " means P implies Q and " P o Q" means P implies Q and is implied by Q. ]

Proper Subset. Definition. Set A is called a proper subset of B i f f A is a subset of B and at least one element of B is not an element of A.


We write this relationship as A c B and read it as "A is a proper subset of B

Thus A C B iff A C B and 3 b e B s. t. b <£ A. [We shall use the abbreviation " s. t. " to denote ' such t h a t '

and the symbol ' g ' f o r ' there exists. ' ]

( i ) A c t B ( i i i ) AC B (•ii) A qt B Fig. 3

We can represent the proper subsets by Venn diagrams. If B is not a subset of A, then three possibilities are to be considered :

( i ) A is a proper subset of B i. e. every element of A is an element in B and Aj=B. [ See Figure 3 ( i i i ) ].

( ii ) There are some elements that are common to A and B and there are also some elements not common to A and B. [ See Figure 3 ( ii ) ].

( iii ) There are no common elements i. e. the sets A and B are disjoint. [ See Figure 3 ( i ) ].

Illustration. If A = ) 1, 2, 3 ( and B = | 1, 2, 3, 4 \ then A c B since

every element of A is also an element of B and there exists an element 4 in B which does not belong to A.

Equal sets. Definition. The sets A and B are said to be equal i f f both of them consist of exactly the same elements.

We write this relationship as " A = B " and read it as " A is equal to B ''. Equal sets can be represented by drawing two coincident closed figures as in Fig. 2 ( i ),

Illustration. Let A = ) x | x is a letter in the word " sister " \;

B = j x 1 x is a letter in the word " resist *' f ; C = ) e, i, r, s, t<\.

SETS : 9

We find that A = B = C as each set has the same elements namely e, i, r, s and t. The example shows that even if the rules -forming the given sets are different, yet they produce identical elements, of course, irrespective of the order and hence the sets are equal.

It can be shown that the sets A and B are equal iff A is a subset of B and B is a subset of A. i. e. A = B <=>• A C B and B c A.

Equivalent Sets. Definition. If the elements of one set can be put into one-to-one correspondence with the elements of another set, then the two sets are called equivalent sets,

For example, if A = ) Monday, Tuesday, Wednesday j and B = ) a, b, c \ , then the sets A and B are equivalent.

One can easily observe that if two sets are equal, then they must be equivalent. However two equivalent sets need not necessarily be equal.

For example, if A = ) a, b, c, d B = \ 1, 2, 3, 4 j, C = \ d, a, c, b, c, a, d then A and C are equal and A, B, C are equivalent.

We now show that : "" The null set (or the empty set ) is a sabset of every set". In order to show this, let A represent any set. Then there are only two possibilities namely 0 is a subset o t A or <j> is not a subset of A. The condition for a subset viz. " xe A, if xe<j>" is satisfied since there is no element in 0. Thus one has to remember that a null set is a subset of every set.

Similarly one can easily see that every set is a subset of itself.

The number of subsets of a finite non-empty se t : We now show tha t :

" From a set containing n elements 2" subsets can be formed. " ( i ) Consider a singleton set A = j a j. It has two i. e.

( 2 1 ) subsets namely A and ( i i ) Consider a set of two elements say B = $ 1, 2 It

has four ( i. e. 22 ) subsets, namely, } 1 1, } 2 ( . n » 2 ( a n d * .


( i i i ) Consider a set of three elements say C ( a , b , c ) . 3

It has eight ( i. e. 2 ) subsets, namely, ) a ) b ) c I ) a, b\, ) a,c ) b, c ) a, b, c( and <f>. By Induction it can be shown that a set containing n elements

has 2" subsets. We now summarize some of the above results about the

subsets. ( i ) Every set is a subset of itself i. e. A C A. ( ii ) Every set has the empty set as its subset i. e. <t> C A.

( i i i ) From a set containing n elements 2" subsets can be

formed and the set consisting of these 2M subjects is called the Power set of the set A and is denoted by P ( A ), in the context of set A.

The definition of equal sets gives us the following results : ( i ) A m A. ( i i ) If A = B and B = C, then A = C. ( i i i ) A = fiiff^Cfiand BCA. The first two are obvious. A proof of ( i i i ) is given below

as ( i i i ) is very often used to establish the equality of two sets. First we shall assume that A = B and prove that ACB and

BCA. Then we shall assume that ACB and BCA and prove that A = B so that the proof will be complete.

( i ) Let A = B. This means by definition that A and B have the same elements and if x e A then x e B and vice versa. This proves that A C B and BCA.

( i i ) Let ACB and BCA. Here ACB means if xe A, then x e B. Also BCA means if x s B, then x e A. But both these results must hold simultaneously and hence

the sets A and B must have the same elements. Therefore A = B.

Disjoint sets. Definition. Two sets A and B are called disjoint sets if they have no common elements. [ See Fig. 3 (i) J.

For example, the sets A and B given below are disjoint, ( i ) A = )1 ,2 . . 3, 4 ( a n d B = } 5, 6, 7, 8$ .

SETS : If

( ii ) A = ) a, e, i, o, u } and B = ) 1, 2, 3, 4, 5 ( i i i ) A = J a | a is an even integer ( and

B = ) b | is a prime number > 2 (

6. Universal Set. Complement of a Set. Whenever we speak of sets or subsets we usually think of

them to be subsets of some given ( o r fixed) set called the Parent set or the Universal set. In a particular context when the Universal set is obvious, it need not be explicitly stated The universal set is also sometimes called the universe of discourse and is generally denoted by the letters X, U, SL or simply by S.

Illustrations. ( i ) A set of real numbers may be considered as the

universal set for any set of integers.

( ii ) A set of Indians may be considered as a universal set for any set of literate Indians.

( i i i ) For a set of colleges in Bombay, a universal set may be taken as a set of colleges in the Maharashtra State or a set of colleges in India.

( iv ) For a set of books on Mathematics in your library, a universal set may be taken as a set of all books in your library.

Complement of a Set. Definition. The complement of any set A which is a subset of some universal set X 4s the set of elements of X which do not belong to A.

The complement of a set A is denoted by A'. It follows from the definition that

A' = )x\xeX;xiZA\

Set A, Universal set X, A' complement of A Fig. 4.


In a Venn diagram, a universal set is represented generally by a Tectangle, The Figure 4 shows the set A, its complement A' and the universal set X.

Illustrations.

( i ) If J = | n | n is an integer X = J x | * is a real number (, then J ' = } x | x is a real number, x(£J

( ii ) If A = ) x | .x is a literate Indian (, then A' = J x | x is an Indian, x A },

( i i i ) If A = j x is a college in Bombay (, then A' = \ x \ x is college in Maharashtra State x £ A j.

[No te that , here, we have taken the universal set as the one formed by all colleges in Maharashtra State].

( i v ) If A = ) x | x is a book on Mathematics in your library | then A' = ] x | x is a book in your library, x £ A Remarks.

( 1 ) The set of elements of A and A', taken together form the parent or universal set.

( 2 ) ACX; also A' c X. But A and A' are disjoint. ( 3 ) If A = X, then A' = For, as A = X, there is no

element of X which does not belong to A or in other words A' has got no element i. e, A' = <fi, the null set.

( 4 ) A, B are the subsets of the universal set X, then if ACB, then B' C A'.

P r o o f : If x e B' then x £ B, A c B and hence x ^ A . xeA'.

Hence * e B' => x e A'. B' C A'. ( 5 ) (A')' = A. This means that the complement of

A' is A. P r o o f : xe(A'y=>xg;A'andxZ;A'=>xeA.

(A')'CA ( i ) Again x e A => x £ A' and x £ A' => x e (A')'.

ACiA'Y ( i i ) From ( i ) and ( ii ), it follows that {A')' = A. [ The method employed in proving the result may be carefully

noted. ].

SETS : 13

Exercise 1 (a)

1. Explain, by giving an illustration, the concept of a set. If A = \ 1,3, 5, 7, 9, 11 describe the set A (i) in words and ( ii ) in set builder notation ( i. e. by the rule method ). Express in symbols the fact that 5 is a member of the set A and that 4 is not a member of the set A.

2. If N = J n | « is a positive integer (, describe the set N ( i ) in words, ( i i ) in tabular form and express in symbols whether the number zero is an element of the set N.

3. Express by ( i ) tabular method and ( i i ) rule method ( i e. in set builder notation ) the set of integers.

4. Explain what you understand by ( i ) a finite, ( i i ) an infinite set. State whether the sets given in examples 1, 2, 3 above are finite or infinite.

5. Define an empty set and express it by a symbol. Is the set ) 0 | an empty set ? If it is not an empty set what particular name can be given to this set ?

6. Express the following set by ( i ) the Roster method ( i . e. tabular method ), ( i i ) the rule method (i. e. in set builder notation ) and state whether they are finite, infinite or null. ( i ) A set of all even positive integers less than 16, ( ii ) A set of all odd positive integers, ( i i i) A set of positive numbers x satisfying the equation

x2 + 3x + 2 = 0.

7. State whether the following relations are correct. If they are not correct write them correctly. ( i ) ) x | ( x - 3 )i = 0 { = 3, ( ii ) ) a, a,a, b,b, b \ = \ a, b

8. State when the sets are said to be ( i ) equal ( i . e. iden-tical ), ( i i ) equivalent. Find which of the following sets are ( i ) equal, (ii ) equivalent. A = ) x | x2 = 0 I B = ; x | = 1 C = i 0, 1, - 1 {, D = )l, - 1 ( , E = J H , F= ) 0 | .


9. ( i ) A is a set of 5 red balls and B is a set of 5 red chalks. Is = B ?

( ii ) Which of the sets given below are equal ? <M o ) <M.

( i i i ) A = ) x \ xe A;x^A (;. What are the elements of the set A ? Is A = 0 ? Is A a subset of 0 ?

10. If N is a set of positive integers and A = \ x | x e N\ x2 - 5x + 6 = 0 B = ) x | xeN, 1 < * < 4 [, find whether A = B.

11. State when ( i ) A is a subset of B, ( i i) A is a proper subset of B. Express both the relations symbolically. Is 0 a subset of itself ? Is 0 a proper subset of itself ?

12. Let A = )l, 4 \,B = }1 ,0 , 1 , 2 , 3 , 4 j. ( i ) Is A a subset of B1 ( i i ) Is A a proper subset of B? (ii i) Is 0 a subset of B ? (iv ) Is 0 a subset of A ?

13. Is it true that for every set A : ( i ) 0 C A i. e. 0 is a subset of A? ( ii ) A C A i. e. A is a subset of A ?' ( i i i ) A c A i. e. A is a proper subset of A!

14. Fill in the gaps :

( i ) The set which has no members is called the set and it is denoted by the symbol

( ii ) A set having only one element is called ( i i i ) X is a subset of Y iff ( i v ) X is a proper subset of Y iff ( v ) When all the elements of two sets A and B are

identical we say that the two sets are ; they are also

( vi ) When there is one-to-one correspondence between the elements of two sets, but the elements are not identical, we say that the sets are

(vii) A and B are disjoint sets iff

SETS : If

15. State the disjoint sets among the following :— A = ) 1, 3, 4 }, B = j 2, 0, 3, 1 {, C = \ 4, 5, 6 E = } 5, 6, 7 F = J 2, 4, 6, 8 f.

16. If ^ C S and S c C , show that A C C. 17. ( i ) Write the subsets of a singleton j a j .

( ii ) Write the subsets of \ x, y, z \ .

18. What is meant by a Power Set of A ? Write it when A = j 1, 2

19. Show that the number of proper subsets of

A = J aua2,a3, , aH f is 2" - 1. 20. Show that every subset of a finite set is finite.

21. If A = ) 0 , 2 then state whether the following state-ments are correct or not

( i ) \2\ C / f , ( " ) \ 2\eA,(m)i>eA, ( i v ) 2 c ^ ,

( v ) O e ^ , ( v i ) 0 C J 4 .

22. Let A — \ Q\,B = )G, l\,C = 4>, D = \ <t> E = ) x | x is a human being walking more than 200 miles per hour (, F = J x | x e A and x e B j. State which of the following are correct : ( i ) ACB, ( i i ) B = F, ( i i i) C C D , ( i v ) C = E, ( v ) A = F, ( vi ) F = 1, ( vii ) E = C = D.

23. Give one illustration of

( i ) a singleton or unit set; ( ii ) an empty set; (iii ) a finite set; ( i v ) a n infinite set; ( v ) a subset of }0, 1 (, ( v i ) a proper subset of } 1, 2, 3 ( ; ( vii) a super set of j 0, 1 (; ( viii) the power set of J a, b (; ( i x ) two equal sets; ( x ) two equivalent ( but not equal ) sets; (xi ) two disjoint sets; (x i i ) the complement of a set of women members of a club ; state also its universal set.

24. What do you mean by saying that two sets are ' equi-valent ' ? Give an example of each of the following cases:—


( i ) a set A which is equivalent to a proper subset of itself ;

( ii ) a set A which is not equivalent to any proper subset of itself.

25. Explain what we understand by ( a ) a universal set, ( b ) complement of a set. Give answers for the following : ( i ) What is the complement of a universal set? ( ii ) Is universal set, a subset or a proper subset of

itself? ( i i i ) Can A = A'1 Can A C..A' or A' c A?

(iv ) Can two universal sets be equal if the number of elements in them are equal but different in character ?

( v ) If [ / ( t h e univereal set) = ) 1,2,3, write the complements of the following :— ( 1 ) ) 0 ( , ( 2 ) 0 , ( 3 ) ) 2, 4, 6, 8, j, ( 4 ) ) x | x is a + ve integer < 12 (.

Answers. Exercise 1 ( a )

1. ( i ) A is a set of all odd positive integers less than 13. ( i i ) A = J * | * is a positive odd integer * < 13 ( ;

5eA, 4&A.

2. ( i ) N is a set of all elements n such that n is a positive integer; ( this is also called a set of natural numbers).

( i i ) N = ) 1, 2, 3, 4, 5 0 N since the number zero is not a positive integer.

3 . ( i ) / = } - 3, - 2, - 1, 0, 1, 2 . 3 . | ; (11) J = * * I * is an integer

4 . Set A in example 1 is finite and sets in examples 2 and 3 are infinite. 5 . ^ 0 ^ is a set having an element zero. It is not an empty set since an

empty set contains no element at all. Since this set contains one element, it can be called " a singleton ".

6 . ( i ) A = J 2. 4 , 6 , 8, 10, 12, 14 A = J * | * is an even positive integer < 1G This set is

finite, ( ii ) B = J 1 , 3 , 5 , 7 (j B = J * | * is an odd positive integer}

This set is infinite.

SETS : 17

(iii) The relation is ( * + 2 ) ( * + U = 0 . This is satisfied by x = — 1 and * = — 2. There is no positive number satisfying the equation. The set has no element and hence it is a null set This set may be described as $ = ) x \ xe A ; x ^ A

7. ( i ) The condition in the braces gives a set having only one element 3. The relation is not correctly written since a set cannot be equated to its element. We can write it a s : ) x | ( * - 3 ) a = 0 $ = ) 3 This is " a singleton ".

Ci i ) The relation is correct since the repetition of the elements while describing it does not alter the set.

8 . The sets A and F have the same element namely zero The sets B and D have the same two elements viz, 1 , - 1 . The sets A, E and F have one element, the element in £ however is different from that of A and F. Thus, A and F, B and D are equal whereat E, F and A are equivalent.

9 . ( i ) A ^f B. A and B are equivalent sets. ( i i i All the three sets are different from each other <fi contains no

element, j 0 J contains one element and is a singleton. ) 4> i is a set con-taining the null set as an element. However jo( and )4>[ can be considered as equivalent.

( i i i) No element, yes, yes.

10 . A = 5 2, 3 B = \ 2, 3 ( . Hence A = B.

11. Every set is a subset of itself. Hence i> is a subset of itself. One set can be a proper subset of the other only when the two sets are not equal; and hence no set can be a proper subject of itself. Hence 0 can not be a proper subset of itself.

12. ( i ) Yes, ( i i ) yes, ( i i i ) yes, ( i v ) yes,

1 3 . ( i 1 Yes, ( i i ) yes, ( i i i ) no .

14. ( i ) Empty, null or void ; ( i i ) A singleton or unit s e t ; (iii) Every element or X is an element of V ; ( i v ) Every element of X is an element of V and there is at least one element of Y which is not an element of X. (v) Equal; equivalent, ( v i ) Equivalent ( v i i ) There is no element common to both of them

15. A and E ; B and C ; B and E.

16. I f * f A, t h e n x e B since A S B ... ( i ) Since * € B, it follows that * E C since B S C ( i i ) From ( i ) and ( i i ) it follows that if * tA, then xf C.

:. A^C. 17. ( i ) 0 a n d M , ( i i ) <p,) x\, )y ) z \ , \ x , y \ . \ x . z ) .

) y. " \ * \ There are in all 2s i e. 8 subsets. C. A.—2


18. A set formed by the subsets of A is called a power set of A and is symbolically denoted by P ( 4 ) ; and this is read as power set of A. If A = j 1, 2 P [A ) = } } 1 f , j 2 i

19. For A there are n elements. It has 2n subsets. Hence proper subsets will be 2n - 1 ( the one which is subtracted is the set A itself ).

2 0 . A finite se^ is one of which all the elements can be counted and count-ing comes to an end. A subset has elements either equal to or less than its superset. If the superset is finite then its subset must be finite.

2 1 . ( i ) " j 2 \ £ A " is correct since j 2 j is a subset of A.

( i i ) J 2 \ is a set and not an element. Hence the statement is wrong. Correct statement will be " 2 e A ",

( i i i) The statement is not correct since <j> is not an element of A, Correct statement is " A ".

( i v ) The statement is not correct since 2 is an element and not a set. Correct statement is " 2 e A " or " j 2 j E A. "

( v ) Correct, since 0 is an element of A. (vi) Correct, since <fi is a subset of A,

2 2 . ( i i ) , (vi ) and Cvii) are not correct; others viz. ( i ) , ( i i i ) , ( i v ) , (v) are correct. In ( v i i ) E = C ; but B ±= D.

23 . ( j ) ) 0 j , ( i i ))x]x is a human being having 10 thousand hands j, I i i i ) \ 0 ( i v l j ..., - 3 , —2, - 1 , 0, 1, 2, 3, ... <, ( v ) j 0 (, I v i ) J 1 ( , ( v i i ) ) 0, 1, 2 [ , (v i i i ) )(p,) a \ b J a b [ ( ix i $ 0, 1 j 0 , 0 , 1, 1 {, (x) ) 0. 1 ( , ) a, b J, i x i ) j 0 | , ] 1 \ , ( xii) The set of men members of the club. The universal set is the set of all members of the club,

24-. ( i ) 4 = ^ 1 , 2 , 3 , 4 \ and B = J 2, 4, 6, • ••[; A is equivalent to B.

( i i ) If A is a finite set it cannot be equivalent to a proper subset of itself.

25 . ( i ) 0 is the complement of a universal s e t ; ( i i ' universal set is a subset of itself; (iii) no no, no ; ( iv ) n o ; ( v ) ( 1 ) no complement since \ 0 J is not a subset of V,

( 2 ) 1 / itself (3 , ) 1. 3, 5, ... ( 4 ) ) x 1 x f U ; x > 12

7. Union of Two Sets.

In Arithmetic we learn the operations of addition, sub-traction and multiplication of numbers; that is, to each pair of numbers a and b we assign the numbers a + b, a — b and ab which are called the sum, the difference and the product respectively. In this and later paragraph we shall consider the operations of union, intersection and difference of sets; that

SETS : If

is we will assign new sets to pairs of sets A and B. These operations are in a sense somewhat similar to the operation on numbers given above.

Consider the set of all students in a college. Let us consider this set as the universal set X. Some of the students from the set X know Marathi; let us say that they form the set A. Similarly some of the students from X know Gujarathi; let us say that they form the set B. It is quite clear that there may be some elements common to A and B; for there may be students who know both the languages, Marathi and Gujarathi. The union of these two sets A and Bis defined as the set of students who know Marathi or Gujarathi or both these languages. This set of the union of the sets A and B, is denoted by the symbol " A U B " ; we read the symbol as " A union B " or " A cup B '' In the set notation we write

A = j x | x is a student of the college; x knows Marthi £; B = ) x | x is a student of the college; x knows Gujarathi (;

/HJj9 = ) x | x i s a student of the college; x knows either Marathi or Gujarathi or both Marthi and Gujarathi

Thus for the sets A and B we may write AU B = \ x \ x z A or xzB or xe both A and B j. Breifiy, A U B is the set of elements which are common and

uncommon in A and B, i. e. in A and/or in B. Definition. The union of sets A and B, denoted by A U B, is

the set of all elements which belong to A or to B or to both A and B.

Illustrations. ( 1 ) If A = J a, b, c, d, e ( and B = ) c, p,q, d\, then

AUB = ) a,b, c, d,e,p,q\. ( 2 ) A = ) x | x is an integer; 1 < x < 50 (

and B = ) x | x is an integer; 31 < x < 70 ( than A U B = ) x | x is an integer; 1 < x < 70

( 3 ) If A is the set of all Ration Shops in the city of Bombay and B is the set of all Ration Shops in Dadar, then A U B is the set of all Ration Shops in the city of Bombay.

It may be noted in this example that S c A and we get AU B = A.

2 0 : COLLEGE ALGEBRA.

( 4 ) If is the set of all even positive integers 2, 4, 6, ... and B is the set of all odd positive integers 1, 3, 5, 7 then A U B is the set of positive integers 1, 2, 3, 4,

Here, it may be noted that the sets A and B are disjoint sets i. e. there are no elements common to A arid B. Also the sets are infinite.

Representation of the union of two sets by Venn diagrams ;

The figure ( 5 ) shows how we can exhibit union of two given sets A and B. The set A and the set B are shown below.

W ( « )

Whole portion of A and B together shows A U B in each case.

In figure 5 ( i ) A c£ B. However some elements of A and B are common. The two circles taken together will show all elements in A and B. Hence AU B will be represented by both the circles taken together.

In figure 5 ( i i ) A c. B. Hence the whole portion of B (which obviously includes the portion of A ) will represent A U B. In this case All B = B.

In figure 5 ( iii ), no portion is common. Obviously A and B are disjoint sets.

The following standard identities on set union can be easily obtained from the definition of the union of two sets A and B.

( i ) For every set A, A U A = A. For if x e A U A, then x t A and/or x s A i. e. x t A. This means that A U A C A. ... ... C i ) Similarly we can prove ACZ AD A. ... ... ( ii) From ( i ) and ( ii ), we have A U A = A. ( ii ) If A C B, then A U B = B.

SETS : 21

Proof : xzA\iB=t-xzA, and/or xzB and since Ac. B, we must have x z B.

AU BCB. ... ... ... ( i ) If x z B, x z A U B, since A U B is a set of all common and

uncommon elements of A and B. BCAU B. ... ... ... ( ii )

From ( i ) and ( ii ) , it follows that A U B = B.

Similarly if B c A, then A U B = A. ( iii ) A U <£ = A. ( iv ) A U A' = X. ( v ) A U X = X. ( v i ) A<z ( A U £ ) and B C (A U B) if A # B. ( vii) If A C B and B C C, then A U B C C.

(viii) A U B = B U A, this is a commutative property i n set union.

Proof : xzA\JB=>xzA and / or B. xzBVA.

:. (AU B)C(BU A). ( i ) Similarly x z B U A=>xzB and/or A.

x s A u B. ( B U ^ ) c ( y ) U B ) . ( i i )

From ( i ) and ( i i ) it follows that A U B = B U A. 8. Intersection of Two Sets.

Consider the example in the paragraph 7 where we have defined the union of two sets A and B. In this example the universal set X was the set of all the students in the college; the sets A and B consisted of all the students from this set X, who knew Marathi and Gujarathi respectively. As was stated there, it is quite possible that there might be some elements common to A and B; i. e. some students might be knowing both the languages, Marathi and Gujarathi. This set of elements common to both the sets A and B is called the intersection of the sets A and B ; we denote this set of intersection of the sets A and B by the symbol " A fl B " ; and we read it as " the intersection of A and B", or " A cap B ".

2 2 : COLLEGE ALGEBRA.

In set notation, we write / t n f i = J x | x i s a student of the college ; x knows both

Marathi and Gujarathi Or having defined the sets A and B as given in that paragraph,

we can write AO B = I x l x e / l a n d x e f i j . Thus A 0 B is the set of elements common to A and B.

Definition. The intersection of two sets A and B, denoted by AO B, is the set oj elements which belong to both the sets A and B.

We can now consider again the illustrations given in paragraph 7.

Illustrations. ( 1 ) If A = )a, b, c, d, e $ and B = ) c,p, q, d,\ then AOB = ) c, d

( 2 ) If A = j x | x is an integer: 1 x < 50 ( and B = ) x | x is an integer; 31 < x < 70 j ;

then A0B = } x \ x is an integer; 31 < x < 50 f.

( 3 ) If A is the set of Ration Shops in the city of Bombay and B is the set of Ration Shops in Dadar, then A fl B is the set of Ration Shops in Dadar.

As in the paragraph 7, we note here that Be . A; and it follows that if B C A then we have A fl B = B.

( 4 ) If A = ) 2, 4, 6, 8, • • • • ( and B = ) 1, 3, 5, 7, • • • • ( then there is no element common to A and B, thus AO B =<j>.

In general, it is clear that if A and B are disjoint sets A n 5 = 0.

Representation of the intersection of two sets by Venn diagrams.

The figures given below show how the intersection of two sets A and B can be exhibited in Venn diagrams.

In figure 6 ( i ), there is no common portion so that A and B do not have any common elements and hence they are disjoint sets so that we have A fl B = 0 as in illustration 4 given above.

SETS : If

In figure 6 ( ii ), we have a common portion which is shaded and has also vertical lines in it. This shows that the two sets

( i ) = 0 ( i i i ) 4 fl B ( i i ) A O S ( Common Portion)

Fig 6

A and B have common elements but neither A C f i n o r JSC A. In figure 6 ( i i i ) , Ac B and hence obviously all the ele-

ments belonging to A will also belong to B and hence Af\B = A.

As in the case of the union of two sets, we can obtain the following standard identities on the intersection of sets.

( i ) AflA = A. Proof : x t A fl A =>Je t A and x e A i. e. x e A.

A C\ AC A. Again x t A =s- x e A and x e A i. e. x s A fl A.

AC Ad A. Hence from ( i ) and ( i i ) A fl A = A. ( ii ) AC\ A' =<t>. Proof : x s A H A ' ^ x s A a n d x e A ' . But A and A' are compliments of each other and have no

common element and hence this set is A PI A' = <t>.

( i i i ) Ad<t> = (t>. Proof : xeAD0=>xsA and x e <f>. But <p has no element. Hence there is no element which

belongs to A and 0 both. .*. a n * = 0.

( iv ) A fl B C A and A fl B C B.

( i )

2 4 : COLLEGE ALGEBRA

Proof, xe Af\ xe A and A; e 2? and hence the result follows.

( v ) If CCZA and C C f i , then CCLAftB. Proof, xe C x e A and xe B, from the given data;

and this =>xeA H B. CCA OB.

( vi ) A fl B = 2?f) A, i. e. the intersection of two sets is

commutative.

Proof, xe A fl B => xe A and xeB=>xeBC\ A. A r i B C B n A . ... ( i )

Similarly xeB(lA=>xeB and xeA=^xeAf\B .'. B H A C A D B ... ( i i )

From ( i ) and ( i i ), it follows that Ad B = B(1 A. ( vii ) ) A fl B ( fl C = A fl ) B fl C ) i. e. the intersection

of three sets is associative. xe J ^ H B C ^ - x e J .4 0 B $ and x e C ,

=> x e A and x e B ; and x e C x e ,4 and ;c e J B?\C\

j f l f l C j -\ AViBinCC-Aft) BC\C\. ... ( i )

Similarly we can prove that ^ r u B n c i c M n ^ n c . ... ( H >

From ( i ) and ( i i ), it follows that ) AV\B ( 0 C = Af\ | B f l C (•

9. Extension to Three Sets. As a result of the definitions of union or intersection of

sets, we have if A, B, C are three given sets, A U B and A D B are sets ; we can, therefore, easily extend these operations as indicated below :

C , ( ^ U B ) R C and ( ^ O B ) U C , ( A t \ B ) ( \ C . Since A U B is a set of all elements common and uncommon

to A and B, we can write , 4 U . B = | * | ; c e a t least one of A and B j and A n B = j x ] .X e both A and B

s e t s : 2 5

Then it follows that

( i U B ) U C = the set of elements x such that x e at least one of ( A U B ) and C;

i. e. = ) x | x e at least one of A, B, C f ; and ( A U B ) l~l C = the set of elements x such that x e both ( <4 U B ) and C;

i .e . = j .t | .x e A and C or x e B and C ( ;

Similarly ( A D B ) U C = the set of elements common and uncommon to ( A D B ) and C i. e. = ) x \ x e A, B, C ; x e both A and B and x C; x e C and x £ both A and B ( ; and < A PI B ) fl C = the set of elements x common to ( A fl B ) and C i e. = j x | x e A, B and C (•

The meanings of these operations as indicated in the above symbols are made clear by Venn diagrams as shown below :

( ^ U B ) U C is the whole of the portion shaded, (either by lines, vertical, hori-zontal or inclined) in the figure 7.

( A U B ) fl C is the por-tion shaded in the figure 8 ; part (1) is common to A and C ; part (2) is common to C and B and part (3) is common to A, B and C.

Fig. 8

2 6 : c o l l e g e a l g e b r a

( A n B ) u C is the portion shaded in the figure; it is the portion common to A and B and the whole portion of C.

( A D B ) (1 C is the portion ( 7 ) shaded in the figure 10 ; it is common to the sets A, B and C.

Fig. 10

Example. Let A = set of men, B = set of graduates-and C = set of teachers. Interpret P U B L I C , / f f l B f l C , ( A U B ) fl C and ( A D B ) U C.

Let the sets A, B, C be denoted by the circles as shown in the figure 10. The interpretation is simplified by the parts of the figure marked ( 1 ) , ( 2 ) , ( 7 ), where part ( 1 ) = all those elements in A which do not belong to B or C and similarly for parts ( 2 ) and ( 3 ) . Part ( 4 ) = all those elements in A and C but not in B and similarly for parts ( 5 ) and ( 6 ) . Part ( 7 ) = all those elements common to the sets A, B and C.

Then A U B U C = the set of elements * which belong to either A, jB or C, to either A and B, B and C or C and A or to all the sets A, B, C; i. e. it = all the parts ( 1 ) , ( 2 ) , . . . ( 7 ); or A U B U C is the set of elements who are men, graduates or teachers, men graduates, graduate teachers or men teachers or are men graduate teachers.

sirrs : 27

A fl B fl C = the set of all elements common to the sets A, B, C i. e. the set of men graduate teachers. It is shown by part ( 7 ).

( A U B) (1 C = the set of elements x which belong to C and to A or B or both A and B; this set is shown by parts ( 4 ), ( 5 ) and ( 7 ) . The set may be said to consist of those men (not graduates) who are teachers, those graduates ( not men ) who are teachers and also of men graduate teachers.

( A fl B ) U C — the set of all elements x common to A and B, which belong or do not belong to C and also of those elements x which belong to C but are not in A fl B i.e. are not in A and B\ i. e. this set = the elements in parts ( 6 ) , ( 7 ) , ( 4 ) , ( 5 ) and ( 2 ) i. e. it consists of the men who are graduates but not teachers ( 6 o f men graduate teachers ( 7 ), of men teachers who are not graduates ( 4 ), of graduate teachers who are not men ( 5 ), and of teachers who are neither men nor graduates ( 2 ).

10. Operations on Sets.

Students are familiar with operations on numbers. For example, addition, subtraction, multiplication, division etc. are operations on numbers. We can similarly think of operations on sets. If the operation is done only on one set, it is called a unitary operation. Complementation is a unitary operation since it operates on a single set only. If the operation involves two sets we call it a binary operation. The union of two sets and intersection of two sets are binary operations. These operations of union and intersection may be extended to more than two sets also. These operations satisfy some properties which are similar to those satisfied by operations no numbers.

( i ) Commutative Property. Students are familiar with commutative property for addition and multiplication of two numbers, namely, m + n = n + m and mn — nm, i. e. the effect of the addition and multiplication of two numbers remains the same even if the order of numbers is changed. This commutative property remains true for the binary operations of union and intersection. In using the commutative property only two numbers or sets are used. Thus if A and B are two sets then we have A D B = Bl) A and A C\ B = B C\ A.


( ii ) Associative Property. If there are more than two numbers the operations of addition and multiplication are associative i. e the order in which the numbers are associated is immaterial. For example, {m + n) + r = m + (n + r ) and (mn)r = m (nr ). This property holds for the operations of union and intersection of sets also. For example,

AUBUC=(AUC)UB*=(CUB)UA etc. and AC)BnC = (AnC)HB = (BnC)DAetc. ( i i i ) Distributive Property of Union and Intersection.

Students know that the operation of multiplication is distributive with respect to addition. For example, we have

m(n + r ) = mn + mr. In the case of sets we have union of sets is distributive over intersection of sets. For example,

^ u ( s n c ) = (y4u B>n(^iu c>. This can be illustrated by the following Venn diagram.

Shaded portion indicates A U ( B f l C) .

Shaded portion also indicates ( ^ U B ) f l ( ^ U C ) .

^ U ( J S n C )

Fig. 11 I

Similarly intersection of sets is distributive over the union of sets. For example

This can be illustrated by the following Venn diagram.

Shaded portion indicates A(\{BV C).

Shaded oprtion also indicates {Aft B)U ( A f t C ) .

^ n ( s u c )

Fig. U 2

SETS : 2 9

( i v ) We now consider here the operation of difference on two sets.

If A and B are two sets then ( i ) A. — B is a set of all elements of .A which do not

belong to B and ( ii ) B — A is a set of all elements of B which do hot

belong to A. Thus A - B = } *e>4; x e B ( and B - A = ) x\xe B-,x<£A\.

( i ) Shaded portion A — B. (ii j Shaded portion B — A . Fig. 12

Example. If A = ) 0, 1, 2 j and B = j 2, 3, 4 f, then A — B = set of elements which belong to A but not to B.

A - B = ) 0, M; and similarly B — A = J 3, 4

Following identities on difference of sets may be noted. ( 1 ) A - B = A 0 B' = B' - A'.

Proof : We have A - B = \x\xeA \ x<ZB\ = j x j x e A; X e B' ( ... ( i ) = A 0 B'; this proves the former

part. Again A - £ = ) x f y 4 , x e f i ' j , from ( i ) above

= j xC A'; xe B' | = ) x e B -, x<Z_A' f — B - A'; this proves the latter part.

( 2 ) A - B = <p, iff A C B. Let A - B = <t>. We have A — B = A Pi B' from ( i ) above. Aft B' = <f> since AO B' = A - B = <p.


But A n B' = <j> means that A and B' have no common element.

A and B' are disjoint sets, ... <#> ( j ) Again B and B', being complements are disjoint sets. ( ii ) From ( i ) and ( i i ) it follows that A £.B. Conversely if A S. B, then A fl B' = <p and hence

A —.8 = 0. ( 3 ) One can easily see that

A - B = B - A, only when A = B\ and A - B = A, only when A fl B = <f>.

11. Illustrative Examples. E x . i , Write in set notations (tabular or set builder notion ) the follow-

ing sets. ( 1 ) A is the set of letters in the word " college ", ( 2 ) B is the set of letters c, e, g, I, o. (3 ) C is the set of letter? i n the word " calculus ". ( 4 ) D is the set of letters a, c, I, s, u. ( 5 ) £ is the set of letters in the word "leg". ( 6 ) F is the set of letters in the word " cell '.

Answers — (1 ) A = ) x j x is a letter in " college " ( 2 ) B = i c,e.g.l,o ( 3 ) C = ' x | x is a letter in " calculus " |. (41 D = ) a,c,l, s, it ( 5 J B = J x \x is a letter in ' kg" ( 6 ) F = ) x\x is a letter in " cell " j.

Ex. 2, With reference to the sets in Ex. 1 above, answer the fol lowing:

( 1 ) Which of these sets are subsets of the sets in the example ?

( 2 ) Which of these sets are equal sets ?

(3 ) Which of these sets are finite sets ?

( 4 ) Write down the sets A (J B. A U C, C (J D, C U E and A U E.

( 5 ) Write down the sets ,4 f l B f ] C. C f l D, C f | E and C f~l F .

Answers —

( 1 ) It is evident that A <= B , B = A; and E and F C l o r B. This follows from the fact that the letters in the word college'' are " c , o,le,g" which are the same as in B ; i e. 4 = B. As the letters " I e, g " are also to be found in A or B, E is a proper subset of A or B. F is also a proper subset of A or B.

SETS : 31

Similarly, C S Z 3 and C £ C ; hence C = P.

(2) As seen in the above answer, A = B and C = P. These are the only equal sets. It should be noted that u A or B ; s ^ or B. Also g C or D. Hence A C or D and B 4= C or P.

Note : The elements of a set can be written in any order ; thus the elements in A and B are the same ; but the order is different. Similar remarks hold for the sets C and D .

If the elements in a set are repeated, they are to be counted once only. Thus the letters I and e are repeated by the rule for the formation of A ; yet A - B.

( 3 ) It is evident that all the six sets are finite. There are five elements in tbe sets A, B, C and D and there are three elements in each of the sets E and F.

( 4 ) It is obvious that A [ j B = /i or B, as A = B ; similarly C IJ D = C or D.

Since A = B and C — D, A |J C = B (J D ; the conmon elements in B and D are c, I; the uncommon elements are e, g, o and a, s, u, :• i l ) C = B U D = j c, I, e, g, o, a, s, u,

Similarly C U E = D U E = j I a, c, s, u e, g Since E C. A or B , A U-B = - 6 L ) E = 4 o r B .

Similarly A U F = B l J f = - A o r B . ( 5 ) Since A f ) B = the set of elements common to A and B = A oiB.

B f ) C = B f | O, Since C = D,

Similarly C (~) £> = C or D ; and C f | £ = O f | ® ('•' C = D) ; i . e . = ) l \ .

Finally, c n f = £ > r i F ' = I c . 1 t h i s s h ° w s that

EX. 3. A = \ 0, 1, 2, 3 Write down all the subsets of A. How many subsets has the set A ?

( i ) <t>, the empty set, is a subset of every set A ; ( i i ) Subsets of A with one element only are ) 0 ( , ) 1 | , ) 2 ^, ) 3 { ; (iii) Subsets of A with two elements only are ) 0, 1 j 0, 2 j ;

) 0. 3 (, J I, 2 j 1, 3 and \ a, 3 ( ; ( i v ) Subsets of A with three elements are ) 0 1 , 2 ^ , j 0, 1 , 3 j ,

\ 0, 2, 3 [ and j 1, 2. 3 ( v ) Finally A, tbe set of four elements is a subset of itself.

There are thus sixteen ( = 2 ' ) subsets of A and there are fifteen proper subsets of A.

Ex. A. The sets A, B, C, D, E, F are given as follows : A = | * | * is an even integer : * > 0


B — J * | x is an odd integer ; * > 0 j.

C = j * | * is a multiple of 3 ; * > 0 \ .

D = J * | x is a multiple of 5 ; * > 0 j .

E = j x | x i s a multiple of 3 ; 1 < * < 50 (.

F =') x | * is a multiple of 5 ; 1 < * 50

Find ( i1) which of the sets are finite and which are infinite ; ( i i ) A U B and / I f l B , ( i i i ) A U C and A f ) C ; ( i v ) A (J E and A E ; and ( v ) E U F and E f ) F.

Answers —

( i ) A = \ 2, 4 6, 8, J and B = \ 1, 3, 5 \ ; similarly C = \ 3, 6, 9 \ and D = ) 5, 10, 15, 20 These are thus infinite sets. However E = J 3, 6, 9, 48 ^ i e. there are sixteen elements in E ; and F = j 5. 10, 15 50 \ i .e. there are 10 elements in F. Thus E and F are finite sets.

( i i ) There are no common elements in A and B as seen above ; hence A U B = J 1, 2, 3 4 , 5 , 6 ( ; i e. A U B is the set of all posit ive integers.

As there are no common elements in A and B, B is an empty set ; or a n B = <t>.

( i i i ) A is the set of multiples of 2 ( * > 0 if x e A) and C is the set of moltiples of 3 ( * > 0 if * e C ); hence A U C is a set of multiples of 2, multiples of 3 and multiples of both 2 and 3 ( i. e. multiples of 6 ) ; i e. A JJ C = j 2, 4, 6, 8, .. and 3 , 9, 15, ... . ( or A U C = j x | x is an even integer or an odd multiple of 3 ; * > 0

A f") C •* the set of common elements in A and C i. e. = { 6 , 1 2 , 18,

Or j 4 f l C = ^ * | * i s a multiple of 6, * > 0 { .

It may be noted that both A U C and A (~) C are infinite sets.

( i v ) A U £ = ! 2 , 4 , 6, ; and 3, 9, 15, . . . . 45 | i. e. A (J E is the set of all even integers ( > 0 ) and odd multiplies of 3 between 1 and 50.

Similarly it is evident that = ) 6 12, 18, 24, 48 ( ; or A f \ E = $ * | * is a multiple of 6 ; 1 < * < 50 (.

{ v ) W e leave it to the students to show that E U F = ) 3, 6, 0 48 ; 5, 10, 20, 25, 35, 40, 50 $ = ) * | * is a multiple of 3 and x > 0 ; * is a multiple of 5 and 1 < * < 50

B f ) F = ] 15- 30, 45 j.

E*. 5. (a) If A C.X, BCLX and A f ) B 4= 0 . draw Venn diagrams to show ( i ) B'.

(ii) A UB and (A U ( i i i ) A H B and (A f l B)'. (iv) A ' f l B ' . X ' n B T . l i ' U B ' ) ' -( v ) A — B and {A — B)'. (v i j B - A and (B - A)'

SETS : 33

(6) If any two of the sets A, B, C have common elements and they are proper subsets of X, draw Venn diagrams to show

( i ) A n ( B u C ) a n d / l u ( B u C ) (ii) (A OB) n (A DC) and (A f lB) u ( A n c )

Answers:— ( a ) ( i )

Rectangle indicates universal set. Plain area indicates B; shaded portion indicated by horizontal lines indicates B'.

Fig. 13 ( « ) ( i i )

Rectangle indicates universal set. A u B = 5 * | * e A and or * e B |

= unshaded area. (A j B ) ' = shaded area.

Fig. 14 («) (Hi)

Rectangle indicates universal set. .A D B = } * | * e A and e B f,

shaded area. (A 0 B )' = unshaded area.

(a (iv)

F i g . 16 formed by the two types of inclined

C. A.—3

Fig. 15 Rectangle indicates universal set.

Portion indicated by straight line* inclined towards the right hand side ( i . e. having positive slopes) indi-cates A'.

Portion indicated by straight lines inclined towards the left hand side (i. e. having negative slopes ) indicates B'.

A' n B' = I » xe A' and x "-B' j. This is indicated by crossed area

34 : COLLEGE ALGEBRA

( A' n B' y = A u B, indicated by two circles taken together.

( A ' U S ' ) ' is indicated by plain area i . e . common portion of two circles.

(«) ( v )

Rectangle indicates universal se t . A - B =) x e A, x f c B , j plain area

( A — B )' = shaded area. («) (vi ) am

Fig. 18

(6) (>) A is indicated by right slant lines.

B |J C isindicated by drawing left lines.

A f ) ( B (J C ) is indicated by cr area formed by the two types of lines.

A U < B (J C) is indicated by the shaded area.

(b) (n)

A n B is indicated by drawing lines inclined towards the right having positive slope.

A f l C i s indicated by drawing lines inclined towards the left having negative slope.

( A n BJ n ( A n C) is indicated by crossed area formed by the two types of , inclined lines.

(A 0 B ) u (A 0 C) is indicated by the whole shaded portion i. e., by the two

types of lines and the crossed area.

Fig. 17 Rectangle indicates universal set. B - A = ) xe B; x £ A shaded portion. (B - A )' = plain area i. e. whole unshaded portion.

Fig. 19

Fig. 20

SETS : 35

Exercisc 1 (b>

1. Define the operation of union between two sets A and B. Explain its meaning by an example. Illustrate by means of Venn diagram, the union of two sets.

2. Define intersection of two sets and explain the defini-tion by means of an exampie and with the aid of Venn diagram.

3. if A = ) a, b, c,d B = \ c, e j \ and \ X = J a, b, c, d, c , f , g, h write in the tabular form the sets ( i ) A U B and ( ii ) A 0 B and indicate them by a Venn diagram. Also write down the sets which are equal to ( i ) A U X, ( i i ) A fl X, ( iii ) A U A', ( iv ) A fl A'.

4. Let X = j — 4, — 3, — 2, — 1, 0, 1, 2, 3, 4 (, A = } - 4, - 3, - 2, - I f , B = J - 1, 0, 1 C = j 1, 2, 3, 4 j, express in the tabular form ( i .e .

by Roster method ) the following sets ( i ) A', ( i i ) A'ClB, ( i i i ) AUC, ( i v ) (AU C ) \ ( v ) A fl x ( v i ) A U 0. Verify that AC\X = A\J<f>.

5. If A = ) a, b, c B = J c, x, y, z (, find ( i ) AuB, £ U . 4 ; ( ii ) Aft B, Bft A.

Is ( i ) /4 U 5 = B I) A, ( i i ) At\B BViAl

6. if X = } 1, 2, 3,4, 5, 6 { and its subsets A = \ 1, 2, 3, 4 B = j 2 ,4 j, C = j 3, 5, 6 write in tabular form, the sets ( i ) A U C, ( ii ) C — B, ( iii ) A' - B, ( iv) (A- C)', ( v ) ( A - B' )', ( vi ) .Bfl A, ( vi i) B\ (vii i) B' U C, ( ix ) C' fl A, (x) ( A fl A' )'.

7. If X - set of all students in F. Y. Arts class, A = set of all students who have taken Mathematics

in the class, B = set of all male students in the class.

Describe in~words the sets ( i ) A' U B', ( ii ) ( A fl B )'. State whether .4' U B' = ( A fl B )'. illustrate with the aid of Venn diagram.


8. If the universal set X = j x | x is a positive integer less than 10 j, and A - J 2, 4, 7, 9 j, B= J 1, 5, 7 verify that ( /I U B )' = A' fl B'.

9. When are two sets said to be equal ? Define empty set. Give an example of empty set and show that A C 0 A = 0.

10. If A = set of all fat people, B = set of all short people,

describe in words ( i ) A' fl B\ ( ii ) ( A U B ) and state whether A' fl B' = ( A U B )'. Illustrate by means of Venn diagram.

11. If X = )a, b, c, d, e,f\, A = \c,d,e\, B = ) a, d j find ( i ) ( A U B )', (ii ) A' fl B' and state whether {AUB)' = A'VlB'.

12. If X is the universe with A and B as its subsets, show that ( i ) ( A U B)' = A' fl B' i. e. the complement of the

union of A and B is equal to the intersection of the complements of A and B ;

( ii ) ( A f t B)' = A'U B' i. e. the complement of the intersection of A and B is equal to the union of the complements of A and B.

Illustrate by means of Venn diagrams the equality of these sets.

[The results given above are known as De Morgan's laws. ].

13. If the universe X = ) 0, A , ! , ?, a, 5, x [ and subsets A = J A , ?, • B = ) o , 0, A i then find ( i ) ( A D B)', ( ii ) A' U B' and state whether ( A fl B)' = A' U B'.

14. With the usual notation, show that

( i ) A ^ B o B'CA';

( ii ) if A C C and B C C, then ( A U B ) Q C and ( A f ) B ) C C .

SETS : 37

15. Given two sets M and N with NC- M, which of the following is true ? ( a ) M f \ N = N or M C\ N = M.

(Z>) M U N = N or M U N = M.

16. Prepare a Venn diagram to show that A n B + <*>, A n C = B f l C * <t>.

17. Given subsets ,4, B, C of the universe AT, write the sets which are equal to ( i (ii)Af)(f>, (iii )AUA, ( i v ) A r \ A ' , f v ) 0 ' , ( v i ) ( i 4 ' ) ' , ( v i i ) ^ n j f , ( v i i i ) A \J X, (a)AC\A, (x)AUA\ ( x i ) X', (xii)A-<fi, ( xiii ) A - A, ( xiv ) A 0 B'.

18. If A = )4, 5, 6, 7 = J 1 , 2 , 3 C = } 8 , 1 0 { express in tabular form the sets ( i ) AU{B\JC), ( i i ) ( / t U B) U C and verify that

X U ^ B U C ) = U U 5 ) U C.

Draw a Venn diagram to show this result. ( This is called associative property for union of sets A, B,C.)

19. Taking A, B, C as subsets of the universe X, show by means of Venn diagram or otherwise that ( i ) = ( B f l C ) ( H ) ^ n ( B n c ) = ( y i n J B ) n c .

[ This is called the associative property for intersection of sets A, B, C. ]

20. If X = } 1. 2> 3> 4< 5> 6> 7' 8- 9» 10' H' 12 * and subsets

A = \ 1 ,9 , 1 0 j , f l = J3, 4 ,6 ,11, 12 t and C = ) 2 , 5 , 6 | , find ( i ) AU ( B f \ C), ( i i ) (A\jB)r\(A\jC) and verify that ^ u ( 5 n c ) = ( ^ u 5 > n ( ^ u c ) .

[ This is called distributive law of union over intersection. ]


21. With the sets in example 18, find ( i ) Aft ( 5 U C ) , ( ii ) ( Aft £ ) U ( ^ f l C ) and verify that - 4 n ( j 3 u c ) = ( / < n s ) U ( y < n c ) .

[This is called distributive law of intersection over union. ]

22. If A, B, C are three non-empty sets, state giving reasons whether the following statements are true. ( i ) If ,4 U £ = ,4 U C, then B = C; ( i i ) I f ^ n s = Af\C, t hen f i = C.

23. Sets A and B are such that A has 32 elements ; B has 42 elements and A U B has 62 elements. Find the number of elements in the set A D B.

24. Show that (AU 5 ) D B' - A only if A D B = <f>. 25. S h o w by using Venn diagrams or otherwise that

( i ) AUB = BVA, ( i i ) Aft B = BV\ A,

( v ) ^ 0 5 = Biff fiC/1, ( v i ) ( ^ U B ) ' = / n 5 ' , (vii) (At\B)'= A'UB\ (viii) A D (BflC) = ( ^ D B / l C , ( i x ) A D ( B U C) = ( ^ U B ) U C, ( x ) ^ U ( B n C ) = ( ^ U B ) n ( 4 U C ) , (x i ) ^ n ( B u c ) = ( / < n s ) U ( ^ n c ) , (xii) A- B = B'-A'.

1. A ( J B = )x\xe A and/or B j . If A = * 1, 2 B = ) 2, 3, 4, (, then A U B = ) 1, 2, 3, 4

2. A f \ B = ] x i x e A and B j. If A = j 1, 2, 3, 4 j and B = j 3 ,4 , 5,6, 7 | , then A f l B = ) 3 , 4

3. A (J B = ) a, b. c, d, e / A f | B = ) c (.

( i i i ) ( A ' ) ' = * A , ( i v ) X l l £ = A iff BCA,

Answers. Exercise 1 (b)

X

Fig. 21 a

SETS : 39

( i ) A U-Y = x, ( i i ) A n x = A.

( i i i ) A I U ' = X. ( i v ) A f l A' - <t>.

4. ( i ) A' = ] 0 , 1 . 2 , 3 . 4 $ . ( i i ) A ' n B = | 0 , l j ,

( i i i ) A u C = j - 4 , - 3, - 2 , - 1 , 1, 2, 3, 4, ' ( , (iv) (A (J C)' = j 0 \.

( v ) A O X = A = J - 4 , - 3 . - 2 , - 1

(v i ) A u<t> - A = ) -4, -3, - 2 , - 1

5 . ( i ) A u B = \ a, b, c, x, y,'z B U A - ] x, y, z, a, b, c\.

A\JB = BUA.

( i i ) A n B " \ c \ , BU A = ) c \ ,

6 . ( i ) A u C = S 1. 2, 3, 4, 5, 6 j,

( i i ) C - B = } 3, 5, 6 $,

(iii) A' - B 5, 6 [ - | 2, 4 { = j 5, 6 j,

( i v ) First let us write A - C = ) 1, 2, 3, 4 ( - $ 3 , 5 , 6 ( = J 1 , 2 , 4 } ,

••• ( A - C ) ' = ) 3, 5 , 6 ' . ( v ) First we write A - B' = \ 1, 2, 3 , 4 ( - \ 1, 3, 5 6 [ = ]2. 4

A (A - B ' ) ' = (1,3,5,6). ( v i ) B H i = j 2, 4 $ , ( v i i ) B' = ) 1 , 3 , 5 , 6

(viii) B ' u C = J 1, 3, 5, 6 $ u i 3, 5, 6 $ = \ 1 , 3 , 5 , 6 5,

. ( i x ) C' 0 A = j 1 , 2 , 4 ( f l j 1 , 2 , 3 , 4 | = ] 1 , 2 , 4

( x ) A f l A ' ^ 0 , :. (A D A ' ) ' = </>' = X = J 1 , 2 , 3 , 4 , 5 , 6

7 . ( i ) A ' u B' = set of all students who have not taken mathematics and or who are not males.

( i i ) A H b = set of all male students who have taken mathematics.

( A O B ) ' = set of all students who are neither male nor have taken mathematics.

the two sets are equal and we have

A' u B' = ( A O B ) ' ,

i. e. the union of complements = the complement of intersection.

10 . ( i ) A'0 B' = set all people who are not fat and who are not short.

( i i ) 4 u B = set of all people who are fat and/or short

( A u B ) ' = set of all people who are not fat and who are not short.


/. (A'n B') - u uB)'. i. e. the intersection of the complements = the complement of the anion.

11. ( i ) We have A u B = ) c, d, e, a, f ( A u S ) ' = } M « ( i i ) We have A' = ) a, b, / B' = \ b, c, e (.

( A u B ) ' = A ' n B ' = j 4 | .

12. ( i ) Starting from the left hand side set, we have xe (A nor * g ; B

x e A' and xe B' ^ x e (A' 0 B' )

(A U J B ) ' S ( A ' n B ' ) . ... ( i )

Similarly starting from the right hand side set, we have x e (A' 0 B'} => xe A' and B', both

=> x e A x B simultaneously

^xe(ADB)'. :. ( A' n B' ) S ( A uB)'. ... ( i i )

From ( i ) and I i i ) we conclude that

(A (jB)' = (A' DB').

( i i ) Arguing out as in ( i ) . show that ( A H B )' £ (A' u B' ) and ( A ' U B') £ ( A D B)'

and hence deduce the result.

13 . ( i ) (A n B)' = \ 0, * 1, ?, 5, * \ ( i i ) (A' u B ' i = \ 0, !, ?, 5, * I

The two sets are equal.

14. ( « ) " ( M O W ) = N " is true. ( b ) "(SI (j AT) = M " is true.

1$.

Fig. 22

SETS : 41

17. ( I ) A U <t>=A. I. ii ) A n <t> = 1>. ( HI) A U -A = , (iv ) A 0 A' = 0, I v * 0' = X, (vi) (A'Y=A, (vii) A F|-X = A , (vii i) A U X = X . ( i x ) A (1 A = A, ( * ) A L M ' = X, ( xi ) X' = ( x i i ) A — 0 = A>

(xiii) A - A = 0 , (xiv) A f l B ' = A - B .

18. ( i ) B 1JC = ) 1, 2, 3, 8, 10

••• A U< B U C ) = j 4, 5, 6. 7. 1, 2. 3, 8 . 10 ( = ) 1, 2, 3, 4, 5, 6, 7, 8, 10 |

( i l A U B = J 4, 5, 6. 7. 1. 2 , 3 (A U - B ) U C = \ 4. 5. 6, 7, 1, 2, 3, 8. 10 [

= } 1. 2. 3. 4, 5, 6, 7 . 8 , 10 Thus A U(B\JC) = (A{JB)\JC.

19. ( i ) Consider the set (A D B ) f l C.

* e ( A f ) B > n C ^ i E A f l B and C, both s - i f A, B, C simultaneously

*f A and ( B f | C) => xt A (1 ( B O C ).

( A n B ) n c E A n ( B n c) ... (1) Similarly we can show that

A F ) ( B N C ) = ( A D B ) D C . ... ( 2 )

From ( 1 ) and ( 2 ) , we deduce that ( A n B ) n c = A n ( B n c

( i i ) Similar proof as in ( i ) can be given.

iO . ( i ) ( B F L C ) = ) 6 / . A U(BOC) = ) 1. 9, 10, 6

( i i ) W e have A (j B = } 1, 9, 10. 3, 4. 6, 11. 12 [•

A (_|C = ) 1, 9, 10, 2, 5, 6

( A U B ) n (A U C > = J i, 9, io, 6

From ( i ) and ( i i ) we see that A u(snc) - (A UB)D M UC).

2 1 . ( i ) W e have B U C = ) 3, 4, 6, 11, 12, 2, 5 J.

A H { B U C ) = (II) Similarly A 0 B = 0 , A L"L C = 0 .

( A f ) B ) U ( A f l C ) = 0 and hence the result.

2 2 . ( i ) Not true. There may be more elements common to say A and C than the elements common to B and C (or vice versa). For example let A = J 1 , 2 , 3, ( , B = } 2, 3, 4 $ and C = j 1, 2, 3, 4 (. Then A U B = A U C = \ 1, 2, 3, 4 But B 4 = C .


(i i ) Not true. There may be more (or different ) elements in C which are not in A ) than those in B (which are not in A) or vice versa Let A = J 1, 2,1 B = > 2, 3, 4 ( and C = J 2. 3. 5 j.

Then A f l B = A |") C = ] 2. 3 '<• But B 4= C.

2 3 . A ( j B will have maximum elements if A and B axe disjoint Since A has 32 and B has 42, A (j B can have 32 + 42 i. e. 74 elements. But we are given that A y B has 62 elements Hence the sets A and B are not disjoint. They intersect and must have 74 - 62 i- e , 12 elements in common Hence A 0 B has 12 elements

2 4 . Plain area B ; Shaded area B' .

Fig 23

To show that ( A u # > n B' = A only if A D B = 0 . Proof : We have the property that intersection of sets is distributive over

the union of sets. .'. (A UB )f1 B = ( A 0 B') u ( B O B' )

= UnB')u<*>. (s ince B and B' do not have any common element.)

= i f l B ' . If A n B = (j>, then A and B are disjoint sets and hence A and B do not

intersect and A C S and hence A f ) B' = A-

Chapter 2

Real Numbers

( l ) Natural numbers. ( 2 ") The number corresponding to an empty set. ( 3 ) Integers. ( 4 ) Rational numbers. ( 5 ) Representation of rational numbers by points on a straight line. ( 6 ) Irrational numbers. ( 7 ) Corre-spondence of irrational numbers with points on a straight line. ( 8 ) Approxi-mation of irrational numbers by rational numbers, ( 9 ) Real numbers. (10) Imaginary numbers. (11) Illustrative Examples. Exercise 2.

1. Natural Numbers.

The students are already familiar with the positive number s 1, 2, 3,..., n,...These positive numbers are called natural numbers as they arose from the natural process of counting. The set o f natural numbers, which we shall denote by N, is an infinite set. The process of counting can be considered as asso;iating a set o f objects with a set of natural numbers N = ) 1, 2, 3,... } in such a way that one and only one object of the set corresponds to one and only one natural number. This type of association is called one-one ( or one to one or 1-1 ) correspondence from one set to the other.

Let us consider a finite set of natural numbers 1, 2, 3, , k. This set has k elements ; denote this set by Nk so that

Nh - ) 1 , 2, 3 k \ .

If A is a given set and the elements of A can be put in 1-1 correspondence with Nk , then obviously A has k elements. We shall write the statement that " the set A has k objects corre-sponding to k natural numbers " as " n ( A ) =k." For example, if there are 15 objects in a set A corresponding to the natural numbers 1, 2, ,15 we shall describe the correspondence by the equation it ( A ) = 1 5 . Similarly if A ~ \ a, b, c, d. .. j \ and Ni0 = j 1, 2, 3,..., 10 j, then the two sets can be

43


put in one-one correspondence and we may write this relation as n (A) = 10.

Consider the two sets N = | 1, 2, 3

and A = J 2, 3, 4,..., n, n + 1, .. . . j. Observe that

( i ) N and A are infinite sets; ( ii ) A is a proper subset of N, since 1 e N and 1 g; A ;

(iii ) there is one-one correspondence between the infinite set N and its proper subset A; and

( iv) the two sets N and A are equivalent.

Thus, an infinite set can have one-one correspondence with its proper subset whereas a finite set cannot have one-one corre-spondence with any one of its proper subsets.

If n ( A) = r and n ( B ) = s i. e. if the number of elements in A is r and the number of elements in B is s, and if r > s, then process of counting the elements in B comes to an end first.

Suppose that there are 15 elements in the set A and 30 elements in the set B, such that the 30 elements in the set B are distinct from the 15 elements in the set A. Here, two sets A and B are disjoint since they do not have any common elements and hence A n B = 0, but n ( A U B ) = 45. This illustration will help us to define the addition in N.

1-1. Addition in AT. If n ( A ) = r and n ( B) = s and A n B = 0, then the sum of two natural numbers r and s is given b y r + J = n ( ^ U B ) .

We shall now derive some fundamental properties of the natural numbers.

( i ) Closure Property. If disjoint sets A and B have r and s elements respectively then A U B has r + s elements; hence r + s is also a natural number. This shows that the addition of two n a t u r a l numbers is also a natural number. This fact is expressed by saying that the set of natural numbers is closed for

REAL NUMBERS : 4 5

addition. A set is said to be closed ( or is said to have the closure property ) for an operation if the result of the operation is an element of the same set. Thus,

if re N, se N, then r + s e N and hence N is closed for addition.

( i i ) The Addition is Commutative. Let A and B be two finite, disjoint sets such that

r = n ( A) and s = « ( B ).

By definition, r + s = n(AU B) — r} ( B U A ), since the commutative

property holds for the union of two sets.

= s + r.

This shows that the addition of two natural numbers is commutative i. e. the sum of two natural numbers remains the same even if their order is changed.

( iii ) The Addition is A ssociative. Let A, B, C be three disjoint sets such that

r = n(A),s=n(B), t =n(C).

:. by definition, ( r + s ) + r = n j ( ^ U f i ) U C | = n \ A U ( B U C ) ( as the

union of more than two sets is associative. = n(A) +n(BUC) — r + (s + t ) = r + s + t.

This shows that the addition of three natural numbers is associative, and hence we have the property that in the addition of more than two natural numbers, the order in which the numbers are associated is immaterial.

It follows from this result that if r, s, t e N and if r + s = r + t then s = t.

This is called the Cancellation Law for addition.

1-2. Multiplication in the Set N. If there are s disjoint sets, say A,, A2, A3...As, any two of them being disjoint and


n ( Ax ) = n ( A2 ) = ... = n(As) — r, then the product of two natural numbers r and s is given by

rs = n j At U A2U A3 ... U A, j.

We can see from the associative property of addition that n Mi U A2 U A3... U As ( = 11 ( Ax ) + n ( A2 ) + ... + n(As)

= r + r + r + s times = r s.

We shall now derive some fundamental properties of the operation of multiplication of natural numbers.

( i ) Closure Property in N. If r and s are two natural numbers then the disjoint sets Av A2,...AS can be found such that n( Ai) = n( A2) = ... =n( As) = r. The number of elements in the union of the sets is given by n ) A^UA2 U A3... U As j and is some natural number ; hence the product rs is also a natural number. This means that the set N is closed for the operation of multiplication.

Thus if r, s t N, then rs t N.

( ii ) Commutative and Associative Properties in N. Let there be J sets, say, At, A2, A3...AS, any two of them being disjoint and n ( A\) =/? ( A2 ) = ...=«( As) =r.

Hence the set, say, C = ] At U A2 U A3 U As ( contains rs elements. From these, we can form the sets

Bu B2, B3,...Br, any two being disjoint and n(Bi) = n(B2) = ...=n(Br) = s.

Hence the set, say, D = B^ U B2 U B3 • • • • U Br contains sr elements.

Since Au A2, ..., As are disjoint and £u B2,...Br are also disjoint and B - sets are formed from the union set C, the union sets D and C contain the same number of elements and hence we must have rs = sr\ and this establishes the commutative property in N.

Let, as usual, r = n ( A ), s = n ( B), r = n ( C ) .

By definition, ( r j ) t = « | ( ^ U 5 ) U C j = n ) A U ( f i U C ) ( = r(st).

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This establishes the associative property in N. For example, ( 3-5 ) -9 = 15-9 = 135;

and 3- ( 5-9) = 3-45 = 135 and hence ( 3-5 )-9 = 3 ( 5 - 9 ) and. this verifies the associative property.

( i i i ) Distributive Property in N. Multiplication is dis-tributive over addition and we can show that

r ( s + t) = rs + rt if r, s, t e N. ( iv ) Identity Element in N. If r e N, we know that

r X 1 = r = 1 X r. Thus, if any natural number is multiplied by 1, the product

is the natural number itself. Hence unity i. e. 1 is called an identity element for multiplication.

1-3. Order in N. If we are given two different natural numbers r and s and it is known that there is another natural number t such that r + t - s, then we say that s is greater than r and write it as s > r or that r is less than s and write it as r < s. We can thus see that

( i ) for two different natural numbers r and s we have either r > 5 or r < 5.

If A = ) 1, 2. 3, ..., r then n ( A ) = r, B = ) 1,2, 3, , 5 then n ( B) = s, C = ) 1, 2, 3. ..., t j , then « ( C ) =

then ( 1 ) if r < 5, A <= B and (2) if 5 < t, B c c . It is clear that in the process of counting 1, 2, 3 , . . . any number is less than its following and greater than its preceding.

All natural numbers being positive integers, there is no natural number less than one. Hence 1 is the smallest natural number and it has no predecessor but it has a successor ; every other natural number has a unique ( i e. one and only o n e ) predecessor and a unique successer.

If q, r, s, t e N. then we also have ( i i ) i f r > 5 , r + t>s + t, ( i i i ) if J">s, rt > st, ( i v ) i f r > s and tyq, r-\-t>s + q, ( v ) if r > 5 and tyq, rt>sq. These are known as order relations in N.


2. The Number Corresponding to an Empty Set. Let Ax = { 1 \,A2 = ) 1, 2 ),A3 = M- 2, 3 (, ,4, = } 1, 2,...r j so that » ( ) = 1, n ( /42 ) = 2 , n (A3) = 3,..., n = r.

Thus, a natural number r can be assigned to every finite non-empty set by one-one correspondence with the set. No natural number is associated with the empty set 0. We, therefore associate the number zero with the empty set and have the relation n ( 0 ) = 0.

The system of numbers is thus enlarged to the set M J ) 0 ( = jO, 1,2, 3 {.

2-1. Operations in the Set NU J 0 (. The operations of addition and multiplication are performed in the set N U ) 0 j by the following rules.

( i ) If r and s are non-zero elements of NU ) 0 (, then r + 5 = s + r and rs = sr.

( ii ) If r is a natural number or zero, then r + 0 = 0 + r = r .

( i i i ) If r is a natural number or zero, then r x 0 = 0 x r = 0 .

From ( i i ) , we observe that if zero is added to the element of NU J 0 (, then the result is the same element. The number zero, therefore, is called the identity element for addition.

From ( iii ), we see that r X 0 = 0 y r eN U ) 0 ( .

[ The symbol ' v ' looking like an inverted A means ' for all ]

Hence, even if r ^ s, r x 0 = 0 = s x 0 and therefore we can not cancel zero from the relation r x 0 = s X 0.

Thus, the concellation law " r x s = rt => s = t" is valid in the set N U \ 0 j only if r 0.

We can also see that the set N U ) 0 j is closed for the opera-tions of addition and multiplication and that these operations are commutative and associative We can also see that the distri-butive property of multiplication over the sum is also true in this set. It may also be noted that the set N U ) 0 { has both the identities-the additive identity " z e r o " and the multiplicative identity " one ", ( also called unity ).

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3. The Integers.

The set N U ) 0 j is closed for the operations of addition and multiplication. This means that if r and se NU ) 0 then we can always find the elements, t and u e NU ) 0 $ such that r + s = t and rs = u. If s = 0, then r = t and if s # 0, then r < t. Conversely, if we are given two elements, r and t e NU such that r = t or r < t ( written as r < t ) then we can find an element s e NU } 0 such that r + s = t. In this case we say that it is possible to subtract r from t and that s is the difference t — r, written as 5 = t —r. Thus, subtraction is the operation inverse to that of addition. We can, however, observe that the operation of subtraction is not closed for the set N U } 0 J, since we can not find an element s e NU } 0 f such that s=t — r if r>t. We, therefore, enlarge the number system by introducing a new set of numbers. This is done as follows. Let r e N, then we know that there is no element zNU J 0 j, which when added to r gives zero. We now assume that there is such an element ( or a number ) denoted by — r such that r + ( — r ) = 0. Thus for each element se N we have an element — s. Thus we have now two equivalent sets

) 1, 2 ,3, ,n,... | a n d ) - 1 , - 2 , - 3 , , - » , . . . Since we denote the first set by N the second set will be

denoted by N. The elements of N are called negative integers since the elements of N are called positive integers.

The set consisting of negative integers, zero and positive integers is called the set nf imagers and we shall represent it by

Thus we have J= N U J 0 $ UN. Obviously JVC J and.

N^J. If r and t are two natural numbers ( positive integers ) such

that r > t, then the difference r — t is a positive integer.

In this case the difference t — r can be defined as the negative integer — (r — t).

Addition in the set J .

We define the operations in J so as to make them consistent with the operations in A^U ) 0 j and N.

C. A—•


Thus, we define addit ion in J in such a way that it remains commutative and associative and that the number zero remains as the additive identity. We should remember that if teN, then the negative of — / is t i . e. — ( — t) = t.

If r e J, s e J we now define the sum r + s. ( i ) If reN U seN\J J 0 r + s is already

defined.

( ii ) If r e N U j 0 j, 5 e N, then - s e JV and r + 5 is defined as the difference r — ( — s).

( i i i ) If r e N and 5 e N U \ 0 then - r e N and r + s is is defined as — ( — r ) + s.

( i v ) If reN and s e N, then - r e N , - s e N and r + s = - ( - r ) - ( - 5 ) i . e . - [ ( - r ) + ( - s ) ] .

With these definitions it can be seen that the sum and dif-ference of two integers is an integer ( closure property ) and that the addition is commutative and associative with zero as the identity element.

Multiplication in J. We define multiplication in J in such a way that the

multiplication is commutative and associative in the set J and the set J is closed with respect to multiplication. For these reasons, if r and s s J, we define the product r X s as follows :

( i ) I f r , s e J V U ) 0 { , then r x s is already defined. ( i i ) If r = 0 ands e •/ then r X s = 0=s Xr,

( i i i ) If r e Wand s e N, then r X s = - [ r X ( - s ) ] .

similarly if s s N and r s N, then rXs <= -[(—r)Xs].

( iv ) If r t N and s s N, then r X j = ( — r ) X ( — J ) . The product defined in this way has all the required properties

stated above. We can also see that unity ( i. e. 1 ) is the identity element for multiplication in the set J. We also observe that, as in the case of N, the distributive property of multiplication over addition holds in / .

Order in J. If r and s s / , then we have either (i) r - s = 0 or (ii) r - seN

or ( iii ) r-s s N. In these cases we say that ( i ) r = s or ( ii ) r>s or ( i i i ) r<s respectively.

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With this definition of order, the set J can be described as J = \ ... , - ( n + I ) , - n, - 1, 0, 1 n,

< n + 1 ),... The element which is to the left of another element is less than it. Thus, - 2 0 < - 5 , - 2 < - 1 , - 30<6 etc.

For r, 5, t, e J the order relation satisfies the following properties :

( i ) if r < s, then r + t < s + t. ( ii ) If r < s and 0 < t, then r X t < s x ( i i i ) If r < s, and t = 0, then r X t — s X t. ( iv ) If r < 8 and t < 0, then r x t > s x t. 4. Rational Numbers. If r, s, t e J and r # 0 and r x s =t, then we define the

number s as the quotient of t and r and write it as s = -y or

s — t + r. This operation of finding the quotient is called division. Just as subtraction is the inverse operation of addition, we may consider division as the inverse operation of that of multiplication. The set N is closed for the operation of addition and multiplication but not for subtraction. In order to generalize the process of subtraction we extended the number system from N to J. However the set J is closed for the opera-tions of addition, multiplication and subtraction but not for division. For, although 1 5 - 4 - 3 = 5, ( - 28) -s- 7 = - 4, ( — 18 ) -7- ( — 6 ) = 3, we do not have any no- re J such that 5 -r 3 = r o r ( - 2 0 ) - i - 7 = r . We, therefore extend the number system to generalize the operation of division and form a new set of numbers, in which division is closed except in the case of division by zero. The new set is of numbers of the type p/q, where p,qef, such a numberpjq , where and q # 0 is called a fraction or a rational number. We assume that the rational number pjq is always given by the equation q (piq) = p, q * 0.

While considering the operation of division of p by q we have always to exclude the case in which q = 0; for as we have seen in para 2 above, 5 x 0 = 0, ( - 18 ) X 0 = 0 etc., and thus if the operation of multiplication is to be consistent, we cannot have an appropriate meaning for pjq when q — 0; for, by the definition of division, "we will get, in the above illustrations, 5 = 0 /0and; ( - 18 ) = 0/0 or 5 = - 18 which is absurd.


We DOW define a rational number.

Definition. A number which can be expressed in the form

~ ~ , where p and q e J and q 0 is caljed a rational number.

We shall denote the set of all rational numbers by Q. If

p = 0, then -£- = 0 and if q = 1, A = p q q

Thus the set of rational numbers includes the number zero as well as + ve and — ve integers; i. e. 0e Q and NC1Q,

Nc. Q. The rational number S - is usually written so as to q

have a positive sign for q.

— 2 2

Thus - 4, —^— , 3, - j etc. are some rational numbers

We thus have, JclQ.

Two rational numbers — and — are said to be equal if q s ps — qr = 0.

2 8

For example, 3 " = j y s i n c e 2 x 12 - 3 x 8 = 0.

Addition in the set Q. Let — and — be two rational numbers so that q s

p, q,r, s e J and q # 0 s and hence qs # 0.

".* P. <7, r, s e J, the numbers ps and qr and their sum ps + qr are all integers since J is closed for addition and multiplication.

Ps ^ r is a rational number. This is defined as the qs

sum of rational numbers — and —• q s

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We see therefore, that the set Q is closed for the operation of addition and we have

— + 1. = p s V q s qs *

the number on the right hand side of the equation being also a rational number.

Multiplication in Q.

If pjq and rjs are two rational numbers then pr and qs are integers and qs 0.

— is also a rational number This number is said to qs be the product of plq and rjs. Thus the set Q of rational num-bers is closed for the operation of multiplication defined as

S- x = q s qs

Properties of Rational Numbers.

By using the definitions and properties of integers it is now possible to state in brief the properties possessed by the set Q of rational numbers. Since similar properties have been discussed once for N and secondly for J, we propose only to "state them and the students can verify them for themselves.

Let a = plq, b = r/s where p, q, r, s,e J and q=h 0, s 4* 0 be two rational numbers.

I. Addition.

( i ) Closure property : If a and b are rational numbers, then a + b is a unique ( i. e. one and only one ) rational number.

( ii ) Addition is commutative : a + b = b + a. ( i i i ) Addition is associative : ( a + 6) + c = a + ( 6 + c). ( i v ) Additive Identity : The number zero is additive

identity i. e. a + 0 = a . ( v ) Additive Inverse : For every rational number a there

is a rational number ( — a ) such that a + ( — a ) = 0 . — a is called the additive inverse of a.


( v i ) Cancellation law for addition . If a b, c, are rational numbers such that a -f-c = b + c, then a — b

II. Multiplication.

( i ) Closure Property : If a and b are rational numbers, then a x b or ab is a unique rational number.

( i i ) Multiplication is commutative : ab = ba. ( i i i ) Multiplication is associative : (ab) c—a (6c). ( iv ) Multiplicative Identity : The number 1 which belongs

to Q is such that a X 1 =a y ae Q; 1 is called the multiplica-tive identity. -

( v ) Multiplicative Inverse : To every rational number a ( 0 ), there is a rational number 1 /a such that

ax = 1. Ma is called the multiplicative inverse of a.

It should be noted that zero has no multiplicative inverse ( v i ) Multiplication is distributive with respect to addition

( and subtraction ) : a { b + c ) = ab + ac. (vii) Cancellation Law for multiplication: If ac = be

then a=Z> iff

III. Order.

Let a, b, c, be rational numbers and a = p/q, b = r/s where p,q,r,sej and q # 0 # s.

and <7>0, s > 0

then a — b = ^ — 2 1 . qs We say that a=.b, a>b or a<b according as ps — qr — 0, > 0 , or <0 . It should be noted that a may be equal to b although

pj=r and q=£s. For example,

L = as 3 x 2 0 - 4 x 1 5 = 0 .

( i ) Order relation is transitive : If a > b, and bye then a > c.

Similar properties hold for = and < relations.

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( i i ) If a > b, then a + c > b + c and similarly for = and , < relations.

(iii) If a > b, ac > be if c > 0 and ac < be if c < 0.

IV. The product of a rational number a and zero is zero. As a result of this property we have if ab = 0, then either a — 0 or b = 0.

V. Denseness.

If a and b are distinct rational numbers; then the number

a b is a rational number lying between a and b~

We give below, a proof for this result.

Let a > b. We have a — ° "t = a , b > 0, a > fc.

Similarly, ~ b = " ^ T ^ > V a > h '

a>«±±>b.

Similarly if a < b, we can prove that

a + b , " a < 2 <

Hence between any two rational numbers we can always find a third rational number.

If we write ax = a b , we have a > at> b, assuming that

a > b. We can similarly find rational numbers a2 and a3

between a and a\ and between a\ and b\ and we, then, have a > a2 > ai > a3 > b.

As this process can be repeated as many times as we like, we can get an unlimited group of rational numbers ( o r infinitely many rational number s ) between a and b. We express this property by saying that the set Q of rational numbers is dense.


Remarks :

1. The operation of subtraction is not commutative, for in general a — b 4 b — a. For example 7 — 3 =j= 3 — 7.

2. Similarly the operation of division is not commutative; for a-i-b 4 b + a in general. For example 8/5 4=- 5/8.

3. The operation of addition is not distributive over multiplication; for 3 + ( 5 x 7) = £ ( 3 + 5 ) X ( 3 + 7 ) .

Eiampla : Obtain s o y three rational numbers between - 1/2 and 1/3.

i s one rational number between - 1/2 and 1/3.

Again - J - ( - - J - - = ( - j - ) — i f !

and - f ( - -/2- + - } - ) = \ = -J - • Thus the rational numbers — and are between - 1/2 and - 1/12

24 8 and between - 1/12 and 1/3 respectively.

Hence - 1/12, - 7/24, 1/8 are three rational numbers between* - 1/2 and lj3. \

Alternatively, we may write

- 1 - 1 2 . 1 8 — - ^ - a n d — = —

and hence . . . , — , are a few rational numbers between 24 t4 24 - 1/2 and 1/3 »

Or we may write

- = - -5 and = -3333

and hence - -4, - -3. - -2, -1, -2, -3 are also some rational numbers between - 1/2 and 1/3.

5. Representation of Rational Numbers by Points on a Straight Line.

Let X'OX be a straight line. Let a point O on this line be taken as origin. Let A, B, C, ... be points at distances of

REAL NUMBERS : 5 7

1,2, 3, ... units and to the right of O. Similarly let the points A', B\ C", ... be at distances of 1, 2, 3, ... units, and to the left

B ' K> A R B " 2 ^ +1 ^ / n + 2 X

Fig. 24

of <9. Then the points O, A, B, C... correspond to the numbers 0, 1, 2, 3, ... and the points A', B', C' ... similarly correspond to -1, -2 , -3,. Similarly, if the point P is the mid-point of BC, it corresponds to the rational number 2 i or 5/2 as OP = 2J units and P is to the right of O; if Q is a point on B' C' such that B'Q = 2/5 then the point Q' corresponds to the rational number - 2f as OQ = OB' + B' Q - 2f units and Q is to the left of O. In general, to respresent any positive rational number m/n, we divide the unit segment OA into n equal parts and take off a length OR on the line equal to m such parts to the right of O. The point R would then represent the positive rational number m/n. If we cut off a length OR' from X' OX in such a way that R' is to the left of O, then the point R' would represent the negative rational number, — m/n. Thus every rational number can be represented by a point on the line X'OX. This naturally raises the question : Is every point on the line a representation of a rational "number ? We shall show in the following paragraphs that the answer to this question is in the negative.

6. Irrational Numbers.

The students are already acquainted with the simple equations of the type 2x +• 3 = 0, x2 — 5x + 4 = 0 etc. When we want to solve these equations, we obtain the roots i. e. numbers which when substituted for x, the two sides of the equation are equivalent. Thus the root of the first equation above is — 3/2 for, if we substitute — 3/2 for x in the equation we get

2 ( — 3/2) + 3 = - 3 + 3 = 0 . Similarly for the second equation, we have ( x — 1 ) ( j c — 4 ) = 0; i. e. x = I or x = 4, The roots of the equation are thus 1 • and 4. In general, for an equation of the first degree viz. ax + b = 0 where a, b e J


and a = — b/a or the rational number — bja is a roo( of the equation.

Consider now the equation x* = 2. It is shown below that

there is no rational number pjq such that there = 2.

Hence, we write the root as 2 and — yf 2 where (\[T )2 = ( - V T )2 = 2. These numbers \f~2 and - \f~~2 are called irrational numbers. In general, we can say that if a is a positive rational number, and there is no

rational number b such that (b )" = a, we write " a root of the equation x» = a " as the irrational number " Thus ty 7 , ^ 1 9 1 are irrational; however the numbers

T , -^21, $ 625 are rational as they are 2 , 3 , 5 respec-tively.

As stated above, if a is a rational number and the num-ber x satisfies the equation -xn = a, then we write x = a • The process of finding the number x or the nth root of a is called evolution A rational number x statisfying the equation x3 = 8 or x* = 16/81 can be found. But as seen above a rational number cannot be found to satisfy the equation x3 = 7; hence is not a rational number. In other words the set Q is not closed for the evolution. We therefore extend the number system and introduce " irrational numbers ".

The number \f 2 is not a rational number; the students can easily verify this by proceeding to extract the square root of 2 in decimal notation. It will be found that the value of

is an endless (non-terminating) decimal. The value of V 2 can only be written correct to some places of decimals, say, as 1-1414, 213,6 Thus the number cannot be actual ly„aadcorrect ly expressed as a terminating decimal and hence cannot be expressed "In the fogm^fiLft fraction Im/n. "Therefore the number is jyot a . j p i fona l number? "In order tP prove this rigorously, we require the following lemma.

Limma: If 2 is a factor of *»*, than 2 is a factor of the integer m. m is an integer; hence it is either an even or an odd integer.

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m — 2n or m = 2n + 1, where « is any integer. m 1 = 4K* or m* = 4 « ' + 4 * + I.

If 2 is a factor of w ' then m 1 cannot be equal to 4n* + 4» + 1 which i s not divisible by 2.

fn* = 4« s and hence m — 2n.

•'• 2 is a factor of m.

Alternatively we may argue that 2 is a factor of i. e. of m. m .

.". 2 must be a factor of either of the two factors and hence of m.

We now prove that V2 cannot be a rational number. We have 1 < 2 < 4.

1 < ^ 2 < 2.

1 and 2 are consecutive integers and sf2 lies in between them,

n/2 cannot be an integer.

If possible, letV2 = a fraction, say, mjn where m and n are integers having no common factor and n 4 0.

2 = m2/n2 m2 = 2n\ /. 2 must be a factor of TO2 and consequently of m.

4 must be a factor of m2 and consequently of In2. 2 must be a factor of «2 and consequently of n.

.'. 2 is a factor common to both m and n. " This contradicts our supposition that m and n have no

common factor. V2 cannot be a fraction of the form m/n.

cannot be a rational number.

Similarly, it can be seen that 'sT^T $ 1» ^ 1 1 are numbers which cannot be expressed as the ratio of two integers such as mjn and hence are not rational numbers. Such numbers are called irrational numbers.

7. Correspondence of irrational numbers with points on a straight line.

At the end of paragrah 5, we had raised the question : " Is every point on the line X'OX a representation of a rational number ?" The answer is in the negative. Although the system of


rational numbers is dense — i. e. although we can find infinitely many rational numbers between any two given rational numbers-there are gaps in this system. We will illustrate this in this paragraph .

The figure of paragraph 5 is shown again here. Let us draw a right, angled triangle OAM having AM = OA = 1 unit and having a right angle at A. Then by the Pythagoras theorem, we have

OM2 = OA2 + AM2 = 1 + 1 i e. 2.

Therefore OM = \[2 units.

Let a circle with O as centre and radius = OM, cut X'OX in points P' and P.

Then OP = OM =\[2, OP' = - s f l . Thus the points P' and P on the line X'OX correspond to the numbers — \[2 and \[2 respectively. Similarly any irrational number of the type y[2 can be represented by some point on the straight line X'OX. This shows that the gaps in the system of rational numbers are filled by irrational numbers. We therefore have the following result.

A number which cannot be rep resented as ratio of two integers alb, ( fc =£ 0 ), but which corresponds to some point on the number axis is an irrational number.

The irrational numbers are of two kinds. The irrational numbers of the types j l^ V 1 + \ /3 are the roots of the algebraic equations x2 — 2 = 0 and x* — 2x2 — 2 = 0 . Such

REAL NUMBERS : 6 1

irrational numbers which are the roots of equations are called! algebraic irrational numbers. The number 7r, which the students know, as the ratio of the circumference oTany circle to its diameter is another type of irrational number. The value of 7r which the students take as 22/7 or 3-14 is only its approximate value. Attempts which were made to find the value of 7r failed to express it either as a terminating or as a recurring decimal. The number e which is taken as the base of Napierian logarithms (Refer to § 12 Chapter 5) , is also an irrational number. The numbers such as -n and e which are not the roots of equations and are irrational, are called Transcendental irrational numbers.

The following illustration will now show how irrational numbers between any two rational numbers can be found.

Illustration. State three irrational numbers between - 5 and — 4.

Let us write — 5 = - V 25 and — 4 = — V l 6 . We now see that

- V24, - V23 , — V l 7 are irrational numbers between - 5 and - 4. In (act if N is any positive rational number, which is not a perfect square of any rational number and which lies between 16 and 25, then -for different values of N will give the Irrational numbers lying between - 5 and - 4.

Enmfln : 1. Express the natural number 3 in the form of a rational nu mber. 2 . If a and b are two rational numbers, prove that a + 6 is also a

rational number. [ Hint : Take a = m/tt and b = plq where m, n, p, q are integers and

« , 5 4= 0. Use also the fact that the sum and product of two integers is also an integer. 3

Further show that a - b, ab, a/b ( b =)= 0 ) are also rational numbers.

3 If m and n are any positive rational numbers, show that 2 m + 3 * m + n

is a rational number between 2 and 3. [ H int : S ince m and n are rational, tn + tt, 2m + 3n and hence

a r e r a t j o a a | n u m b e r s . Further observe that m + n

2m + 3« „ , , 2m + in . 2 and 3 ; are both positive. ]

m + » m + n

4 . State the smallest integer greater than 6 and the greatest integer greater than 6.


[ Smallest integer greater than 6 is 7. Observe that the greatest integer greater than 6 can not be given. Every integer is followed by the next greater integer and hence the system of integers is endless ]

5 . State ( i ) three rational numbers between - 2 and — 1 ; ( i i ) three irrational numbers between — 2 and - 1.

6 . The numbers 5 + V 3 and 5 - -J 3 are irrational.

I s their snm also an irrational number ?

Is their product an irrational number ? 7 . State two irrational numbers whose sum is a rational number.

8 . State two irrational numbers whose sum is an irrational number.

9 . Show that J i + V 5 is not a rational number- ( Assume that V l 5 is not a rational number) .

[ Let if possible V3 + V 5 be a rational number equal to m/n , where m and n are integers and 0.

by squaring, we get 3 + 5 + 2 «/ l5 = I/.'/M*.

.'• V l 5 = • , — which is a rational number. We thus arrive at an 2n absurd conclusion.

Ji + <J 5 cannot be rational number. ]

1 0 . If a is rational and Jb is irrational, show that a + V b i s not a rational number.

[ If possible, let a + -J b = a rational number.

/ - m m p b — « = —; where

n n q q

... j j = 2SJL2* . nq

But the result is absurd since I. h. s. is irrational and r. h. s. i s rational since in, », p, q are alt integers where n, q ^ 0. Hence the given result. ]

8. Approximation of Irrational Numbers by Rational Numbers. As will be seen in paragraph 9, the real numbers form an

ordered set. AU the real numbers can be represented in decimal notation? The rational numbers form either termi-nating o7 recurring ( i. e. same digils being repeated ) decimals; the" irrational numbers form non-terminating and non-recurring ( infinite ) decimals.

Thus ( 1 ) the rational number 2/5 = 0-4, a terminating decimal; ( 2 ) the rational number 1/3 = 0-3333 ... a recurring

REAL NUMBERS : 6 3

decimal ( denoted as 0-3 ); and ( 3 ) the irrational number\(~2 = 1*414.216, ... a non-terminating, aon-recurring decimal.

Elementary arithmetical methods might be used to find an approximate value of an algebraic irrational number. Thus the method of extracting the square root of a number enables us to know the value of say a (especially when a is a positive integer and not a perfect square) correct to any number of places of decimals. In addition to the methods suggested in illustrative examples given below, logarithmic tables can be conveniently used for finding approximate values of irrational numbers. Use of the Logarithmic Tables will be explained in the chapter on Logarithms.

Ex. 1. Find tbe value o f V 5 7 correct to three places of decimals.

145 800 5 725

1504 7500 4 6016

15089 148400 9 135801

150988f 1259900 'f

Hence the value of «/57 correct to three places of decimals is 7-550.

Ex. 2 . Obtain an approximate (rational va lue ) of \ /2 .

By using method of finding the square root, as is done in Ex am pie above we can find the approximate value to any number of places of decimals One may however find the value by inspection and successive trials as indi-cated below. W e have 1 < 2 < 4. [ Considering the preceeding and following perfect squares ] .

L < V2 < 2. ( i )

Again taking the numbers 1-1, 1-2, 1-3, • • • we see that ( 1-1 )» = 1-21 < 2, ( 1 - 2 ) 1 = 1-44 < 2, (1-3)® = 1.69 < 2 , (I-4)5 = 1-96 < 2. (1-5)» = 2-25 > 2.

1*4 < « / ? < 1-5. ... . . . ... C«)

Again taking the numbers 1-41, 1-42, 1-43

-we see that

( i - 4 1 ) 1 = 1-0881 < 2. (1*42)> = 2-0164 > 2


1-41 < J 2 < 1-42. ... iii>

Proceeding in this way we can find better and better approximations to the value of V2.

In a similar way rational approximate values corresponding to the numbers V3~, v r etc. can be found. W e illustrate below by one more example.

E*. 3. Find the value of V l 1 correct to three places of decimals.

By successive trials we can verify that

23 < 11 < 3 s ;

(2-2 )3 = 10-64 < 11 < 12-16 = (2-3 )3 ; ( 2-22 )3 = 10-94 < 11 < 11-09= (.2-23 Y ; (2-223 )s = 10-99 < 11 < 11-01 < (2-224)®.

We also have

( 2-2239 )8 = 10-9988 and ( 2-2240 )8 = 11-0003

Hence the value of v n correct to three places of decimals is 2-224.

Ex. 4 . Find the value (correct to two places of decimals) of

As Sn Ex. 1 aboveV57 = 7-550 approximately. H e n c e "J 9 + V 5 7 = + 7.550 = Vl6-550. If we extract the square root of 16-550 as in Ex. 1

above we get the required value as 4-068, i. e. 4-07 correct to two places of decimals; or we may work as in Ex. 2 above and have the following figures :

4 » = 16 ; ( 4 - 1 ) ' = 16-81 ; ( 4 - 0 5 ) J = 16-403 ; (4-06)* = 16-483 ; and ( 4 0 7 ) ' = 16 565.

Hence V9 + J57 = 4-07 correct to two places of decimals.

E*. 5 . Find two rational numbers between 2 and V H -

As in Ex. 3 above 3 / u = 2-224. Hence the numbers 2-10, 2 -15 , 2-20 lie

between 2 and v n . They can be written as 210 215 220 21 43 U_ 100 ' 100 ' 100 ° r a S 10 ' 20 ' 5

and are therefore rational.

9. Real Numbers. We have considered the irrational numbers in the para-

graph 8 above; we shall denote the set of all the irrational numbers by Q t .

The set of all rational and irrational numbers is called the set of real numbers. In other wards every real number is either rational or irrational.

r e a l n u m b e r s X 65>

We shall denote this set of real numbers by R; thus, Q U = R. We can thus write e R, -^TT <= R, n e R, 27/11 e R, — 3/4 e R and so on. It may also be observed that Q C R and Qx c R. As N c Q, N c r: similarly J C R.

We can define the operations ot addition, subtraction multiplication and division ( except by zero ) in the set R in a suitable manner so as to make them consistent with the opera-tions in N, J and Q. These definitions therefore must be such that in the set of R ( 1 ) there exists a closure property, ( 2 ) addition and multiplication are distributive and ( 3 ) multi-plication is distributive over addition. In other words the set of real numbers possesses all the properties enumerated earlier for iV, J and Q.

Thus the numbers 0 and 1 are the identity elements for addition and multiplication in the set R of real numbers also. With the exception of zero, every real uumber has its additive and multiplicative inverse. The additive inverse of zero is zero and it has no multiplicative inverse. The set R is closed for addi-tion, subtration, multiplication, involution ( i - e. raising the number to any power), evolution ( i . e. extracting the root of a number ) and division ( except division by zero ) .

We can consider the order in R by representing them on the number axis.

Fig. 26

Order in R. We can see that every real number—rational or irrational—can be represented by a point on the number axis X' O X; conversely every point on X'OX corresponds to a real number—rational or irrational. The straight line X'OX is therefore called axis of real numbers.

If a and b are any two real numbers and the points corre-sponding to a and b on X'OX are R and S we say that a > b,

C. A—5


a = b or a < b according as R is to the right of S, R coincides with S or R is to the left of S. Thus, in the above figure 3 > \(T as the point C representing the number 3 is to the right of P representing the number \ [ 2 ; similarly — 1 < 2 as A' is to the left of P. Following this procedure we can show that the properties of order for rational numbers, given in paragraph ( 4 III ) hold also for real numbers. Thus the set of real numbers is an ordered set.

We now summarize the fundamental properties of Real Numbers to emphasize their basic importance and to enable the students to remember them.

I. Addition. If a and b are real numbers then a + b ( called the addition ) is a unique real number.

Addition obeys the following rules :

( 1 ) Addi t ion 's commutative : a + b = b + a. Thus, the order of addition is not material.

( 2 ) Addition is associative : ( a + Z>) + c = a+ ( b + c ) .

Thus, in the addition of more than two numbers by the repeated use of the first rule, the order in which the numbers are associated is immaterial.

( 3 ) Additive Identity : The number zero belonging to the system of real numbers is such that a + 0 = a, for every real number a. Zero is called the additive identity.

( 4 ) Additive Inverse : For every real number a, there is a real number —a such that a + ( - a ) = 0 ; — a i s called the additive inverse of a.

Subtraction can be defined in terms of addition : a + { - b) = a - b.

It can be seen that subtraction is not commutative, i. e. a - b b - a.

II. Multiplication. If a and b are any two real numbers then a x b (also written as ab ) is also a unique real number. The Multiplication obeys the following rules :

( 1 ) Multiplication is commutative : ab = ba. ( 2 ) Multiplication is associative ( ab ) c = a {be)

r e a l n u m b e r s X 67>

( 3 ) Multiplicative Identity : The number I, belonging to the system of real numbers is such that a x 1 = a for every real number a . " 1 " called the multiplicative identity.

( 4 ) Multiplicative Inverse : To every real number a (aj* 0 ) , there is a real number 1 ja such that a x 1/a = 1. 1 la is called the multiplicative inverse of a. It should be noted that the number zero has no multiplicative inverse.

( 5 ) Multiplication is distributive with respect to addition: a (b + c) = ab -f ac.

Division can be defined in terms of multiplication. Thus a + b = a x lib provided b ± 0.

III. Order. If a and b are any two real numbers, then either a > b, a --- b or a < b. The system of real numbers, therefore, forms an ordered set of numbers.

( 1 ) The order relation is transitive. If a > b and b > c then a > c.

Similar relations hold for the = and < relations. ( 2 ) If a > Z> then ( i ) a + c > b + c if c is + ve or — ve;

(ii) ac > be, if c is + ve\ but ac < be, if c is — ve.

It may be noted that the above relations of-order satisfied by the Real Numbers are not obeyed by Imaginary Numbers introduced in the next article. It can also be realized that there is no greatest real number since there is always a number greater than any given real number. Similarly there can not also be any least real number. In fact the real number system is endless on both the sides.

IV. Property of Denseness. We may also observe that between any two real numbers there always exists a third real number. For example, f (2 +\fs ) is a real number between 2 and Similarly J ( ^ 2 + ) ' s a r e a ' number between >sj2 \[T. fact> a s already explained, we can find an unlimited group of real numbers between any two given real numbers. This property is known as the property of denseness of the Real Number System. Due to this property it is not


possible to obtain the greatest or the least real number between any two given real numbers. However one can get an unlimited group of real numbers between two given real numbers.

It may also be noted that there does not exist a greatest or -a least number in the system of real numbers.

AS stated above, between any two real numbers a and b there exists a third real number, say c. Conversely any number c which satisfies all the properties of real numbers stated above and is such that a < c < b, then c is a real number. This property of real numbers is called the axiom of completeness.

It should be noted that the system rational numbers although dense does not possess this property of the axiom of completeness. For if a, b are any two rational numbers there exists a third rational number c between a and b; but a number d between a and b may not be rational ( i. e. may be irrational ).

V. In addition to the above rules and characteristics, the real number system has the following properties :

( i ) The product of any real number and the number zero is zero.

( ii ) If a and b are real numbers and ab — 0 then either a = 0 or b = 0.

( i i i ) The product of two real numbers, one positive and the other negative is a negative real number.

( iv ) The product of two negative real numbers is a positive real number.

( v ) The square of any real number is non-negative.

We are giving below the proofs in support of the above five properties. First year students however, are not expected to go through the proofs

; i ) Let a and b be two real numbers.

We have ab + 0 = ab = a (6 + 0) = ab + a-0

0 = a-0.

( i i ) Let a 4= 0, so that — exists.

REAL NUMBERS X 69>

Now 6 = 6 X 1

= (ab ) • — - 0 • X - = 0 . a a

Thus if a i= 0, then 6 = 0 . Similarly if 6 =4= 0 ; then a = 0.

(iii) Let a and 6 be two positive real numbers, so that a and - 6 wil l be two real numbers, one positive an d the other negative.

Assume that a is a positive integer ; then *•( - b) = ( - b)-a

= { - 6 ) + ( - 6 ) + a times = - b-a = a negative real number of magnitude ba.

( i v ) Let a and 6 be two positive real numbers so tuat - a and - 6 become two negative real numbers.

Now { — a ) ( - & ) = ( - « ) ( - 6 ) + 0 = ( - a) ( - & ) + «•<) = ( - «) { - b) + a ( - b + b) - ( - a) ( - b) + a ( - b) + ab = ( - « + «) (~b) + ab = 0 ( - b) + ab = ab.

( v ) If a is any real number then a > 0 or a < 0 or a = 0,

If a > 0, then a-a > 0. Henoe a* > 0.

If a < 0, then let b = - a so that b > 0.

Now a* = a<a = ( - 6 ) ( - 6 ) = 6-6, due to ( i v ) = 6* which is + ve since 6 > 0.

If a = 0, then a ' = a-a = 0-0 <= 0.

Example. State to which proper subsets N, J, Q or Qi of the set of rea 1 numbers the following numbers belong.

( i ) - 1 0 , ( i i ) - 3 / 2 , (Hi) 0, ( i v ) 1. ( v ) 13/5, ( v i ) W, (vii) e, (viii) VTT.

Auk, ( i ) - 10 e / a s w d l a s t o Q ; { i i ) - 3/2 e Q ; ( i i i ) O e Q ;

( i v ) l EN.J.Q; ( v ) " f Q; ( v i ) IT E Q,: ( v i i ) e f Q . i

(viii) « / H e Qi.


Students should also remember that the number zero is considered as an integer. No sign is attached to it. Further 0 x 0 is considered as zero. Again f. js considered as indeter-minate and that the division of a real number by zero has no meaning since zero has no multiplicative inverse by the axioms.

10. Imaginary Numbers.

We have seen that \J'f, V/kT ^Tl a r e examples of irrational numbers. We also know that ^ \ ( 2 5 / 4 , -fylT • • • are rational numbers. It follows that the roots of only positive numbers are either rational or irrational numbers i. e. real numbers. Consider now the square root of a negative number — 4. Since the square of any positive or negative real number

gives a positive number, therefore _ 4 cannot be a real number. Such numbers which are the roots of negative numbers are called imaginary numbers.

Thus -v/Try; \ f ~ ~ 9 , yf - 25/4 are examples of imaginary numbers. In order to have a meaning for the opera-tion of extracting the square roots in all cases we are further required to introduce imaginary numbers. We shall consider these imaginary numbers in the next Chapter.

We now classify in the following chart the several types of numbers considered so far :

Numbers

Real I

Imaginary

I Rational Irrational

L

I ± Integers Zero ± Fractions Algebraic Transcendental

( I r ra t ional )

I

REAL NUMBERS X 71>

11. Illustrative Examples.

Ex. 1. Which is greater + 3 or — 7 ? By how much ?

On the number scale + 3 is to the right of - 7.

+ 3 is greater than - 7. Starting from — 7 and then moving 10 units to the right we get the number + 3.

.". + 3 is greater than — 7 by 10.

E*. 2. Prove that 73 is not a rational number. We have 1 < 3 < 4.

1 <«/5" < 2. Now 1 and 2 are consecutive integers and *J3 lies in between them.

.'. J i cannot be an integer.

Let, if possibleV3 = mjn, where m and » are integers having no common factor.

.. 3 =- m'/w* ; ••• m* = 3«*. 3 must be a factor of m* and hence of m.

.'. 9 must be a factor of m* and hence of 3n*. 3 must be a factor of » ' and hence of n. 3 is a factor common to both m and ft.

This contradicts our supposition that m and » have no common factor.

J 3 cannot be a fraction of the form tn/n.

.'. V3, since it cannot be an integer nor a fraction, cannot be a rational number.

Ex. 3 . Is 1-3 ( i . e. 1-333 a recurring decimal) a rational nnmber? If so, express it in the form mln where m and » are integers.

Let, in the limit, * = 1 3 ( i ) 10* = 13-3 ( i i ) 9x = 12, subtracting ( i ) from ( i i ) . x = 4/3.

1-3 is a rational number and is 4/3.

[ For alternative method refer to the Chapter on Progressions].

Ex. 4 . Express 0«234 in the form mjn. Let, in the limit, x= -234 ( a recurring decimal -2, 34,34, 34 )

/. 10* = 2-34 ( i ) Also 1000* = 234-34 ( i i ) .'. 990* — 232, subtracting ( i ) from ( i i

_ 232 . 116 " * 990 495'


Ex. 5. State whether the following statements are true or false giving resons in support of your answer.

( i ) The product of two integers is a positive integer. ( i i ) The product of two equal integers is a positive integer. (iii) There is a greatest real number less than 2. ( iv) There is a greatest integer less than 2. ( v ) There is a least integer greater than 2. ( vi) There is a least real number greater than 2. (vii) Zero is a positive integer.

(viii) — and -325 are both rational numbers

Ans. ( i ) False. The product of two integers, one positive and the other negative, is a negative integer.

( ii ) False. The number zero is considered as an integer. We further have the result 0 x 0 = 0. Here the product of two equal integers is an integer ; but it is not positive since no sign is attached to the number zero.

( i i i ) False. It is not possible to state the greatest real number less than 2, since the system is dense.

( iv ) True. 1 is this integer. ( v ) True. 3 is this integer. ( v i ) False. No such real number exists. (v i i ) False. Zero is an integer. Numbers greater than zero are positive

and those less than zero are negative. No sign is attached to the number zero

(viiil False. is not a rational number, as 7T is irrational. M

However -325 i. e. 325/1000 is rational.

Exercise 2

1. Which is greater + 5 or — 7 ? By how much ? 2. What are the following numbers in the number scale ?

( i ) 10th (un i t ) number before zero. ( ii ) 8th ( unit ) number after zero.

3. Define a rational number and show that \ [ 7 is not a rational number. Represent it by a point on the number axis.

4. Prove that s f T l is an irrational number and find a point representing it on the number scale.

5. Explain why division by zero is meaningless. 6. Explain why % is indeterminate.

REAL NUMBERS X 73>

7. Show that V 2 + \f 5 is not a rational number, assuming that sf 10 is not a rational number.

8. Is 1-2 ( i. e. a recurring decimal 1-2222 ) a rational number ? If so, express it in the form mjn.

9. Is 1-6 a rational number? If so, express it in the form mjn.

10. Express the following recurring decimals in the form m/n:

( i ) 5-23, ( i i ) -234, (iii ) 1-125.

11. Express 1/7 in the decimal notation. Calculate to 14 decimal places and show that it is a recurring decimal.

12. Express \[~5~ in decimal notation. Is\j~5 a terminating decimal ? Can you express yf 5 in the form mjn, where m and n are integers ? Hence state whether it is a rational or irrational number. Represent it by a point on the number axis.

13. Is the set of natural numbers closed under subtraction ? 14. Is the set of real numbers closed under evolution ? 15. Does the associative law hold for subtraction ? 16. Does the commutative law hold for division of rational

numbers ? 17. What is the additive inverse of — \f 2 ? State also its

multiplicative inverse.

18. If a is rational and sf b is irrational, show that a — yf b is not a rational number.

19. If a, b, c, d are real numbers, such that a > b and c > d show that ( i ) a + c > b + d, (ii) ac > bd, if a, b, c, d are all positive and ac < bd, if a, b, c, d, are all negative.

20. Show that if a and b are unequal real numbers, then ( i ) ( a + b ) 2 > 4 ab, ( i i ) a2 - 6a + 14 > 0. [ Use the fact that square of a real number is non-negative. ]

21. State whether the following numbers are rational or not and if they are rational, express them in the usual form plq

( i ) 3 . 7 2 , ( i i ) - ? ~ , ( i i i ) v / T + 5, ( i v ) -3.


5 \ f T 2 ( v ) , ( v i ) r — (vii) $ 243 , (viii)^245.

V 12 5 - ^27

22. Give ( i ) three rational and (ii) three irrational numbers which are greater than 3 and less than 5.

23. State giving reasons whether the following statements are true or false : ( i ) Every positive integer is a rational number. ( ii ) Every negative number is a rational number. ( i i i ) Every real number is a rational number. ( iv ) Every integer is a natural number. ( v ) a; is real where x2 + 1 = 0.

24. State how ( i ) 3/5, (ii) - 7/5 and (iii) a rational number pfq can be represented by points on the number axis.

25. ( i ) Between n and ( n + 1 ) there is no natural number. Hence (. n + 1 ) is called the successor of it and n, a predecessor of ( n + 1 ). If n is a natural number does it always have a predecessor ? ( ii ) Show that between any two rational numbers there is at least one rational. Hence show that no rational number has a successor. ( i i i ) Is there a first / last integer ? Does every integer have a successor and a predecessor ? ( iv ) Is there a first natural number / a last natural number ? ( v ) Give examples o f :

( 1 ) two rational numbers between 1 and 3/2, ( 2 ) two irrational numbers between — 8 and — 3, ( 3 ) two rational numbers between \[ 2 and 7 , ( 4 ) two natural numbers m and n such that m — n

is a natural number but mjn is not a natural number.

26. Rewrite the following statements, duly corrected where required. ( i ) Any fraction is an irrational number. ( i i ) Since ir = 22/7, ir is a rational number.

REAL NUMBERS X 75>

( i i i ) The smallest real number greater than 6 is 6-1.

( i v ) , , are consecutive rational numbers.

( v ) 4-2 is the only rational number between 4-1 and 4-3. 27. Answer the following as required :

( i ) State the smallest real number greater than 5. ( ii ) State the greatest real number less than 5. ( i i i ) State the smallest integer greater than 5 and the

greatest integer greater than 5. ( i v ) If m and n are positive rational numbers, show

than — is a rational number between m + n

5 and 6. ( v ) If m is any positive real number, show that

m + 1/m can never be less than 2. ( vi) State the property of real numbers from which it

follows that the roots of x2 + 4 = 0 are not real numbers.

(vii) State three real numbers between 1 and 2.

28. ( i ) Obtain two rational numbers and" one irrational

number between 2 and 7.

( ii ) Show that 2 ( f 37 - 1 ) lies between 2 and 3.

(iii) Show that the number 3 + 2 - ^ 5 lies between 6 and 7.

29. State whether the following statements are true or false. If a statement is true, prove it ; if you consider a state-ment to be false, give an example in support of your answer.

( i ) The product of two rational numbers is rational. ( i i ) The sum of two irrational numbers is irrational. (iii) The product of two odd integers is an odd integer. ( iv) If x is less than y then .vJ is less than y1. ( v ) Given any real number x, we can find a real

number y sueh that xy = 1.


30. ( a ) If R, N, N, J, Q, Qu C denote the sets of real, natural, negative integral, integral, rational, irrational and complex numbers state which of the following statements are correct.

( i ) 9eQ, ( i i ) - 12eJV, ( i i i ) 0 e Q, ( i v ) i r e Q

( v ) Q> (v i ) - - ^ - e j v U j 0 { UJV, (vii; - j eQi .

( b ) If A = J 1 , 3 . 5 , 7 , . . . 5 = J 2, 4, 6, f, find whether the sets A and B are closed under the operations of addition and multiplication.

Answers. Exercise 2.

1 . 5, 12. 2. ( i ) - 10. ( i i ) + S.

5. We have the property that there is no multiplicative inverse for the number zero. However, if we assume that the multiplicative inverse of zero exists we come to absurd conclusions as shown below in this and the next example and hence division by zero excluded.

a gf Let 6 = 0 in - y • Let - y = x so that a — bx i. e. a 0-*. ... ( i )

Now R. H. S of ( i ) is equal to zero for every x.

equation ( i ) is impossible if a. 0.

If, now a = 0, then ( i ) is satisfied by every value of x.

Thus, if 6 = 0, then the operation is either impossible or indeterminate,

division by zero has no meaning.

6. We have 2 x 0 = 0, 3 x 0 = 0, 4 x 0 = 0.

0 „ 0 „ 0 , - "T = 2' "cr=4-o

•'• — has no definite meaning and hence is indeterminate

a 11 n 5 , „ . . . 5 1 8 . . . 232 1124 8. 9. — • 10. (,) - , m ) j f i . (mi —

11. 0142857,142857,14

12. V5 = 2-23607 It is not a terminating decimal, no, irrational. t3 . No. 14 . No. 1 5 . No. 1 6 . No.

17. JT, 2 ) . The numbers . ( i i i ) , ( v i ) and ( vi i i ) are «Ot V 2 _

rational. 2 2 . ( i ) 7/2, 4, 9/2 ; ( H ) « / l 0 . VTl and / l 7 -

REAL NUMBERS X 77>

False. Sign only does not determine the rational number; Number such as - J 2 , - V3 are negative but irrational F^ise. Real number may be irrational also ; False. Natural numbers are only + ve integers : False. Square of a real number is never negative. N o . The natural number, i has no predecessor since 0 is not a natural number, ( i i ) It a and 6 are two rational numbeis,

— i s also a rational number. Between any two rational

numbers any number of rational numbers can be inserted and hence the result, (iii) No. Every integer has a successor and a predecessor, (iv) 1 is the first natural number. There is no last natural number, (v) 1 1-1, 1-2; (2) — VlO. - A/I3u Any numbers of the form - where p is not a perfect suqare and 9 < £ < 6 4 ; ( 3 ) 1-5, 16; ( 4 )m = 5, n = 3.

26. ( i ) Any number mjn where m, n are integers and » =t= 0 is a rational number.

(ii) Since 7r = 22/7, approximately and since its ej^act value can not be expressed as a ratio of two integers, it is not a rational number.

(iii) The smallest real number greater than 6 can not be given. (iv ) 1/2, 1/3, 1/4 are not consecutive rational numbers. (v) 4-2 can not be the only rational number between 4-1 and 4-3.

27 . (i ) Smallest real number greater than 5 cannot be stated. (ii) The greatest real number less than 5 cannot be stated. (iii) Smallest integer greater than 5 is 6 ; but the greatest integer

greater than 5 can not be stated. ' 1 V ) Since m and » aTe positive rational, m + » and 5m + 6n and

hence ( 5m + 6n ) / ( m + n) are rational. 5m + 6n , 5m + 6n , , .. ... ,

Also 6 ; and ; - 5 are both positive and m + n m + n hence the result.

(v) m + — > 2 if m - 2 + — > 0 m m

i. e. if [ V m - —^ ) which is true since the square of a real

number is always non-negative. Equality sign is possible if m = 1Jm.

( vi) The square of any real number is non-negative,

(vii) - y ( 1 + V"2), + + V 2 ) J .

t { 1 + T[1 +

23. ( i i )

(iii) ( iv) l v )

2 5 . ( i )


2 8 . ( i ) We have 4 < 2 - < 7, .\ 2 < - f - < V 7 • *T 2 .*. 2*5, 2-6 are required rational numbers

- y ( 2 + V7>. - J - [ 2 + - | - ( 2 + / 7 ) J

are required irrational numbers,

( i i ) Show that 3 - 2 [ V 3 7 - 1 ] > 0 and 2 ( V37 - 1) - 2 > 0.

(iii> Show that 7 - [ 3 + 2 V J ] > 0 and 6 - [3 + 2 V 5 ] > 0.

2 9 ( i ) True. ; If p. q. r, s e J and q £ 0, s ±0;

br then p , qs, e J and hence -— € Q.

(ii) False (a + V 6 ) + (a - -Jb I = 2a, a rational number (iii) True. (2m + 1 ) ( 2« + 1) = 4 m n + 2m + 2« + 1 is odd. (iv) False. - 4 < 3 but 16 > 9. (v ) False. O x y ^ l for any y.

3 0 . ( a ) ( i ) Correct, since 9—9/1 and hence is a rational number. ( i i ) False- - 1 2 (jT N since JV contains only positive integers.

6 (iii) Correct, since 0 is a rational number — when > = 0and g=£0.

P> qc J-( i v ) False T is an irrational number, 7r e ( v ) FalSe. 7r/2 is an irrational number and hence Q.

( v i ) False. N u \ 0 [ U # = \ . . . - 3 , - 2 , - 1 , 0, 1, 2 ,3 . . . | = /

and - 3.'2 GL J. In fact y f g .

(vii) False. -3 = -3333 ... = ~ - e Q .

(b) The set A is not closed for addition but it is closed for multi-plication. The set B is closed for addition and multiplication.

Chapter 3

Complex Numbers

1. Complex numbers, 2. The imaginary unit i. 3. Algebraic operations with complex numbers 4. Properties of complex numbers. 5 Geometrical representation of a complex number. 6. Modulus and Amplitude of a complex number. 7. Illustrative Examples. Exercise 3.

1. Complex Numbers.

We have already referred to imaginary numbers as different from real numbers in Chapter 2. We will again consider them here in this chapter.

The students can easily see that the root of a linear equation of the type 3.x + 7 = 0 is a rational number. In general, the root of the equation ax + b = 0 is — bja which is a rational number. Again the roots of the equation x2 — 7 = 0 are + \jl which are irrational. In both the above cases, the roots are real. But the roots of the equation of the type x2 + 4 = 0 cannot be real. For, the equation can be written as x2 = — 4, and hence the two roots are + N/ — 4. Since squares of all real numbers, whether positive or negative, are positive, it follows that the square root of — 4 cannot be a positive or a negative real number. Numbers which satisfy equations of this type, have their squares negative, and are called imaginary numbers. Consider further the quadratic equation

x2 - 4x + 13 = 0.

We may solve this equation by the method of completing the squares as follows :

x2 - 4x + 13 = 0. .'. x2 — 4x + 4 + 9 = 0 , ( x - 2 ) 2 + 9 = 0. ( x — 2 )2 = — 9. x - 2 = ±\J~^9. X = 2±yf-^9.

79


Here, it is seen that the two roots are partly real and partly imaginary. The roots of such quadratic equations are called Complex numbers \ Thus the complex numbers are partly real

and partly imaginary. Thus it is seen that the idea of the number system has to be

extended in order to enable us to solve completely equations in general

It should be noted that the word " Imaginary " is Hot used in the usual sense of the term. We have already seen that real numbers can be taken to measure quantities and can be represented by points or in terms of line segments taken on a straight line. Since the numbers of the type \f — 4 or 2 ± V — 9 can not be taken to measure quantities in this sense, they are called imaginary numbers. The term " imaginary number" is sometimes used to denote the number of the type \ [ — 4 as well as the number of the type 2 ± V - 9, which we have called above a complex number.

2. The Imaginary Unit " i ".

The roots of the equation x2 + 4 = 0 referred to above are given by

x = + \ f ^ 4 = + \/" 4 ( - 1 ) = ±\j~4 X \ J ~ 1 = ± 2 \ f ~ l . Similarly the roots of the equation x2 -}- 7 = 0 are given by

* = = + V 7 ( - 1) = ± > / T , f ^ T . In a similar way, the square root of any negative number, say — a, ( a > 0 ) may be written as

\1 - a i. e. \f a — 1 ) i. e. \Ja s[ — 1, where \fa is a real number.

Therefore, any imaginary number can be written as a real number multiplied by \J — 1. Thus, every imaginary number can be expressed in terms of \f and hence is taken as the imaginary unit and has the characteristic property that

V ^ l X = - 1.

It is customary to use the first letter i of the word " imaginary" to denote this imaginary unit.

Thus i = — \ and i2 = - 1.

COMPLEX NUMBERS : 81

The number ai, where /, the imaginary unit, is multiplied by a real number a, is called a purely imaginary number. Thus, 7/, — 2//3, \[5i are examples of purely imaginary numbers.

A number represented by a + ib, where a and b are real is 3 4 r—

called a complex number; 2 + 3/, — — —/, + 5/',

— "v/2 + n/3. I, are examples of complex numbers.

If in the complex number a + ib, a = 0 then the complex number reduces to a purely imaginary number. Similarly if b — 0, then the complex number reduces to a real number. It can be seen that 7 + 0-/ is a real number written in the form of a complex number.

We, therefore, have the following definitions :

If a ER, b e R and b 0, then a + bi(or a + ib ) is called a complex number where i =\f - 1.

If in the complex number a + ib, a= 0, the number 0 + ib or ib is called an imaginary number.

The two complex numbers a + ib and a — ib are called conjugate complex numbers.

In the complex number a + ib, a is called its real part and b is called its imaginary part ; hence conjugate complex numbers differ in the sign of their imaginary parts.

2 + 3/ and 2 - 3 / , ( - \ f f ) + 5i and ( - \ f 3 ) - 5/ are examples of conjugate complex numbers. We will denote the •set of complex numbers by the symbol C. 3. Algebraic Operations with Complex Numbers.

With the above definitions we can assume that the complex numbers obey the usual rules of algebraic operations. Thus we may have

( 3 + 4 / ) + ( 5 + 7 i ) = ( 3 + 5) + (4 + 7 ) / = 8 + 11/. Similarly

( 7 + 8/ ) — (5 — 4/') = (7 — 5) + (8 + 4) f = 2 + 12/. We now define the algebraic operations with complex

numbers as follows : C. A.—6


Examples. 1. Find the value oi

( 2 + 3») — ( 4 - 5 i ) + ^ — * ^ . Ans. - | L

2. Find the product of ( 2 - 3i) and ( 4 + 5i ). Ans. 23 - 2/.

9 — 7 i 3. Simplify —_ ^ • Ans. 3 + ».

4. Evaluate ( * + iy ) s when * = 2 and y = - 3. Ana. - 46 - 9 i ,

5. Geometrical representation of a complex number. We have seen in chapter 2, how real numbers can be

represented by points taken on a straight line X'OX. In order to consider the representation of a complex number in a similar manner, we shall first take the simple case of the imaginary unit«'.

Let the points A and A' represent the numbers + 1 and - 1 respectively on the axis X'OX.

We know that i2 = — 1. i. e. l X t ' X i = - l , . '. the symbol i must represent an operation such that

when repeated twice on the number + 1, gives the result as — 1.

Y

B

X '

- 1 —i—

A* 0

B'-

+ 1 — t -

Fig, 27

the operation by the symbol i can be considered as the revolution through a right angle in the counterclockwise direction by convention.

COMPLEX NUMBERS : 8 5

.*. the number i can be, represented by a point B at a unit distance from O on the jp-axis namely Y'OY. Similarly the number — i can be represented by a point B' at a unit distance from O on OY'.

Therefore, in the light of the above discussion, it will be convenient to take the points on the *-axis to represent real numbers and the points on the y - axis to represent purely imaginary numbers.

We will now consider how a complex number a + ib in general can be geometrically represented ( Refer to Fig. 28 )

Let X'OX and Y'OY be the axes of reference, Let OA represent a units and AB represent b units. In order to represent a complex number a + ib, we have to take a units on OX and operate by the symbol i on b units. The operation will result in the revolution of AB through a right angle in the counterclockwise direction to the position AP. Thus the point P can be taken to represent the complex number a + ib.

P (a + ib)

Thus any complex number a + ib can be represented by a point P having co-ordinates (a, b) in the plane of axes of reference.

Many times, the line OP as well as the point P are taken to represent the complex number a + ib. Such a diagram representing the complex number by a point in the plane of the axes is known as the Argand's Diagram.


If , in the number a + ib, the real part a = 0, then the purely imaginary number bi will be represented by a point on the j'-axis, hence the j-axis is called the axis of imaginaries. Similarly if b = 0, the number becomes a real number a and is represented by a point on the x-axis and hence x-axis is called the axis of reals.

Example. Represent the following complex nambers in Argand's Diagram.

3 + 5/. - 2 + - * - « - . - - 4 / . 3 - « .

6. Modulus and amplitude of a complex number.

I. Modulus. From the Fig. 28, we have OP = \fa2 + b2.

Definition. If a + bi is a complex number, + \f a2 + b2

is called its modulus and is denoted by | a + ib |. We, therefore, have ( 1 ) | 3 + 4f [ = + + 16 = + 5.

(2 ) | «/3- 5 i | = + V 3 + 25 = + 2 J f . ( 3 ) | 5 | = | 5 + 0/ 1 = V 2 5 + 0 = + 5.

( 4 ) | - 3 | = | - 3 + 0/ | = A/9 + 0 = + 3 . ( 5 ) | — 2 | = + 2 : Also | 2 | = + 2.

In examples 3, 4 and 5, b = 0 and hence the numbers are real. We also have the moduli of the corresponding real numbers obtained in the same way.

n . Amplitude. In Figure 28, if OP makes an angle 9 with the positive direction of £-axis, then the angle 9 i» called the amplitude of the complex number P = ( a + ib).

If OP = r and XOP = 9, then from A OAP, we have OA = a = r cos 9...( i ) and AP = b = r sin 9 ... ( i i ) From ( i ) and ( i i ) , we have

r = \ja2 + b2 and 9 = tan~i (:b/a ) • • • ( iii ) We thus have the following definition.

Definition. The angle 9 which satisfies the equations

„ a a . n b b cos 9 — — = — , sin 9 =— = -r V a2 + b2 r \fa2 + b2

and — T < 9 < -n is called the amplitude of the complex number a + ib.


We can also write a + ib = r ( cos Q + i s i n # ) . Thus by writing a = r cos 9 and b = r sin0, we can find both the modulus and amplitude of the complex number a + ib.

Illustration. Find the modulus and amplitude of the complex numbers

(O 1 + (i 1 - V T + i. ( i ) Let 1 + i = r (cos 9 + i sin 9)

1 = r cos Q and 1 = r sin 9 by squaring and adding, we have t® = 2-the modulus = -J 2 •

Now the amplitude 9 is given by

1 = J 2 cos 9 and 1 = V 2 sin Q

•'• cos 9 ~ J^ a Q d sin 0 ~

the amplitude is 45° or 7t/4 radians

( i i ) Let - V 3 + i = r (cos 9 + i sin Q), — V3 = r cos 9 and 1 = r sin Q.

by squaring and adding, r8 = 4 and hence r = 2. cos 9 = — Vii" 12 and sin 9 = 1/2. the amplitude 9 is 150° or 5 TT/6 radians.

7. Illustrative examples. Ex. 1. Express as complex numbers in the form a + ib.

i m 1 .. \ 3 + 2.' . . . . . 3 + 2i 3 - 2 ( 0 7=r~. (10 r — T " ("0 +

( J ) ^ = 3 + • / 2~- t = 3 f V2~. t

3-^2-i (3 - J z - , 1 (3 • ' )

= 3 + VT- i = 3_ VT 9 + 2 11 11

3 + 2t _ (3 + 2Q ( 5 + 3i) 1 5 - 6 ) 4-t (9 + 10) ( I I > 5 - 3i~ (5 - 3i) (5 + 3») (25) — (9i®)

= 9 + l9 '̂ = jL I 19. •

25 + 9 34 + 34

3 + 2; 3 - 2» _ (3 + 2i W 2 + 5 . ) + (3 - 2H (2 - 5}M ( l l l ) 2 — 51 2 + 5i — ( 2 - 5 < ) ( 2 + 5<)

- [ ( 6 —10) + ' (4 + 1 5 ) ] + [ ( 6 — 10) + »' - 4 - 15 ) ] 14) - ( 25t*)

= [ ( ~ 4) + i ( 1 9 ) ] + [ ( - 4) + i{ - 19) ] = _ 4 + 25 29


Ex, 2. Find the square root of 6 + 8 -J-1 in the form a + b.

Let-j6 + 8i = a + ib 6 + 8/ = (oa - b') + 2 ab. ;. a* - 6 2 = 6 ( i ) and ab = 4 ( i i )

From ( i n , = ^ (i i i) a'

Substituting in ( i ) , = 6.

a* - 6a' - 16 = 0. (a' - 8) (a» + 2 ) = 0. a 1 = 8 or c? — — 2,

The value — 2 is inadmissible since a is real and hence its square must be always non-negative.

.\ a' = 8 and hence from ( i i i ) , b3 = 2.

/. J 6 + 8; = ± ( V T + « -,/2 ).

E*. 3 . Find the cube roots of unity. Show that each of its imaginary roots is the square of the other.

1I? Let * = ( ! ) ' , * s = l . *8 - 1 = 0. /. (* - 1) (*2 + * + 1) = 0.

» - l = 0 or ( ' + » + l = 0 - 1 + - 4 - 1 ± » V T

. . * = 1 or * = 2 1 e- 2 '

Thus, the three cube roots are

- 1 + »VT - 1 - iJJ 2 ' 2

If w and w' denote the imaginary cube roots,

- l + i ' V I , - i - J'VF then w = j- and w = g '

We observe that w2 = ( X - 2i'V3 - 3) i. e. — e •

Similarty tw'a = w>.

Exercise 3

1. Multiply :

( i ) 4 - 3 / by 5 + Ii.

( i i ) by +

( i i i ) 2 \ f^~3 + 3 by 4 \ J ^ 3 - S s J ' ^ .


2. Express with rational denominator :

1 3\[~2 + 2\T—5

3. Simplify and show that

, . . 3 + 2/ , 3 - 2 / . , , ( i ) r»—^ + ~ , is a rational number . 2 — 5 j 2 + 5/

. . . . n/3~ + 1 %/T , i N/2" . , , ( n ) ! — + ^ is a rational number.

•n/3 — I -S/2 n/3 + / 'V2 4. Express in the form a + the square root of

( i ) 5 - 1 2 / , ( i i ) - 11 - 60/.

5. Express in the form a +jb : . . . 3 + 5/ .... (2 + 3i)2

( D ( n ) ;

( i i i ) - i ( 9 + 6 » ) ( 2 - »)-». Represent these numbers by points in a plane.

- 1 + «V3 y : 6. Prove that ( 2— — j is a positive integer.

7. Find the values of ( i ) x4 - 5x3 + 14*2 - 21x + 25,

when x = 1 + 2 V - 1; ( i i ) x3 - 3x2 - 8* + 31, when z = 3 - 4 - 1.

8. Write the values of : ( i ) |V2 - 3/1, ( i i ) | 10 |, ( i i i ) | — 10 f -

9. Represent the complex numbers 4 + i2, 4 — il and 4 + t by points in a plane, referred to co-ordinate axes, and verify that they are collinear.

10. Find the real values of- x and y satisfying the equation ( i ) ( 2 * - 3y ) + i(x' - j 2 ) = 4. ( i i ) ( 3 + i ) * + ( 1 - 2i ) y + li = 0.

11. Show that the points representing the numbers 3t, 3 + 3i, 3 and the origin 0 are the veritices of square.


2 — i 12. Express -—;—. in the form * + iy and hence show that 1 + t

2 -

1 + i 1 + Verify this result by representing the three complex

2 ' numbers . ,2 — i and 1 + i by points on a plane

1 + i referred to the co-ordinate axes.

13. If x = k + 3t satisfies the equation x2 + 4x + 13 = 0, find the value of k.

14. If 1, a and to' are the roots of x3 — 1 = 0 , then show that ( i ) 1 + <o + a>' = 0, ( i i ) a2 = <o'.

15. Solve, by the method of completing the squares, x2 — 2x + 5 = 0 and represent its roots on the Argand's Diagram.

16. Answer the following as required : ( i ) Classify the following into real and complex

numbers: 5 + \[~-37 - 5 + V37 1 + i y f ^ T , 3 + i3.

( ii ) State why \f — 1 is not a real number. (iii) Correct the following statement, if required, giving

reasons in support of your answer. ( Assume that a and b are positive real numbers.)

s ! ^ x s f ^ b = \f ( - a ) ( - b ) ( i v ) Give examples of complex numbers whose sum

and product are real numbers. ( v ) Give examples of two complex numbers whose

product is of the term ia where a is real. ( vi) State the value of ( - i )4"+ 3 , where n is a positive

integer. (vii) Express ( x2 + y2 ) as a product of two factors.

Answers. Exercise 3 1. ( i > 41 + 13/, (ii! -13, (iii) 6 - 2 V<f.

2. ( i ) ~ ( 5 - (ii) - 1 9 - 6 VlO. 3 . ( 1 ) - = ^ . (UJ - § - -


A. ( i ) ± ( 3 - 20, ( i i) ± (5 — 6/ )-

. . . . - 9 , 19 . . . . . 2 , 29 ..... 21 12 . 5. ( . ) -13-+-13- ' - ( » ) " 3 - + - r ' - (»•) "5 T ' '

6 . 1. 7 . ( i ) 10; ( i i ) 1.

8 . ( i ) + V n ; ( i i ) 10; ( i i i ) 10. 4

10. ( i ) x = y = - 4 or x = - y — -g-, (ii) x = - 1, y = 3.

12. i - l i = JL _ J - . 1 3 . fe=_2. 1 + » 2 2

16. ( i ) - 5 + «/3 and 1 + i J - 2 are real, others are complex.

( i i ) Square of every real number is non-negative. However

( V - 1 ) (V - 1) is ia and is equal to - 1 . Hence V - 1 is not a real number.

(iii) J -a X *J - b= i *J a X.i *J b = i* *Jab = - *fab-( i v ) If a: and y are real, then x + iy and x — iy are the required

numbers. Such numbers are called conjugate complex numbers.

( v ) If x and y are real, then x — iy and y — ix are the required numbers.

(vi) = =

( vii ) (* + ! } ) ( * - iy).

Chapter 4

Indices (Exponents)

1. Index Notation. 2-3. Laws of Indices. 4. Definitions of a " for any rational number m. 5. Exponential Equations. 6- Illustrative Examples. Exercise 4 ( « ) • 7. Laws of Indices for all rational values. 8. Illustrative Examples. Exercise 4 ( b ) .

1. Index Notation. The students already know that the repeated multiplication,

of the same number can be expressed as a power.

Thus a2 = a X a, a3 = a X a X a, a1 — a X a ... multiplied 7 times.

In general, if m is a positive integer, we can write am= a x a X a m times.

Thus am is a symbol and is used to represent a power, m is called the index or the exponent, while a is called the base.

It can be seen that the symbol am, for the present, ceases to have any meaning if m is not a positive integer. For example, a'5 cannot be defined as a X a X a ... mulitplied ( — 5 ) times, a5n cannot be interpreted as a X a X a X ... multiplied ( 5/7 ) times ; nor can a0 be taken to mean a multi-plied by itself zero number of times. Hence the usual mean-ing which we are giving to the symbol am fails if m is not a positive integer. It is, however, found convenient to assign meanings to the symbol am when the index m is not a positive integer. The meanings which we are going to assign to these

92

INDICES ( EXPONENTS ) : 9 3

symbols in this chapter will be found useful in developing the subject further. We shall also study in this chapter the general rules of algebraical operations with such numbers expres-sed in index notation.

2. Laws of Indices. The general rules for multiplication, division and involution

with numbers expressed in index notation are given below. They are known as the Laws of Indices.

I. am x an — a"1* II. am + aK = am~n,

III. ( a m )" = a m n = ( a " ) " ' ,

IV. ( ab )"' •=am-bm,

V (-"X = ^ . \ b J b«>

Here it is assumed that a # 0, b 0 and me Q and ne Q. The students are already familiar with these laws when

m and ne N.

For example, ( i ) 22 x 2* = 2S.

( i i ) ( 3 B ) 2 = 3 3 X 3 3 = 36 .

( i i i ) ( 4 X 5 ) 2 = 4 2 X 5 2 . -

3. These laws can be easily proved for positive integral values of m and n.

I. To prove tha t am X an = am+n it me N and ne N.

We have, am = aXaXaX m times, [ since m is a positive integer. ]

Similarly, an = a X a Xa n times, [ since n is positive integer. ]

a*» x a" = ( a X a x a m times ) x (a X a ... n times )

= a x ax a ( m + n) times.

= am + ». [ by definition]

.*. a'" X aH = am+\


Similarly, am X a« X a* = ( am x a" ) x a* r= Qtn + ii y , a t ) — g m + n + ii

and in general, am x a" x a<> x x a'' =

II. To Prove that am 4- an = am'n if m > n, 1 .. = if m < n a"'"'

= 1 if m = n, if m e N and n e N.

We have am — a X a X a x m times ; an = a X a x a x w times ;

_ a X a X m times a" a X a X n times '

( i ) If m > n, n factors each equal to a can be cancelled from the numerator and denominator; and there will remain m — n factors each equal to a in the numerator.

am . . — = aX ax ... (m — n) times

a n ^ >

= am~». ( b y definition > ( i i ) If m < n, m factors each equal to a can be cancelled

fiom the numerator and the denominater and there will remain n — m factors each equal to a in the denominator.

a"' 1 1 ^ = axax...(n^myimef = def ini t ion>

(iii) If m = n, then a>» _aX aX ... m times a^ ~ aXaX . . .m times < a s n = m - '

= 1. ( cancelling m factors each =a. > a1 , . , aH 1 1 , a9

Thus = a 7 ' 4 = a 3 = - T 2 - 5 = a n d = a4 a1-* au 8 a4 a

III. To prove that ( a « ) " = ainn = ( a» )m. We have ( am )« = am X am X am x ... to « factors

— qm + m + t o n terms

INDICES (EXPONENTS) : 9 5

Anlernatively, it can be proved as follows :

(a"1)" = amx am X am X ... to n factors = (axax ...m t imes) {aXaX ... m times ) ...

... to « factors = ax axa x ...mn times = am».

Similarly we can prove that (an)m = a'""-

IV. To prove that (ab) m = a»'bm• We have (ab)'" = (ab)-(ab)-(ab) to m factors.

= ( a-a-a...to m factors) X ( fox bx ... m factors ?

= ambm-

V. To prove that j , ( b + 0 ).

We have = . . ^ ... to m factors

_ a-a-a m factors ~~ b-b-b m factors _ am

~ bm

4. Definition of a'", a e R, m e Q.

Tn article 1, we have stated that it is convenient to assign meanings to the symbol am for all rational values of m. It is evident that these meanings should be consistent with the laws of indices given in para 2 and proved in para 3 for positive integral indices. These meanings are now given in the following definitions :

Definition 1. If a e R, and m t J V then am = aX aX ax ... m times.

Definition 2. If a e R, a ± 0, then a° = 1. [ If a — 0, we consider the symbol 0° as indeterminate.]

Definition 3. If a e R, a > 0 and p, q e N, then a^o = [The restriction on the value of a > 0 is necessary to

restrict the discussion to a set of real numbers only.]


Definition 4. If a e R, a > 0 and m e Q, then a~'n = •

The students are already familiar with definition 1. From definition 2, we have 3° = 1, ( 127 = 1, ( -\JT)° = 1 and

The definition 3 shows that 5® 52 =i/25; ( 7/3 )3/4 = ^ { I f y f = ^343/27. It should be noted here that a is any positive real number and p, q are positive integers.

~ 3 1 1 From the definition 4, we see that ( 5 ) = —j- = J^J ;

( 6 >'"2'3 = ( 6 * * = W = ; a n d ( 2 / 3 ^ = ( 2 / 3 ; -

= 1_ _ 1 _ v t w t * ~ &s]Tf

The meanings given in the above definitions are such that the laws of indices proved for positive integral exponents remain true for all rational values of the exponents ; this is being discussed in article 7.

Note. ( 1 ) (16 ) i l 2 = V i 6 = ± 4. Thus (16 )1 /2 has two values. But

( 1 6 ) " shall be taken to mean + 4 and not - 4. Otherwise a confusion is likely to arise.

For example, [ ( 4 ) 1 / 2]» = ( 4 ) m x (4 ) V 2 = 2 x 2 = 4 = (4s)1'2. ... ( i ) 1/2 /—

But 4 = V4, and 4> = 16,

from ( i ) , ( - J l ) * = ( 16) 1 / 2 = Jl6. :. ( ± 2 ) ' = ± 4. 4 = ± 4.

We have arrived at a contradiction since l e ^ o r v ^ i e has been taken

± 4 . In order to remove the contradiction, we have to tak'-

16J/2 = + V l 6 = + 4 only.

Thus o*'9 will bs always taken to mean positive gth root of a * .

Again we have, V = + 4 or - 4

/. tyl6 = ± V T or ±

/. V l 6 = ± 2 or ±

Thus has four values ± 2, ± but (16)1'4 will be taken to mean only positive V l 6 which is equal to 2,


Note. ( 2 ) It should also be noted that index for «2* is 23 i. e, S. whereas the index for (a3)Jis 2 x 3, i. e 6. Thus a2S = a6 and ( a

2 ) i = aad hence a-* 4= ( a 1 ) . Similarly the index for 2<s is 4® i . e . 64, whereas the index for (2l 8 is 4 X 3 i. e. 12.

Thus 21 ' = 264 and ( 2 ' f = 213 and hence 2J~ =p ( 24

n n n In genera l ,a m However, if a = ( a ' " ), it follows, that

m n = mn ; m, n are not independent but connected by a relation in" = n

Not . (3 ) . It will be possible to compare the numerical magnitudes

of nu mbers of the type a ^ provided they can be expressed with the same denominator in their fractional indices. The process is explained in the illustrative exa u les 5.

5. Exponential Equations. An equation is called an exponential equation if the variable

in it occurs in the place of the index (exponent). Thus 2" = 8 is an exponential equation and its solution can be obtained by writing it in the form 2* = 23 so that x = 3 is a root of the given equation. In general if a = a , (a # 0 , a ±1 ) , then .v = r is a root of the equation. Thus an exponential equation can be solved by expres-sing both the sides as powers having the same base and then equat-ing the indices If both the sides of the exponential equation cannot be expressed as powers having the same base, then the equation is to be solved by taking the logarithms of both sides; this will be explained in the next chapter.

6. Illustrative Examples. 3 • 27s • 94

Fx. 1, Simplify 3 • (81)4

3>-273-9' _ 3' ( 3" )*•( 3 ' )4 = 3*-39 3»

We have y ^ f j T - 31.(34)4 3'-316

oJ+9+3 5 = — = z n ' v = 3 .

3 1+16

Ex. 2. Simplify - i i i 16* — 2 • (8)*

We will express every term in the expression as a power whose base is a prime number.

C. A —7


32 • C (22) ]3 expression = rrr

<24)* 1 - 2*(23)'v

32 • 24' ' 24* + 4 _ 24.r+l 2* - 21 14

Ex. 3. Find the values of ( i ) ( 81 ) "5, ( i i ) (-001) ' . • 5 1/2 r ~

( i ) ( 81 ) = (81 ) =V81 = 9.

, i i ) (-ooi )1/3 = V-ooi = ^ -555

Ex.4 . Express in the form of 1 0 * : ( i ) -0001; ( i i ) V l 0 0 -

(ii) VlOO = VlO^ = 102/3.

Ex. 5. Arrange in order of magnitude : 21'3 and 31'5.

We will express the rational indices of both the numbers with the same lowest common denominator namely 15.

t i 1/3 15 r^, 15 Thus 2 = 2 = V2" = V32. and 31'5 = 3 3 / 1 5 = 1 V 3 ? = 1 V 2 7 ;

V 32 > 27. ••• , 1SV32 > 1V27, .'. 21/3 > 31/s .

Ex. 6. Prove that it _ + 5 nS . I a T

= 21. 2 [ 2 7 1 + m » 3"'+ 5 + 3" • ( 9 )2 m ] 21 • ( 81 ) m + 1

Let us express every power with the base as a prime number. We shall prove the result for a positive integral value of m. But the students can later on see that the equality is true for any rational value of m.

L. H. S. 23-(3*p + 1

0 2 [33+3m. 3m + 5 + 3 .3imj 2 34m+8 + 34m + 5

2s . 34m + 4 ~ 2* ' 34m+4

S.


Exercise 4 ( a )

1. Find the values o f :—

( i ) 9 ~2, ( i i ) 91 '2 , (iii) g2 '3, ( iv ) ( 6 4 ) 5,

( v ) ( -001 ) 2 ! \ ( v i ) ( 3 | ) "1 / 3 .

2. Express in the form of 10* :—

( i ) -01, ( i i ) s j loooo.

3. Express with fractional indices and simplify :—

( i )

( i i ) x-s/3 X •ty&'x. [ Use the Laws of Indices ]

4. Simplify :—

( i )

(iii) (/>-* + q-i ) (p-* - ).

5. Prove that 1 + xa~b 1 + xb-a = 1.

6. Arrange in order of magnitude :—

9. Prove that

10. Show that :—

(81)" (3)5 - 34""1. ( 2 4 3 ) _ 4 ( 3 " ) = 4. 92"- (27 )

( 27 ) 1 - — show that m=n +1.


12. If X1P + y13 + Z1/3 = 0. show that ( x + y + z ) 3 = 27 xyz.

13. If a is a rational number not equal to 1, show that

[ 2 3 a - 1 11/3

- ( a + £ ) - - 4 - + °a _ t I is rational.

14. If x # 1 and a, b, c are positive integers and,

y x • \ x • V x =1, show that at least two of a, b, c are equal.

15. If a* X a* x a = 1, find the value of x. « C 3/7 2 -2/21

16. If 2x - 5 = a x a x a x a • , find the value of x.

17. If p* x pu* = />2, find the value of x2 + •

18. If ( 81 )* = ( 1/27 f = 9, find the values of x and

19. If " I J 3 m ' - ( 3m) J = 81, find m.

20. Solve the equations :

( i ) 2* + 2*+1 + 2*+2 + 2*+3 = 30;

( i i ) 6* — 3 (2*) - 4 ( 3 * ) + 12 = 0; ( i i i ) 3 2 t + 1 — 3* _ i ,x + 3—9;

( i v ) 4* - 5 ( 2 * ) + 4 = 0; / \ «W W3 • -1/3 \ R-( v ) 2 ( x + x ) = 5.

21. Write down the values of the following numbers arrange them in ascending order of magnitude.

d ) 2* 3 , (22) 3 ; ( i i ) 2 1 / 5 ; 3 , / 7 ; ( i i i ) ( 2 3)2 , ( 2 Simplify and show that :

22. 2a112 - 5a12 + 2a™ = ( 2a - 1 ) {a - 2 ) ^ .

23. 2 (2a - 3)5 2 - 3a ( 2a - 3 )3/2 + a2 ( 2a - 3 )1/2

= 3 ( a - 2 ) ( a - 3 ) V 2 a

INDICES ( EXPONENTS ) : 101

- x'1) (y-x"y - x "y xy + x y

2 2 -2 -2 , -1 W " K 94 -v + y - x - y i (x - x ) Q - y ) _ ^ ' 2 - 2 - 2 -1 -1 —

25. | 1 - £ 1 - ( 1 - a2 | = a.

if J C 1 .

Answers. Exercise 4 (a)

1. ( i ) ( a ) 3. (iii) 4, (iv) 8. (v) j i g . (vi - f - .

2. ( i ) 10"2 , ( i i ) l o \ 3. ( i ) / , ( i i ) 1-

b * ' ( i ) « ( « + »)• 'vH) ablJ+V)'

6. ( i ) 4 , / 3 < 1 0 l r t < 5 ' ' 2 ( i i ) i 1 ' 3 < 2 9 , , 6 < l 0 , / 4 < 31 / 2 . 8 . 2.

15. - 2 or - 1 . 1 - . 3. 17. 2. 18. * = l /2, 2/3. 19. 6. 2 0 . i ) 1; ( i i 1 ,2 ; ( i i i ) - 1 , 2 ; ( i v ) 2 o r 0 ; t v ) 8 o r l / 8 .

2 1 . ( i ) 2> 3 = 2 s = 256, ( 2 s )3 = 26 = 64 ; ( 2> ) 3 < 2*3.

( ii ) 21 '5 < 31'7, since the values respectively are S Vl28 i ~>>/243-

t •• < (2*3 )2. 7. The laws of indices for rational exponents.

With the help of the definitions given in article 4, we can show that the laws of indices proved in article 3 only for integral values of exponents remain true for all their ratoinal values.

We shall not consider these proofs here in detail, but indicate the method of the proofs by proving the first two laws only.

1. To prove that cT X a = if m, Tie Q and aeR, a > 0. ( i ) If m and n or both are equal to zero the result is trivial;

for example o » . n n a x a = 1 X a = a ,

, o+n " .. o w n o+w

and a = a ; thus a x a = a ( i i ) If m and n are positive integers, the result has already

been proved in article 3. ( i i i ) Let m and n be positive fractions, say p/q and rjs respec-

tively so that p, q, r, s e N; further , m n , m+n let x = a X a and y = a


We have x * = ( a " x an )os

= c V a T r x c V o o '

= c ( V « * ) B r x t ( V ^ ) s r = [ « Y x = a x a

Ps + cir = a since p,g,r,seN QS ps + ar

. . x = a Hence x =

t + r-Similarly f =(am+'l)os = (a " f

P ar = [ a as r = ( f = as+c'r . QS As+tfr . . y = a

• m n m+n . . x = J or a x a = a .

( i v ) Let m be a positive fraction pjq and n be a negative fraction - rfs so that />, g, r and se N;

let v x = a'n x a" and y = a'+".

We have x 9 s = (a"'x a" ) 9 S = (a^ 9 x a " r / s ) 9 s

= [ > T x ^ J S

/ — — 1 = ( « / « * ) « . = ^ ^ = x

ate = = X a - w = a^-or . ( A )

Similarly / * = ( am + ")" s

m+n >» + » . = a X a x a X qs times

(«t+n)qs (plg-rlsftjs =a = a

( B ) From ( A ) snd ( B ), x»s = y^ . Taking (qs )th

root ( q and 5 are positive integers ) of both sides, we get x = y; i. e. amx a" = am +«.

INDICES (EXPONENTS ) : 1 0 3

If both m and n are negative fractions, say — pjq and — r/s respectively, so that p, q, r, s e N, we have

"1 n , i i a X a = a'i'1" x a~rls

= - L - x - L - = Qph ar,s apla+rls '

and a m + n = = a-<£'o+Ws>

_ 1 fl^/a+r/s

Thus am X a" = am+n, for all m,neQ.

This completes the proof of the first index law for all rational values of m and n and a > 0.

It is obvious that the second law of indices is just the first law stated in a different form. For. if m, n e Q, then m,— n also e Q; and am -r- an = am X a~» = am'n by the first index law.

The remaining laws of indices can be similarly proved. With the help of the meanings of am as given in article 4, we can take the laws of indices of para 2 as valid for all rational numbers m and n and a>0. We are therefore, free to usg zero, fractional and negative indices in all algebraical operations with which the students are familiar in school Algebra.

8. The meaning of the symbol ax : a > 0 a e R and x irrational.

(This article should be read after the students get introduced to the idea of a limit of a sequence in Calculus ).

The definitions stated earlier give the meanings of the symbol am thus :

( i ) If m is a positive integer, am = axaxax m times when a is any real number.

( ii ) If m = 0, am = 1, a is any non-zero real number.

( i i i ) If m is a positive fraction, say pjq, then am = a^", = "\[a*>, where a is any positive real number.


( i v ) If m is a negative rational number, say — /-, then

alu = a~r = , where a is any positive real number.

The students have already seen ( and they will see it further in the later chapters) how this extension of the meaning of a»» from positive integral m to all rational values of m, is useful in the development of the subject. It has been found useful to extend the meaning of the symbol still further to those cases where m is irrational. A detailed discussion regarding the extension of the meaning of a m to cases where m is irrational, consistent with the laws of indices, is beyond the scope of this book. We propose, however, to explain this briefly in the following lines.

For any irrational number x, we have already seen how a sequence of rational numbers can be found by finding successive rational approximations to the value of x in decjjnal notation.

The table below gives approximations to the irrational number \ f l . The sequence in the first row gives the numbers p to successively increasing number of places of decimals; the sequence in the second row gives the numbers q obtained by increasing the last digit of the corresponding p by 1.

1 2 3 4 5 6

p 1-4 1-41 1-414 1-4142 1-41424

q 1-5 1-42 1-415 1-4143 1-41425

It is to be noted that the successive differences between q and p go on decreasing rapidly and tend to zero in the limit; yet it is evident t h a t g >\fl > p for any corresponding values of p and q. If we. therefore, call the sequence of rational numbers

in the first row p^. p2. Pi, and the sequence of rational numbers in the second row qi , q - i , q $ , then we


see that the sequence of p 's is an increasing sequence and the se-quence ofq 's a decreasing sequence; but both these sequences tend to\f 2 in the limit. This can be seen from the diagram below.

Let us now consider a particular example, where .v is irra-J~2

fional, of a say 3 : whatever meaning we ^assign to the Q p

symbol 3 it must be consistent with the fact that if q>p, 3 > 3 ; if therefore in the above sequences each qr > \f 2 > correspond-

ing pr we must have 3° > 3 > 3 . That is to say we must

assume that as each of the sequences px, Pi, and

qx, q2, qz tend to \[2 in the limit, 3^', 3p!, 3P\ and

39 ' , 1q\ 3? ' , tend to a definite common limit, which can be

defined as the number 3 . It should be noted here that for each

p and q, 3^ or 3" have definite meanings as p and q are rational.

The interpretation given here is made further clear from the rough figure drawn without scale where the graph of y = 3* is shown. We can plot the points (0 , 3°), ( 1, 31) (1 - 2, 31'2) and


say (1-6, 31-6 ). If we draw the ordinate at x = sf 2 it must have VT.

the length 3

Fig. 30

Graph of y = 3* The definition of a when a > 0 and x is irrational can now

be given. Definition. If xi, x2, x3, ... , x„, be a sequence of rational

numbers and if this sequence tends to an irrational number x as

the limit, then there exists a sequence a1, ax\ a*3, , ax",..., which tends to a as its limit.

If the nth number of the sequence x\, x2, is denoted by xn and the nth term of the sequence n a*2, a*s is denoted by ax" we write

a = Lt a*", if Lt x„ = x, with the meanings given n — n —

above for a, xn and x.


We have assumed here that a sequence of rational numbers Xu x2, x3, ,x„ ,... can always be found by considering the rational approximations to the irrational number x in decimal notation.

It is interesting to note that l'" = 1 for all rational values

of m and as such with the above definition 1* = 1 for any irrational

jc; for any sequence of the above type namely ( 1 )*', ( 1 )*a, ( 1 ) x \ is just the sequence 1, 1,1,

and as such tends to 1. We shall see in the chapter on logarithms, how approximate

values of ax, where a > 0 and x is irrational, can be calculated. [ See Illustrative Example 2 ( i i i ) of article 10 and Ex. 5 (v)

of Exercise 5 (b). ]


Ex .1 . Simplify *a + * 2 + 2 x3 +x 3 + 3 ( * + x'1)

. (*+*~1)A 1 X The expression — - = = • . ( x + x ^ f x+x-1 *S + 1

Ex. 2. Multiply 3X'V3 + X + 2X2'3 by - 2. Arrange in descending powers of x and multiply.

{X + 2X2I3+3X-113) (XV3-2)

= (*4/3 + 2 * + 3 ) - 2 * - 4 * 2 / 3 - 6 * - 1 / 3

4/3 i/3 -1/3 the product is x - 4x~ + 3 - 6 *

C Q T U VI 4 / 3 2 / 3 1 / 2 , V - 1 / 2 J - ""1/3_I_ - 1 1 4 Ex. 3. Divide * + x y + y by x y + x + y Arranging dividend, divisor and each remainder as it occurs in descending

powers of x, we have, 1/3 -111 , -1/4 , -1/3 \ 4/3 , 2/3 1/2 , / 1/2 2/3 3/4 1/3 x y +y + x ) x + x y + y \ x y - x y + x y

4/3 , 1/4 , 2/3 1/2 x + xy -f x y 1/4 , — xy + y 1/4 2/3 1/2 1/3 3/4 - xy — x y - x y

2/3 1/2 , 1/3 3/4 , x y +x y + y 2/3 1/2 , 1/3 3/4 , x y + x y +y

i U .. * . 1/2 2/3 3/4 , 1/3 the quotient is xy — x y + x y.


Ex 4 . F ind the square root of

4 * - 1 2 * + 1 3 - 6 * + *

The expression is already arranged in descending powers o f * .

4 « 4 / 5 - 1 2 * 2 / 5 + 1 3 - 6 * 2 / 5 + *

4, 4 / S

4 * 2 / 5 - 3 - 12*2fo + 13

- 12*2 '1 + 9 . 2/5 -2/5 4* - 6 + X

. , '2/5 -4/5 4 — 6* + *

. , -2/5 -4/5 4 — 6 * + *

, „ 2/5 -2/5 ( 2 * — 3 + *

.'. square root is ± ( — 3 + x 2'5).

I n many cases fractional indices may be changed by proper substitutions and the operations can be conveniently carried out by applying the wellknown algebraical identit ies We wi l l i l lustrate few cases of such a type.

u <r „„ w i 2/3 , 1/3,1/3 , .2/3 . 1/3 1/3 Ex. 5. Mul t ip ly a + a b + 6 by a — b . r i J .1/3 Let a — x and b = y

-)/3 2/3 so that a" = 6 «• y ' , a = *® and b = y8

•V product = ( * * + * y + ys) (x - y) = *s — ys ~ a — b. 3/4 1 '2 1/4 Ex. 6. F ind the square root of a + 4 a + 10a ' + 12a + 9.

r i l^4 , 1/2 . 3/4 . . Let a = * , so that a = * * , a = a8 and a =

the expression reduces to x' + + 10*' + 12* + 9. B y an inspection of the expression i t can be seen to be eqnal to ( * ' + 2* + 2 )*.

square root = ± ( * 2 + 2x + 3 ) = ± { al2+2aL,i ->- 3 ).

Ex. 7 . I f m = a11* + « " 1 / 3 , then snow that m" — 3m = a + — • a We wil l wr i te the given relation in the form x + y + z 0 and make

use of the ident i ty , x° +y' + z3 = 3 x y z .

Thus we have m + ( - a 1 ' 3 ) + ( —a 1 /3 = 0

ffl' + ( — a 1 ' 3 ) 3 + ( - a ' 1 1 3 ) 8 = 3 ( m ) (—a*'3) ( - a ~ V 3 > .

m 5 — a — a = 3m.

f«8 — 3m = a + -J— a '

INDICES (EXPONENTS) : 1 0 9

We may, otherwise, use the identi ty ( * + y = *s + y5 + 3xy ( * + y).

1/3 , -1/3 V in = a + a

B , !'3\3 . / "1/3XS , 1/3 "1 /3/ l/3 I -1/3.

.". «ts = a + a " 1 + 3m.

m s — 3m = a H—— • a

[ i J 1 3 1 h'1 / 2 v I 312r vi r 1/2 n 1/4

\a -b ) -I and find its value when a = b = 81.

r (g'6-3/3) X I g - M - b ^ n ^ EXP" = L a^b-V* J

= [(«• ( fc -^ - f t^ .d^) ] 1 ' 4

= [«» 6 " 1 ] 1 / 4 = a 3 ' 4 - * , ' 1 ' 4

I f a = 6 - 81 = 34,

then the value of the expression = (3 ' )3 '4- ( 3 4 ) " 1 / 4 = 3s 3 " 1 = 3a = 9-

Ex. 9 . I f a * = V = c®, and 6a = «c, then prove that x~1+z~1 = 2y"1.

Let each of the given ratios be equal to ft.

so that ( i ) ax = ft, :. a = kVx.

( i i ) 6" - ft, :. b = ft1'*.

(i») c* = ft,

Substi tut ing these values of a, b, c i n terms of ft, x, y and z i n the second given relation, ft can be eliminated and a relation in , * , y, z can be obtained. Thus, we have

(ft1'*)' ^ ft1'* ft1''. ,. k2 h = J1*"1*. 2 1 . 1 . n "I . — 1- . . 2y = x + z y x z

I n problems where two or more quantities or ratios are equal, i t is con-venient to assume all of thequanti t ies equal to some unknown quant i ty, say, ft. The required result, then, can be established by eliminating ft.

Ex. 10' Solve 2X + 3" = 17 and 2 * + 3 - 3 y + 1 = 5. The equations can be wi j t ten as

2* + 3* = 17 ( i ) 4-2*-3-3Y 5. ( i i ) Mul t ip ly ing ^ i ) by 3 and adding to ( i i ) , we get

7-2* = 56. / . 2X = 8. i . e. 2. :. x = 3.

1 1 0 : COLLEGE ALGEBRA

S u b s t i t u t i n g t h e value of x in ( i ) , we ge t

3V = 9 i. e. 32 . y = 2.

x = 3 a n d j) = 2 a re t h e requi red r o o t s .

Exercise 4 (b)

1. Express as powers of 8 : ( i ) | , ( i i ) -125, ( M ) t y Y .

2. Express x as a power of :

( i ) x 3 , ( i i ) \[x, ( i i i ) x".

3. Expand the following :

( i ) U 1 / 3 + l ) ( * 2 / 3 - * 1 / 3+ i ) .

( i i ) ( x / 2 + / 1 / 2 ) ( x 1 / 2 - / 1 / 2 ) . , . . . . , 1/3 -1 /3 2/3 , 1/3 -1 /3 -2 /3 ( i l l ) ( X - J ) ( X + X y +y ' ).

( i v ) ( x 1 / 3 + / / 5 ) ( ^ - , 1 % 2 ' 5+ / ' 5 ) .

( v ) ( , 1 / 2 - x 1 / 4 + l ) ( . 1 / 2 + , 1 / 4 - l ) .

Cvi ) ( 1 + x1'2) ( 1 + xVi) ( 1 + x'3) ( 1 - x1 / 8).

4. Express as a product of two factors :

( i ) x + 2 , - 3 ^ % 1 / 2 ; ( i i ) x + y + z - ^ y ^ z 1 * ; , ••• 3 . 2 , „ 2/3 1/3 . 4/3 2/3 (in) * + j — z + 3xy z ; (iv) x ' + x + 1.

5. Find the values of : 3/2 3/2 3/2 , 3/2

• ( i i ) * . I 1 ; 1/4 1/4 ' ^ ^ 1/2 1/2 T 1 / 2 , 1/2 >

x - y x - y x + r 4/2 „ 1/3 3)2 , 1/2 2/3 „

t;;; \ a — 8 a b . ^x + x y -2y I 1 1 1 ^ 2/3 , ^ 1/3 ,173", ~ J 7 T ' 0 V ) 1/2 T / 3 7 - ^ *

a + 2a - o + 4 6

6. Find the square rools of :

( 1 ) x - 2 + A - ; (11) 1 — 2 + 4 ; 4 5/2 1/2 2 ( i i i ) X - 6x - 30x + lOx + 9x + 25; . . . . 4/5 2/5 -2/5 -4/5 ( i v ) 4a - 12a + 13 - 6a -:- a


7. ( i ) If X = 32'3 - 3 "2/3, prove that 9x3 + 27x = 80-2/3 -2/?

( ii ) If y = 3 + 3 , prove that 9f ~ 27y = 82. , . . . . T „ -2/3 . -213 2/3 -2 /3

( in) If x = 3 — 3 and y = 3 + 3 prove that 41 ( x3 + 3x) = 40 (y 3 - 3y ).

8. Prove that : /- a112 . A1'2 , , , j.1'2 !'2 x v ( 4 - o + c ) (a + b - c ) X

1/2 1/2 1/2 1\2 1/2 1/2 ( a — b + c ) ( — a + b + c )

= 2bc + 2ca + lab - a* - b2 - c2. 9. Evaluate :

- 2 -1/2 5/6 , . - 3 -1 /2 +2/3 . - 1 / 4 , „ 2 3 a + 2 3 a — 10 (27 a ) where a = 64. Simplify and show that :

10. ( X2 + y2 ) X (x2 -y2 ) = X2 - y2 . i i_ _1_

( f t + c \ a " b / c + a / a + 6

JCC" j x l x a b \ x a;b"c ) = 1.

if a, Z>, c are unequal quantities. ( V m+» / \ ri+J / , \l+m -(5) •(£) •>•-

( '+T)"( ' -T)

x , i _ *2 /3 _ i _ - r 2 / 3 + 2

1 . 1 15. 1 + X"~m + X^-m 1 + xm~n + X*"n

+ ! = i 1 + x™-* + Xn~t>


16. \ f p q + r = 0, then prove that

- + I + xr + x-* = 1.

[ Hint : Multiply every term of 2nd and 3rd fractions by x'o and xt> respectively].

17. If a" , a" = a""1, then show that m (n - 2) + n(m - 2 ) == 0.

18. Jf a = l ( f , b = 10'and ( a* b*f = 100, show that xyz = 1.

19. If a1" = ( a V , then show that . v ^ V ' ' " 1 = / . 20 If xy = z, yz = x and 2

V = v, show that xyz = xy y~ z* = 1.

21. If m"x = n'v = {ni • prove that xyz = 1.

22. If in = a\ n — a and a = (in • n 'v)\ show that .v v r= 1. 23. If a" = by = c~ — d" and ab — cd,

show that ——(- — - = ——h — . x y : w

x z 24. If — = — , prove that

in . m , in , in x +y +z +iv in

( i ) 8-22*-1 = 16-2*-1; ( ii ) = ( x \ f x )

(i i i) 27* = 9 " and 8 f = ( 2 4 3 ) 3 ' ;

( iv) 2A = 4 5 = 8= and ~~ + j -

and hence find its value if x = y = 8,

and hence find its value if x = , y 8 '

INDICES (EXPONENTS) : 113

a. Show that L 4 x ^ 2 " J

cannot be 8 for any value of rt.

29. Simplify f ^ " , , ^ 2 Q Q ) . l / 3 • (\ 3/2 / \ 2/3

y 1 + =

31. If 4* = 8V = 16z,

prove that : H — = — • x z y

t Particular case of Illustrative Example 9 ].

32. If 4* = 3" = 2* = 6W,

• 1 1 1 . 1 prove that : 1 — = — f - — • x y z w

[ Particular case of example 23 in this exercise ].

33. If 2* = 3* = 12*, prove that xy = z ( x + 2 y ).

34. If 6 * = 3 y = 2, prove that — + — + — = 0 -x y z

35. If 2* = 4" = 8* and xyz = 288,

,« , 1 . 1 . 1 11 prove that - + - + _ = _ .

Answers. Exercise 4 (b).

1 / • l o~1/3 / •• \ o"1 I ... > .W8 1 . ( 1 ) 8 , (n) 8 , (in) 1

2. ( i ) ( ^ V ' 3 . ( i i ) ( V i T \ ( i i i ) ( * " ) 1 , B .

;1/2 + >'-1/3-

3 . ( i ) * + l , X ( i i ) x - - y , ( i i i ) * - i ( i v ) * + / / 5 .

( v ) x - xV2 + 2xVi - 1. ( v i ) 1 - * .

C. A.—8

: COLLEGE ALGEBRA

/ . > , 1/2 1/2. , 1/2 . 1/2. 4. ( i ) (* - y )(X ~2y ), . . . . . 1 / 3 , 1/3, 1/3. . 2/3 , 2/3 2/3 1/3 1/3 1/3 1/3 1/3 1/3. (ii) (* +y +2 ) (* +y +z -x y -y z -z x ) , , , 2/3 1 /3 . , . , 4/3. 2/3 2/3 2/3 1/3, 1/3 . (iii) (x + y -x )(*' + ? +« -*y + y « +« *).

(iv) ( ,2 '3 + ^ 3 + l ) ( * 2 , 3 - / 3 + l ) .

5. ( i ) ( * W + , 1 / 4 ) («) 2 (* + y ),

(iii) a '3 ( a1'3 — 261'3), < iv ) * + ^ / 3 + 2 / 3 .

6. ( i ) (* 1 , 4 -*~ l M ) . (ii) ( * " - ! ) . (Hi) **-3* 1 / 2 + 5

(tv) 2A2LS - 3 + A~2I\ 9. 51^3.

25. ( i ) («) X-9(*. (Hi) * = ' . y = 3/2 (tv) * = 7/16, y - 7/32, * - 7/48.

26. 1. 27. ^ V 3 : 3)32. 29. 7/4.

Chapter 5

Logarithms

1. Definitions. The identity N = a 2. The restrictions on the base a and the number N . 3. The Laws of Logarithms. 4. Il lustrative Examples Exercise 5 (a ) . 5-7. Common Logarithms. Rules to find characteristics and Mantissae. 8. Four-figure Logarithmic Tables. 9. Anti-Logaritbms. 10. Use of Logarithmic Tables. Illustrative Examples. 11. Problems on Compound Interest. 12. Napierian Logarithms. Exercise 5 ( b ) .

1. We will consider in this chapter an important application of the theory of Indices, namely Logarithms. Logarithms are very useful in simplifying complicated numerical calculations! If N is any positive real number and N = a , (where a > 0, 1) then x is called the logarithm of the number N to the base a, and we briefly write this as x — log,, N. Thus, we can define the logarithm of a number as follows .

Definition. The logarithm of a number N (> 0 ) to any base a (>0,¥=1) is the index of the power to which a must be raised to make that power equal to N.

It can be seen from the definition that a logarithm is merely an index. It should also be noted that

are only different forms of expressing the same relation between a, x and N. The first equation expresses the relation in the index or exponential form and the second in the logarithmic form.

Thus, as 24 = 16, 4 = log2 16.

a = N and x = log^ N

-2 1 Similarly, as 5 = — , — 2 = log,

115

'S


Thus, the two equations a = N and JC = log,, N are only transformations of each other and should be remembered to change one form of the relation into the other.

1 1 . The Identity : N =aoi"N •

Consider the two equations

N = a* ( i ) and x = log,, N (ii).

The equation ( i ) expresses the value of N in terms of x and the equation ( i i ) expresses the value of x in terms of N, Substitut-ing the value of x given by ( i i ) in ( i ) , we get the relation.

. logfl N ... N = a (111). Similarly, if we substitute the value of N given by ( i ) in ( i i ) ,

we get the relation

x = loga ( a ) ( iv ) .

The relation ( i i i ) and ( i v ) can be called identities. For example, we have

2 1 o ^ = 4 , = 3 l o i 3 < 5 ) ! = 5* = 25.

2. Restrictions oo the Base a and the Number N.

Let us again consider the two equations

N = a . . . ( i ) and x = log„ Ar ... ( i i ) .

Restrictions on the base a. ( 1 ) We observe that if the base a in ( i ) is taken as 1,

X it is satisfied for all values of x if N - 1. For 1 == 1 for all x.

However, ( i ) is not satisfied for any value of x if N # 1; hence from equation ( i i ) , we see that x i. e. the logarithm of N to the base 1 is not properly defined.

Similar arguments hold if the base a is taken as zero.

We, therefore, assume that the base ' a ' of a logarithm is not to be taken equal to either one or zero.

( 2 ) If we take the base ' a ' in the equation ( i ) as a nega-tive number, say — 2, then,

LOGARITHMS : 117

( a ) for values of x, such as, 2, 4, 2/5... etc. N is a positive number, namely, 4, 16, ^ 4 , . . .etc.;

( b ) for values of x, such as 3, 5, 1/3, ... etc. N is a negative number namely, — 8 , - 3 2 , — ... etc. and ( c ) for values of x such as, 1/2, 1/4, etc. N is an imaginary number, namely, \ [—2, -ty—2, etc.

Thus if the base ' a ' is taken as a negative number, certain values of x give imaginary values of N. We are considering real numbers only for the present, and hence we shall assume that the base ' a ' of a logarithm is not to be taken as a negative number.

We shall, therefore, choose the base ' a ' as a positive number 1).

Restrictions on N.

Now if we take the base ' a ' as a positive number 1) in the equation ( i ) , then N is always positive, whatever be the value of x ( i . e. positive or negative). Thus N cannot be a negative number for any value of x, if a > 0 and=£l. Hence from equation ( i i ) , we conclude that x i. e. loga N does not exist if N is negative and a > 0 , # 1. Hence, we shall assume that logarithms of nega-tive numbers are not to be considered.

Two important results.

( 1 ) If log, 0 = x, then 0 = a". This is not possible for any x. Hence the logarithm of zero to any positive base does not exist.

( 2 ) Since a0 = 1, we have 0 = log^ 1.

Similarly, since a1 = a, we have 1 = loga a-

Hence we notice that the logarithm of one to any positive base 1 ) is zero and that the logarithm of any positive number (=£ 1) to the same base is always one. We should therefore, remember the two important relations :

logB 1 = 0 and loga a = 1.


3. The Laws of Logarithms. We shall now proceed to prove the laws of logarithms which

are valid for any base a ( > 0 b u t # 1). Since the term " logarithm " is merely a substitute for the term " index," the laws of logarithms can be easily deduced from the laws of indices- The laws of logari-thms are given in the following theorems.

Theorem I. The logarithm of a product of any two numbers is equal to the sum of the logarithms of the two numbers to the same base.

i. e. loga (mn) = leia m + loga

Let logam = x and log0« = y.

Then by definition, a = m and a = n.

x+y

. . a = mn. by definition, x + y = loga ( mn).

Substituting for x and y, we get loga m+log 0 n = log,, (mn). Cor. The logarithm of a product of any ( finite ) number of

factors is equal to the sum of the logarithms of these factors to the same base.

For, logfl ( mnpq) = loga ( mnp ) + loga q

Ex. log^ 110 = log,, ( 2 X 5 X 1 1 ) = loga 2 + loga 5 + loga 11.

Note : It should be noted that log ( m + n)i=\og m + log/i unless m + n = mn.

Theorem II. The logarithm of the quotient of any two numbers is equal to the logarithm of the numerator minus the logarithm of the denominator to the same base,

m X n = a x a = ax+v [by the Index Law ].

= loga (mn ) + log„ p + log„ q = loga m + log,, n + logap + loga Q-

i. e. loga = m ~

Let loga m — x and logan = y.

by definition, a* = m, and a = n.

LOGARITHMS : 119

.". — = — — a " [ by the Index Law 1. n av

m ax~y = — • n

by definition, x — y •= loga ^ f j '

Substituting for x and y, we get

loga m ~ loga n = log,,

Cor. loga = loga 1 - log„«.

= - loga n. [ V loga 1 = 0 ]

Ex. loga ^ = loga 110 - loga 21,

= loga 2 + loga 5 + loga U - logo 3 - log,, 7. Theorem III. The logarithm of a power of a number is equal

to the product of the index of the power by the logarithm of that number,

i. e. loga ( m* ) = n loga m.

Let loga m = x] by definition, m = a .

;. mn = ( a

x f = a"x. [ by Index Law]

ax=m. by definition, n-x = loga (in).

Substituting for x, we get loga (rn") = n-logam.

Cor. log„ C-J™ )=loga(wi1,n) = _Llogam. n

Let logam = x. :. m = a. 1/n , x.lln . 1 In xtn . . m = (a ) : . . m = a •

X , , ll« . . 1 In xln •'• — = loga (m ). . . m = a .

1 i i « Vn x . . lOga m = loSa ( m ) '


E x . 1 . l og a 81 = l o£ a ( 3 4 ) = 4 logB 3.

E x . 2 . loga ^ ^ J - l o g a ( 5 - 4 ) « - 4 1 o g a 5 .

E x . 3 , loga ( ) = - loga 7.

Change of Base The theorem for the change of base of a logarithm enables

us to find the logarithms of numbers to a new base, say b, if loga-rithms of numbers to some base a are known.

Theorem IV. To prave that logb m = l°ia m. • loga O

Let log6 m — x.

by definition, m — b*. Taking logarithms of both sides to the base a, we get

loga m = loga ( b* ) = X log„ b. [ Theorem III.J

- * = ]°Sa m .

loga b

Substituting for x, we have logbm = ^ .

Ex. tofcai-l^f1. logio 5

Thus, it will be seen that if the logarithms of the numbers 31 and 5 to the base 10 are known, then the logarithm of 31 to the base 5 can be found.

Cor. ( 1 ) . If we put m = a in the above theorem, we get

. lo^a a 1

.'. logja X loga fc = 1. Cor. ( 2 ) . log6 m = loga m X !og6 a.

4. Illustrative Examples. Ex. 1. Evaluate ( 1 ) log2 32. { i l ) l o g „ (-04] and (HI) log ( 6 4 ) .

2^ 2 ( i ) Let log, 32 •= x.

by definit ion, 2 * - 32. i . e. 2* = 2\ * = 5. .*. log2 32 = 5,

LOGARITHMS : 1 2 1

( i i ) Let logj» ( -04 ) =

... ( 2 5 ) * = . 0 4 = J _ ^ = 2 5 - \

* = - 1. log»s (-04 ) = - 1.

( i i i ) Let log _ (64 ) = * . / . ( 2 - J l ) x = 64. 2^2

(2mf = 2. 2 W 2 =2 8 .

.'. — * = 6. .'. x = 4. log ( 64 ) = 4. 2

E x . 2 . S o l v e : ( i ) 5* = — • (ii) l og , ( 8 1 ) = - 4 .

( i > ••• 5* = ± = i r = 5"4; ••• « — 4 .

( » ) V l o g , ( 8 1 ) = - 4 . - 4 1

by defini t ion, x = 81 = 3' = — 4

1

25 SI

3. S imp l i f y : log - log j + log

(225 \ I 64 \ /

32~ / + l 0 g I 729 / l 0 g V

= l o g { ( f * § ) + f i } [ byTheoremI I ]

= log{Sx^xS}=log (2) [ o n s i mPu f i c a t i°n ]-

Ex. 4 . Simplify : ( i ) log ( log a 3 ) - log ( log a ) .

( i i ) loga (625) - M o g a ( V " 5 ) .

( i ) We have log ( log a 3 ) — log ( log a )

= log £ J [ b y Theorem I I ]

" I o g £ i s j r ] ~ l o g 3 - [ b y Theorem I I I ]


( i i ) We have loga (625) - log,, ( V 5 ) = ' ° g " j 5 *1 lOgn ( 5 )

= l j ® i i £ - f = 8. [ by Theorem I I I . ] 4 log,! 5

. Ex. 5. I f x, y, z are all positive, prove that l og* (y ) x l o g y ( z ) x log2 ( i ) = 1.

Changing the logarithms of all numbers to the base a, we get by the Theorem of the Change of base formula

L. H. S. = ^ - ^ = 1. log,, * log,, y log,, z

6 . i f = J£fJL = togj.. v e t h a t a * . b \ 0 ' m q - r r - p p - q

Let —g a = -g— = l o g c =. k q - r t - p p - q

.'. log a = k (q - r ) , log b = k (r - p ) and log c = k ( p — q ) .

I f the common base is, say, e then hKq-r) _ kir-i>) , k(p-Q>

a = e , 6 = e and c = e t> bk<Q-r1 ,o qk(r-p) . cr rUp-q)

a => e , o =» e and 3 1 t> .Q r tk(a-r) a rkib-a) .. a • b • o = e X e X e

_ ik(o-r)'+qk(r-6)+rk(j>-q) — 6

E*. 7 Prove that 1 + 1 + 1 2

logxy (xyz) l o g y z ( x y s ) l o g z x i * y * )

I Making use of the result of the corollary of the change of base formula,

•namely, ^ = log,

UH.S = log ( x y ) + log { y z ) + Jog ( z x ) . xyz xyz xyz

= log (x * y ' z • ) xyz

= log ( xyz )• xyz

= 2 log ( x y z ) = 2 =• R. H. S. xyz

Ex. 8. Show that

log,0 ( l i ) + logu ( l i ) + 1 gio ( 1 I ) + ... to 198 terms = 2.

LOGARITHMS : 123

L. H. S. = log,, ( f ) + log,0 + log,, + + log,. ( f g )

, r 3 4 5 200 "1 - ®10 L 7 ' T • 7 • • • 199 J = log,o [ 100 ] = logw (103) = 2 logio (10) = 2 = R. H. S.

E». 9. Find Ihe positive integer » such that loga 96 lies between » and n + 1. [

We have 3' < 96 < 35

••• log3 O M < log, (96) < log3 (3s ) .*. 4 < logs 96 < 5 ; .•. ft < log3 96 < » + 1, if « = 4.

E* 10. I f log0«i = * and logr t» = f where a , n t , n are positive and a+=1. tnn =j= 1, express log ( a ) in terms of x and y.

We have log a = log0»Mn logam + loga« * + y

Exercise 5 (a)

1. Write the following equalities in the form x = loga N :—

( i ) 54 = 625, ( i i ) 3 "5 = ^ 3 . ( i i i ) 6 4 " 1 , 3 = - l •

2- Express the following statements in the form a* = N :—

( i ) log7 343 = 3, (ii) log.

( i i ' ) log,

3- Find the values of :-

(s-)' (iii ( i ) l o g v _ (125), (ii) log4 ( ^ r ) , (iii) logs (-008).

4. Find the values of .v which satisfy the following equations:— ( i ) log2 v ,- ( 144 ) = x, ( i i ) log. ( - 0 9 ) = 2,

( i i i ) log8x = - -y- •


5. Write down the values o f :—

, ^ J o g a 5 ">> — 2 log 7 ( 3 ) ( i ) 2 , ( H ) p , ( i i i ) 7

6 . Assuming the base as 10 throughout, prove that ( i ) log 25 = 2 - 2 log 2, ( i i ) log 800 = 2 + 3 log 2, ( i i i ) log 12-5 = 2 - 3 log 2.

7. If logay + 4 = 2 logax + 5, express y in terms of x and a.

8. Assuming the base as 10, prove that

log (20) + 7 log + 5 1 o g ( i £ ) + 3 1 o g ( g . ) = 1.

9. Show that

10. If a = log (10/9 ), b = log ( 25/24), c = log ( 81/80), prove that log 2 = la — 2b + 3c.

11. Prove that to any base

( i ; log

( i i ) log ( 1 + 2 + 3 ) = log 1 + log2 + log 3.

12. If log ( a + b ) = log a + log b, find a in terms of b and hence prove that log b — log a = log ( b — 1 ) .

log b log a 13. Prove that a = b

14. Show that log ) V a2 + 1 + a \ = - log) \ /a 2 + 1 - a { .

16. if log ( ^ r ^ j = ( I o g a + l o g p r o v e t h a t a = K

17. If log { ^ T ^ j = \ ( l 0 S a + l o S b ) ' P r o v e t h a t

a2 + b2 = 11 ab.

LOGARITHMS : 125

18. Simplify the following :— ( i ) ( log x> — log .v5 ) -7- (log x3 — log x2 ); ( i i ) log ( x? + 3*2 + 3x+ 1) log ( JC2 + 2x + 1 ) .

19. Solve : -

I ; \ lQg * _ lQg C 6 4 ) K ' log 4 ~ log ( 1 6 ) ' ( i i ) log1 0x + log10 ( x + 3 ) = l.

( i i i ) logio (7.v — 9 )2 + logto ( 3 x — 4) 2 = 2.

20. Find the values of : ( i ) l 0 g 3 (4) x log4 (81); ( i i ) log2 (3) x log3 (4) x log* (5) x logs (2)-( i i i ) log3 (bed) if a — b2 = c3 = di(j= 1") where

a, b, c, d are positive real numbers. 21. If p, q, r, s are all positive, prove that

( i ) logpq X loga r X log, s = log^s; ( i i ) log r p X logpq = log, 5 X logs q.

22. If ( 3 • 7 )* = ( • 37 )" = 1000, show that - l - l - i

x — y = 3 •

23. If = . , prove that b — c c — a a — b

xyz = x6+= • • z i+6 = xa • yb • ze = 1.

24. If log x _ log y _ log z b + c - 2a ~ c + a - 2b ~ a + b - 2c '

prove that xy: = 1. 25. Show that ( i ) logy (.v3) X log, ( y 3 ) X logx ( z 3 ) = 27;

( i i ) logy2 (x) X log,» 00 X log.,.. (z) = y •

26. Without using logarithmic tables, show that

( i ) "fij < log102 < - y ; ( ii) - y < logi03 < - L •

27. Show that ( i ) .v = fl1/logvW

;

log1 (6-logjc-log£( i log^.v ( ii ) x = a


28. I f a2 = b3 - c* = ds, prove that logj (bed) = 47/30,

OQ TF I ° g a l°g b log c . ,,, , , 29. If ^ • = —|— = — , prove that a5b3c'2 = 1.

30. Tf b2 = ac, prove that + 1 2

logax log,.x logjX


1. ( i ) 4 = logs 625. ( i i ) - 5 = l o g s ( ^ ) . ( i i i )

2 . ( I ) 7s = 343, ( i i ) 4 " 3 / 2 = i , ( i i i ) a ' 2 = \ . o a 3. ( i ) 6. ( i i ) - 3 . ( i i i ) - 3 . 4. ( i ) 4, ( i i ) -3, ( i i i ) 1/4.

5. ( i ) 5, ( i i ) >•. ( i i i ) j - 7. y = ax\

12. 6 = ~ T 1 8- C fi > 2- («)

19. ( i ) * = 8, ( i i ) x = 2, the other root * = - 5 is to be rejected as

the logarithm of negative numbers are not considered (iii) 2 or 13/21.

20. ( i ) 4 ; ( i i ) 1 ; ( i i i ) 13/12.

Common Logarithms 5. Logarithms to the base 10 are known as Common Logarithms. Common Logarithms are useful in simplifying numerical calcula-tions. They were first introduced in 1651 A. D. by Henry Briggs. We now proceed to consider how the common logarithms can be used in practice.

When we are referring only to common logarithms, the base 10 is usually omitted to avoid repetition. Hence hereafter when-ever the base is not mentioned, it should be assumed as 10, as we are now considering common logarithms only.

6. Consider the following relations— Y 10° = 1 , 0 = log 1. Y 101 = 10, . '. 1 = log 10.

102 _ 100, 2 = log 100. • 103 = 1000, 3 = log 1000.

-.•10° = 1 , 0 = log 1. Y l O - 1 ^ , /. - l = log (-1). Y l 0 - 2 = -01, A - 2 = log (-01) Y 10-3 = -001,

- 3 = log (-001)

LOGARITHMS : 127

We observe that—

( i ) the common logarithm of a number greater than one is positive, and of a number less than one is negative; while the common logarithm of one is zero.

( i i ) the common logarithm of a number which is an exact power of 10 is either a positive or a negative integer;

( i i i ) the common logarithm of a number which is not an exact power of 10 is partly integral and partly fractional;

( i v ) the logarithm of a number, say, between -1 and -01 lies between — 1 and — 2 and hence can be written as — 1 — ( a fraction ) or — 2 + ( a fraction). It has been found convenient to write the fractional part of the common logarithm of a number with always a positive sign, and hence the above logarithm is always written as — 2 + ( a fraction) and not as - 1 - ( a fraction).

7. Definitions. When the logarithm of a number is written so as to have the positive sign for the fractional part, then the fractional part is called the mantissa and the integcral part is called the chara-cteristic.

We shall now proceed to see how the characteristics and man-tissae of the common logarithms can be obtained.

I. Characteristic of a number greater than one.

From the first four relations of the previous article, we observe that

( i ) a number lying between 1 and 10 has one digit in its integral part and its logarithm is = 0 + / j ;

( i i ) a number lying between 10 and 100 has two digits in its integral part and its logarithm is = 1 + /2 .

( i i i ) a number lying between 100 and 1000 has three digits in its integral part and its logarithm is = 2 + f3 ; and so on.

Here the integers 0, 1, 2, denote the characteristics and ]\,fi,fz ( the + ve fractions < 1), denote the mentissae. Thus we observe that the characteristic of a number ( > 1 ) is positive and is one less than the number of digits in the integral


part of the number. We, therefore, have the following rule to find the characteristic of a number greater than one :

" If a number N is greater than 1, and has p digits in its integral part when expressed in decimal notation, then the characteristic of its logarithm is (p — 1 ).

II. Characteristic of a number less than one. From the last four relations of the last article, we observe

that ( i ) a number lying between 1 and • 1 has no zero between

the decimal point and the first significant digit and its logarithm is = - 1 + / ' .

( i i ) a number lying between 1 and -01 has one zero between the decimal point and the first sinificant digit and its loga-rithm is.= - 2 + / " ;

( i i i ) a number lying between -01 and 001 has two zeros between the decimal point and the first significant digit and its logarithm is = — 3 + / " ' ; and so on.

Here the integers - 1 , - 2 , — 3, are the character-istics and / ' , / ' ' , / ' " ( the + ve fractions < 1), denote the mantissae.

Thus we observe that the characteristic of a number ( < 1 ) is negative, and is one more than the number of zeroes between the decimal point and the first significant digit of the'number. We, therefore, have the following rule for finding the characteristic of a number which is less than one.

" If a number N is less than one and has q zeroes between the decimal point and the first significant digit when expressed in decimal notation, then the characteristic of its logarithm is — (q + 1 ). "

Thus the characteristics in log -32, log ( -02) and log (-0032) are - 1 , - 2 and — 3 respectively. When the characteristic is negative, the negative sign is not prefixed but is written over the integers and is read as ' bar ' . Thus the above characteristics are

written as 1, 2, 3. This method of indicating the negative sign of the characteristic over the integers avoids the association of negative sign with the mantissae.

LOGARITHMS : 129

lit. Mantissae in logarithms of numbers having the same sequence of digits.

Consider log ( 1234 ), log ( 12 -34 ) and log ( • 1234 ).

1234 We have = 103. 1-234

log ( 1234) — log ( 1 -234) = log (10 3 ) = 3 log ( 1 0 ) = 3.

.-. log ( 1234) = 3 + log( 1-234).

Similarly, it can be shown that log ( 12-34 ) = 1 -f log (1-234). and log ( • 1234 ) = - 1 + log ( 1-234).

It follows, therefore, that the mantissae of log (1234), log (12 • 34 ) and log ( • 1234 ) are the same as that of log ( 1 • 234 ).

In general, the logarithms of numbers having the same sequ-ence of digits have the same mantissae. We thus have the following rule :—

" If the numbers N and N' have the same sequence of digits and differ only in the position of the decimal point, then log ( N ) and log ( N') have the same mantissae. "

This can also be seen easily as follows. We have, by data,

= 10^, where p is an integer.

log ( N / N ' ) = log ( 10* ). = log ( 10 ) = p. ( Y log 10 = 1)

log ( N ) — l o g ( N ' ) = / > . log N and log N' differ only by an integer p and hence

their mantissae are the same.

Thus the numbers having the same sequence of digits have the same mantissa, whereas their characteristics vary according to the position of the decimal point in the arrangements of the digits of the numbers.

8. Four Figure Logarithmic Tables. To find the logarithm of a number we have to find the characteristic and the mantissa. The first two rules of the previous article enable us to write down the

c. A.—9


characteristic by inspection of the number. The mantissae for numbers with different sequences of digits have been tabulated and are given at the end of this book for reference. In these tables the mantissae for four digit numbers are only tabulated and hence these tables are called four figure logarithmic tables.

We shall now show how the logarithm of a number can be obtained by the following illustration.

Suppose we have to find the logarithm of 763-5. The number under consideration is greater than one and has 3 digits in its integral part. Hence by the first rule the characteristic of log ( 7 6 3 - 5 ) is 2. To find the mantissa of the logarithm of the number, we refer to the logarithmic tables given at the end of the book. Following is the required extract of the table for ready reference.

! • 1 2 3 4 5 6 7 g 5 j Mean Difference 1 1 1 2 3 ( 4 5 6 ( 7 8 9

75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 2 2 3 3 4 5 5

76 8808 8814 8820 8825 8831 8837 8842 8843 8854 8859 1 1 2 2 3 3 4 5 5

77 8365 8871 6876 8882 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 5

Choose a row from the first column containing the first two digits of the number, namely 76. Proceed along this row until we come to the column headed by the third digit 3 of the number. We get the figure 8825. We next add to this figure the number 3 to be found in the same row in the mean differences column headed by the fourth digit 5 of the number. Thus we get the number 8828. This number with the decimal point prefixed gives the mantissa of log (763-5 ) as -8828.

Hence we have obtained that log ( 7 6 3 - 5 ) = 2-8828.

The students can verify that

log ( 7 - 5 2 9 ) = 0-8767; log ( -0077 ) = 3-8865;

log ( -7605 ) = 1-8811 and log (75 ) = 1-8751.

9. Anti-logarithms. In many cases we are required to find the number, the logarithm of which is known.

If y = logio x, then x is called the atitilogarithm of y to the base 10; and this is shortly written as x = anti-log y.

LOGARITHMS : 131

When we have to find the number whose logarithm is given it is convenient to make use of the table of anti-logarithms given at the end of the book. The method of using this table of anti-logarithms, is much the same as in the case of the table of logari-thms. From the mantissa of the given logarithm, we first find the sequence of digits in the number. Then the characteristic of the given logarithm determines the position of the decimal point in the number. We illustrate this process by the following example.

Example. Find the numbers whose logarithms are ( i ) 2-3456, ( i i ) 2-1576.

( i ) To find the sequence of digits in the required number, we look up the table of antilogarithms given at the end of the book. Since the mantissa is -3456, we look out for the row in the first column containing the first two digits of the mantissa • 34, and in this row we select the number 2213 in the column headed by the third digit 5 of the mantissa. We, then, add to this the figure 3 in the same row and of the column of mean differences headed by the fourth digit 6 of the mantissa.

We, thus, get the required sequence of digits in the number as 2216. As the characteristic of the given logarithm is 2, there must be 2 + 1 i. e. 3 digits in the integral part of the number.

Hence the number whose logarithm is 2-3456 must be 221-6.

( i i ) As in case ( i ) , from the table of anti-logarithms, the sequence of digits in the number, corresponding to the mantissa •1576, is found to be 1437. Now the characteristic is 2. There-fore, there must be one zero between the decimal point and the first significant digit in the number.

Hence the number whose logarithm is 2-1576 must be •01437.

10. Use of Logarithmic Tables. Illustrative Examples.

We now proceed to illustrate the use of logarithmic tables to simplify the numerical calculations.


„ . „ , , . , 3871 x 4-53 Ex, 1. Find the value of • - „ —— •

58-9 x 73-01

3871 x 4 53 Let x -

58-9 x 73 01

log x = log 3871 + log 4 53 - log 58 -9 - log 73-01 = 3-5878+ 0-6561 - 1-7701 - 1 8624 = 4 2439 - 3 6335 = 0-6104.

log * = 0-6104. .-. * = anti-log (0-6104) = 4 078.

Ex.2. F ind the values of ( i ) V T l . ( i i ) ( -7)-'. ( U l ) 7 ( i ) Let * = V l T = (11 )1 '7.

taking logarithms of both sides, we get

l o g * = log (11 )1 /7 = y l o g (11) = y x 1-0414 = 0 1488.

.". .r = anti-log ( 0-1488 ) = 1-409.

( i i ) Let * = ( -7) - 7 , / . log * = log ( -7 ) ' 7 = -7 x I03 ( -7 .

= ~ X ( f -8451) = x ( - 1 +0-S451)

= To { ~ 7 + 5 9 1 5 7 ) " ^ x < ~ 1 0 + 8 ' 9 1 5 ? >

= - 1 + -8916 = 1 8916.

* = anti-log (1-8916)= -7791.

( i i i ) L e t * = 7 ^ 5 '

log x = J5 • log 7

= J5 (0 8451) = y say. 1/2

y = 5 -(0-8451 ).

•'• logy = - i log 5 + log (0-8451 )

= 6990) + T-9270

= 0-3495 + 1 -9270. log y = 0-2765.

y = 1-890. .-. log * = y = 1-890. * = 77-62;

or = 77-62.

LOGARITHMS : 1 3 3

The student is advised to note carefully the various steps in mult ipl ication and division in the case of logarithms of numbers wi th negative characteristics.

Ex. 3 . F ind the value of

(-06235)* x s/ (-7273 )

6-7 x (• 0726 )5 x V 7 ^

N

Let * when N and D denote the numerator and denominator of the

given fraction.

Then log N = 4 log (-06235 ) + f l o g ( -7273 )

= 4 x 12 7948) + 1(7-8617)

= 4 x ( - 2 + -7948 ) + J ( - 1 + • 8617 )

= - 8 + 3-1792 + 4 ( - 2 + 1.8617)

= - 8 + 3-1792 - 1 +-9308

= - 9 + 4-1100 = 5-1100.

Similarly log D = log 6-7 + 3 log ( -0726 ) + H o g ( -95)

= 0-8261 + 3 x (2-8609 ) + I X { 1-9777)

= 0-8261 + 3 x ( - 2 + -8609 + J ( - 1 + -9777)

= 0 8261 - 6 + 2 5827 + H - 3 + 2-9777 )

= 0-8261 - 6 + 2-5827 + ( - 1 ) + -9926

= - 7 + 4-4014 = F-4014.

Now log x = log N - log D = 5~1100 - 3 4014 = 3 7086.

.-. * = anti-log (3"-7086) = -005112.

Ex. 4. Solve correct to two places of decimals the equation

7 * + 1 = 2 4 * - 6 ,

We have 7 = 2

Taking logarithms of both sides to the base 10, we get

l °g(7*+ 1 )=l°g(2 4*- 6 ) . .-. ( * + 1) log 7 = ( 4 * - 6 ) log 2. / . * ( log 7 - 4 log 2 ) = - 6 log 2 - log 7.

+ 61og2 + log 7 6 x -3010 + -8451 1-8060 + -8451 " * - 41og 2 - log 7 4 X -3010 - -8451 1-2040 - -845l"

= 2 , 6 5 1 1

* 0-3589'

l o g * = log (2-6511 ) - l o g < 0 - 3 5 8 9 ) - 0-4234 - ( f -5550) = 0'8684


* - anti-log (0 -8684 ) = 7 -386

E * . 5. F ind the number of digits in 31 ' .

Let * = 3 " .

log * = log 315 = 15 log 3 =« 15 x 0-4771 = 7-1565.

* = anti-log (7 1565 ).

As the characteristic is 7, the number of digits in the integral part of * is 7 + 11. e. 8.

.*. the number of digits in 315 is 8.

E x . 6 . F ind log71 (731-5).

By the ' change of base • formula log-, (731-5 ) = ' ° g l 1 7 3 1 5

Iogi0 71 Let this fraction be denoted by x. : . log * - log ( 2 -8642 ) - log ( 1 -8513) = '-4570 - 0-2674

* = anti-log (0 1896 ) = 1 • 547. logn (731-5) - 1-547.

This example illustrates the method of finding the logarithm of a number to any base by using the common logarithms.

* ! 1 , Problem« on Compound Interest. The common logarithms can be con-veniently used to solve problems on compound interest. Let P denote the principal, r the rate of interest per cent per annum, n the period i n years and A the amount of P in « years.

f The interest on P for the first year is P - ——• 100

.'. the amount at the end of the first year is

P + P - ^ i . e . p ( l + r ^ ) .

Similarly the amount at the end of the second year is

P ( 1 + i 5 o ) x ( 1 + i o o ) i " e ' p ( 1 + 100

tne amount at the end of n years is P ^ 1 +

-'• A = P - ( 1 + l 3 o ) " - ••• A = P - R " w h e r e R - l - f . ^ .

From this formula, if any three of the four qnantities A, P, r and n are given i t -will be possible to calculate the remaning four th quant i ty wi th the help of logarithmic tables. We wi l l explain the method by the fol lowing examples.

= 2 S 6 i 2

1-8513'

= 0-1896.

LOGARITHMS : 1 3 5

Ex. 1. F ind the amount of Rs. 150 in 10 years at 5 p. c. per annum com-pound interest.

Here P «= 150, r = 5 and » = 10, we have to calculate A,

We have A = P - R n where R = 1 + ^ •

" a = p ' ( 1 + i5O) " p ( 1 '0 5 ) 1° = 150 { 1 - 0 5 )I0-

log A = log 150 + 10 log 1-05 = 2-1761 + 10 (-0212) = 2-1761 + -212 = 2-3881.

A = anti-log (2-3881) = 244-4.

Therefore, the required amount is Rs. 244 correct to the nearest rupee.

E* . 2. F ind the principal which amounts to Rs. 2000 in 12 years at 3J p. c. per annum compound interest,

3-5 Here A = 2000, R = 1 + — i . e. 1-035 and n = 12, we have to find P.

100

We have A = P R * where R = 1 + ~ •

2000 = P- (1 -035)12 .-. log 2000 = log P + 12 log ( 1 035).

.'. log P = log 2000 - 12 log (1-035) = 3-3010 - 12 ( -014940). = 3-3010 - -179280 = 3-3010 - -1793 = 3-1217.

.-. P = anti-log 3-1217 = 1323. the required principal is Rs. 1323.

[ W e have taken log 1-035 as -014940 correct to six places of decimals to obtain more accurate result ]

Ex, 3. I n how many years wi l l a given sum double itself at 4 p. c, per annum compound interest ?

4

Here, let the principal be P, so that A = 2 P ; R = l + j ^ = 1'04 ;

a n d we have to find n.

We have A = P R " . .-. 2P = P (1 -04) " . A 2 = ( 1 - 0 4 ) " . log 2 = n log 1-04.

- M = J £ « i _ = = 1 7 . 7 " " log 1-04 0-0170

.'. the required period is approximately 18 years.

Examples

1. F ind the amount of Rs. 1500 in 8 years at 3% per annum compound interest. [Ans . Rs. 1900 ]


2. F ind the number of years in which the sum of Rs. 2500 wi l l amount to Ks. 3625 at 4^% p. a. compound interest [ Ans. 8 years. ]

3. F ind the sum which wil l amount to Rs. 1500 in 7 years at 5% per annum compound interest. [ A n s . Rs. 1066. ]

4 . At-what rate per cent p. a. compound interest wi l l Rs. 750 amount to Rs. 912-5 in 5 years ? [ Ans. 4% p. a. ]

12. Napierian Logarithms. For numerical calculations we have seen how the common logarithms are used. The base for the common logarithms is 10. However, it has been found convenient to obtain the values of logarithms of numbers to a base which has been designated by the symbol V . The number e is irrational, and its value is 2-71828...The logarithms of numbers to the base e are called Natural ( o r Napierian) logarithms. The method of calculating these loganthms is beyond the scope of this book. The common logarithms given at the end of this book have been calculated with the help of these natural logarithms.

The students must have noticed the advantages of using the common logarithms or logarithms to the base 10. These advantages are—

( i ) The characteristic can be found by inspection.

( i i ) The numbers having the same sequence of digits have the same mantissae.

The logarithmic tables given at the end of this book give the results correct to four significant figures. If more accuracy is required seven figure logarithmic tables are also available.

Exercise 5 (b)

1. Write down the logarithms of the following numbers. ( i ) 2788, ( i i ) 2-304, ( i i i ) -0089, ( i v ) • 707, ( v ) 345700.

2. Write down the anti-logarithms of :—

3.

4.

LOGARITHMS : 137

Find as accurately as the tables permit, the values o f : —

83-71 x -0783 5. ( i ) 3759 X -057 x -1783; ( i i ) 102-7

( i i i ) \f 59-8; ( i v ) V-0082 ; (iv) f where c =

6. ( i ) ( -7592 )3<4; ( i i ) ( 0-053 ( iii ) ( 0-73 )°'7

7 / : < > (23-45)2 X (5-432)3 f l 7 - 8 X ( - 5 9 7 ) 0 7. ( i ) ^ ^ ^ (2-35)2 J

1/5

8. Solve, as accurately as the tables permit, the following equations :

( i ) 5* = 23, ( i i ) 9 2 * + 7 2 ^ 3 2 x + * .

9. According to the census of 1961, the population of Bombay is 41 lakhs. Estimate of the growth of population is 25 % every ten years. Show that the population of Bombay after 40 years i. e. in 2001 will be approximately one core and ten-thousand.

10. A Saving Certificate of Rs. 100 gives Rs. 150 after 12 years. Show that the interest is earned at 31 % per annum com-pound interest.

11. Show that : logw (logi0 x5) - l o g w (log10 x ) = 1 - log l0 2.

12. Prove t h a t : - = - 2. logs ( 7 ) l°gl6 ( 3 )

13. Simplify and show that logw 343 3

1 + i l og l 0 (49 /4 ) + | log,0 (1 /125)

14. Prove that : log6 7 = 3 •

15. Find the number of digits in ( i ) 323, ( i i ) 2S0, ( i i i ) 513 x 1" .

16. Find the number of zeros between the decimal point and the first significant digit in

( , \15 / \100 A A , ( i i ) (-079 ) 1 8 , ( i i i ) { \ y


19.

17. Find the number of digits in N if ( i ) log2 ( log 2 N) = 8, ( i i ) log2 1 log3 ( l o g 4 N ) | = 2.

18. When do you say that log, x = m ? State, giving reasons, the restrictions generally put on x and a.

State when ( i ) m > 1, ( i i ) m = l, (iii ) m < 1.

If log = - j - ( l og * + log y),

prove that — + 2 - s 47,

y x '

20. Prove that : log„ (\fx)- lag, ( / ) •log* ( ^ z 2 ) = 1.

21. If = a + x -show that : x log ( b / a) = log a. 22. Find the value of x if

logiaZ ( 1 ) x 10*. ( i i ) l og 4 ( j t )+ Iogi«4 = 5/2.

23. If log,.? = log,z = logzx, show that x = y => z. 24. If p = abc and a, b, c are all + ve real numbers 1,

prove that 1- T—̂ — + — = 1. loga P ^gbP logcp

25. Answer the following as required :

( i ) Arrange the following in ascending order of magni-tude.

l o g a ( 1 2 8 ) , l o g v _ ( 2 1 6 ) s l o g 3 4 3 ( 4 9 ) , 3'<>*<7>-

( i i ) If log6x = m and log6y = «, write, in terms of m and n the values of : l o g 6 ( * / j ) , logs ( X2y), l o g ^ ( i 2 ) .

(iii) State the values of the following : 1

5log;,* log37 i o c „ 81

(iv) If x = logba, show that log, ( a« ) = .v, if c _ b«

LOGARITHMS : 1 3 9

( v ) If log,o 2 = -3010, logic 3 = "4771, logi0 7 = -8451. find the values of : logio 5, logic 9, logio 42.

( vi) If log2 ( xy) = 5 and log2 (x/y) = 1. find the values of x and y.

(vii) Express x in terms of y if ( i ) 2 logio x - 3 logio y = 2. (ii) logic x = 2 logio y + 3.

(viii) Without using logarithmic tables, prove that 2 , 3

- y < logi0 5 < - j - •

( i x ) In the equation logic ( 1 + y ) = 2 logi0 x + C, when x = I, y = 4. Find y when x = 2.

26. Without using tables show that :

log 3\/"3 + log 2\[2 - l o g 5\/5" _3_ . log 1-2 2

27. Solve the following equations : ( i ) log3 x + log9 x = 6; ( i i ) log, 3 + log, 9 + log, 729 = 9.

28. If p — log,o 2, q = logio 3, express in terms of p and q ( i ) logio 108, ( i i ) log10 10-8, ( i i i ) \og10sfYjs.

29. If log ba + logc a — 2 logj a • log,, a, prove that a2 = be. 30. State, giving reasons, whether the following statements

are true or false, ( i ) log ( a + b ) = log a + log b, ( i i ) log a — log b = log a log b, ( i i i ) logt (1) = 1 , (iv ) logi a = 1, ( v ) loga 1 = 1, ( v i ) loga 0 = 0, (vii ) loga 1 = 0- Also lo&, 1 = 0. Hence loga 1 = logb 1. a — b.

Answers. Exercise 5 ( b )

1. ( i ) 3-4453, ( i i ) 0-3624, ( i i i ) 39494, ( i v ) f8494, (v) 5'5387. 2. ( i ) 3-723, ( i i ) 52-71, ( i i i ) 01043. ( i v ) 007923, (v) 0-6310.

3 . 1-5682 and 4-5682, 4. ( i ) 5"1 < 2 " < 3»5, ( i i ) 10" < 2 " < 7*°.


5. U ) 38-2, ( i i ) -06386. ( i i i ) 1-272, (iv) 0-449, (v) 8177. 6 . 0-S132, ( i i ) 5-826, ( i i i ) 0-7947. 7 . ( i ) 0 02519, U i ) 0-8364. 8 . ( i ) 1-948, ( i i ) 0-5 or 1-446.

l i . ( i ) 11, (ii) 16, ( i i i ) 24. 16. ( i ) 5 , ( i i ) 19, (iii) 30. 17. ( i ) 78, (Hi 49.

18. II a m " x, then we define log,, x = m. Restrictions are: x > 0 and a ~ l and a > 0,

I f a > 1. then tn >1 , = I , < 1 according as * > a , = « < a.

}{ a < 1, then m > 1 , = 1, < 1 according as * < a, = 0, > a.

22. ( i ) * = 100, ( i i ) * = 2 or 16.

25 . ( i 3rd number < 1st number < 2nd number < 4th number;

( i i ) m — n, 2m + n ; 2/ ( m + n ) ;

( i i i ) * , 0 , 3 , (v) -6990.-9542, 1-6232; ( v i ) * = 8 , y = 4 r

3/2 (vii) * = 10y , * = 1000y'. ( i x ) y = 19.

27 . ( i ) 81, ( i i ) 3.

2 8 . ( i ) 2/> + 3<?1 ( i i ; 2p + 3 q - l , ( i i i ) (P + q - 1 ) .

30 . Al l statements except ( i i i ) are false. However in ( i i i ) , l i s not the only value of log 1. Reasons are left for the students to give. I n ( v i i ) log 1 = log 1 is only a relation of equality and a, 6 have no relation. I n fact a b log 1 to any positive base is always zero. Hence no relation between the bases, can exist.

Chapter 6

Surds

1 . 2 . Surd Numbers. 3. Rules for operations with surds 4. Mult ipl i-cation and division of surds, 5. Definitions. 6. Simplification of monomial sards. 7. Addit ion and subtraction of surds. 8, Comparison of surds. 9. Simplification of surd expressions. 10. Rationalizing factors. 11. Illus-trative examples. Exercise. 6 ( a ) . 12. Properties of quadratic surds. 13. Il lustrative examples- Exercise 6 ( 6 ) . 14. The square and square root of a binomial quadratic surd. 15 Method of inspection- 16- The sqnare o f a trinomial quadratic surd. 17- Illustrative examples. Exercise 6 ( c ) .

1. Surd Numbers. We have seen that a number, which can be expressed as a

ratio of two integers in the form mjrt, is known as a rational number. The numbers which are not rational, such as V3, ^ S , are known as irrational numbers. Irrational numbers are of various

types. For example \ [3 and + \ [3 are both rrational

numbers, but they are of different types. First number is a root of a rational number 3, whereas the second number is a root of an irrational number 2 +\J 3. The number of the first type is called a surd.

Definition :—If the irrational number x satisfies the equation

x" = a, where a is any positive rational number and ne N, then x is called a surd.

Y x satisfies x" = a', .'. x = \/a;

.'. x ( otsja ) is a root.

A number, therefore, will be a surd if, and only if, ( i ) it is irrational and (ii J it is a root and ( i i i ) a root of a rational number.

141


We notice that if a is any positive prime integer then \/ a is a surd.

[ y a i s a + ve pr ime i n t e g e r . '•J a c a n n o t b e an i n t ege r .

If possible, let "Ja = p'q, a f rac t ion where p and q have no common fac to r .

.-. « = .'•. p» - aqn.

a is a factor of p" and hence of p.

:. a" is a fac tor of p" a n d h e n c e of aq".

: . a is a fac tor of q" a n d h e n c e of q.

a i s a fac tor common to bo th p and q which is con t ra ry to o u r hypothesis.

V a 4= . a ra t ional n u m b e r . 9

"yfa is i rrat ional a n d h e n c e a surd number .

W e have assumed above tha t a is a positive prime in teger and hence we could use t he result ( w h i c h the s tuden t s should only n o t e at this s t a g e ) tha t " If a i s a f ac to r of ji", then a is also a factor of p whenever a is a p r ime n u m b e r . "

However, if a is not a prime integer but any positive rational number which is not the perfect nth power of any other rational

number, t h e n \ / a is a surd, since in this c a s e \ / a can be expressed

in terms of other surds of the type \ / b where 6 is a positive prime integer.

For example : ^ 1 6 = 2 ^ 2 , ^ 6 = ^ 2 X \f%

J T = T ^ = T I V 5 X V 3 ]•

2. In a surd a, n is called the order of the surd or the surd index and a is called the radicand. In order to see whether a number is a surd number or otherwise, we have to find the nature

of its value as well as the form. For instance, has a form of

a surd number; but its value 2 is rational and hence -^8 is not

a surd. Again " J 2 + \ f 2 is irrational but is the square root of

SURDS : 143

an irrational number (namely 2 and hence is not a surd.

B u t a p p e a r s to be a root of an irrational number; yet it is

really equal to ^ 2 and therefore is a surd.

It should be noted that every rational number can be expressed in the form of a surd of any required order. A rational number 2 may be written as-v/^, tyl', ^ 2 5 i. e. as a surd of 2nd, 3rd and 5th order respectively. In general, it can be seen from the laws of indices that V a " = ( a n )Vn = a, and hence any rational number a can be put in the form of a surd of order n. However, a surd can-not be expressed in the form of a rational number.

The students may be interested to note that the number -n-which they know as the ratio of the circumference of any circle to its diameter is irrational but not a surd.

Examples,

1. State which of the following are surd numbers :

( i ) 49". ( i i ) V 64, (iii) V 2 5 ; ( i v j V r + T T ( v ) V j f . Ans. ( i i i ) , (v).

2. When is a real number said to be ( i ) rational, ( i i ) irrational, ( i i i ; surd ? Rewrite the following statements, duly corrected where required.

( j ) -Jib I 25 is an irrational number and also a surd,

( i i ) V 8/25 is both an irrational number and a surd.

( i i i) Every irrational number is a surd,

Ans. ( i ) V16 125 = 4/5 and hence is a rational number ;

, 2 «/2 ( i i ) Vo/25 = — is irrational and is also a root of a rational

number and hence is also a surd. (ii i) Every irrational number need not always be a surd. For example

>J 3, v etc. are irrational but not surds.

3. Rules for Operations with Surds. Surds can always be expressed as numbers with fractional indices. Therefore, the laws of indices which we have studied in the fourth chapter can be applied to surd quantities for algebraical operations. Thus we have the follow-ing rules for operations with surds.


If m, n, are positive integers and a, b are positive rational num-bers, then

( i ) Va~ V r = V a T ,

( H ) VT

( iii) m J ( \ f a ) = mnJ a = ^ ( y Z b

( i v ) = ""K/^S

tn ( v ) ( V T ) = V " m -

Examples. ( i ) V I X VI = 31'3 • J l f i - ( 3-5 )1/3 - ( 15 J1* = VlS. . . . . V I 5W / 5 V'*

( i i i) v ^ « = p / 2 ] , / 3 - ( « ) 1 / 6 = v r .

also / V ^ = [ a 1 / 3 ] 1 / 1 = t « ) , « - Va.

• = y i r r = V a

( i v ) V 2 » = ( 2° )1 / 3_ I_22 / 3 = 24'6 = v ?

VT» = 3XV23X2 = V ? . Thus the order of the surd can be changed by mult iplying the surd index

and the index of the radicand by the same integer.

(v) ( W ) 5 = (4V3 )6 = 4S/3 = W . 4. Multiplication and Division of Surds. By the first rule it can be seen that surds of the same order can be multiplied. Thus, we have ' '

V 7 X y T = a ' n - b V H = ( a b f ' ^ M a f . In general, if the surds have rational coefficients, we have

P W X q \l~b = P<1 "\[ab Similarly with the help of the second rule, division of surds

of the same order can be simplified as

"/~b = bvn = ^ J " « nJ<L.

SURDS : 145

In general, if the surds have rational coefficients, we have

P \]~a <7 "/~b = J L n J ° . m

If the surds are not of the same order, we should first reduce them to the same lowest ( possible) order and proceed as before. The surds can be reduced to the same order by the fourth rule viz. ahlm = = " j ^ / ^ ; we have only to multiply the surd index and the index of the radicand by the same convenient integer in order to produce the required order for the surd.

E x . 1 . Multiply 5 V~2 by 2 V 5~.

T h e first is a surd of t h e t h i rd order , whereas the second is a surd of second order . T h e lowest o rde r t o which bo th of them can be reduced is six. T h u s the p roduc t = 5 l/~2 x 2 2 V T = 5 x 2 %/¥.

= 5X2 V?T s .

5 \l~2 x 2^/7 = io fyioo. /

Ex. 2 . Div ide 3 V l O by 6 \ ! z .

T h e lowest o rde r to w h i c h bo th the surds can be reduced is twelve.

3 V l O 3 1 2 V l 0 3 1 1 2 / T T 3 , required q u o t . e n t - — - - ^ J _ _

5. Definitions. Here, we will introduce a few terms associated with surd quantities.

Surds of second, third and fourth orders are called quadratic, cubic and biquadratic surds respectively. Surds of the same order are called equiradicai surds.

Surds are also named according to the number of terms. Surd expressions involving one, two, and three terms are called monomial, binomial and trinomial surds respectively. Expressions containing two and three surds are called, as a matter of convention, binomial and trinomial surds respectively instead of specifically mentioning them as surd expressions. In general expressions containing two or more surds are called compound surds.

C. A.— 10


If the coefficient of the surd is ± 1, the surd is called an entire surd; otherwise it is known as a mixed surd. For example j j

are entire surds, whereas - 5-^/4" -^ / (T are mixed surds. It is obvious that any mixed surd can be transformed into a surd of entire form. For example, the above mixed surds can be written in entire forms as

Surds which can be expressed as rational multiples of the same surd are called like or similar surds; and those which cannot be so expressed are called unlike or dissimilar. Thus ^27V~18 a n d 5 0 / 4 are similar surds as they can be expressed as 3 \f~2 and whereas & ~ tysT, are unlike surds.

6. Simplifications of Monomial Surds.

For purposes of calculating approximate values of expressions containing surds, much tedious work can be saved by first simplify-ing the surd expression.

For example, if it is required to find the numerical value of y[5~ / V~3"we have to find the value (i. e. 2-236 ), and the value of«/~3t (i- e. 1-732 ), and then obtain the quotient

0.01/'

J—732"-' --. Much of this labour can be saved by multiplying the

numerator and denominator of yj "5" / \J~3~ by y[JT Thus V5 V J V J \l 15" 3-87298 . - Q n o , . . -j^ = — x -r=. = = ' — 1-2909... The device of V3 V3 V3 3

removing the surd quantity from the denominator of the surd fraction is very convenient and is known as the the process of rationalizing the denominator. We shall return again to this process later.

W e say that a monomia l surd is in its simplest fo rm when the number under the radical sign is the least possible positive

SURDS : 1 4 7

integer. We illustrate the simplification of monomial surds by the following examples.

( i ) \[300~ = VlOO X 3 = VlO2 X 3 = 10 V3T

(«> «vf - - H =4^ = 2yf2X.

(iii) vs~= V27 7, Addition and Subtraction of Surds.

The addition and subtraction of like surds can be simplified by taking out the common surd factor. Thus,

n r " r 11 /

\a + q \ a = ( p ± q ) ' \ l a. Ex- 1. Simplify 3 V l 6 2 - 7 V » + V5 l2 -

The expression = 3 V 8 1 7 x 2 - 7 V l 6 x 2 + V 2 5 6 x 2 -

= 3 x 3 - V5" - 7 x 2-V2 + 4-%/J = ( 9 - 14 + 4 ) V 2 = - V 2 .

Ex. 2 Simplify 3 V I47 - S V I +

7vT V3~ /~~3 The expression = 3 V49 x 3 j ^ - x — + 7 y ' 2 7 x 3

3 V3 V3 y 7 . , = 1 3 x 7 ) 7 3 " " - — V 3~ + ~ V 3

The addit ion and subtraction of unlike surds cannot be further simplified . Thus _ _ _ _

Vl8+Vl2 -2V2 = 3^2 +2V3 - 272 = (3 - 2)J2 + 2V3"

and th is cannot be far ther simplified.

8. Comparison of Surds. Since we cannot find the exact values of surds in rational

forms, we can not directly compare their magnitudes. We can, however, compare the magnitudes of surds by expressing them as entire surds of the lowest possible common o r d e r .


Ex. 1. Arrange in ascending order.

The lowest common order to which all the three surds can be reduced i s 12.

We have = = 13V729, V ? = 12V73 = 2 ^ 3 4 3 ,

and

Bu t 343 < 625 < 729. .'. K/T < V J <-JT. Ex. 2. Which of the following is the greater : 3 1/S" or 2 . / I ? The lowest common order to which both the surds can be reduced is 6.

Thus 3 = 3 VT2 = V 3 ' 7 T a = V5 5 = V6561

and 2 -J 5 = 2 V =M Z6~¥~= VsOOO.

But 8000 > 6561. .\ V8000 >V6561 , 2 ^ 5 > 3 i f f .

9. Simplifications of Surd Expressions.

In order to simplify surd expressions, we first put each surd in its simplest form. For addition or subtraction we group like surds just as we collect like terms in algebraic addition or subtrac-tion. The multiplication of compound surds is performed like the multiplication of compound algebraic expressions. In the case of division we rationalize the denominator by multiplying the numerator and denominator by a suitable factor.

3 Ex. 1. Simplify V250 + V l 2 8 + — 1 / 2 - l / i .

Simplifying ear.h term separately, we have

X/250 = 2/125 x 2 = 3v'5 r2 = 5 Vi".

V l 2 8 = V64 x 2 = = 4 1/2-

> / ' J ^ ' n / t x 7 ~ V ¥ = l ^ 2 '

.". the given expression = 5 i j t + 4 U2 + ~ VI" V2

= ( 5 + 4 + l - i J 'V2 =10V2.

E x . 2 . Multiply *JT - J F + v T by ^ 3 + ^ 2 -

The required product

= -J3 +J1) +V2 (V5 -V3 )

SURDS : 1 4 9

= Vl5 - 3 + ^10 - *J6 + 1

« Vl5 - 3 + VT~X I + JlO - V6 + 1 = Vl5 - 3 + J </ 6 + VlO - V 6 + 1 = Vl5 + VlO - J V6 - 2.

E x . 3 . Divide V98 - V50 by « / l2

We have the required quotient :

S8 - -J50 l4z - 5 V 2 SS - • = • - • — • — — —

J12 2 ^ 3 2j / f = j _ ^ 2-Js 3

10 Rationalizing Factors. Definition : When the product of two surd expressions is rati-

onal, then each one of them is called the rationalizing factor of the other.

We shall use the abbreviation R. F. for " rationalizing factor." We discuss few simple cases.

Case 1. Monomieal or simple surds.

In this case, the rationalizing factor can be found by inspection. Thus

( i ) v a\[x x \ f x = ax;

sjx is R. F. of a\[x.

( i i ) y 3 x 4/2 = 3 ( 2 3 ) W (2) i /4 = 3 • 2*/4 = 3-2 = 6.

/. </2 i s R . F . o f - ^ 8 . . . . . . . 3/2,-113 215. w , 1/2 -2/3 3/5, ( i l l ) v ( a •b -c ) x ( a •b ~c )

cflc = a2Z>_1 • c = •

1/2 , -2 /3 3/5 „ c. 3/2 , - 1 / 3 2/5

. . a -c is R. F. of a b c .

Case 2. Binomial Quadratic Surds. Y ( a\[x + 6 Vy ) ( a\/x — b\{y )= a2x — b2y.

a\fx + b \[y and a'sfx - b \(y are R. F.'s of each other.


These two binomial quardratic surds differ only in sign between their two terms. Two binomial quadratic surds which differ only in the sign connecting their terms are called conjugate or complementary surds. Therefore, the rationalizing factor of a bino-mial quadratic surd is its conjugate. Thus

V ( 3 \[5 +2y[3) X ( 3 \[5 — 2 s f j ) = 45 - 12 = 33,

3 \ f 5 + 2\[3 and 3\[~5 - 2 ^ 3 are R F.'s of each other.

Thus the R. F. of a binomial quadratic surd can be obtained by simply changing the sign connecting the two terms of the given surd.

Case 3. Trinomial Quadratic Surds.

The general trinomial surd is of the form a\[x + b\fy + c\J z. The general process to rationalize such a surd is to reduce it to a binomial quadratic surd by grouping any two terms together and then follow the same procedure as is followed in the case of binomial quadratic surds. The process of rationalization is a completed in two steps.

We will illustrate the process by working out some examples.

Ex. 1. F ind the ra t ional is ing f ac to r of : V 2 + V 3 + J 5 .

Here , the three rad icands are 2, 3 and 5. If we choose the surd whose rad icand is the sum of the o ther two rad icands to be removed first then the first step would reduce the surd expression to a monomial quadra t i c surd which can be easily rationalized Otherwise, the first s tep would reduce t h e expression to a binomial quadra t ic surd which will have to be rat ional ized again by mult iplying by i ts con juga te binomial quadra t ic surd . T h u s in such examples much work can be saved by a judic ious choice of the surd to be removed first.

T h u s [ ( J J + V T ) + V~5] [ ( V ! + V3") - V s ]

= (V2~ + V3~)- - (VJ)3 = 2 + 3 + 2 Vfi" — 5 = 2^/(7. the expression can be rationalized by multiplying fur ther by a

f a c t o r J 6 which would give the p roduc t , 12, a rational n u m b e r .

H e n c e J 6 ( ViT + V 3 - -J~5 ) is a R . F . of J 2 + + J 5 7

Ex. 2 . Express with a rat ional denomina tor — • •J 7 + V 3 - V 2

SURDS : 151

We reduce the trinomial quadratic surd to a binomial quadratic surd by grouping the first two terms together and then multiply by its conjugate surd

XI,, 1 - 1 V ^ + ^ + J2 Thus — = = =— = - = = — X — p ^

V7 + V3 - V2 V7 + V3 - V2 V7 + 73 + V2 ~Jj + 73~ + -J2 J 7 + 72

(V7 + Vf)S !- (V2)a _ 7 + 3 + 2 V2T-2 77 + y f + 2 _ 1 77 + 7 ^ + 7 2

8 + 2 721~ ~ ~2~ ' 4 + V2l" Now the denominator is reduced to a binomial quadratic surd and hence

can be rationalized by multiplying i ts conjugate surd.

1 J7 + J3 + 72 4 - 72T T h e e x p ' = 4 +V2T X

1 4 y/7 + 4 VJ" + 4 7 2 - J7J2I - V3~ V2l" - ^2 721 = T ' 1 6 - 2 1

1 4 ^ 7 + 4 y r + 4 ^ 2 " - 7 V f - 3 J7 - V42" = T ' - 5

= J _ (3 73 - 77 - 4^2 + 742).

Case 4. Rationalizing Factors of Higher Order Surds.

Many times standard identities in factorization can be used to obtain the rationalizing factors of higher order surds. We will illustrate a few simple cases.

v3 1/3 E«. 1. Find the K. F . of x" + x + 1.

Let a so that x* 3 = a^, and * = a ' , which is rational. The given expression = a% + a + 1. Now we want to find a factor which when multiplied by the given expression will give terms in ax or its powers. By inspection, obviously, I a - 1) is such a factor.

Thus ( x ~ 3 + * 1 , 3 + 1 ) n = (« 2 + a + 1 ) ( a - 1) = a 5 - 1 = x — 1, which is rational.

1 ! i ~ 3 1 x - 1 is the K. F. ol x + x + 1.

t : 1 .'4 Ex. 2. Find the ! '. F. cf .r + .T + 1.

•1 ' ~ Let x' = re, so that x' "= a- and * = «4 which is rational. The given

expression reduces to a2 + a + 1. We have, therefore, by inspection to find such factors which when multiplied by the given expression would give terms in a' or its higher powers.


W e have ( a1 + a + 1 ) ( a 1 - a + 1 ) = ( a 1 + 1 )* - a* = a' + 2«a + 1 - aJ = a* + a2 + 1.

Still the product conta ins t e rms in a 2 which are no t rat ional . Again by inspect ion , we find a factor which when multiplied by a* + a 2 + 1 would render the result to conta in t e rms in a ' or its higher powers only.

W e a g a i n h a v e ( a ' + a1 + 1 ) ( a 4 - a2 + 1 ) = ( a* + 1 ( ) » = a 3 + 2a 1 + 1 - a 1 = a* + a 1 + 1, which is rat ional .

T h u s to rat ional ize the expression we were required to multiply by (a1 - a + 1 ) and (a4 - a 2 + 1 ) .

Therefore the two rat ional izing fac tors are ( x ^ - * 1 / 4 + 1 ) and ( x - x m + 1 ) .

Note. I n order to obtain the rat ional izing factors of h igher order surds following types of ident i t ies in factorizat ion will be found very use fu l .

( i ) *s + )>3 = (x+y) (*2 -xy + y*) (ii) x1 - y% = (x - y ) (x1 + xy + y2) ( i i i ) + * V + y* = U 2 + xy + j>2 ) ( - xy + y>) ( iv ) x6 - y6 = (x s - y3) (*» + y » )

= ( * - y ) ( x2 + xy + y1) ( * + y ) ( *2 - xy + y2) {v ) x3 + y' + a3 - 3xyz — ( i + y + j )

( *a + y% + 2 2 - xy - yz - ).

11. Illustrative E\ampies.

E x . 1 . Simplify s y - T l j f - ^ ~ '

Let us first rat ional ize each denomina to r . Le t 3j5 = a so tha t i/25 = a1

and 5 = a1 which is rational. Similarly let = 6 so tha t \[9 = b2 and 3 = ft1 which is rat ional . W e shall, therefore , multiply the denomina to r s by such factors tha t the p roduc t will conta in a z and 63 or the i r h igher powers af ter ra t ional iza t ion.

8 2 _ 8 ( a + b) _ 2 ( a - 6 ) Thus exp. - a2_ jb + &2 a>.ab+ a3 + bi ai _ bi

= {a + 6) - {a - 6), 7 a3 + S3 = 8, a ' - 6 ' = 2. = 26 = 2 X/i-

Ex. 2 . Wi thou t finding approximate values, find which is greater

V n + V 7 or V35-

V n W ~ 7 > -J 35 according as (J 11 + -J~7 V > 35

i. e. according as 18 + 2 V77 > 35.

i. e. accordi ng as 2 V77 > 17,

i. e. according as 308 > 289

SURDS : 153

we conclude that 7 l l + Jl > V 35, since 308 > 289.

Ex. 3 . If -/To = 3-162277 find the value o[yJ"E- correct to three

places of decimals.

We have first to simplify the surd before finding its value.

Now y f l = 1 V40 = - f 7 1 5 = - J - (3-162277)

= 1-265 correct to three decimal places.

E t 4 . Show that —

3 V 2 V6~ 4 7 3 '

V 3 + V 6 7 2 + 7 3 7 6 + 7 2 3 V~2 ( 7 3 - 7 e ) 7 6 ( 7 2 - 7 3 ) 4 7 F ( 7 1 - 7 T )

L. H . s . - 3 - 6 + 2 - 3 ~ 6 - 2

= - 7 l i - 7 6 + 7l~8 ~ 7 l 2 - 718 + V 6 = 0 = R. H, S.

Ev. 5. Simplify — p i — — 7 l 0 + 7l4 + V15+ 721

The exp = •„ , — — - 1 ^ = V 2 x 5 + V 2 - 7 + V 3 - 5 + 7 3 - 7

1 ( 7 7 + 7 5 ) ( V 3 W ~ 2 )

= i j j ^ - 7 1 ) ( 7 3 - 7 ~ 2 ) = a (V21 - V 1 4 - V T 5 + V 1 0 ) .

Ex. 6. Simplify —

12 _ _ 4 "7 6 + \ / 2 4 7 3 - 7 5 + 7 2 7 2 - 7 3 + 7 5 7 5 - 7 2 + 7 3 It will be convenient to rationalize each denominator before simplification.

12 12 ( V 3 + 7~2 + 7 3 ) We have — ,i_ — , - p :— 7=

^ 3 - 7 5 + 7 2 ( 7 3 + 7 2 ) 3 - ( 7 5 )2

= Vet 73 + V2 + 7 5 ) ; 4 7 e _ W i ( 7 2 - 7 1 - 7 1 )_

72 - 7 3 + 7 5 ( 7 2 - 7 3 - ( 7 3 )-= - 2 ( 7 2 - 7 3 - 7 1 ) ;

724 2 7 6 ( 7~3 - 7 2 - 7 1 )

7 1 - 7 1 + 7 3 ~ ( 7 3 - 7 2) 2 - . 7 1 ;2

= - 1 7 3 - 7 2 - 7 1


exp. = ( 3 - J 2 + 2 V 3 + 7 3 0 ) + 2 ( 7 2 - 7 3 - 7 5 )

- ( 7 3 - 7 3 ~ 7 3 )

= 7 3 ( 3 + 2 + 1 ) + -s/3 ( 2 - 2 — 11

+ 7 5 ( ~ 2 + 1 I + 730-= 6 - 7 2 - 7 3 - 7 1 + 730-

Exercise 6 ( a )

1. State which of the following are surd quantities :—

2. By placing the surd coefficient under the root sign, express the following as entire surds :—

3. Express as surds of the 12th order : —

( i ) ( i i i ) \''4.v-\ir .'

4. Express in the simplest form :—

( i ) \[l92, ( i i ) -^56, ( i i i ) \ / 3 6 ^ , ( i v ) 27720.

5. Express as surds of the lowest common order :—

( i ) 75T V u , f / 1 3 ; ( a ) w , yf3,

6. Arrange the following surds in ascending order of magnitude :—

( i ) 7 5 7 W V W ( i i ) 7"<T ^ 2 2 5 . ^ 3 4 3 .

7. Which of the following is the greater surd ?

( i ) m . ( i i ) J \ +\[3\ ( i i i ) ^ 3 .

( i v ) V32. ( v ) Tr. (Vi )

( i ) 5 ^ 2 , ( i i ) A ( i i i ) (a+b) J \ a-b a + b

SURDS : 1 5 5

9. Multiply and simplify :—

( i ) ( $2 + 1 ) ( ^ 4 - #2 + 1 )•

( i i ) ( a2 + l ) ( a 3 + V 2 7 a + l ) ( a 2 - 1 ) ( a 2 - N / ^ a + 1).

( i i i ) (\[a + x +\[a - x) (\[a + x -\(a - x).

( i v ) ( V ( v / T T - \[5~+sf2).

( v ) (yfa+sfb) (Va+Vb)(Va - W -

( v i ) ( ^ 2 + ^ 4 ) ( 2 ^ 2 + ^ 4 - 2 ) .

10. Find the values of :—

( i ) 4^63"+ 5\ /7 — 8 ^ 2 ^

( i i ) 3^162 - 5 ^ 3 2 + ^ 1 2 5 0 ?

( i i i ) 3^147 +

( iv ) V C ^ + T ? - \[x*~+~x*y - \ [ x f T y K

11. Given that 2 • 449489 find the values of the following to three places of decimals:—

l ^ J - T * 7 = ' ( i i i ) V™-** 2 V600

12. Without using approximate values, determine which of the expressions is the greater

( i ) 2-v/6~ orV5~ + \ fT, ( i i ) s f T ' + y f T ' or V3~ + \ [5 .

13. Find the rationalizing factors for each of the following :— ( i i ) ^ T - l .

(iii) \[5 +\f2 - \ f f . ( i v ) i/~5+

( v ) <^3 + V^2. ( vi) 2\[T ~ 3\f2.

14. Rationalize the denominators and simplify :—

7 + 7 ~ • i 1 ) .— + ,— > 3 +N/5 3 - N/5

( i i i )

( V )

(v i )

ALGEBRA

1 1- 2

^ 9 + ^ 3 + 1 ' V9 - ^ 3 + 1

1 1

1 + \ / 2 - V 3 1 + V 2 + V 3 '

A + b

Va + Vb

5 • -\J~6

15V3 - 2%/32 + 2^50 - 8V12 7

\[2 + i/2 + 1

15. Prove that :—

( i ) _ 3 _ = - 1 ; ^ 1 6 + ^ 4 + 1

( i i ) - _ 5 _ 3 = 2 ; ^ 1 6 - ^ 4 + 1 ^ 1 6 + ^ 4 + 1

( i i i ) + L _ ^ 1 6 - ^ 1 2 + ^ 9 ^ 1 6 + ^ 1 2 + ^ 9

( i v ) __ L + - = = ^ 4 a

^ 2 5 - ^ 1 5 + ^ 9 ^ 2 5 + ^ 1 5 + ^ 9

Simplify and show that :—

16. ( 7 ~ 2 ^ 5 ) (31_+ 13v/5~) = 2 g

3 + \ f 5

17. _ 1 _ | 5 _ I _ 2 3 _ = 3 ^ 3 • 2\f5 - 3\[2 3\f2 - 2 ^ 7 4 \ / 7 - 2 ^ 5 2

18. + _ 6 _ 4\[3 _ =

3 + V 6 2 ^ 3 + ^ 6 V 2 + V 6

19. — = ^ = 17\[3 - 2\f32 + 3^18 - 4\/48

SURDS : 157

20. Show that the following is a surd.

2\f2 _ 1 _ 7 5 + 7 3 \ f 2 + ^ 5 \ [ 3 - \ [ 2

21. Answer the following as required.

( i ) Show that "*" ^ ^ is an irrational number

between 4 and 5.

( ii ) State an irrational number which is not a surd.

( i i i ) Write the following numbers in the descending order of magnitude.

• y - , - V7/2, 2-9, 2\f2, - 2-1.

(iv) Is "79/169 a surd number ?


1. ( i ) , ( i i i ) and (iv ).

2. f i ) 7250 . m J ~ o r V - i , v 45 * 13

3. ( i ) ( i i ) ^Vz ' * 8 y 1 6 . ( " i ' ^ V ^ T ^ y K

4. ( i ) Ss/3 , (ii 2 V 7 , ( i i i ) 6 « 7 « , ( i v ) 24

5. (i) Vl25, V m . Vl3, ( ii ) VST V V& 6 ( i ) V 9 , M l , (ii ) 7(5, V343 V225.

7. ( i ) V3~ ( i i ) - 4 ( i i i ) V 5 V 2

8 . ( i ) 2 V t 2 , ( " ) - J - V 2 4 , (in ) Ja.

9. ( i ) 3, ( i i ) < ? s - 1, / i i i ) 2*. ( iv ) 14 -2^55, ( v ) a - 6, ( vi) 6.

10. ( i ) 7 7 . (ii) 4 $2. • (iii)2l<j3, ( i v ) 0. / i ^ 4 £/£

I t ( i ) 1-225 , ( i i ) -122, ( i i i ) 4-899.

12. ( i ) 27<T ( i i ) 7 2 + Vt".

1 3 . ( i ) V J - V y , ( i i ) 3 7 9 + 7 T + l .

(iii) ( - 7 T + V ~ - 7 7 ) ( 7 7 - J T — j T ) ( 7 7 + 7 T + 7 f )

or 7 I o (73" + 7 2 + 7 7 ).


(iv! ( V 5 - J 2 ) ( 5 V 5 + 2 V25 + 4 ).

[ Hir.t :—First remove t h e quadra t ic surd ]

( v) V~9 - V 6 + V 7 . ( v i ) 2 V T + 3 VT.

i4. (i) 3. (H) yr, (iii) We, (iv) y ^ - i / ^ - v p ,

( v ) - i . (7>/2 + V 3 ). ( v i ) I + 2V2 - 3 Vz + V<$ •

20. 2>/i0.

2 1 . ( i ) On rat ional izat ion the n u m b e r is 7 - -J6. No te tha t J 6 l ies between 2 and 3 and hence the result ,

( i i ) 7r, e,V1+Jf etc

(iiil Express the number s in the decimal notat ion u n d e r the root s ign.

2-9 > 2V2 > 14/5 > - V7/2 > - 2-1. (iv> T h e number is 3/13 and hence not a surd .

12. Properties of Quadratic Surds.

As the quadratic surds are the simplest surds, we consider some of their properties in this article.

Theorem I If a + \!b = x + sjy where a and x are

rational and \fb and \ f y are surds, then a — x and b = y.

If a is not equal to x, then let a — x +- m where, since a and are rational, m is a rational ^ 0.

Since a + \fb = x + \ f y we have

x + m 4 - \fb = x + \ f y .

. ' . m 4- \fb = yfy .

Squaring, in1 + 2m \b + b = y .

Or 2m\fb = y — b — m2.

As m --i- 0, wc can divide by 2m; we then get «

_ v - b — in2 , . •sjh fm ( ' }

Sincc yjh^y are surds, b and y are rational; also m is rational.

Hence the R. H. S. = a ratioaal number.

SURDS : 1 5 9

Therefore the equation ( i ) implies that \[b = a rational which is impossible.

Hence our assumption that a x can not be valid. Hence a = x. If a = x we get from the equation

a + \fb — x + sJy, that \[b = \[y or b = y, Thus, a = x and b = y.

Note : It is absolutely necessary in using this theorem, to

remember that y[b and \[y are really quadratic surds and do not have only the form of quadratic surds; otherwise by equating the rationals and surds of both sides in the equality of the theorem we may get absurd results.

For example, 2 + si 16 — 3 +\J 9, each being equal to 6.

But 2 3 and 16 9 as \ [ 16 and \ [9 are rational numbers 4 and 3 put in the form of quadratic surds; and hence the theorem cannot be applied to equate the rationals and irrationals of both sides in the equality.

Thus in any equality involving rational and a quadratic surd numbers on both sides, the rational parts on both sides can be equated so also the surds.

Cor 1. If \fa =-- b +\[c where b is rational and \fa, \fc are surds, then 6 = 0 and a = c.

VVe may deduce this result from the theorem 1 or may establish it independently as follows :

We have \j 'a = b 4 Vc.

Squaring we get, a = b2 + 2b \Jc + c.

On transposition, 2b \[c = a — b2 — c. Nov/ either b = 0 or b 0.

If we suppose b 0, we get = ^

This is impossible since \ f c is a surd and hence can not be equal to a rational number.


Hence b — 0 and consequently it follows that a = c. Thus, a quadratic surd can not be equal to the sum or diffe-

rence of a non-zero rational number and a quadratic surd.

Cor 2. If a + \fb = x ± \ f y , then under the same con-ditions of theorem 1, we have

a — \fb — x — \[y. Theorem II. If ( i ) a, b, x, y are all positive rational numbers,

sfb a quadratic surd,

and (ii) ^Ja + \[b = \[x

then ^Ja - ^ b = + (s/x - \ J y ) .

V J a + a/6 + \ f y ,

we have by squaring both sides, a + \fb — x -+- y 2sfxy.

. ' . by theorem I, a ~ x + y a n d \ f b = 2 \fxy.

.". a - \[b - (x -f y ) - 2 \ [ x y = ( \ f x - sfy )2-

•'• J a ~ ^ ( V * ~ >/">)•

It is customary to wirte only the positive square-root;

i. e. if x > y, AVC take the square root as (\[x — 4y )

if y > x, we take the square root as {\Jy — \ f x ).

13. Illustrative examples.

Ex. 1. If x = 2 - find the value of 3z» - 14*2 + 11* - 1.

I t will be tedious to find the value of the expression by actually subst i tut-i n g the value of the variable x. I t will be found convenient to find the value of the expression as fo l lows:—

W e have x = 2 - J3. x - 2 = - JT.

: . by squar ing both sides, x' — 4* + 4 = 3, * 2 - 4 * + l = 0. ( i )

SURDS : 161 Now the given expression

= 3*s - 14x2 + 11* - 1 = 3x (x' ~ 4* + 1) - 2*a + 8x - 1 = 3x (x' - 4* + I ) - 2(x3 - 4x + 1 ) + 1 ' = 3* [ 0 ] - 2 [ 0 ] + 1 = 1. [ b y ( i ) ]

Ex. 2. If 2x = A / ~ r - J — , prove that ' - jl - « + &. V 6 v a x + J l + x '

Exp. =

2 a -Jx + x%

2a Vl -+ x3 x- -Jl + xa

x+JT+x* x- VFTi3

2j s/l + *2 [* - Jl + x1 ]

a Now * =

2 J ab

4 ab 4ab

2~J ab

Exp. = — 2a 2 J ab t a — b _ a + b I

2 Jab 2 Jab J 4ab ( a + b ) , ,

= — T a b = a + 6 '

Ex. 3 If a j z + b J 3 = 0 , where both a and 6 are rational, then show that a = 0, 6 = 0.

On squaring the given relation, we have_ ( 2®a + 3Z>2) + 2ab Je = 0

This is possible if, and only if, the rational and irrational parts are each equal to zero.

/ . 2a2 + 362 = 0 and 2ab J6 = 0.

Now 2aa, 36a are both non-neggtive and hence their sum can not be equa to zero unless each is equal to zero Hence a = 0 and b — 0

The product involving ab, then, automatically reduces to zero Ex. 4- If a and 6 are rational and

3 + n/20 - V45 = a + b V 5, 7 W 3 2 0 - 125 find a and 6.

3 + ^ 2 0 - V45 3 + 2 J5 - 3 Jl W e have • ^ r - = 7 = ~

7 + 320 — v 125 7 + 8 ' 5 - 5 V 5

_ 3 ~

J + 3 V 1 C. A —11

1 6 2 : COLLEGE A L G E B R A

> (7 ~ 3 -J 5) 4

(21 + 15 > - 16 7 5 4

36 - 16 7 1 4

= 9 - 4 ^ 5 .

.'. by given data , 9 - 4 - V 5 = a + b -J5.

.". a — 9 and & = - 4 .

v J" + 1 1 Ex. 5. If x = -—; = — , show that * s + xy + r 2 = 35.

V2 - 1 y

™ u + 1 f 7 2 > 1 )! We have x = —: = , 0 n rationalization

V 2 - 1 1

= 3 + 2 2

c . ., , 7 2 - 1 ( 7 2 - 1 Similarly j> = -j.- = 3 - 2 V 2 v2 + 1

.'. x -r y — (> and xy = 1

Now + + y s _ ( ^ + y y _ ^ _ 3 6 _ j _ 3 5

Exercise 6 ( b )

1. If A- + a 7 y = 0, where ,v and a are rational and \f~y is a surd, show x — 0 and a = 0.

2. If a \[ 2 + b\[ 5 = 0 , where a and b are both rational numbers, show that a and b must both be zero.

3. If a \f~2 + £ 7 3 + c 7 5 = 0 , where a, b, c are rational numbers, show that a b = c - 0.

4. State the conditions under which the statement

" If a + \f b = c + 7 d, then a = c and b = d ". is true. Give an illustration to show that the result is not true when the conditions are not satisfied.

5. If a and b are rational numbers and

5+2sfi

7 +4\[ 3 find a and b.

= a •b\ l 3,

SURDS : 1 6 3


l + s f ^ M a a + b y l x

7 + V192 - V 4 8 find a and b.


( v r + l f = a + w 3, find the values of a and b and hence state the value of

( V 3 - 1 )3-8. If a and b are positive rational numbers and

(2 + y[3 yZ = a-b\l~, find a and b.

9. if " { ^ t ^ / J A = xV i , find the value of x.

\[5 + \f 3 10. Prove that :

( i ) x3 - 9x - 12 = 0, i f x = ^ T

( ii ) + is a root of the equation x3 — 6x — 6.

11. I f * = ^ T T 7 B + , 4 + \( \ 5

show that x3 — 3x + 8.

[ H i n t : Express x = ( 4 + s / l 5 ) 1 / 3 + ( 4 + \ [ lS ) " v ' ' ] .

12. Find the values of ( i ) a3 - 6«2 + 7a + 8, when a = 3 - 2 \J~T;

( ii ) 2a3 - 3a- - 5a + 3, when a = 3 + ^ 5- •

13. Simplify the expression yfl + x - \] I - x

V T and find its value when x = •


14. If a = , find the value of a2 — 10a + 1. 5 + 2 ^ 6

15. If a = \[~2 + \(T, show that ^ J - + = a — 1 a - f - 1

16. If a = 1 — _ and b = -——, 2 - y[3 2 + V ?

find the values of

( i ) 3a2 + Sab + 3b2 and ( i i ) az~bi.

17. I f a ^ l - ^ l and b , V3 + V"2 \f3 ~ yj2

find the value of a 3 + b3.

18. If _ a + find the values of a and * 3 - \[2

assuming that a and x are rational.

19. I f a + U T - \Sb-la~ ( b + 4 ) Vl25, where a and b are rational, find a and b.

20. Show that the real number lies between N/3 + \[2

5 and 6.

21. If x = show t h a t * 2 ( x - 1 4 )2 = 1. V 7 - 4 \ U

22. Find x and y if ( \ f x + yfy" f = 5 + ^24 , it being given that J x,-J y are unlike surds. Hence find k if

( 5 + \ f 2 4 f 2 + ( 5 - V24 f 2 = k yjT.

2 3 .

show that L = 4. x + V1 + *2

SURDS : 1 6 5

24. If a, b, c are rational numbers and

— J = = 1 + W3 + b \[S + c -s/TF. \[\5 - \(5 - N/3 + 1

show that a, b, c are all equal to 1.

25. If (x + *Jx2-bc) • (y-yly2-ca)'(z-4z2-ab), = ( x - \[x2-bc)-(y+\/y2-ca)-(z + \fz*-ab) prove that each expression = ± abc.

Answers. Exercise 6 ( b )

4. The condi t ions are : a, c are rational and Jb.-Jd a re surds .

I l lustrat ion : 2 +*Jl6 = 3 + *J9, each be ing equal to 6. Bu t 2 =F 3

and 16 =f= 9. since V l 6 , *J9 are no t real surds as requi red .

5 . (1 = 1 1 , 6 = 6. 6. a = 3 3 . 6 = - 1 9

7. <t = 10, 6 = 6. and (V3~- 1 )s = - 10 +

8. <r = 26, 6 = 15- 9 . * = 15.

12 . ( i ) 26-12*/!, ( i i ) 3 + V 5 " . 13. J 3. 14 . Zero.

16 . ( i ) 47, 30J3. 17. 970. 18 . a = y and * = y -

— 2 2 19. a = 11. 6 = ——. 2 1 . Hin t , Rational ize the denomina tor of the

expression unde r the radical sign ; we get * = 7 + 4V 3 -2 2 . fc = 18.

14. The Square and Square-root of a Binomial Quadratic Surd.

Let x and y be both positive rational numbers so that yjx + \ f y is a binomial quadratic surd. Let us assume that \ /x and yfy are unlike surds, so that sjxy is not rational.

We have ( \ /x + n/}02 = x + y + 2\{xy = p + 2 \[q, where x + y=p, and q=xy = a + yfb, where a = p, b = 4 q.

Thus, we observe that the square of a binomial quadratic surd \[x + Jy may be expressed as another binomial quadratic surd a +\[b. We now consider whether the square


root of a binomial quadratic surd a + \[b can be expressed as another binomial quadratic surd in the form \ f x + \ f y and if so, under what conditions, in the next theorem.

Theorem : The square root of a + \[b, where a and b are rational and by 0, can be expressed in the form of a binomial quadratic surd \[x +\[y where x and y are positive rational numbers i f , and only i f , a is positive and a2 — b is a perfect square.

Let \[cT+Jb=\fx + \fy, ... ( i ) squaring both sides, we get

a + \fb = x + y + 2\fxy. ... ( ii )

Now sfxy is not rational as otherwise \[x and \ f y will be like surds and r. h. s. of ( i ) will not be a binomial surd as given.

by equating rational and irrational parts from ( i i ) , by Theorem I of the article 12, we have

x + y = a ... ( i i i )

and 2 Jxy=\[b i. e. 4xy = b. ... ( i v )

In order to solve equations (iii ) and ( iv ) simultaneously, we may make use of the identity

( x - y ) 2 = (x + y)2-4xy.

( x - y ) 2 = a2-b.

.'. x-y= ± \ja2 — b. ... ( v )

Solving equations ( iii ) and ( v ), we have

* = \ -ja± V f l F ^ J - ; y = \ | a T - J a 2 - ... ( v i )

Since x and y are given to be positive rational numbers, therefore by ( i i i ) , a must be a positive rational number.

Again since x — y is rational, a2 — b by ( v ) , must be a perfect square.

S U R D S : 1 6 7

Thus, if a is a positive rational number and a2 - b is a perfect square, then by taking either the upper or lower signs in ( vi ), we have

J a + \[b = J ajW^HZ5 + j J a ~ •

Thus, the square root of a + \Jb where a and b are rational and b > 0, can be expressed in the form \ f x + \[y where * and y are positive rational numbers if, and only if, (i ) a > 0, and ( ii ) a2 — b is a perfect square.

Corollary *•

Ja~\[b = J a + ^c^b - J a - \l'a2-l),

if a > 0 and a2 — b is a perfect square. Note : In this chapter as per convention, we will only

write the positive square root of a — \fb.

Ex. Find the square root of 47 + V2200.

Let V47 + V 2 W = + J y .

squaring both sides, we get

47 +J22Q0 = x + y + 2 Jxy. .'. equating rational and irrational parts, we have

x + y = 47 . . . ( i ) , 2-jxy = V2200 i e. 4xy = 2200, (ii ) ( x — y )2 = ( x -f- y )* — 4xy.

:. (x -y)*= (47 )'2 - 2100 = 2209 - 2200 = 9, .'. x — y = ± 3. l iii)

Solving ( i ) and ( i i i ) , we get x = 25, y = 22 or x = 22, y ~ 25.

.'. the required square root is V23 + V 2 5 i. e. 5 + V22.

15. Method of Inspection. In many numerical cases the square root of binomial

quadratic surds can be found more conveniently by the method of inspection based on the identity

(^Jx±\Jy)2 = (x + y) ± 2\fxy-For (\fx ± \ f y ) 2 is of the form p + 2 \[q

[ if p = x + y, q = xy ].


(\/3c ± \[y )2 is of the form a ± \fb [ i f a = p , b=4q].

in order to find the square root of a ± \fb by inspec" tion, we have first to write it in the form p ±2\jq. We, then, have to factorize q into two factors whose sum is p. If x and y are two factors, then the square root of a ±\fb would be \Jx ± \[y. We will illustrate the method in the following examples.

Ex. 1. F ind the square root of 7 + 7 4 8

W e have 7 + ^ 4 8 = 7 + 2 V l 2

= ( 4 + 3 ) + 2V4 X 3

= (V4 +V3 )2.

V7 + 7 4 ? = V 4 + J F i e 2 + V 3 .

W e have only wri t ten the positive value of the sq. root .

Ex. Fir.d the square root of 2

W e have 2 - V3

Ji - 1 - W i - y / i j \ v i

/— . VF- 1 • Jg -V2 the square root of 2 — V 3 is ——— i. e. ^ •

V2 ^ V3 - 1 Note The square root really is ± — — — • But , by conven t ion , we V2

v r - 1 only write the positive square root and hence we have taken — — as the

V2 1 - VF

square root and not V2

E». 3 . F ind the four th root of 3 — + ~ > / 5 . 2 2

W e have 3 ~ + = y ^ 7 + V 4 5 j = ~ ^14 + 2 ^ 4 5 j

= i ^ 9 + 5 + 2V 9 ~ 5 j = -i- + J~5

SURDS : 1 6 9

v the square root of 3 — + — 7 5 is - - ( 3 + J 5 )

•'• the required f o u r t h root will be the square root of

H - ^ B « , ( , + V R ) - / L ( . W I

Note. In the above examples it can be easily verified tha t the condi t ions I i ) a > 0 and ( i i ) a"- — b is a perfect square, a re satisfied and hence the square roots are of the form 7 x ± 7 y If . however, these condi t ions are no t satisfied then the square root may take a different form as can be seen f rom the following example

Ex 4. F ind the square root of 6 + 4 7 3 .

W e have 6 + 4 7 3 = 6 + 7"48".

In this example a = 6 and is positive ; and a1 — b = 36 - 48 = - 12 and is negative. Hence the square root cannot be expressed in the form of a

binomial surd x + *Jy. W e can, however, proceed as follows :—

c + 4 7 F = 73~l 2 7 ^ + 4 ) = V3*(4 + 2 73") = J i ( -73 + 1 )a.

•'. the square root of 6 + 4 is ( 7 3 + 1 ) . " Ex. 5. F ind the squre root of 9 -76 - 6 V l 2

Sometimes a common factor of both the te rms of the binomial quadra t ic surd can be taken out before proceeding to find the square root.

Thus, 9 7 6 - 6 7 1 2 = 7 6 [ 9 - 6 7 2 ] = V 6 [ 9 - 2 7 } 8 ]

= 7 6 [ (6 + 3 ) - 2 7 < T x 3 ] = 7 6 [ 7 6 - 7 3 ] "

square root of 9 7 6 - 6 -7l2 is V 6 [ 7 6 - - 7 ! ] .

13 4 Ex. 6. F ind the square root of -r- + — • 3 7 3

W e have j + A = j ( 13 + 4 7 3 " ^ = y ^ 13 + 7 4 8 j

= j ^ 13 + 2 712 ^ - j ^ 712 + 7 F j •


.-. / » + « i ( v s + i ) . 3 V3 \ /

•'. t h e required square root is 2 + • V 3

16. The Square of a trinomial quadratic Surd.

We have the identities —

( V T + « / y + J ~ y = x + y + z + 2-Jxy + 2-Jyz + 2 Jzx.. ... ( i )

vy-W)2

= x + y + z + 2-Jxy - 2-Jyz - 2 Jzx. ... (ii)

( j ~ x _ v y - -Tz >2

= x + J' + z - 2\fxy + 2 y f y z - 2\fzx. ... (iii)

Thus, we observe that the square of any trinomial quadratic surd (\f x ±\[ y + V 2 ) is one of the forms

a + 2 \[~b + 2 + 2 \[~d or a + 2 \f 'b - 2 \f~c - 2 Jd where a, b, c, d are rationals. ( iv )

Conversely it may be possible to express the square root of the expression of the form ( iv ) in one of the forms of ( i ), ( ii ) and (ii i) . The problem of finding the square root of the expressions consisting of four terms one of which is rational and the three remaining terms, quadratic surds can be best illustrated by the following examples.

Ex. L Express as tr inomial surd the square root of

10 + JZA + Jwi + J 60.

T h e expression can be wr i t ten as

= 10 + 2 J 6 + 2 JlO + 2jl5. If the square root is + Jy + J z then

x + y + z + 2 Jxy + 2 Jyz + 2 JZx.

= 10 + 2 J~6 -f 2 JlO + 2 J l J .

.'. we have x + y + z — 10, ( i ) xy = 6. ( i i ) . yz = 10. (iii! zx — 15. i iv l .

SURDS . 1 7 1

Any three of t he fou r equa t ions are sufficient to de te rmine the three u n k n o w n s y, z. W e will solve i i i ) , I i i i ) and I iv ) and use the equation ( i ) to verify the values.

Mult iplying ( i i ) , ( i i i ) and (iv ) we have x Y z * = 6-10-15 = 2-3 x 2-5 x 3-5.

/. x i * = 2-3-5. . . . ... . . . ... ( v )

Div id ing ( v ) by ( i i ). we have z = 5.

Div id ing ( v ) by ( i i i ) , we have x — 3 .

Dividing ( v ) by ( i v ) , we have y = 2.

Since x + y + z = 3 + 2 + 5 = 10, the values of x, y, z satisfy ( i ) .

.'. the requi red square root is V 3 + V 2 + V 5 .

Ex . 2 . Express as a t r inomial surd the square root of

21 - V l 9 2 - V l 6 8 + V 2 2 4 -

T h e expression c a n b e wr i t t en as

21 - 2^48 - 2^42 + 2-756. Let the square root be J x - *Jy +*Jz so tha t

x + y + z — 2*J xy — 2 s / y z + 2«J zx

= 21 - 2«/4S - 2>/42 + 2^56 .

we have x + y + z - 21 ( i ) , xy = 48 ( i i ) , yz = 42 (iii) f

xe = 56 ( i v ) .

Fo r solving these equa t ions s imultaneously, we mult iply ( i i ) , ( i i i ) a n d ( i v ) .

W e have * V z 2 = 48-42-56

= [ 2 X 2 X 2 X 2 X 3 ] [ 2 x 3 x 7 ] [ 2-2-2-7 ] = 2 8 -3 s .7 a .

.*. xyz — 24-3-7 = 16-3-7. ... . . . ... ... ( v )

Div id ing ( v ) by ( i i ) , we have z = 7.

Div id ing ( v ) by ( i i i ) , w e have * = 8.

Div id ing ( v ) by (iv), we have y = 6.

S ince x + y + z = 8 + 6 + 7 — 21, the values of x, y z satisfy ( i ) .

.'. t h e requi red square root is = V8 - J b + V 7 .

= 2J2 - J6 +


Ex. 1. Simplify (4 + V l 5 )3 / 2 + ( 4 - V l S ) 1 ' 2 -

We will first find (4 + V B ) 1 ' 2 i. e. V4 W l 5 -

4 + V l 5 = I ( 8 + 2 V l 5 ) = -b ( J 5 + S ) ' -


W e have to evaluate a" + b'.

W e observe that a + b — JlO and ab = 1.

Now® 1 + 6 s = {a + ft)3 - - 3 a 6 (a + 6 )

- (JlO)s - 3-1 (VlO) = xoVio - 3 Vio = 7 JlO

:. ( 4 + V l 5 ) 3 / 2 + ( 4 - V l 5 ) 3 , 2 = 7 V l 0 -

E*. 2 . Prove that

1 1 2

Jl2-Jl40 VfT- V60 ViO+a/84

W e have 12 - V l 4 0 = 12 - 2 ^ 3 5 = [JJ - V 5 )a;

8 - J60 = 8 - 2jl5 = (Vs" - V3~)'; 10 + V84 = 10 + 2 V 2 I = (V7 + J3~) a .

V7 - V 5 V5 - V 3 vV +V3

V F + V s V i + V i 2 ( V t - V £ ) 2 2 4

1 2

W5 -Vs -J3 -Ji + J3 = 0.

= R . H . S.

Ex. 3 . Given Js = 2-23607 find the value of

10V2 V10+V18

Exp. = 10 Ji Jio + 3J2

3 J2 — (J 5 + 1 ) 2 J2 + ~ (J 5 - 1 ) Jz J2

SURDS : 173

= 10 x 2 _ 2-J 5 + 6 = 10 x 2 _ 2 ( 3 + J 5 ) 6 - V I - 1 4 + V5 —1 5— J 5 (3+ J~5)

20 (5 + J 5 ) . . = 25~- 5 ~~ • ' 0 n r a t l 0 n a " z a t l 0 n

= 3 + "v/ 5 —5-23607, ( correct to 5 places of decimals ).

Answers. Exercise 6 (c)

Express the square roots of the following as binomial surds:

( i ) 1 1 + 2 ^ 3 0 , ( i i ) 17 — 2 \[70,

( i i i ) 4 - ^ 7 7 (iv) 16 + 5 \ / T , ( v ) ~ - 2 V~2 •

(v i ) 5 3 - 1 2 sflO, ( vii) 62 + 20 V~6\ (viii) 3 - i - + \[T.

Find the square root of :—

( i ) 5 *J~2 + 2 \{l2. ( i i ) 15 V T —

( i i i ) 11 \fT + 28. ( i v ) x + V*2 - y2-

( v ) 2a — \JTa2-2ab-b2.

( v i ) a + Z> + c + 2Va6 + Z>c. ( v i i ) \[96 - \ f 7 2 .

Find the fourth root of :—

( i ) 56 - 24 -JT. ( i i ) 124-32\f l5~.

( i i i ) - ^ - N / T + 3 - i . ( i v ) l | - v T

Express as trinomial surds the square roots of : -

( i ) 37 + 1 2 ^ / T + 6 \ T I 4 + 4 x / 2 T

( ii ) 49 + 6-f6 -4\[l0 -12-Jl5.

<m> 2 i - 2 J i r - 2 J i + 2 J i r '

Show that ( 2 8 - 1 0 ^ 3 )1/2 - ( 7 + 4 \[3" )"1 / 2 = 3.


* t i . t . • I 6. Prove that — = = = = = = = —— • \[2 + V 7 — 3\[ 5 \j5

7. Find correct to three places of decimals the values of

( i ) if x / T = 2-236068 ; 7 - 2 V 5

( ii ) if J T = 1-414213

V17 - 2\[288

8. Show that J 3 + -IT + ^ 3 - \f~5 = Via

Hence or otherwise prove that ( 3 + V5")3 '2 + ( 3 - si5 f 2 = 4\/l0.

9. Simplify and show that

VlO - 2 + V 7 - 2 V l 0

10. + ^ 1

^ 2 + ^ 3 + ^ 5 V 2 - V 3 - V 5 2

n . = V 1 1 - 2 V 3 0 V 7 - 2 V 1 0 V8 + 4 V 3

12. Show that

( * } T [ 7 19 - 4 VI2 + 7 7 + 4 V T ]

is a rational number ;

( » ) J ^ + + V T - ^ 3 0 - 2^56 - VT2

is a surd. 13. Prove that

/ 4 - VT ~ j

2J~2+y/t-3j~f

SURDS :

14. Prove that

_ \[T 5\j5 _ >

V 4 T V T 5 V 7 2 ^ 2 \ / 3 5 4A/3 - 2\[7 2

15. Given -J3 = 1-732050

find the value of * V 2 ~ V6 —3\f3

16. Prove that

6 __ _ y f j + 2yj5 ^ ^6-^35 V 4 - V I 5 n / T ^ N / 2 1

17. Find the value of ( 3+\fJ )3/2 — ( 3 — \ / 5 )312

( 2 + V 3 ) M - ( 2 - V 3 ) 3 ' 2 ' Simplify and show that

19. t z i L . « 6VI4 . V2 — V4 — V 7 V2 + V4 + V 7

20. Find the value of 6 if

10 V2 _ _ _ y r o ~ + V l T = 2 + s f t .

-sfTF- n / 3 T N / 5 \f%+yf3-\f~5

21. If a, c are rational and

( ; ) 5 + 3\/3_ = a + b s f s h o w that a + 6 = 0. 7 + 4 \ /3

( i i ) J __ = a + /> v/2~+ c\T67 show that I - \ [ 2 - \ j 3

a + b + c = 0. 22. Prove that

7 + 3N/T _ 7 ~ 3 ^ . _ _ = 2N/27

V2 + V7+3VF V2 + V7 — 3\[ 5


23. Show that

J _ - / T + + \[5 + "irT^ 4y[l5 is rational.

24. Prove that

- 2 / 2 + 4 / 2 + >f5 - V8 - \jW = V T + n/T~

25. Prove that

3 + 1 ? = 1

1 - ^ 2 + ^ 3 1 - V 2 - N / 3 1 + ^ 2 - ^ 3 \ f 5

Answers. Exercise 6 (c)

1. ( i ) J6+J5. (ii) V l O - J T . (iii) 4 ( > / i 4 - a / 2 ) .

(iv) £ (5*j2+*jl4). ( v ) - g - [ 3 V 6 - 4 -v/T].

(vi) 3 J J - 2 J2. ( vii) 5 V2 -I- 2 V3~7 (viii) J [2 V 3 + 1 ] .

2 . ( i ) V ~ 2 [ V 3 " + V 2 ] , ( i i ) [2 V2 - V7 ] .

(iii) V 7 [ 2 + V 7 ] (iv) J L + J -

(v)—— + b - V a - b J - (vi) 4a~+~c + Jb.

(vii) 6 [V~3 - 1 ] ,

3 . ( i ) v ' l - l , ( i i ) J s - V f . (iii) i ( V l + 1). (iv) i ( V3~-J).

4 . ( i ) 3 J 1 + 2 V 3 + ' 1 . ( i i ) V j + 3 V3~ — 2

7 . ( i ) 11-472. ( i i ) 5-828. l 2 - ( » ) 6/5. ( i i ) J~2. 15 . 3-1732050. 17 . 8/5. 20. 5. 21. 3/4. 2 3 . 1.

Chapter 7

Quadratic Equations 1. Int roduct ion. 2 General method of solution of the quadratic equation.

3. Nature of the roots of the quadratic equation. 4. Factors of the quadratic expressions 5-6. The factor theorem and its applications. 7. Relations between the roots and the coefficients. 8 Symmetrical expressions of the roots. 9. The signsof roots 10 Irrational and Imaginary roots. I I . Cubic and Biquadratic equat ions 12. Illustrative examples . Exercise 7 ( a ) 13. Common roots. 14. Equat ions reducible to a quadrat ic . Exercise 7 ( b ) .

1. Introduction. The students are already familiar with simple equations

of the type 2x + 3 = 0 . This equation is of first degree in x and is generally called a linear equation. The general form of a linear equation, in one variable, is ax + b - 0, [ a = £ 0 ] Solving this equation we have x = — b j a. This equation has, thus, one root viz. — b/a. Any equation of the first degree in one variable can be put in the above linear form ax + b = 0 and hence can be easily solved.

The students are also similarly acquainted with equations of the type 2x2 — 5x + 2 = 0. The left hand side of this equation is a second degree expression in x and the right hand side is zero The second degree expression on the left hand side can be facto rized and the equation can be written as ( 2 x - l ) ( x — 2 ) = 0. Thus x = 1/2, x = 2 satisfy this equation and hence 1/2, 2 are the two roots of this equation. Any equation of this type, when the highest degree of x is two, is called a quadratic equation. Every quadratic equation must contain a second degree term in x, and may contain a first degree term in x and a term without x. Thus the general form of the quadratic equation can be taken as ax2 + bx + c = 0, where a, b, c are constants and a 0. In the above numerical quadratic equation, we have a - 2, b = — 5 and c = 2. C. A.—12 177


When the equation is so written that all the terms are on the left hand side and the right side is zero, then the left hand side expression is called the characteristic of the equation. The term c iii ax2 + bx + c = 0 is called the absolute or the constant term of the equation.

The students aie already aware of the process of finding the roots of the quadratic equation by factorizing the characteristic of the equation as shown in the above numerical case. We shall now proceed to consider in the next article the solution of the quadrat ;c equations in general.

2. General method of solution of the quadratic equation.

We first illustrate the method by the following numerical example.

Let us solve the equation 5x2 + 2x + 3 = 0. W e canno t easily factorize the characterist ic. Such an equation can be solved by a process known as completing the square. Transpos ing the abso lu te term to the right hand side and dividing by the coefficient of x a , we get

„ . 2 3 x + —x — — — •

5 5

We complete the square on the left hand side by add ing 1/25 to both sides.

, . 2 1 1 3 Thus, we have * + — x + —— = — — •

— ± -4- V -14; ••• - i

14 25

5 5

T h s method will now be used to Gad the roots of the quadra t ic equat ion in general.

• To solve the quadratic equation ax2 + bx + e = 0.

U is assumed that a + 0, as otherwise the equation would become a linear equation. Transposing c to the right hand side, we get

a.x2 + bx = — c. Dividi^g,by a, the coefficient of x2, [ a 0 ], we get

b c

QUADRATIC EQUATIONS : 179

Completing the square on the left hand side, we get

, , b . b2 b2 c X2 + .V + ~r -y = ~ •

a 4 a2 4 az a

b2 — 4 ac 4a2

b V / \[b2-4ac x2

X+2 a ) = 2 —

v -|. JL- 4- V - 4 a c

2 a ~~ 2 a

— b , Jb2 - 4ac x = -X— + r • 2a 2 a

- h ±\Jb2-4ac . . . * = nZ ( 1 )

By giving suitable values to a, b, c in the formula ( i ) , we can write down at once the solution of any quadratic equation ; and hence ( i ) is known as the general solution of the quadratic equation ax2 + bx + c = 0, where a, b, c are any real constant numbers { a 0 ) .

We illustrate the use of the formula ( i ) by the following examples.

Ex. 1. Solve 10*a - 13* - 77 = 0.

He re a = 10, 6 = - 13 and c = - 77.

- 6 ± v V - 4 ac W e have x = 2 a

_ - ( - 13) ± V( - 13) ' - 4 ~ 1 0 ) ( - 77 ) * 2 x 10

_ 13 ± J 1 6 9 + 3080 _ 13 ± ^ 3 2 4 9 13 ± 57 - 0 20 20

70 44 * = ^ r o r —^r ' 20 20

7 11 the requi red roots are and — •


Ex. 2. Solve 10*5 - 13* + 2 = 0,

He re a = 10, 6 = - O and c = 2.

^ - 13) +_V ( - 1 3 ) ' - 4 ( 1 0 ) 2) 2 x 1 0

13 t V i f r T - ' s o 13 ± J&9 20 20

. J 13 + N / ~ 1 3 - ^ 8 7 . . the required roots are ^ — and •

Ex, 3. Solve 10*a - 13* + 4 -9~ = 0. 40

Here a = 10, b = - 13 and c = 4 - 4 r = - - • 40 40

" ( - 13) + 13 )J — 4 ( 10 )/ M j

2 X 10

13 f V l 6 9 - 169 _ 13 f 0 20 _ 20

• 1 3 1 3 t he requi red roots are and —

Ex. 4. Solve 10*a - 13 c + J = 0.

H e r e « = 10, 6 = - 13 and c = 5.

_ - ( - 13 ) ± V ( - 13 )• - 4 (10 ) ( 5 ) -•• * - 2 x 10

_ 13 ± V169 — 200 _ 13 ± V ^ l l ~ 20 _ 20

13 + V ^ T T 13 - V ^ T F the requi red roo ts are ^ and j o '

Examples. Solve the following quadra t i c equa t ions :

( i ) 3 * ' + 2* - 4 = 0. ( i i ) x% - 2* + 10 = 0 .

( i iU * 2 - 2 * + 3 = 0. ( i v ) 48*a = 104* - 51. ( v ) 4*2 + 2a a = 3ax. (vi) abx3 -(a — b,'x-(a- 6)J = 0.

Ans. ( i ) — ( " ) 1 ± 3«. (iii) l ± - J T - j .

. . 3 17 , . 3 a±aj^23 , a - b b-a (.v) - r . j j - ( v ) g •

QUADRATIC EQUATIONS : 181

3. Nature of the roots of the quadratic equation.

The roots of the quadratic equation ax2 -f bx + c = 0 are given by

— b±\fb2 -4ac , „ 2 a

The nature of the roots depend on the value of b2 — 4ac i. e. on the numerical value of b2 — 4ac. Since b2 — 4ac discrimi-nates the nature of the roots, it is called the Discriminant of the quadratic equation. The discriminant is usually denoted by the symbol A , which is a greek letter pronounced as delta. In all the quadratic equations that we deal with, in this book, a, b, c will be rational. Assuming that a, b, c are rational, we obtain the following results about the nature of the roots.

( i ) If A > 0 and is a perfect square, then \ [ a is rationa and both the roots are rational and unequal.

(ii ) If A > 0, but not a perfect square, then V A is irrational and both the roots are irrational and unequal.

Thus from ( i ) and ( ii ) we see that if A > 0 then the roots are real and unequal.

( i i i ) If A =s 0, then = 0 and both the roots are rational and equal, each being equal to — b{7a.

Thus if A > 0, the roots are real.

( iv ) If A < 0, then \J A is imaginary and both the roots are complex.

The roots will be purely imaginary if 6 = 0 and b2 - 4ac < 0 i . e . if ft = 0 and a, c have the same signs.

Remark. If the coefficients a, b, c are assumed to be real then we have the following results.

( i ) If A > 0 and is a perfect square, the two roots are real and distinct. The roots will be rational or irrational according as

— b ±\[ A b^QjQg r a t ional or irrational. 2 a ( ii ) If A > 0, but not a perfect square, the two roots are

irrational and unequal.

1 8 2 '. COLLEGE ALGEBRA

7 - 1 1 the roots, - — , —— are rational and unequal-

( i i i ) If A = 0, the two roots are real and equal, each being equal to — b'la. They will be rational or irrational accor-ding as b[a is rational or irrational.

( iv ) If A < 0, both the two roots are imaginary. Obviously the two roots wiil be purely imaginary if a and c

have the same sign and b = 0. In general, the roots of the quadratic equation are real and

unequal, real and equal or imaginary according as A > 0, A = 0 or A < 0.

W e may now consider the nature of the roots of numerical quadratic equations given in the four illustrative examples of the last article,

( i ) 10*' - 13* - 77 = 0. Here, the discriminant A , ft1- 4ac = 169 + 3080 = 3 249. Hence A is a perfect square of 57.

7

the roots, - — ,

( i i ) 10*2 - 13* + 2 = 0.

Here, the discriminant A , 6a - 4ac = 169 - 80 = 89.

Hence A is a positive number, but not a perfect square. 13 + V89 13 - V89

.'• the roots ^ o " — • 20 "— are irrational a nd unequal.

( i i i ) 10*J - 13* + 4 ~ = 0.

Here, the discriminant A = 169 - 169 = 0. 13 13

The roots are found to be — and — which are real and equal,

( i v ) 10*' - 13* + 5 = 0. Here, the discriminant A. 6a - 4ac = 169 - 200 = - 31 Hence A is a negative number

13 + \ f ~ T l 13 - V - 31 The oots are found to be -q a n c ' 20 which are

imaginary ( or complex) .

Note. Sometimes the standard quadratic equation is taken as ax' + 2hx + 6 = 0. Proceeding on similar lines as § 2, the roots can be

- h i J h" - ab found as These roots, further as in 5 3 can be shown to

be real and distinct, real and equal, or imaginary according as h2 - ab > 0, K1 — ab = 0 or h1 = ab < 0 In this form of the equation h1 — ab becomes the discriminant of the equation. Student should obtain these results from first principles as in articles 2 and 3, when the standard equation of the qua-dratic is taken as ax1 + 2hx + b = 0

QUADRATIC EQUATIONS : l B 3

Examples.

1. Ascertain with the help o f the disciiminant the nature of the rccts of the following equations.

( i ) 3 * » - 4* - 1 7 = 0 , ( i i ) + 2x + 10 = 0.

Ans. ( i ) Real and unequal, ( i i ) Imaginary.

2. Answer the following questions for the quadratic equation

x* + 3ht + 3 6 = 0. ( i ) State the cond i t i ons satisfied by h' so that the roots of the

equation become real; ( i i ) State the possible values of h which satisfy the condit ions

In ( i ) .

Ans. <\) h1 > 16, ( i i ) h > 4 or h < - 4.

3 . Find the condit ions to b e satisfied by k, to that the roofs of

+ * + k = 0 become ( i ) complex, ( i i ) equal.

Ans. ( i ) k > 1 / 4 , ( i i ) * = 1/4. 4 . Find the value of k in x> - 6x + k = 0 if

( i ) both roots are equal ; ( i i ) both roots are real. Ans ( i ) k = 9, ( i i ) k < 9.

5 . Find the condit ions to be satisfied by h* and hence by h so that the roots of + 2/i.v + 4 = 0 become

( i ) real and unequal, ( i i ) real and equal, ( i i i ) complex. Ans. ( i ) h1 > 4 i. e. h > 2 or h < - 2, „

( i i ) = 4 i. e. A " ± 2 , (iii) h* < 4 i . e . \ h \ < 2 or - 2 < h < 2.

6. Find the possible values of h for which the roots of the equation x% + hx + i = 0 are real.

Ans. h > 2 or h ^ - 2-7. If the equation 3 * + k = 0 has equal roots, show that the

value of k is 9/8.

8 . Find the condit ions satisfied by k1 and hence by k so that the roots of the equation 3*' + 2k c + 3 = 0 become ( i ) real and unequal, ( i i ) real and equal, ( i i i ) imaginary. Ans. ( i ) k* >9 1. e. k > 3 or k < - 3, i i i ) k* = 9 i. e. k = i 3,

(iii) k' <9 i . e . , < 3 o r - 3 < f t < 3 .

9 . Solve the equation - 6x + 13 = 0 and represent the roots by points on a plane referred to co-ordinate axes,

Ans. Roots are 3 ± <2, Required points are ( 3 , 2 ) , ( 3 , - 2 > . 10 . If one root of kx' + 2x + 6 = 0 is 1. find the value of k and deduce

the other root. Ans. k - - 8, other root = - 3/4.

184 '. COLLEGE ALGEBRA

4. Factors of the quadratic expression. An algebraical expression, where the highest degree of the

variable is one, is called a first degree or a linear expression in that variable. The general linear expression in x is taken as ax + b. Similarly an algebraic expression, where the highest degree of the variable is two, is called a second degree or a quadratic expression in that variable. The general quadratic expression in x is taken as ax2 + bx + c. Here a, b, c are constants and x is a variable. The students are familiar with the factorization of such an expression in certain simple cases. The general method of factorizing the quadratric expression ax2 + bx + c will now be considered. We have

ax2 + bx + c = a a2 + — x + ~ L a «J r x 2 + i . x + i i - 1)2 + - £ . 1

a 4al 4a2 a J

( completing the square. )

where A = b* — 4ac.

• 11 L R r , b+ si A I r , b - \ T A I I .max>+bx + c = a [ x + — s r - J- ...I

As in the case of a quadratic equation the discriminant A ( i . e. b2 — 4ac) determines the nature of the factors of the quadratic expression.

Thus ( i ) If A is positive, the factors are real and different, ( i i ) If A is zero, the .factors are real and equal and the

expression is a perfect square.

QUADRATIC EQUATIONS : lB3

( iii ) If A is negative, the factors are imaginary.

( i v ) If A is a positive perfect square, the factors are rational and different, assuming that a, b, c are rational.

Since the quadratic expression when equated to zero gives the general quadratic equation, we can deduce the general solution of quadratic equation from the factors given in I.

Note : In obtaining the factors of any quadratic expression it is convenient to proceed ab initio instead of using the formula I.

Examples Resolve into two linear factors : —

( i ) x% — 8x + 11. Ans. t * - 4 + J i ) ( * - 4 - V T ) .

T - 1 + ^ 3 4 1 T - 1 - V34"l ( ii ) — 2x - 11. Ans. 3 * + ^ * + 3 I •

(iii) 2* + 3* + 5. Ans. 2 * + * + } I

5. The Factor Theorem.

Theorem I f « is a root of the equation ax7 + bx + c = 0, then x - « is a factor of ax2 + bx + c, and conversely, if x — * is a factor of ax2 + bx -f- c then « is a root of the equation ax2 + bx 4- c = 0,

( i ) Since * is given to be the root of the equation, it satisfies the equation ; and hence we have, a*2 + b« + c = 0.

ax1 + bx + c = ax2 + bx + c — ( a«J + b« + c) = a(x2 - «2) + b(x - «) = ( * - « ) I > ( * + « ) +*>]•

Hence x — * is a factor of ax2 + bx + c.

( ii) Conversely if x — <x is a factor of ax2 + bx + c, then <* is a root of the equation ax2 + bx -+ c = 0.

We have ax2 + bx 4- c = ( * — < * ) ( px + q) where p and q are some constants.

x = o< satisfies the equation ax2 + bx + c = 0. * is a root of the equation.

6. Following applications of the factor theorem should be noted.

I. To form the quadratic equation whose roots are given.


Let « and 0 be the given roots. Hence x - « and x — 0 are the factors of the characteristic.

the equation is (.r — <*) (x — 0) = 0. i. e. x 1 - ( « + 0 ) x + = 0. i. e. x2 — (sum of roots ) x -f product of roots = 0.

Thus the equation whose roots are 3, — 5 is ( * - 3 ) ( * + 5 ) = 0 .

i. e x2 + 2x - 15 = 0. i. e. x2 — ( sum of roots) x + product of roots = 0.

Example. F ind the equation whose roots are 3 + -Js, 3 - J57

W e have the sum of the roots = 6 ;

also the product of the roots = (3 + -JI) ( 3 - V J ) = 9 - 5 « 4 the equation ! s - ( 6 ) * + ( 4 ) = 0 .

i. e. v * ' - 6 * 4 4 = 0.

Thus if the sum and the product of the roots are known, the equation can be found.

II. A quadratic equation cannot have more than two roots.

Let the given quadratic equation have two roots « and 0. By the factor theorem, the equation can be written as

( x - « ) ( x - / 3 ) = 0 . If this equation has a third root y, distinct from x and 0,

then it must satisfy the equation i. e. ( 7 — « ) ( 7 — /3) = 0. But this is impossible since, 7 J= and y 0. Hence there cannot be a third distinct root for the quadratic

equation. Similarly it can be shown that a cubic ( i . e. third degree)

equation ax3 + bx2 + cx + d = 0.

has three roots and cannot have more than three roots and that a biquadratic fourth degree ) equation

ax4 + bx3+ cx2 + dx + e = 0 has four roots and cannot have more than four roots. See para 11.

In general, it can be shown that an equation of the nth degree cannot have more than n roots.


Exa nple. Form the quadratic equations whose roots are :

( i ) 13 + V s 7 13 - </5~; Ii i ) 2 + 7». 2 - 7i \

3 + V - 111 3 - s/ - 111 . 2 + V T 2 - V T (...) • — . — (iv) — f .

Ans ( i ) x' - 26* + 164 = 0, ( i i ) * ' - 4 * + 53 = 0.

( i i i ) 2*' - 3* + 15 = 0. ( i v ) ** + 18* + 1 = 0.

7. Relations betweeen the roots and the coefficients. If « and /3 are the roots of ax2 + bx + c — 0, then

<* + /3 = — — and « S = — • a a

The roots of ax2 + bx + c = 0 are — b ± \jb 4ac 2 a

, * -b + yjb2 ~4ac , -b - \fb2 -4ac Let <* = sr and /3 = ^ 2a 2a Hence, writing A for b2 — 4ac, we have

- ~b+\fX - b - \ ( X 2 a

_ -2b _ -b ~ 2a a

Similarly „ =

= & ~ A ^ b2~ (b2-4ac) 4 a2 4a2

_ 4 a c _ <r

~~ 4 a 2 — a

( i ) the sum of the roots = ~ b - c o e f f i c i e n t o f x

( i >

( U )

and ( ii ) the product of the root =

a coefficient of x2 '

c _ absolute term a ~ coefficient of x2


Example, Write down the sum and product of the roots of the equation 2*a - 3x + 7 = 0 and hence deduce the sum of the squares of i ts roots.

.Aus. The sum of the squares of roots = — 19/4.

8 Symmetrical expressions in the roots of the quadratic Equation.

An expression in « and 0 is said to be symmetrical if it remains unchanged by the interchage of <* and 0.

Thus «2 + 02 becomes 02 + «2 by the interchange of <* and 0.

Therefore *2 + 02 ;s symmetric in « and 0. Other examples of symmetrical expressions can be given as

« + /3, cc/3, + - 1 + . 1 , - L + i . .

The students will note that the expressions like o<3 + 02 a r e n o t symmetrical in * and 0, as they are altered by the interchange of * and 0.

If «, 0 are the roots of the quadratic equation

b c ax2 + bx + c - 0. then we have « + 0 = and <*Q = — • a a

Any symmetric expression in «, 0 can be expressed in terms of « + 0 and «0 and hence can be evaluated in terms of the con-stants a, b, c of the equation.

Thus ( i ) « a + j 8 » s ( * + j 8 ) » - 2 « 0 = 4 - 2 — = v ' a2 a a2

( i i ) «3 + 03 = (* + 0)3-3«0 (« + £ )

_ - b% + 3 b c - 3 a b c - b * _ ~~ a3 a2 a3

1 -t- JL - - b!a)2-2cla (U1) 1? + 0* ~ «202 ~ {c!a)2

— b2 ~ c l _ ~ ~ a2 ' a2 ~ c2


Tn a similar way any other symmetric expression can be expressed in terms of the constants a, b, c.

It should be noted that the symmetric expression can be evaluated without actually finding the roots of the equation.

Ex. 1. <x, /3 are the roots of 2x- - 3* + 5 = 0 ; find the values of

( i > <*" ( « ) i + Y '

Since P are the roots of 2 * 2 - 3* + 5 = 0, * + 0 = y . = - y - •

( ! ) ( « ' + - f - 5 = - Z - " .

We have ( « + / 3 l = ( « + / 8 ) - 2 * 0

, •• , ™ u 1 , 1 «8 4- & ' ( u ) We have —r- + — = „ . ' «» fj,

v o<S + = (« + £ ) ( - */3 + 3*).

[- 21 "1 = 63 4 J «

- 11

3 I - 21 I 63 = i r * L — J

• JL + J - „ « s + ft1 = - 63 x = ~ 63 _

«» ft /8» 8 125 125 The method of constructing the quadratic equation whose roots are

symmetric functions of the roots of the given equation is illustrated in the following examples.

Ex . 2. If 3 are the roots of 3 c' + 4* + 7 = 0. find the equation whose

« * P roots are —pr and • (j <*

Since *, 3 are the roots of 3*' + 4* + 7 = 0,

The sum of the roots of the required equation

= JL + A + _ (« + £ ) ' - 2«P P « ~ <*/3 «/3


16 14 16 ~ 9 — r x • 1 4

y 3 3 - 2 6

7 7 21

The product of the roots - — • — = 1. 0 «

Now the required equation is x2 - ( sum ) * + product = 0. ( by § 6 )

' e ' ** ~ ( ) * + 1 = e" 21*" + 2 6 * + 2 1 = 0

Ex. 3 . If <x, 0 are the roots of 2*J + 5x + 7 = 0, find the equation

whose roots are 2* + 30 and 3 x + 20

V 0 are the roots of 2*s + 5x + 7 = 0.

The sum of the roots of the required equation = {2* + 30) + {3* + 20)

Similarly, the product of the roots = ( + 3 3 ) ( 3 x + 2 3 ) = 6 (<«.» + fi') + l3*/3 = 6(« + 0,* + <x0

: required equati Now the required equation is x ' - ( sum] * + product = 0. ( by § 6 )

i. e. x' - [ ) * + 41 = 0 i e. 2x> + 25* + 82 = 0.

Ex. 4 . Find the equation whose roots are the reciprocals of the roots of ax2 + bx + e = 0.

Let <*, be roots of the given equation a .* + 6k + c = 0.

« + 0 = rJ, «/3= a The sum of the roots of the required equation

= _L + — x JL n.'b

<x 0 a c c

1 1 1 - a The product of the roots = — • — — — = — . 0 0(0 C Now the required equation is x1 - ( sum) x + product = 0 by § 6 )

i. e + ~ = 1 e ' cS + bx + a = 0.


If should he observed that the coefficients in th i s equat ion are reversed.

Two such equations are called reciprocal equations. Ex, 5 . Find the condit ion that the roots of the equation axJ + bx+c - 0

may differ by 3. Let « and « + 3 be the two roots.

sum of the roots = 2x + 3 = — ; ... . . . . . . C I J a 0

and the prodixt of the roots = «a + 3 * = ... .. ( i i )

The condition can be obtained by eliminating * in ^ i ) and ( i i ) . We shall obtain the value of « from ( i ) and will substitute it in ( i i ) .

/ . v b + 3a From ( i ) , * = - • 2 a

, . . , ( 6 + 3 a \ \ , f 6 + 3 « \ c substituting in ( i i ) , we have I 1 + 3 1 1 = — .

6a + 6a6 + 9a"1 - 6ab - 18aa = 4ao.

.'. 6a - 9aa = 4ac is the required condit ion.

9. The signs of the roots of the quadratic equation. If /3 be the roots of ax1 + bx + c = 0, then

— b c * + B = and «/3 = a ' a ( i ) When ft are both positive, « -f ft and- «/3 are both

positive. a, b have opposite signs and a, c have same signs

i. e. cr, c have a sign opposite to that of b. ( i i ) When <x is positive and ft negative, « being numerically

greater than ft, then « + ft is positive and «ft is negative. a, b have opposite signs and a, c also have opposite

signs. i. e. b and c have a sign opposite to that of a.

( i i i ) When « is positive and ft negative, °c being numerically less than ft, then « + ft is negative and «/3 also is negative.

a, b have same signs, and a, c have opposite signs, i. e. a, b have a sign opposite to that of c.

(iv) When * and ft are both negative then * + ft is negative and <*ft is positive.

.'. a, b have same signs, and a, c have also same signs, i. e. a, b, c have all the same signs.


10. Irrational and imaginary roots of a quadratic equation. Theorem. If one root of the equation ax2 + bx + c = 0 is

P + V<7. the other root is p — \[q (where a, b, c, p, q are rational

and W a surd ).

Since ( p \[q) is a root of ax2 4 bx + c = 0, we have a (P + \A7)2 + b (p + \[q ) + c = 0 ;

i. e. a (p2 + 2p\fq + q) + b(p + V? ) + c = 0.

( ap2 + aq + bp + c ) + \(q (lap + b) = 0.

This is possible if, and only if,

ap2 + aq + bp + c = 0 and 2ap + b = 0. ... ( i )

Now ( p - \[q ) will be a root of ax2 + bx + c = 0

if a ( p - yfq )2 + b(p - \Jq ) + c = 0,

i . e . if a (p2 - 2p\fq + q) + b(p -\Jq ) + c = 0, l. e. if ( ap2 + aq + bp + c) - \[q ( 2 ap + b ) = 0

which is true because of ( i )

p — sfq is a root of ax2 + bx + c = 0. Similarly, the following theorem can be proved. Theorem. If one root of ax2 + bx + c = 0 is p + iq, the

o ther root is p — iq where a,b,c,p, q are real and i is the imaginary unit.

Examples.

1 If 3 + i 2 is a root of ** - 6* + * = 0, find k if k is a real number. Ans. k = 13.

2. If - 1 - i J 3 is a root of the equation + 2hx + 4 = 0, state the other root and the value of A.

Ans. Other root is - 1 + i */3, h = 1. 3. Find a quadratic equation, with rational coefficients, one of whose

roots is 2 + ^3. Ans. x' - 4* + 1 = 0 4. Find a quadratic equation with rational coefficients, one of whose

roots is ( i ) 7 + 4 V f ; ( i i ) 3 - »' 4. Ans. ( i ) * 1 - 1 4 * + 1 = 0, (i>) - 6r + 25 = 0.

5 If one of the roots of x" - 4* + k = 0 is 2 + <3, find k. Ans. k = 13. 6 Find the values of b and c if one of the roots of the equation

* ' + &; + c = 0 i s ( i ) 2 + ^ 3 , ( i i ) 5 - 3 < . Ans. ( I ) b = - 4, c = 1 ; ( i i ) 6 = - 10, c = 34


11. Cubic and Biquadratic (quartic ) Equations. We have seen that if and ft are the roots of the quadratic

quation ax2 + bx + c = 0, then

<x + ft = — — and oc/3 = — • ... ( i ) a a v

Similar relations between the roots and the coefficients hold for equations of higher degree. In this paragraph we shall consider such relations for cubic i. e. equations of third degree and biquadratic or quartic i. e. equations of fourth degree.

Cubic equation. Every cubic equation when so written that the right side is zero and all the terms are on the left side must contain a term of third degree in x and may contain terras of second and first degree and a term without x ( i . e. a constant term). Thus the general form of the cubic equation can be taken as ax3 + bx2 +- cx + d = 0 where a, b, c and d are real number constants and a 0. Thus 2x3 + 3x2 — 4 = 0 is a cubic equation ; here a = 2, b = 3, c = 0 and d «= — 4.

A quadratic equation has two roots; we can generalise and say that a cubic equation must have three roots. Let us therefore assume that the cubic equation

ax3 + bx2 + cx + d = 0 ... ( ii ) has the roots «, ft and y.

On dividing by a ( a j= 0 ), the equation ( ii ) can always be

written as J;3 + — ~ x + — = 0. ... (iii) a a a

The equation whose roots are «, ft, y can be written as U - « ) ( * - ft) (x - y) = 0. v

i. e. - ( « + ft + y ) x2 + ( «ft + fty + y<* ) x - «fty = ... ( i v )

Equations (iii) and (iv) have 1 as a coefficient of x I f the equation ( i i ) has the roots ft, y then equations ( i i i) and ( iv ) must be identical. Hence by comparing coefficients of equations ( i i i ) and ( iv ), we have

^ a C. A . — 1 3


c «/3 + + 7« = _ ,

and a.By = — — • a

These relations between the coefficients a, b, c, d, of the cubic equation and the roots «, /3, 7 of it are similar to the relations ( i ) for the quadratic equation.

Let us write 2 <x (read as sigma alpha) for <* + /3 + 7 and 2 « 0 for <x/3 + fiy + 7 * so that the relations between the constant coefficients and the roots of 'the cubic equation ( ii ) can be expressed as

2 « = —, 2 «/3 = , and «/3 7 = - — • ... ( v ) a a a

Illustration 1. If °c, 7 are the roots of the equation 2x° - llx3 - * + 30 = 0, find 2o<> 2 x / i , «/3y and verify their values by actually solving the cubic equation. Find also the value of 2* a .

__. _ ^ b ( - 1 1 ) 11 We have 2. * — 5 = - - . a 2 2

2 x p = = - 1 H a 2

and 'xfi~Y = — = — 15. where •*, 8, y are the roots of the given

equation

We can however factorize the left side of the given equation since x — 2 is one of its factors ; we have, then,

( * - 2 ) (2*a - 7* - 15) = 0 ;

or (x - 2") [2x + 3) ( * - 5) = 0.

.'. the roots are 2, - 3/2 and 5

Hence 2 « = 2 + ( - 3 / 2 ) + 5 = ^ - ,

2«/3 = ( 2 ) ( - ' - { - ) + ( - T ) ( 5 ) + ( 5 ) ( 2 )


and « / 3 y = ( 2 ) ( ~ - f ~ ) ( 5 ) = - 15.

The relations are thus verified.

Now 2<v> = <*.» + yS" + >a = (<x + ft + y )a - 2 ( + fry + y * )

A1SO2«» = ( 2 ) 2 + ( - " I " ) =

Illustration 2. If /3( y are the roots of the equation x3 + x — 2 = 0,

show that = - <xfiy.

Here a = 1, 6 = 0, c = 1, d = - 2.

• 2* = _ = 0. 2«/3 = -L = 1, «/3y = - _ 2.

Hence = ( 2<x )a - 2 2 * / 3 = 0 - 2 = - 2 and «/3y = 2.

2 « a = - «fty.

It may be noted here that two roots of the equation are imaginary ; this can be verified by the students by factorizing the left side. ( * — 1 ) is one factor of the left side.

Biquadratic or Quartic Equation. It is evident that the procedure followed above for obtaining the relations between coefficients of a cubic equation and its roots can be generalized to quartic equations, i. e. equations of fourth degree. If the right side of a quartic equation is zero, the left side must contain a term in fourth degree in x and may contain terms in third, second, first degree in x and a constant term. Thus the biquadratic or quartic equation can be written as

ax* + bx3 + cx2 + dx + <? = 0 ... ( v i ) where a 0.

This equation, since a 0, can always be written as

x* + — x3 + — *2 + — * + — = 0. ... ( vi i ) a a a a

If the roots of the equation—four in number—be taken as <x, fi, y, S we can write the equation ( vii ) as

0 - « ) 0 - / 3 ) 0 - y ) ( * - 5 ) = 0 ;


i . e . as -+ « 0 y 8 = O ... (v i i i )

where 2 « = ° < + /3 + y + o, 2«/3 = oc/3 + «y + «s + Py + PS + yS,

and 2«/3y = «Py + + «yS + pyS. Since the coefficient of x* is equal to 1 in equations ( vii) and (viii), equations (vii) and (viii) must be identical.

by comparing coefficients of the equations (vii) and (viii), we get

2 « = — , 2»;/3 = — , 2 « / S y = — — a a a

and «/3y5 = ~ - ... ( ix )

where 2« = sum of the roots taken one at a time, 2«/3 = the sum of the products of two roots taken at a time

2oc/Sy = the sum of the products of three roots taken at a time.

Illustration 3 . Verify the relations ( i x ) for the equation

2x' - 3** - 27*' + 62* - 24 - 0. ... ( 1 )

If P> y , & are the roots, we have

5 ( - 3 ) = _3_ 2 0 , f l = _ £ _ = _ 2 7 _ a ~ 2 = 2 ' a 2 '

2 * / 3 y = - - c - = - = - 31 and «/3yS = ± = - = - 12. a 2 ' a 2

The students can check and find that ( x - 2 ) and ( * - 3 ) are factors of the left side of ( 1 ). Thus the equation is

( * - 2 ) ( * - 3 ) ( 2 * ' + 7* - 4 ) = 0 ; or (2x - 1) { * - 2 ) (x - 3) <* + 4 ) = 0.

Hence the roots are - ^ - , 2 , 3 , and - 4 ; and 2<x = - i - + 2 + 3 - 4 = 3/2.

The other results can be similarly verified.

Illustrat on 4 . For the equation 2x' + 3*a + 2A + 5 = 0, ... ( 1 )

show that =• 3 ~2<xdy. where « , P, y, d are its roots. Show also that the sum of the reciprocals of the roots is two-fifteenth times the sum of the squares of the roots.


We have 2<x = 0 as 6 = 0 ; 2<x$ = c /a - 3/2,

2<x/3y d - = - — 1 and = 5/2. at '• :. 2 « » t» ( 2 « ) * - 2 2 « / 3 = 0 - 2 (3/2) = - 3.

2ocJ = - 3 = 3 2«/Qy. Also the sum of the reciprocals of the roots

« « B y S

_ fty$ + «yS + <x/3S + <xfiy

_ _ n 1 L

- ~ 5/2 = 5

Hence = - - f - = £ ( - 3 ) = £ 2 * . Examples.

1. For the equation 2*"' + 3*a + 4* + 5

find the values of 2 « - , 2 — and 2 • « oc p

2 . For the equation 3** - 6* + 4 = 0 show that the product of the roots is one-third of the sum of the squares of the roots.

3 . For the equation 3x* + 6*1 - 9* + 5 = 0

find the values of 2 « « 2 — - and 2 • « «p

4 . For the equation

2xl + 3xs + 4*- + 5* + 6 = 0

find the values of 2 « a and 2 — - •

£ Hint. For the latter part, find 2 and 2 • J

5 . Find the equation with roots • , -jg- . and - y - if <x, Q, y, S

are the roots of the equation in Ex. 4.

Answers. 1. ( - 7/4), ( - 4/5 ) and (3/5). 3 . ( - 4 ) , 9/5, 4/5. 4 . ( - 7 / 4 } , ( - 23/36). 5 . 6x' + 5x* + 4*« + 3x + 2 = 0.


12. Illustrative Examples. Ex. 1. If the roots of the equation

J . + _ _ L +

x x + a m m + a

are equal in magnitude but opposite in sign, show that a 3 = 2m1,

. .. 2 x + a 2m + a The given equation i s — j — = —-. -* (x + a ) m (m + a)

x {x + a) [2m + a) - (Zx + a) m (m + a) = 0.

(2m + a ) x3 + [ a (2m + a) - 2m (m + a ) ] * - am (m+a) = 0. The root of this equation will be equal in magnitude but opposite in sign

if the snm of the roots is zero. a ( 2m + a) — 2m ( m + a ) = 0. a3 = 2m'.

Ex. 2 . Show that if x* + px + q = 0 has two distinct real roots, the equation + px + q + k (2x + p ) = 0 has also two distinct real roots for any non-zero real value of k.

V + px + q = 0 has two distinct real roots. its discriminant p3 - Aq > 0. ( i )

Now the second equation i s + * ( p + 2k) + ( q + pk) = 0. This equation will have two distinct real roots if i ts discriminant

(p + 2k)3 - 4 ( q + pk)>0 1. e. if p3 + 4pk + 4feJ - 4 q - 4pk > 0,

1. e. if p' — 4q + 4fea > 0 which is true, since p3 - 4q is positive due to ( i ) and 4&a is positive, being a square of a non-zero real number.

Ex. 3 . If one root of the equation ax3 + bx + o = 0

Is the square of the other, prove that

63 + a'o + ao3 = 3 aba. If one root is <*, the other root will be «"

W e have the sum of the roots x + «a = - — ( i ) a x

and the product of the roots i . e . , * 3 = ( H )

In order to obtain the required condition, we have to eliminate « from ( I ) and , i i ) It may be found mors convenient to eliminate x as follows:—

From ( i ) , we have <xa + <x + — = 0. a

:. («»)» + ( « ) a + ^ - j - j = 3«a-<x • •


o<e -p-* ' + ~ = 3 . — . ar a

c' c d1 c 6 T c r 1 ••• ~ r + — + —r = 3 — ••• = — from («)

a a a a J

ffc2 + a'c + f>® = 3abc.

Ex. 4 . If the roots of the equation ax* + ax + c = 0 be in the ratio of p : q, prove that

Ji+JT-tJT-* the ratio of the two roots is j> : q, they can be taken as !>"*, gx .

We, then, have the sum of the roots, (p + q) « = = — 1 ... ( 1 )

and the product of the roots, pq <xJ =» — • ( i i )

We have to eliminate <x in ( i ) and ( i i ) to get the required condition. In order to eliminate « we find the value of <** from ( i ) and ( i i ) and equate the two.

— 1 o From ( i ) , <* = — — and from ( i i ) , «a = — r • P + q ap q

, ( A + , )» = . (P + q? "Pq ' IP " c

!a P + q=

japq^ a

Vi+Vf-W-f-+/F */?-<••

Ex. 5 . If the two values Of y obtained from the equations ax1 + by3 = 1 and ax + by = 1 are equal show that a + b = 1.

We have, ax + by = 1. ... ( i ) . ax' + by3 = 1. . . . ( i i )

We will eliminate * from ( i ) and ( i i ) and get a quadratic in y whose discriminant will be equated to zero to obtain the required condition.

1 — by From ( , i )»* •'• substitutin a

tituting in ( i i ) , a ^ — V j + by* = 1.

(1 - by)3 + a by' - a = 0.

b i a + b ) y> - Zby - a - 1) = 0.


The discriminant of this equation in j>, is zero if

4 6 ' + 4 6 ( a + b) ( a - 1 ) = 0 ;

i. e. if 6 + {a + b ) (a - 1 ) = 0 ;

i , e. if b + a(a + b)~a — 6 = 0 ;

i . e. if a + b = 1.

Exercise 7 ( a )

1. Solve, by using the formula or otherwise, the following equations:— ( i ) ( x - 1 ) ( x - 3 ) = 9800. ( ii ) x + x"1 = a. ..... 1 . 1 2 ( i n ) —r 1 x + a x—a a ( i v ) (a - b)x2+ (b - c)x + ( c - a) = 0.

( \ -J— 4. _ L _ - _ L 1 v ; x + l + x + 2 ~ x + 3 ' (vi) fc ( x2 —10 ) = x ( 2 — 5k2 )

2. Prove that the roots of the following equations are real, ( j ) ( x - a ) ( x - b) = <2

( ii ) x2 - 2ax + a2 - b2 - c2 = 0. ( i i i ) ( a - f c + c ) x 2 + 3 ( f l - Z > ) x + { a - 6 - c ) = 0 . ( i v ) a ( l - x - x 2 ) = 6 ( 1 + x) ,

3. Show that the roots of the following equations are rational. ( i ) (b + c)x2-(a+b + c)x + a= 0. ( i i ) ( a 2 — b2~) ( x2 + 1) = 2 ( a2 + b2) x.

4. Find the values of k, if the following equations have equal roots. ( i ) x2 - 2 ( 1 + 3k ) x + 7 ( 3 + 2k ) = 0. ( ii ) ( 2k - 1 ) ( x2 + 1 ) = ( 2k + 3 ) x.

5. Prove that (_x-a)(x-b) + ( x - b ) ( x - c ) + (x-c)(x-a) = 0 will have equal roots if, and only if, a = b = c.

6. Prove that the roots of ( a - f c ) 2 x 2 + 2 ( a + fc-2c)x+l=0 are real or imaginary according as c deos not or does lie between a and b.


7. Prove that the roots of

(a2 + ft2)*2 - 26 ( a + c ) x + (fc2 + c*) = 0 will be equal if, and only if, a, b, c are in continued proportion.

8. Prove that the roots of 9x2 + 4ax + 4 = 0 will be imaginary if, and only if, a lies between — 3 and 3.

9. The sum of the roots of the equation —J— H j-r- = — x+a x+b c

is zero. Prove that the product of the roots is

- i - 0 2 + Z>2).

10. Find k, if the roots =*, ft of 3x2 - (2Jc+1) x + (fe + 1 1 ) = 0 are such that « — ft = 1.

11. Find h, if one of the roots of the equation

k ( x - 1 )2 = 5x - 7 is .double the other. 12. Guess one root of the equation, and find the other root.

( i ) ( 2x - 49 ) ( 2x - 51 ) = 123 X 125. ( i i ) (b + c-2a')x2 + (c + a-2b)x + (a + b-2c') = 0.

13. If «, ft are the roots of 2x2 + 3x + 7 = 0, find the values of ( i ) «2 + /S2, ( i i ) « 3 + )ff$, ( i i i ) ( i v ) ( v ) ( « 2 - / 3 y + ( P > - « y , ( v i ) ( 2 « - f 3 ) - 2 + ( 2 / 3 + 3 ) - 2 , ( v i i ) « * - / 3 3 .

14. If the sum of the roots of a quadratic equation is 5, and the sum of their cubes is 20, find the equation.

15. Form the quadratic equations whose roots are

J _ rz s~S>fa+\lb \[a - J~b ( 1 ) a + \j b, a — \Jb; (n)y-~ , V a — V b V a +\] b

16. If «, ft are the roots of x2 - 2x + 3 = 0, form the quadratic equations whose roots are

( i ) * + 2, /3 + 2; ( i i ) 2« — 1, 2$ — 1; (iii) ~ , -2-; P *

( iv ) 2« + 3ft, 3« + 2/3;


17. If <* and <3 are the roots of the equation 3x2 + 6x +10 = 0, find the equation with roots « + 1 and /3 + 1. Hence find « and /3.

18. Find k, if ( i ) the roots of 2x2 + 3x + k = 0 are equal; ( ii ) one of the roots of the equation x2 - 4x — k = 0

i s 2 ( l + V " 3 ) ; ( i i i ) one of the roots of the equation x2 — 6x + k = 0

is 3 + i \ j2 . ( i v ) one root of the equation x2 — 6x + k = 0 is double

the other. 19. If /} be the roots of ax2 + bx + c = 0, prove that

the equation whose roots are a« + bS, b<* + a/3 is ( a x + i J ) ( x + i ) + c ( a - 6 ) ^ = 0.

20. If 2X2+kx + 3 = 0 has distinct real roots, show that the equation Sx2 — (k 2 — 24) x + 7 = 0 has no negative root.

21. If the roots of the equation ax2 + 6x + c = 0 are in the ratio p : q, then show that ac ( p + q)2= b2 pq.

22. If the two values of x obtained from the equations kx2 + 4y2 = 1, kx + 4y = 1 are equal, find k.

23. Find the values of the expressions ( i ) 2x* — 12x3 + 3x2 + 9x + 8 if x = H 1 + V3 ). ( i i ) 2x* — 10x3 + 37x2 — 37x + 40 if X = 2+3J".

24. Examine the signs of the roots of the equation ( A : - 2 ) x 2 - 3 x + ( f c - 5 ) = 0

when ( i ) k < 2, ( i i ) 2 < k < 5 and ( i i i ) k > 5. 25. For the quadratic equation x2 + 3hx + 36 = 0, find

( i ) the condition satisfied by h2, so that the roots are real;

( i i ) the values of h, which satisfy the condition in (i) ; ( i i i ) the other root if one of the roots is 3 (1 + iy [3 ) ; (iv ) the value of h when the roots are as in ( i i i ).

26. If « be a root of the equation 4x2 + 2x — 1 = 0 , prove that 4«3 — 3« is the other root.

27. If the ratio of the roots of the equation x 2 +px + q = 0 be equal to the ratio of the roots of x2 + P\X + qt = 0, show that p2qi e= pt

2q.


28. If the difference of the roots of the equation x2-px+q = 0 is the same as the difference of the roots of the equation x2 — qx + p — 0, then show that p + q + 4 = 0.

29. Two candidates attempt to solve a quadratic equation of the form x2 +px +q = 0 . One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and — 9. Find the correct roots.

30. If <x, ft, y are the roots of the equation 2x3 + 3X + 4 = 0

find the values of ( i ) 2 ( i i ) 2 — , ( i i i ) 2 - i -v ' <x v ' o<p

and ( iv ) . K2

31. Solve the equation 2x3 - l l x 2 + 17x — 6 = 0,

it being given that the product of two of its roots is equal to 1.

32. Solve the equation x1 + x3 — 7x2 ~ x — 6 = 0,

it being given the sum of two of its roots is equal to zero. 33. Solve the equation

2x3 + 7x2 + 2x — 3 = 0,

it being given that the sum of two of its roots is ~ .

34. Solve the equation 9 x 3 — 36X 2 + 2 3 x + 1 2 = 0 ,

it being given that one of its roots is half the sum of the other two, [ Hint: Let the roots be a — d, a, a + d ].

35. If ft, y are the roots of the equation x3 + 2x + 3 = 0 find the values of ( i ) ( « + fi) C/3 + Y) (Y + « ) , ( i i ) «3 + ft3 + y\

1 4 - _ i _ 4 . 1


36. «, /3, y, <5 are the roots of the equation * 4 - 1 0 x 2 + 9 * - 2 = 0

and 2 « J = n, <* y % = q and 2 = r. «P

Show that ( i ) p+l0q = 0; and ( ii ) p=4r.

Answers.—Exercise 7 (a)

I . ( i ) * = 101 or * = - 97. H i n t : 9800 = 98 x 100 and hence a ± — 4 <—

dednce one root. ( i i ) * = g ' ( ) x = — (1 ± -J5 ).

( iv ) * = 1, x = ——7" • H i n t : By inspection * = 1 satisfies the equation. a — b c — ct

.*., 1 is a root The product of the two roots is _ ^ , hence dednce the

2 other root, (v) * = - 3 ± V 2 . (vi) * =-7-; * = - 5k. k

4 . ( i ) k = ( i i ) 10. ft = 7, — 5, 9 2 6

I I . ft = 2, - 2 5 . 12 ( i ) * = 8 7 . - 37, ( i i ) 1 . " + 6 ~ - | C -b + a — ?a

13. ( , ) — T - ( „ ) (in) f . v )

( v ) ( v i ) i r - - (vi i ) - l 7 ^ 7 -14. - 5* + 7 = 0.

15. ( i ) * a - 2a* + a2 - b = 0.

( i i ) ( a — 6) *a — 2 (« + &)* + (a — b) = 0 .

16. ( ! ) ** - 6 * + 1 1 = 0 . ( i i ) * ' - 2* + 9 = 0. (iii) 3*" + 2* + 3 = 0. ( iv ) ** - 10* + 27 = 0. ( v ) 3*' - 2* + 1 = 0.

17 . * ' + j « 0 ; x = - 1 +

1 8 ( i ) k = ~ . ( i i ) ft = 8 . (iii) ft = 13 . ( i v ) ft ^ 8.


2 2 . k 3 2 3 ( i ) 4 - > / 3 . ( i i ) 3 ( l + »).

2 4 . ( i ) Both negative. ( i i ) One positive and one negative.

( i i i ) Both positive. 2 5 . ( i ) > 16 . ( i i ) 4 or < - 4 . (iii) 3 ( 1 - » V ? ) .

( i v ) ft = - 2. 2 9 . Roots are - 4 and - 3,

3 0 . ( i ) - 3 . ( i i ) - 3 / 4 . ( i i i ) 0. ( i v )

3 1 . 2, 3. 3 2 . 1 , - 1, - 3 2.

» H ^ - ' - 3 , - 4 - ^ . 3 .

3 5 . ( i ) 3. ( i i ) - 9 . ( i i i ) - J - '

13. Common Roots. If the two linear equations, say,

ax + b = 0 and a'x + b' = 0 are satisfied by the same value of x, then there must be a relation between the constants of the two equations.

The common value of x = — = — • a a

the relation between the constants is. - a'b = - ab' i. e. ab' - a'b = 0.

Similarly, we may expect that the constants of two qua-dratic equations will satisfy a condition when the equations are satisfied by the same value of x. We proceed to obtain such conditions.

I. To find the condition that ax2 + bx + c - 0 and a'x2 + b'x + c' = 0

may have a common root. If oe be the common root, we have

a*2 + b* + c = 0 ( 1 ) and a'«2 + b'« + c' - 0. (2 )

These two equations can be considered as two simultaneous equations in two unknowns «2 and «.


Multiply ( 1 ) by a' and ( 2 ) by a and subtract ( 1 ) from ( 2 ) ; we get

a'a<*2 + a'b« + a'c = 0, aa'«2 + ab'<x + ac' = 0

.*. ( ab' — a'b) <* + ac' — a'c = 0.

« = ( 3 ) ab — ab v '

Similarly multiply ( 1 ) by b' and ( 2 ) by b and subtract ( 2 ) from ( 1 ) ; we get

b'a«2 + b'b* + b'c = 0, ba'ofl + bb'« + be' = 0.

( ab' - a'b)*2+ ( b'c - be ) = 0.

•• ab' - a'b

By eliminating <x between ( 3 ) and ( 4 ) , we get the required condition; thus

O ' ~ fe'c> _ - M2 - ( ca - c q)2 . (ab' - a'b) ~ ~ l ) ~ (ab' - a'b)2

Cancel the factor ab' — a'b from the denominators and cross multiply. We get the condition as

( ca' - c'a )? = (ab' - a'b ) (be - b'c ) . ... ( 5 ) When this condition is satisfied the common root of the

two given equations is CCj, C-y, ( or which is the same as • c a ~ c a ( ab' — a'b \

- b'c \ - c'a j '

be -ca

In solving examples it will be found simpler to proceed ab initio without quoting the above result.

If <x, (8, and /3' be the roots of the two given quadratic equations, then

„ _ c ___ _ _c_ ca — c'a _ c ( ab' — ab ) a ' ~ a ab' — a'b ~ a {ca' — c'a)

, a, c' c ca — c'a c (ab' — a'b) and p = — r - v - « = —r -T- - n rr = ~ r - — , — • a a ab — ab a {ca — c a)


If. To find the conditions that ax2 + bx + c = 0 and a x2 + b' x + c = 0

may have both the roots common.

Let <x, /3 be the common roots, so that

, a —b —b' A O c c'

« -u ft = — = —j— and «/3 = - — • = — , - • a a a a

Therefore, — ~ A- a n d -— = • a a a a

.". — = = — are the required conditions. a b c

When these conditions are satisfied, we have a = ka, b' = kb and c' = kc where k is a constant. Hence the second equation can be written as k ( ax2 + bx + c ) = 0 . Therefore, the two given equations becon.e identical.

Ex. 1. Solve the equations 48*3 - 82* + 35 = 0, 66*a - 67* + 10 = 0,

it being given that they have a common root.

Let x be the common root of the two given equations. 48«2 - 82x + 35 - 0 ; 66 xs - 67* + 10 = 0.

Solve these two relations as simultaneous equations in xa and x ; we have on multiplying first by 11 and second by 8,

528 «a - 902 x + 385 = 0, 528 «a - 536 x + 80 = 0.

.-. subtracting - 366 x + 305 = 0.

= = ~ " * 366 ~ 6

The product of the two roots of first equation = • 48

5 7 Since the common root is the other root = -5- • 6 0

Similarly the product of the two roots of the second equation

10 5 2 = — i. e. ; and hence its other root is ---- • 66 33 11

Ex. 2 . Determine the value k for which the equation 2*a — 5* + k = 0 and 4* ! — 8* + k = 0 have a common root.


Let « be the common root of the two given equations

Then, 2xa - 5* + k = 0 ; ... ... ... ( i )

4x* - got + k = 0. . . . ... ... { ii )

Solve equations (i ) and ( i i ) as simultaneous equations in and x ; multiply (i ) by 2 ; we have

4x2 - 10x + 27: = 0 _ ( i i i )

Subtract ( i i i ) from ( i i ) and we get 2x - k = 0 or x = &/2.

Substitute in ( i ) . Then

7-1 k 2— 5 • — + k = 0 ; or k' - 3k = 0.

/. k ( k - 3 ) = 0 k = 0 or 3.

Ex. 3. Prove that if the equations x1 + bx + ca - 0, x* + cx + ab * 0

have a common root, their other roots will satisfy the equation a2 + ax + be — 0.

Let x be the common root. + 6x + ca = 0 ... ... ... ( i )

and «2 + c< + «6 = 0. ... ... ... ( ii ) Subtracting ( ii ) from ( i ) , we have x ( 6 - c ) + a ( c - 6 ) = 0.

[b — c) ( x — a ) = 0 <x, the common root, must be a. [ 6 cannot be equal to c ; otherwise

the two equations become identical. ]

a satisfies the first equation. .'. a2 + ab + ac = 0. .•. a (a + b + c ) = 0. .'. we have the relation a + b + c = 0. ... ... (iii)

the product of the roots is ca, the other root of the first equation is c.

Similarly, the other root of the second equation is b. the other two roots satisfy the equation

{x — b ) \x — c) = 0 i. e x* - (t + c) x + be - 0. x"1 + ax + be — 0. [ v 6 + c = - a from ^ iii ) ] .

14. Equations Reducible to a Quadratic.

Wc propose fb consider, here, some types of equations, the solutions of whicli can be made to depend upon the solutions of quadratic equations, by suitable devices. The few types, which are given be'ow, should be carefully noted by the students.


I. Equations reducible by inspection.

Ex. 1. Solve ( * + 1 ) ( x + 2) ( * + 3 ) ( * + 4 ) = 120.

"Rearranging the factors

} ( * 1) U + 4 ) $ ) (x + 2) (x + 31 \ = 120.

( x1 + 5X + 4 ) ( x 2 + 5x + 6 ) = 120.

Putting y for x1 + 5x + 4, the equation becomes y ( y + 2 ) = 120. y' + 2y - 120 =-0. (y + 12) (y - 10) = 0,

. . y = - 12 or y = 10.

x3 + 5x + 4 = - 12 or x2 + 5x + 4 = 10.

x1 + 5x + 16 = 0, or x1 + 5* - 6 = 0.

Using the formula for the first equation, we have

_ ~ 5 ± "J 25 - 64 _ - ^ 5 J - - 7 ^ 3 9 * - ' 2 ~ 2

Factorising the left hand side, the second equation is

( * + 6 ) ( * - l ) = 0 ; •"• * = - 6 or + 1 ,

the solution is given by - 5 ± V X i j

x = 1 or - 6 or t = ^ •

In general, an equation of the form ( * + a ) ( x + b) ( * + c ) (x+d) = k where the sum of any two of the four quantities a, b, c, d is equal to the sum of the remaining two, can always be solved in this manner.

II. Equations reducible by suitable substitutidb.

* 2 + 4 Substituting y for , the equation becomes y2 - 3y - 40 = 0.

••• < y - s ) ( y + 5) = o. y = 8 or j = - 5 ,

*2 + 4 0 x 3 + 4 . . — • = = 8 or - = - 5.

* x J

x* - 8x + 4 = 0 or x2 + 5x + 4 — 0.

+ R ± J 64 - 16 * = ^ or (* + 4) ( * + l ) = 0 .

x = 4 ± Jl2 or * = - 4 or * = - 1,

the solution is given by x = 4 ± 2 - 4 or - 1.

C. A.—14


Ex. 3 . Solve 4 1 + * + 4 1 " * = 10.

W e have 4-4* + 4 -4~* = 10. 4-42* - 10-4* + 4 = 0.

Let 4 x = y : the equation reduces to 4-y' - lOy + 4 = 0 .

2y' - 5y + 2 = 0, ( 2y - 1) ( y - 2) = 0.

1 X l X •'• y = y or y = 2. 4 => y or 4 = 2-

22* = 2 " 1 or 22* = 2. 2x = - 1 or 2* = 1.

- 1 1 .. a; = -r- or x = — • 2 2

III. Simultaneous equations.

The solution of simultaneous equations, one of which is linear, and the other of the second degree, can be obtained by solving a quadratic equation in one of the two unknowns.

Ex. 4 . Solve * + y - 2 = 0 . . . ... ... ( i )

+ y"- + 2* + 3y - 7 = 0. ... ... ... ( i i )

The general method of solving such equations is to eliminate one of the two unknowns x or y.

Frou ( i ) , we h a v e y = 2 — x

Substituting this value in equation ( i i ) , we get

x* + ( 2 - *)» + 2x + 3 ( 2 - ) - 7 = 0 .

x1 + (4 - + ^ e 2 ) + 2x + 6 - 3* - 7 = 0.

.'. 7x* - 5x + 3 = 0. ( x - 1 ( 2x - 3 ) = 0.

3 .. x = 1, or x = — • 2

If * = 1, then y = 2 — x i. e. y = 1 ;

and if x = ~ ; then y = 2 — x i. e. y = * • 2 2

ttie required solution is

3 1 x = 1. y = 1 or * = - , y = J '

Students who are familiar with Co-ordinate Geometry know that the pairs' of values of (*, y satisfiyir.g the equations i ) and ii) give the co-ordinates of the points of intersection in which the line ( i ) intersects the curve ( i i i.


Exercise 7 ( b )

1. Solve the equations 28x2 + 71x + 4 5 = 0 , 32x2 + 20x — 25 = 0,

it being given that they have a common root.

2. Solve the equations 104.x2 — 115x + 21 = 0, 7 2 x 2 - 103x + 35 = 0,

it being given that they have a common root. 3. Find k, if the equations

8x2 — 18* + 3fc = 0 and 6x2 - llx + 4k = 0 have a common root; and obtain the common root for this value of k.

4. Find k, if the equations 9** — 36x + 5k = 0, and 9x2 - 30x + 4k = 0

have a common root; and obtain the common root for this value of k.

5. If x2 + ax + b = 0 and x2 + bx + a = 0 have a common root, show that either a — b or a + b +1=0.

6. Show that the equations 2x2 — 7x + 3 = 0, 2 ( 2 + A:)x2 + ( 4 + 3A:) x-(3 + 2k) = 0 have a common root for any value of k. Find the value of k for which the two equations have both roots common.

7. If x2 + ax + be = 0 and x2 4- bx + ac = 0 have a common root, show that their other roots satisfy x2 + cx+ab = 0.

Solve the following equations :— 8. 0 - 5 ) ( j c - 7 ) 0 + 4 ) ( x + 6 ) = 504. 9. 0 + l ) ( 2 x + 3 ) ( x - l ) ( 2 x - l ) = 63.

10. 0 + 2 ) ( * + 4 ) ( x + 5 ) ( x + 7 ) = 55. 11. ( 3 * 2 — 5JT + 4 ) 2 — 5 (3x 2 — 5x~) — 26.

13. 4* -33 -2*" 2 +2 = 0.

14. 9* — 4-3*+2 + 243 = 0.

2 1 2 : COLLEGE ALGEBRA /

15. x + ^ = * 2 - 2 j > J = l . 16. x = 2>> + 5, x2 + 4y2 — 3* + 14>' + 1 4 = 0 . 17. 3x - 2y = 7, xy = 20. 18. x1 + 4 j 2 — 1 — 30y + 80 = 0, xy - 6. 19. jc + j? = 4, xy = - 45.

20. x* - 7x* + 4x2 + 32x - 24 = 0, it being given that

3 + yf~5 is a root of the equation.

Answers—Exercise 7 ( b )

_ 5 _9 5 5 7 3 7 5 1. — I-y- and — . j . 2. - and

3 3. ft = 0 or 3 and corresponding common roots are 0, — "

2

4. ft = 0 or 4 and corresponding common roots are 0, — •

6. ft = - y • 8. * = 3, ~ 2, 8. - 7.

9. * = 2. ~25 . i ( - l ± /V4"7).

- 9 ±V - 19 - 9 ± 3-/5~ 10. - , — ^

1 1 O 1 11. * - 2 . - y , — fi •

- 4 ± VlO 7 ± 12. * =

2 ' 8

13. * = - 2. 3. 14. * = 2, 3.

15, je = 3, 17; y = 2, - 12 16 * = 1, 2 ; y = - 2, ~ 3 •

17. * = 5, ~8 ;y = 4, • 18. * = 6, 2, 4, 3 ; y = 1, 3 3/2, 2.

19. * = 9, - 5 ; y = - 5, 9 20. * = 3, - 2, 3 ± >/J.

Chapter 8

Method of Induction 1. Introduction. 2 Principle of Mathematical Induction. 3. The

sum of natural numbers. 4. The sum of squares. 5. The sum of cubes. 6. Illustrative examples. Exercise 8.

1. Introduction. Many important propositions or formulae depending on the

positive integer n can be established by a method known as the Method of Induction. We denote the formula or the proposition to be established by P ( n ).

2. Principle of Mathematical Induction. If a given proposi-tion P ( n ) involving n is true for n = 1 and if its truth for n = k implies its truth for n = k + 1, then P ( n ) is true for every natural number n.

Thus, proof by Mathematical Induction is completed in two steps :

( a ) The proposition P ( n ) to be proved is verified in the first step, for n = 1.

( b ) The assumption of the proposition P ( n ) for n = k is sufficient to show that P ( n ) is true for n = k 4- 1.

We now use this method to prove some important formulae in the following articles.

3. The sum of natural numbers. We prove that the sum of the first n natural numbers is

n(n + l ) 2

Here, P ( n ) is

1 + 2 + 3 + . . . . + « = ... ( i )

213


( a ) We must first verify that the formula is true tor n = 1. Put n = 1 in ( i ) ; the L. H. S. = I ;

the R. H. S. = 1 ( 1 + 1 } = 1.

L. H. S = R. H. S. and P ( 1 ) is true.

( b ) We must now show that the truth of P ( n ) for n = k* the truth of P ( n ) for n = k + 1 :

i. e. we have to show that the truth of P ( k ) , namely,

1 + 2 + 3 + = ... ( U )

implies the truth of P ( k + 1 ) , namely,

! + 2 + 3 + • . • + k + < * + 1 ) = i i ± m « ± i i t i j . . . (iii)

L,. H. S. of ( i i i ) = [ 1 + 2 + ••• + * ] + (ifc + 1)

k(k+l) = - ^ — - + ( ^ + 1 ) , due to the assumption ( i i )

= U + l ) I ^ + 1

- (k + l ) [ k + 2] ~ 2

_ ( * + i)[(A;+i) + i ] 2

= R. H. S. of ( i i i) . By the step (. a ), P ( « ) is true for n = 1. Therefore, by step

(b), P (n ) is true for n = 2, and hence for « = 3, and hence for « = 4 and so on for every natural number n.

4. The sum of squares. We prove that the sum of squares of first n natural numbers is

n(n + l ) (2n + l ) 6

Here P ( n ) is 1 2 + 2 2 + y + . . . + n 2 = r L ( n ± n i 2 n ± U ( . }

METHOD OF INDUCTION : 2 1 5

( a ) We first verify the formula ( i ) for n = 1.

Put » = 1 in ( i ) ; L. H. S. = I2 i. e. 1 ;

L. H. S = R. H. S. and P ( 1 ) is verified to be true.

( b ) We now show that the truth of P ( k ) , namely,

p + ^ + . - . + ^ W l l ^ , ... ( i i )

implies the truth of P ( k + 1 ), namely, 12 + 22 + 32 • • • • k2 + ( k + 1 V

= (k + \) (k+ 1 + 1 ) (2k + 1 + 1 ) _ (iU)

6

L. H. S. of (iii) = ( l 2 + 22 + 32 + • • • k?) + (k + 1 )2

= A- ( f c + 1 ) (2 fe+ l ) + ( k + J y 6

due to assumption ( ii )

_ k + X

~ 6

k + l

[k{2k + \) + 6 ( * + l ) ]

12k? + + 6 ]

__ {k+D (k + \ + l ) ( 2 f c + I + 1) - — ^

) = R. H. S. of ( iii).

From the two steps ( a ) and ( b ), P ( n ) is true for every natural number «.

5. The sum of cubes. We prove that the sum of cubes of first n aatural numbers is

n2 ( n + 1 )2


Here P ( n ) is

!3 + 2 3 + 3 3 + . . . + „ 3 = ! ! ! W ! . m > ( i )

( a ) We first verify the formula ( i ) for n = 1.

Put n = 1 in ( i ) ; L. H. S. = l 3 i. e. 1 ;

R . H . S . = 1 2 1 1 + U 2 = 1 . 4

L. H. S. = R. H. S. and P (1) is verified to be true for n = 1. ( b ) We now show that the truth of P ( k ), namely,

l 3 + 23 + 33 + - - - + A . . . ( i i )

implies the truth of P ( k + 1 ), namely, l 3 + 23 + 33 + • • • + &3 + (A: + 1 ) 3

= 1 4 ... (m) L. H. S. o f ( i i i ) = ( l 3 + 23 + . . . + ^ ) + (it + l)3

= + l ) 3 ,

by assumption ( i i )

= (k+A

l)2'[t<? + 4(k+ 1 ) ]

- ( & + l ) 2 ( f r + 212

4

- (fc + i ) 2 ( £ + T + l ) 2

T "

= R. H. S. of ( i i i ) .

From the steps (a) and (b), P ( n ) is true for every natural number n.

6. Illustrative examples. The results which are known before hand intuitionally or ;,iven otherwise

and depend only on positive integers can only be proved by Induction. It might happen that a particular proposition holds for all positive integral values of « ^ m ; in this case in the first step ( a ) we verify the result for w = m and


then proceed with the second step ( 6 ) as before. This point is illustrated in the illustrative example 4.

Ex. 1. Prove by the method of induction that P ( » ) : 3 + 7 + l l + + (4« - 1 ) = n (2n + 1) . ... ( i >

The first, second, third etc. terms on the L. H. S . of ( i ) are obtained by putting K = 1, 2, 3 etc. in ( ^n — 1) ; this shows that the last term on the L. H. S. of ( i ) is the ntb term of the series.

( а ) We verify the result for n = 1. Put n = 1 in ( i ) . L. H. S. = 3 ; R H. S. = 1 (2-1 + 1) = 3 .

/ . L. H. S. = R. H. S.; P ( 1 ) i s verified to be true.

( б ) We now show that the assumption of the truth of P ( f t ) , namely, 3 + 7 + 11 + • • • • + (4ft - 1) = ft (2ft + 1 ) . ... ( i i )

implies the truth P ('ft + 1) , namely, 3 + 7 + 11 + • • • • + (4ft—1) + (4 f t+3 ) = ( f t + 1 ) [2 (ft + l) + l ] . (iii)

L. H. S. of ( i i i ) = [ 3 + 7 + 11 + ••• + (4ft - 1 ) ] + (4ft + 3 ) = ft ( 2ft + 1 ) + (4ft + 3 ), by assumption ( i i ) = 2fe* + 5ft + 3 = (ft + 1) (2ft + 3 ) = (ft + 1) [2 (ft + 1) + 1] = R. H. S. of ( i i i ) .

We have thus proved that if the result ( i ) is true for n = k, it remains true for the next integral value of n namely ft+1. Since the result i s verified for n = 1 in step ( a ) , it follows that it is true by step ( b ) for n = 1 + 1 = 2 and hence for n = 2 + 1 = 3 and so on for all positive integral values of «.

Ex. 2 . Prove by the method of induction that

_L + JL + L + . . . . + 1 = . 1-2 2*3 3-4 « ( » + ! ) ( n + 1 )

Let P (n be : + ~ + i - + • -• + . 1 , •--; = r ~ X - P i - - ( 1-2 2-3 3-4 » ( » + 1) ( n + 1 )

( a ) We verify the result ( i ) fdr « = 1

Put « = 1 in ( i ) , L. H . S . = l T r V r ) = - T R ' S> = I T

L. H. S. = R. H. S P ( 1 ) is verified to be true. / ( b ) We now show that the assumption of the truth of P (ft), namely.

I l l 1 ft_ 1«2 2 *3 3*4 ft(fc + l ) — ft + 1 ' "'

implies the truth of P (ft + 1) , namely, J- + J _ + i 1 i 1 - f e + 1

1-2 2-3 ' T ft(fc + 1) (ft + 1) ( f t + 2 ) ft + 2 " " U ' "


L. H. S. of (iii) - + - » 3 + • • • + j ^ ] + [ ( ~ k + l ) \ k + 2 ) ]

= m + (IhTTW)' by assumption (!i >

fc' + 2ft + 1 ( f t + 1 ) i i + 2 )

(fe + i y (ft+l) (ft + 2 t

= t—r-4 = H- H. S of (iii '. k + 2

We have thus proved that if the result ( i ) is true for n = ft, it remains true for the next integral value of n namely k + 1. Since the result is verified to be true for « = 1 in step ( a ) , it follows by step (b ) that it is true for *» = 2, and hence for n = 3, and so on for all positive integral values of »,

Ex 3. Prove by the method of induction that 2 n

3 — 1 is divisible by 8 for all positive integral values of n.

Let P ( « ) be : 3""- 1 is divisible by 8. ... •-• (. i )

( a ) We verify the result ( i ) for n = 1.

Put » = 1 in ( i ) . L. H. S. = 3 2 - 1 = 8 which is divisible by s.

The result ( i ) , thus, is verified to be true for « = 1.

(b) We now show that the assumption of the truth of Pik namely,

3 2 f t - 1 is divisible by 8. i e.

32'1— 1 = 8m, when m is a positive integer, ... ... { i i ) i implies the truth of P (K + 1 ) , namely,

3-'i + 2 _ l is divisible by 8. ... ... ... ( i i i )

We have 3 — l = 3 - 3 - 1 2

= ( 8f» + 1) 3 - 1, by assumption ( i i I = ( + 1)9 — 1 = 72 m + 8 = 8 ( 9m + 1 ), which is divisible by 8.

We have thus proved that if the result ( i ) is true for n — k, it remains true for the next integral value of n namely k + 1. Since the result is verified for « = 1 in step (a), it follows by step (6) that it is true for « = 2 and hence for n = 3 and so on for all positive integral values of n.

Ex. 4 . Prove by induction the ' inequality I 1 + * ) " > ! + «* for « = 2, 3, 4, and * > - 1 , * =fc 0.


Let P( n ) be : ( X + * )" > 1 + nx. .. ... ... ( i )

( a ) We verify the result ( i 1 for » = 2.

Put « = 2 in ( i ) , L. H. S. = ( 1 + * ) 2 = 1 + 2* + x7-

R. H. S = 1 + 2 *

We obviously have 1 + 2x + x2 > 1 + Zx, since * 2 > 0

The result ( i ) is thus verified to be true for n = 2.

< 6 ) We now show that the assumption of the truth of P (k), namely,

(1 + * ) f e > 1 + ft*. ... ... — ( " ) implies the truth of P ( ft + 1 ) , namely,

( 1 + * ) f t + 1 > 1 + (fe + 1 ) * . . . . .» ( i » )

We have ( 1 + * = ( 1 + * )* ( 1 + * ) > ( 1 + ft*) ( 1 + * ) , doe to assumption (ii)

and * > - 1.

(1 + * > 1 + ( fe + 1) * + kx2.

:. ( 1 + * ),i+1 > 1 + ( k + 1 ) * , since fe*2 > 0.

We have thus proved that if the result ( i ) is true for n - k, it remains true for the next integral value of n namely k + 1. Since the result is verified to be true for n = 2 in step ( a ) , it follows by step ( b ) that it is true for tt = 3, and hence for « = 4 , and so on for all positive integral values of » ^ 2.

Ex. 5 . Prove by induction that 2 n > n for every natural number » .

Let P ( n ) be : 2** > « . . . . . . . « » — ( » )

( a ) We first verify that P ( 1 ) is true.

Put n - 1 in ( i ) . L. H. S. = 21 = 2 ; R H. S. = 1.

V L. H. S > R H. S. , the result { i ) is verified to be true for » = 1.

( 6 ) W e now show that the assumption of the truth of P ( fe ), namely,

2& > fe. . . . . . . ... — ( i « )

implies the truth of P ( fe + 1 ) , namely,

2* + 1 > fe + 1 (iii)

L. H. S. of ( i i i ) = 2 / i + 1 , „

= 2 -2 > 2 • fe, due to the assumption ( i i )

We notice that, for each natural number fe, it is true that 2fe > fe + 1, because 2fe = fe + fe and fe + ft > fe + 1 for every fe > 1

L H. S of ( i i i ) which is greater than 2ft i s obviously greater than ft+1, which is R. H. S. of ( i i i ) .


We have thus proved that if the result ( i ) is true for n = k, it remains true for the next integral value of n , namely, fe + 1 Since the result is verified to be true for n = 1 in step (a), it follows by step ( b ) that it is true for n = 2 and hence for n = 3 and so on for all positive integral values of n.

Exercise 8

Use the method of induction to prove the following results.

1. 1 + 3 + 5 + • • • • + ( 2 n - l ) = n .

2. 2 + 4 + 6 + + 2/7 = n ( n + 1 ).

3. 1 + 4 + 7 + - . . . + ( 3 « - 2 )

4. 2 + 6 + 1 0 + . . . . + ( 4 « - 2 ) = 2n.

5. 2 + 22 + 2 3 + f 2* = 2 n + 1 — 2.

6. I3 + 33 + 53 + • • • • + ( 2n - 1 )3 = n (2n - 1 ).

7. 1-2 + 2-3 + 3-4 + + « ( n + l ) = | n ( n + l ) (« + 2).

8. 1-2-3+ 2-3-4+ 3-4-5+ - - - + «(w + l ) (B + 2 )

= T ( " + 1 ) ( " + 2 > ( " + 3>-

9. 1 -3+2-4 + 3-5 + . . . + « ( « + 2 ) = ~ (« + 1 ) (2» + 7).

i n J _ , J _ , _ J _ , , 1 _ n 1U- 1.4 4-7 7-10 (3« — 2) (3» + 1) ~ 3 n + l '

11 1 + 1 -I- 1 I I 1

" " 3-5 5-7 T 7-9 ^ T (2/1 + 1) ( 2 n + 3)

n ~ 3^2n + 3) '

12 1 + 1 + 1 l I 1 A 3-8 T 8-13 T 13-18 ^ T ( 5 n - 2 ) ( 5 n + 3 )

— n ~ 3 (5rt + 3)


13. 24" - 1 is divisible by 15.

14. 5*" — 1 is divisible by 24.

15. 32" + 7 is divisible by 8.

16. 2 2 b + 1 + l i s divisible by 3.

17. a" — bn is divisible by a — b.

[Hint: a fe+1 - b™ = a ( a - b ) + b(a - b* )].

18. an — b2n is divisible by a + b. 19. 10" + 3 ( 4"+ 2 ) + 5 is divisible by 9.

20. n ( n + 1 ) ( 2« + 1 ) is a multiple of 6.

21. n + In is divisible by 3.

22. 2 " ~ 1 < « ! where n\ = n (n — 1 ) ( « — 2 ) - " 2 - l .

23. log x" = n log x. 24. (cos 9 + i sin 9 )" = cos n 9 + i sin n 9.

25 — f x " ) = rix"1. [This example should be attempted ax by students knowing calculus ].

Chapter 9

Progressions 1. Sequences and Series 2. Definition of Arithmetic Progression.

3. »th term of A. P. 4. Sum of » terms of A. P. 5. Sum of first n natural numbers. 6. Sum of first tt odd natural numbers. 7 Convenient representa-tion of terms in A. P. 8. Illustrative examples. Exercise 9 (a). 9. Definition of Geometric Progression. 10. nth term of G. P. 11. Sum of « terms of G. P. 12. Convenient representation of terms in G. P. 13. Illustrative examples. Exercise 9 (b). 14. Definition of Harmonic Progression 15. »th term of H. P. whose first two terms are « and fi 16. Important property of three consecu-tive terms in H. P. 17. Illustrative examples. 18 Definitions of Means. 19 Values of A. M., G. M. and H. M. 20. Theorems on Means. 21. Insertion of n means between two numbers. 22. Illustrative examples. Exercise 9 (c).

1. Sequences and Series : Definition. If to every positive integer n, there corresponds

a number an given by some law then the terms au a2, a3..an... form a sequence.

A sequence is sometimes denoted by bracketing its nth term. Thus (an) means the terms au a3 a„, ... The suffix of a denotes the rank of the term.

If ( i ) aH = n , ( a „ ) is l2, 22, 3 2 . . . , n2, • . •

( " ) «„ = ( - 1 ) " , ( « » ) is - 1 , 1 , - 1 , 1 , - 1 , . . -Generally the rule for forming the terms in the sequence is

given by expressing an in terms of n. From strictly analytical point of view, a sequence is defined as

a correspondence from a set of positive integers to a set of numbers a3, • • • Hence a sequence can be defined as a function

whose domain is a set of positive integers. In fact the numbers a i , a2> a3. ••• foTm the Tange of the sequence. However, by convention we say that the terms a t , a2,... an form a sequence.

222

PROGRESSIONS : 2 2 3

Illustrations. Consider the following sequences. ( i ) 1 , 4 , 9 , 1 6 , 2 5,

( i i ) 5 , 9 , 1 3 , 1 7 , 1 1 1 ± 2, y > 8 , i r

( i v ) 3, - 9 , 27, - 8 1 ,

/ , 1 2 . 1 ± v ' 2' 3 ' 4 ' 5 '

It will be observed that the relations between n and an in the above sequences are given as follows :

In ( i ) a„ is the square of n.

In ( ii ) a„ is obtained by adding 4 to a„-1 and at = 5.

In ( i i i ) the numerator of a„ is 1 and denominator is obtained by adding 3 to" the preceding-dehominator; and a t =

In ( i v ) a„ is obtained by multiplying the preceding term <z»-i by —3; and ax = 3 .

In ( v ) the numerator of a n is n and the denominator is obtained by adding 1 to the numerator.

The above relations can also be briefly stated^thus :

In ( i ) o„ = n* ; in ( i i ) an ~ 4n +1; in ( i i i ) a„ = >'

in (iv ) a„ = ( - 1 ) 13* and in ( v ) a„ = j f - q p f • Thus the

various terms at, c2 , d3 • • • a„ • • • in the above sequences are obtained by giving to n the positive integral values, 1, 2, 3,- • •» • • •

Series. When the sum of the terms of a sequence is under consideration, it is called a series.

Definition. A series is an expression consisting of the sum of terms of a sequence.

Thus if a n is the nth term of a sequence then FLI + A7 + CJ3 H A N

is a series of n terms.


Arithmetic Progression.

2. Definition of Arithmetic Progression.

A sequence ( an) defined by al = a and an = an_x 4- d where a and d are given numbers is called Arithmetic Progression.

The number d is called the common difference. An arithmetic progression is usually referred to as A. P.

Thus ( i ) 3, 7, 11, 15, 19

( i i ) 8, 5, 2, - 1 , - 4 are examples of arithmetic progressions.

In ( i ) , the first term is 3, and the common difference is 4. In ( i i ) , the first term is 8, and the common difference is — 3. Thus if the first term and the common difference are known, the A. P. is completely known.

If a is the first term and d the common difference of an A. P., then the progression can be written as

a, (a + d), (a + 2d), (a + 3d)

3. To find the nth term of the Arithmetic Progression a , ( a + d), ( o f 2d), ( a + 3d), ...

We observe that 1st term ax = a + (0) d, 2nd term a2 = a + (1) d, 3rd term a3 = a + (2) d, 4th term at = a + ( 3 ) d,

It is noticed that the coefficient of d is one less than the suffix of a, which denotes the rank of the term in the sequence.

nth term, a,, = a + ( n — 1 ) d. We can prove the formula for the nth term of an A. P. by

mathematical induction.

Let P (n) be : nth term, a„ = a + ( n ~ l ) d . ... ( i ) ( a ) Put n = 1 in ( i ) , L. H. S. of ( i ) = a,. R. H. S. of ( i ) = a + ( I — 1 ) d = a.


We have at = a. Hence P ( l ) is true,

( b ) We now show that the assumption of P(k), namely, ak = a + (k - 1 )d ... (ii)

implies the truth of P ( k + 1 ), namely,

= a + {k + l — \)d i . t . a + kd ... (iii)

L. H. S. of (i i i ) , ak+1 = ak + d, by definition = [a + (£-1 )d] + d, by ( i i ) = a + kd - d + d. = a + kd = R. H. S. of ( i i i ) .

Thus, from steps ( a ) and ( b ) the formula ( i ) is proved by mathematical induction for all positive integral values of n.

Illustration. The «tb term of the A P. 15,'8, 1, - 6, - 13 is 151+ ( « - l ) ( - 7 ) i. e. 15 - 7n + 7 i e. 22 - In. We can, therefore, write the general i. e. the nth term of this sequence as

a n = 22 - 7«.

By giving to n , the values 1, 2, 3, 5, we can get the various terms of the given A, P.

Ex. 1. Find the «th terms of the following progressions. __

( i ) 3, 8, 13, IS,

( i i ) 1 1 , 7 , 3 , - 1 , - 5 ,

<»*> 4 - 4 > 2 "

Ans. a„ = 5n - 2.

Ans. a = 15 — 4»

28 - 5» Ans. a n = 4 2 4 * 4

, . , - 15 - 7 1 9 8n — 23

( lv> ~ T " • T ' T - T ' - A n s ' a , ! = —

Ex. 2 , Find the 10th term in each of the progressions given in example 1.

Ans. ( i ) 43, ( i i ) — 25, ( i i i ) ( i v ) 7 - | - •

4. To find the ium of the first n terms of the A. P. « . ( « + d ) , ( a + 2cf ) , ( a +

Let S„ denote the sum of n terms ; and / denote the last (i. e. n th) term. Then

S„ = a + ( a + d ) + (a + 2d) + ... + (I-2d) + (1-d) + /• ... ( i )

C A.—15


Writing the series in the reverse order, we have S„ = / + ( / - « / ) + ( I - 2 d } + ... + (a + 2d)

+ ( a + d) + a. . . . ( i i ) Adding ( i ) and ( ii ),

2 S . = ( a + / ) + C« + 0 + ( « + 0 + + ( a + / ) + ( o + / ) + ( a + / ) .

There are n terms on the r, h. s. each equal to a + /. 2S „ = » ( « + / ) .

••• S„ = - £ ( « + O , ( i » )

• / I is the nth term, we have [ = a + (n — I ) d.

S„ = - § [^2« + ( n - l ) r f ] - - ( i v )

Formulae ( i i i ) and ( i v ) are useful and should be remembered.

4-1 Alternative Method The method of mathematical induction can be used to prove this result.

Let P (w ) denote the formula for the sum of n terms, so that

P ( n ) : a + ( a + d ) + (a + 2d) + • - + (a + n - l d )

= ~ [ 2 a + ... ( i )

(а) Put n = 1 in ( i ). L. H. S. of ( i ) = a.

R. H. S. of ( i ) = \ [ 2a + ( 1 - 1) d] = a.

L. H. S. = R . H. S., P ( 1 ) is true. (б) We now show that the assumption of the truth of

P { k ) , namely, a + (a + d)+ • + (a+ k - Id)

= ~ [2a + d] ... ( i i )



a + (a + d) + • • • + (a + k - l d ) + (a + kd)

k+ 1 [2fl + (fc + l - l ) </]. ... (iii)

L. H. S. of ( iii ) = - y [ 2a + k - 1 d] + (a + kd),

by assumption (i i )

= ka + ~ ( k 2 - k ) d + a + kd

= (k + 1 )a + (k2 + k)d

= k-±±[2a + kd]

= R. H. S. of ( i i i ) .

From the steps (a) and ( b ) we see that P ( n ) is true for all positive integral values of n.

Ex. 1. Find the sum of ( i ) 8 + 13 + Is! + 23 + ... to 25 terms. Ans. 1700. ( i i ) 21 + 15 + 9 + 3 + ... to 20 terms. ~ Ans. - 720.

Ex. 2. Find the sum of ( i 1 2 + 4 + 6 V 8 + ... to « terms. Ans. n ( n + 1 ). ( ii ) 21 + 22 + 23 + 24 + ... + 60. Ans. 1620

(iii) 4 + 8 + 12 4- 16 + ... + 100. Ans. 1300.

S. The sum of the first n natural numbers. The positive integers 1, 2, 3 a r e the first n natural

numbers.

Let S„ denote the required sum. Then Sn = 1 + 2 + 3 + + ( n — 1) + «

The right hand side is an arithmetical series where a = 1 I = n and there are n terms in the sum.

S „ = - y U + » ) • [ by (iii ) of para 4. ]


A S n = ~ • n ( n + l ) . U )

Note. We have proved the formula ( i ) by mathematical induction in Chapter 8.

6. The sum of the first n odd natural numbers.

The first odd natural numbers are 1 . 3 , 5 , 7, 9,

They form an A. P. with the first term a = J, and the common difference d = 2.

Let SM denote the required sum. Then

S„ = + ( " - ! ) X 2 ] , [ by ( iv ) of para 4 ].

S „ = - y [ 2 n ] = n\

Therefore, the sum of the first n odd natural numbers in n2. Note. The proposition that " the sum of first n odd natural numbers is

« a c « n also be proved by mathematical induction.

7. Convenient representation of terms in A. P

When 3, 4, 5...terms in A. P are under consideration in any problem, it will be found more convenient to represent the terms in the following form—

( i ) 3 numbers in A P. : a — d, a, a + d. ( ii ) 4 numbers in A. P. : a - 3d, a - d, a 4- d, a + 3d ( iii) 5 numbers in A, P. : a — 2d, a — d, a, a + d, a + 2d.

The advantage of assuming the terms in the above form can bs seen in the following illustrative example 1 of the next article.

8. Illustrative examples

Ex. 1. The sum of 4 numbers in A. P. is 22, and the sum of their squares is 166. Find the numbers.

Let the fonr numbers in A. P. be a — 3d, a — d, a + d, a + 3d.


the sum of 4 numbers is 22, 4a — 22, .'. a = '

the sum of the squares is 166,

(a - 3d)' + (a - d)1 + (a + d)' + (a + 3d)' = 166.

4aJ + 20ds = 166. ... ( i i )

Substituting the value of a from ( i ) in ( i i ) , we get

4 ( T 1 ) + 20D1> = M ' 20A '2 = 45- •'•

11 3 Thus, we have a= — and d = ± —— •

•'. taking positive value of d, the four numbers will be 1, 4, 7, 10.

The other value of d will give the numbers in reverse order.

Note. If some terms of A. P. are given and we have to obtain some result involving such terms, then it will be found convenient to express the given terms in terms of the first term and the common difference of the A. P The following examples will illustrate the convenience of the process.

Ex. '2. If a, b, c, are in A. P prove that a 2 + Aac + ca = 2 ( a b + be + ca).

Let d be the common difference of the given A. P. so that b — a + d and c = a + 2d.

L. H S. = a 2 + 4a ( a + 2d ) + ( a + 2 d

= a3 + 4a2 + Sad + a" + 4 ad +

= 6a' + \2ad + W1. .. .. ... ( J )

R. H. S. = 2 [ a{ a + d ) + I a + d ) ( a + 2d ) + ( a + '~2d ) a ]

= 2 [ a 2 + ad + a1 + lad + 2d2 + a 2 + 2ad ]

= 2 [ 3 a 2 + 6ad + 2d% ]

= 6a» + I2ad + 4dr'. ... (H)

L H. S. = R H. S. and hence the result.

Ex. 3. If a, b, c, d are in A. P , then show that

6 + c + d, c + d + a, d + a + b, a + b + c are in A. P.

Let the common difference oi the given A. P . be x, so that

b = a + x, c = a + 2x, and d = a + 3x.

Now the given numbers will be in A. P. if the successive differences between any two terms are the same.

W e have ( 6 + c + <i) - ( c + d + a) = b - a = x, (c + d + a ) - ( d + a + b ) = c - b = x , (d + a+ b ) - ( a + b + c ) = d - c = x,

Since the difference between any two consecutive terms is the same, namely the given terms are in A. P .


Ex. 4 . Find the nth term and sum to « terms of the A. P. 7, 3 , - 1 , - 5. ...

Here a = 7, d =" - 4,

nth term, an = a + ( n - I )rf = 7-H(*» — 1) ( - 4 ) = u _ 4 „ .

The sum of » terms, S„ = [ 2a + ( n - X ) d ]

••• S„ = - — [ 2 x 7 + ( » - 1 ) ( - 4 ) ] = JL. [ h - 1 « + 4 ]

- " 5 - CIS - 4 « J .

S„ = « [ 9 - 2 « ] .

Thus the 10th term of this A P. a, 0 = 11 - 4 x 10 = 11 - 40 = - 29 and the sum of 10 terms, S,*, = 10 [ 9 - 2 x 1 0 ] = - 110.

Ex. 5. If the 4th term of an A P is 9 ~ and the 9th term is 17, find the

nth term.

[ If any two terms of an A. P. be given the progression can be completely determined ; for from the given data, we can obtain two simultaneous equations In a and d and solving these equations we can find a and d. ]

19

We have a, = a + 3d - — ... ... ... ( i )

and as = a + 8d = 17. ... ... ... ( i i )

Solving these two equations, we get a — 5, and d = — •

.-. the nth term a„ = « + ( » > - l ) < * = 5 + ( » - l ) | - = » — ~ •

Ex, 6 . Find the sum of integers from 11 to 50. Find also the sum of integers from 11 to 50, excluding those which are exactly divisible by 3. Let the required sum be denoted by S, so that S = 11 + 12 + 13 + • • • • + 50-.

= [ 1 + 2 + 3 + + 5 0 ] - [ l + 2 + 3 + - - - - + 10 ]

= 5 [5-51 - 11 J = 5 [255 - 11 J = 5 [244] = 1220.

Now the integers between 11 and 50 which are exactly divisible by 3 are 12, 15, 18, • • • 48.


These numbers are in A. P., where a « 12, d = 3 and n = 13 since 48 = 12 + ( n - 1 ) 3.

Hence their sum, say, S' is given by

S ' - ? [2-12 + ( 1 3 - 1) • 3 ] «

= ^ [ 24 + 3 6 ] = ^ [ 6 0 ] = 390. 2 « the sum of integers from 11 to 50 excluding those which are exactly

divisible by 3 is given by

S - S' = 1220 - 390 = 830.

Ex. 7, Find the first four terms of the series the sum of which to n terms

is (3« + 5) .

[ Note. If the sum of » terms is given in terms of n, then the terms of the progression can be determined.

For S n = a, + th + a3 + a4 + • • • • + + a„

a n d S « - l = + a i + + a»-l, :. S - S . = a . « « —2

This example will illustrate the use of this formula. ]

Hare, we have S j t = ~ ( 3 « + 5 ).

. « - 1 , n 4. , 1 _ ( n - 1) (3*t + 2) •• S «- l = — 2 ~ [ ~ 1) + 51 j

V the /ith term a n = S B — S n _ i ,

the «th term a n of the given series

- " (3n + 5 ] _ (»» ~ l)(3w +2) = 6n +2 2 2 2 "

a = 3» + 1 n By putting n = 1, 2, 3, 4 we get first four terms as 4, 7, 10, 13.

The students should note the formula a — S — S ,

This formula determines the «th term of a series if the sum of « terms is known. By giving particular values to » , particular terms can be obtained.

Ex. 8, The sums of the first n terms of two A. P.'s are as 3n + 5 : 5» - 9. Prove that their 4th terms are equal.


Let the two A. P.'s be a, a + d, a + 2d a + (n - X ) d, a', a' + d', a' + 2d' a + (« - 1) d'.

We are given that

s„ _ a » » + < " - ' > ' ] 3 j l j L 5

. la + (w - 1 ) d = 3» + 5 2a' + (n-l)d' 5n - 9 '

[ by data ]

( i )

We wan+ *o find the ratio of 4th terms ; that is, we want to find a + 3 e i

a' + 3d'

which can be obtained from ( i ) above, by writing it a s j. e . by 2a + 6d

putting « = 7 in ( i ) above,

• a + - d =

2 a + 6 d _

" a' + 3d ~ 2a' + 6d' =

4th terms of the two series are equal.

Students may use this method to solve examples 13 and 14 in the next Exercise.

Ex. 9 . The sum of the first 1! terms of an A:. P. is 19 and the sum of first 19 terms is 11. Find the sum of the first 30 terms.

Let the first term and the common difference of the given A. P be a and d respectively. From the given conditions of the problem, we have

— [ 2 a + 1 0 4 ] = 1 > ; ... ( i )

X-j- I2a+ 18d] = 11 ; ... ( i i )

and we have to obtain the value of 30 — [2a + 2d]. ... ( i i i )

Equations ( i ) and ( i i ) can be solved for a and d and their values, if substituted in (i i i) would give the required result. But it may be found more convenient to proceed in examples of this type as follows :—

Subtract equation ( i ) from ( i i ) as follows:—

[2a (19 - 11) + d (19-18 - II 10) ] = ( H - 19 ),

. [2a (8) + d- (232) ] = - 8.


Cancelling common factor 8 from every term, we get ~ [2a-h29d] = — J. A .'. the required result can be simply obtained by multiplying both the

sides by 30,

.'. [ 7a + 29rf ] = - 30.

Students may use this method in examples of such type, such as examples 16 and 17. in the next Exercise.

Ex. 10. The sum of the first p terms of an A. P. is the same as the sum of the first q terms. Show that the sum of the first (p + q) terms is zero.

Let a and d be the first term and th- common difference of the given A.P.

P We have, sum of p terms, say, St> = [2a + (p - I ) d ] .

Similarly, the sum of q terms, say, S , = [ 2a + ( q - 1 ) d ],

from the given condition of the problem, we get

+ --£-[ta + (q-l)d]. ... ( i )

Now we have to obtain the sum of (p + q) terms, say, S,i+g.

We have Sf,+, = [ 2a + ( p 4- q - 1 ) d ] . . . . ( i i >

From ( i ) , we get

2a (p - q) + d [P(P - 1) - q{q - 1)] = 0-2a {p - q) + d ip1 - q*-p + = 0.

2a (P - q) + d (p - g) {P + q - 1) = 0.

2a f d (P + q - 1) = 0. ... ( i i i )

Substituting the value given by iii in ( i i ) , we have

Exercise 9 (a)

1. Find the nth terra and also the 10th term of each of the following sequences

( i i i ) 1-3, 3-5, 5.7,7-9, • • • ( iv) 31,24,17, 10, • • •


( v ) 2, 5, 10, 1 7 , . . . (vi;

( v i i ) I**' ( v i U ) 2 4 3 ' 2°8 ' 173 ' 1 3 8 ' ' '

2. Find the nth term and the sum to n terms, if ( i ) the 7th term of the A.P. is 30 and the 10th term is 21. ( i i ) the 4th term of the A.P. is 64, and the 54th term is - 61.

3. If p, q, r , s are any four consecutive terms of an A. P., show that p2 - 3q2 + 3r2 - s2 = 0.

4. If the j?th, gth, rth terms of an A. P. are a, b, c respe-ctively, show that a (q — r )+b(r~p)+c(p — q) = 0.

5. I f p , qr r, s, t are in A. P. show that p + t = q + s =2r.

6. If the pth term of an A. P. is q, and the qth term is p, find the rth term.

7. Find three numbers in A. P. such that ( i ) their sum is 15 and the product is 105; ( ii ) their sum is 27 and the sum of their squares is 341-

8 Find four numbers in A. P. such that ( i ) their sum is 44 and the sum of their squares is 564; ( i i ) the sum of 2nd and 3rd is 22 and the product of

1st and 4th is 85.

9. Find five numbers in A. P. such that ( i ) their sum is 20 and the product of the first and the

last is 15; ( ii ) their sum is 25, and the sum of their squares is 135.

10. Find the sum of the series

( i ) 4 - 7 + 10 - 13 + 16 - 19 + . . . to 21 terms ; ( i i ) 2 + 3 + 5 + 6 + 8 + 9 + 1 1 + 1 2 + 14 + 15 + '

. . . to 2n terms; ( i i i ) 2 + 3 - 4 + 6 + 7 - 8 + 1 0 + 1 1 - 12 + ...

to 3n terms. Deduce also the sum of ( I n + 1 ) and ( 3« + 1 ) terms in series ( i i ) and ( i i i ) respectively.


11. Find the sum of all natural numbers between 200 and 400 which are exactly divisible by 6.

12. Find the sum of all natural numbers from 1 to 150,

( i ) excluding those which are exactly divisible by 7 ;

( i i ) excluding those which are exactly divisible by 3 or 7.

13. The sum of n terms of two arithmetic series are in the ratio of In + 1 : 4n + 27; find the ratio of their 11th terms.

14. The sum of n terms of two arithmetic series are in the ratio of In - 5 : 5n + 17 ; show that the 6th terms of two series are equal.

15. The first and the third terms of an A. P. of 40 terms are — 29 and - 22. Find the sum of all its positive terms.

16. The sum of the first 13 terms of an A. P. is 21, and the sum of the first 21 terms is 13. Find the sum of the first 34 terms.

17. The sum of the first m terms of an A. P. is n, and the sum of the first n terms is m Find the sum of the first ( m + n ) terms.

18. The first and last terms of an A. P. a r e a and / respec-tively. If s be the sum of all the terms of the A. P., show that the common difference is

P-a2

2 s - ( / + o ) '

19. If j j , s2, s3 be the sums of p, q, r terms respectively of an A. P., show that

S - j ( q - r ) + f ( r ~ p ) + ^ C p - q ) = 0.

20. ( i ) If the 6th term of an A. P. is 121, find the sum of its first 11 terms.

( i i ) If the 35th term of an A. P, is 30, show that the sum of its first 69 terms is 2070.

( i i i ) Show that the sum of 11 terms of an A. P. of which the middle term is 121 is 1331.


21. If the first, second and the «th terms of an A. P. are a, b and x respectively, then show that the sum of its n terms is

x + a x2 — a2 +

2 2 ( b - a )

22. If a, b, c are in A. P. then prove that ( i ) a 3 + c3 + 6abc = 8b3;

( i i ) a2 + 4ac + c2 = 2 ( ab + be + ca ).

23. If a, b, c, d are in A. P., then show that (b + c+d), (c + d+a), (d+a+b) and ( a + b + c ) are in A. P.

24. Show that the sum of the terms in thsf nth bracket of ( 1 ) + ( 3 + 5 ) + ( 7 + 9 + 1 1 ) + is n3.

25. Prove that the sum of an odd number of terms of an A. P. is equal to the middle term multiplied by the number of terms.

26. Find the «th term of the series of which the sum of n terms is

( i ) 3/j2 + 5n, ( ii ) n ( n + 1 ), ( iii) 3" — 1.

In each case write the 4th term.

27. If s2, s3 sp are the sums, each to n terms, of p arithmetic progressions whose first terms are 1, 2, 3, ....

and common differences are 1,3, 5, 7, respectivelv, then show that

+ +sp=nJLOv±1).

28. Show that the sum of all odd numbers between 2 and 1000 which are divisible by 3 is 83,667 and of those not divisible by 3 is 1,66,332.

Answers—Exercise 9 (a)

, . . n + 1 11 . . . . 2» + 3 23

(iii) (2» - 1 ) (2« + 1 ), 19-21 ; ( iv) (38 - 7 « ) , - 32 ;


( v ) » ' + 1, 101; , v i )

' b" b1(l

(vii) . ; (viii) (278 - 35n), - 72.

2 . ( i ) ( 5 l - 3 „ ) . ^ ( 3 3 - « ) ; ( i i ) 1 - ! 8 - ^ , (291 - 5 » ) .

6. p + q - r . 7. ( i ) 3 , 5 , 7 ; ( i i ) 2 . 9 , 1 6 .

8 . ( i ) 5 , 9 , 1 3 , 1 7 ; ( i i ) 5 , 9 , 1 3 , 1 7 .

9 . 3 , 3 y , 4 , 4 - | , 5 ; ( i i ) 3 , 4 , 5 , 6 , 7 .

10 . ( i ) 34 ; ( i i ) n ( 3 n + 2 ) . S 2 n + 1 ® ( « + 1) ( 3 « + 2 ) ;

(iii) » ( 2» - 1 ) . S 3 n + 1 = 2na +• 3» + 2.

11 . 9900. 12 . ( i ) 9708 ; ( i i ) 6471. 13. 4 ; 3. 1 5 . 1705.

16 . - 3 4 . 17. — (tn + tt)- 20 . ( i ) 1331

2 6 . ( i ) a n = 6n + 2 ; = 26 ; ( i i ) a ( ( = « , a4 = 4 ;

(iii) a n = 3" - " , a4 = 54.

Geometric Progression

9. Definition of Geometric Progression.

A sequence ( an ) which is such that an = r • and = a, r and a being given numbers, is called Geometric Progression.

The number r is called the common ratio of the geometric progression. A geometric progression is usually referred to as G. P.

Thus ( i ) 3, 6, 12, 24, 48,

( i i ) 9, 3, I, 1/3, 1/9, ( i i i ) 1 , - 2 , 4 , - 8 , 1 6 ,

are examples of geometric progressions.

In ( i ) , the first term is 3, and the common ratio is 2.

In ( ii ), the first term is 9, and the common ratio is 1/3.

In (iii ), the first term is 1, and the common ratio is - 2 . Thus, if the first term and the common ratio are known,

the G. P. is completely determined, <


If a is the first term and r the common:ratio of a G. P. then the progression can be written as

a, ar, ar2 ar3, ar4,

10. The nth term of the Geometric Progression a, ar, ar2, ar3,

We observe that 1st term a t = ar°, 2nd term a2 = ar1, 3rd term a3 = ar2, 4th term a4 = ar3.

It is noticed that the index of r is one less than the suffix of a which denotes the rank of the term in the sequence.

nth term a„ — ar"'1. We can prove the formula for the «th term of a G. P. by

mathematical induction.

Let P ( n ) be : nth term, an = ar"-1. ( i )

( a ) Put n = 1 in ( i ) . L. H. S. = ax- R. H. S. = ar°=a. Y L. H. S. = R. H. S. since at = a.

P ( 1 ) is true. ( b ) We now show that the assumption of the truth of

P ( k ) , namely, ak = ar'1'1 ( i i )

implies the truth of P (k + 1 ), namely, a f t + 1 = i . e . ark ( i i i )

L. H S. of ( iii), ak+1 — ak- r by definition h-\ .

= ar • r, by assumption ( n ) = ark. by index law. = R. H. S. of ( i i i ) .

From steps ( a ) and ( b ) , P ( n ) is true for all positive integral values of n, by mathematical induction.

The nth terms of the geometric progressions referred to in the last article are ( i ) 3-2"" 1 , (ii) 9-( 1/3 and (iii l - ( - 2 ) " " 1 - By giving to n, the values 1, 2, 3 we get the various terms of the G. P.'s.


Ex. 1. Find the nth terms of the following G- P.'s.

( i ) 1/4, 1/2, 1, 2, Ans. 2""3 . 1

( i i ) 2 7 , 9 , 3 , 1 ,

(iii) 7 2 , - 1 8 , 9 / 2 , - 9 / 8

Ans. n-i

9 Ans. ( - 1 ) " " 1

2 ,2n-5

6 27 27>/3 3»-<1/2)

Ex. 2. Find the Sth terms in each of the progressions given in example 1.

Ans. ( i ) 32, ( i i ) ~ . (iii) - , (iv) '

11. To find the sum of the first n terms of the G. P. a, ar, ar2, arz,

Let S,t denote the sum of n terms ; so that tt-3

S„ = a + ar + ar2 + ar'6 + + ar + ar«-2

Let r 4= I. Multiplying every term by r, the common ratio, we get

r-S„ = + ar + ar2 + ar3 + . . . . . . + an"'1 + ar".

Subtracting, S„ — rS„ = a - arn.

(1 - r)S„ = 0(1 - rn).

= ( i )

Changing the signs of the numerator and denominator, we may also write

S„ = < L ( Z l l I > . ( i i ) » r — 1

It will be found convenient to remember both the forms ( i ) and ( ii ) for S„. The form ( i ) may be used when r < 1. and the form ( i i ) may be used when r > 1.

If r = 1, the G. P. reduces to a, a, a, a / . S„ = na. ... ... ... ... ( i i i )


11-1 Alternative method. We now prove the formula for the sum of n terms of the G. P. by mathematical induction.

Let P (n) be :a + ar + • . • + a r " ~ 1 = r _ ~ 1 1. ... ( i )

(a) Put n - 1 in ( i ). L. H. S. of ( i ) == a.

R. H. S. of ( i ) = € _ < > i z i ) _ a r — 1

V L. H. S. = R. H. S„ />(1) is true,

(b) We now show that the assumption of the truth of P ( k ) , namely,

a ( rk — 1 1 a + ar+ • • • • + = ^ — . . . ( ii ) r — 1 v y


a +ar + + ar^ + ar" ~ 1 ) _ (iii >

r — 1 v '

L. H. S. of ( i i i ) = [ a + ar + + ar*-'1] + ar*

a (r f t — 1 ) . , , . x = r ~ i + a r " ' by C i i )

= ~ ~ 1 + r H r - \ ) \

- —~r \ r k — \ — r*] r — 1

a ( r;.+i _ i ) = — - = R - H - s- of ( hi )•

From steps ( a) and ( b ), we prove by mathematical induction that P ( n ) is true for all positive integral values of n.

In respect of the G. P.'s U ) , ( i i ) , ( i i i ) , given in the beginning of article 9, we have

3


and in ( i i i ) , a = l , r= 2,:. S#. = 1 • ^ 1 - ( 2 )

Ex. Find the sum of

/ . v 1 . 1 , 2 , - . 2°59 ( ' > X T "9 + t e r m s- u H '

( i i ) 2 - 4 + 8 - 16 + to 10 terms. Ans. - 682.

3 1 3 h ( i i i ) — + 1 —- + 3 + to « terms. Ans — ( 2 - 1 )• 2 4

16 64 ( i v ) 3 - 4 + — — + to 2m terms

12. Convenient representation of terms in G. P. When 3, 4, 5 ... terms in G. P. are under consideration in any

problem, it will be found more conveninent to represent the terms in the following form —

( i ) 3 numbers in G. P . : ~ , a, ar ;

( ii ) 4 numbers in G. P. • , — , ar a/3 ; v / rs r

(iii ) 5 numbers in G. P. : , — , a, ar, ar2 etc. rl r

The advantage of assuming the terms in G. P. in the above form can be seen in the example 1 below. 15. Illustrative examples.

Ex. 1. The product of three numbers in a G. P. is 216, and their sum is 26. Find the numbers.

Let , a, ar be the three number in G. P. r

: the product is 216. a® = 216. '. a = 6. ... ( i )

V the sum is 26. — +• a + ar = 26. r

:. — (1 + r + r1) = 26. r

C . A. —16.


5 + 6r + 6ra = 26r, ( V a =• 6 )

.-. - 20r + 6 = 0. .'. lr% - lOr + 3 = 0.

( 3 r - 1 ) ( r 3 ) = 0. r = - y - or 3

Thus, we have a = 6 and r — - i - or 3.

.'. taking r = 3, the three numbers will be 2, 6, 18.

Taking r = * , we will get the same numbers in the reverse order.

Note If some terms in G. P. are given and we have to obtain some result involving such terms, then it will be found convenient to express the given terms, in terms of the first term and common ratio of the G. P. The follow-ing two examples will illustrate the process.

Ex. 2. If a, b,c d are in G. P. then show that (a — d)* = (6 — o)' + (c - a)* + {b - d)1.

Let r be the common ratio of the given G. P., so that b = ar. c = ar* and d = at

L. H. S = [a - ar*)* = a3 (1 - r5 )a = a 2 ( 1 - 2r* + r6 ). ( i ) R. H. S. - ( a r - a.')1 + (ar2 - a)1 + (ar - ar ' ) a

= a" (r3 2 » + r*) + a 3 (r* - 2ra + 1 ) + a1 (r* - 2r' + r6 > = a' [r2 - 2"' + r* + r4 - 2r' + 1+ r> - 2r* + r6] = aa [ i - 2rs + r° ].

L. H. S. = R. H. S. and hence the result

Ex. 3 If a, b, c. d are in G. P. then show that ( a - b )2, t b - c ),a

( c — d y are also in G. P. Let r be the common ratio of the given G. P. . so that b = ar c = art

t

d — ar3.

Now (a - b)2. (b - cy, ( c - </)' will be in G. P.

•f ( b ~ c)' = (c - d)2 . b - c = c - d (a - b y ( 6 - c y ' e ' a - b b - c

, ar - ar2 ar1 - ar3 . ., ar ( 1 - r ) ar1, ( I - r ) i. e. if = i i. e. if — r = ^ ' • a - ar ar — ar' a ( 1 - r) ar (1 - r)

i. e if r = r which is true and hence the result.

Ex. 4 . If a, b, c are in A. P . and x, y, z are in G. P., then show that b c a c a f> x • y • z = X y • z •

Let d be the common difference and r the common ratio of the given A. P. and G. P. respectively We, thus, have b=-a + d,c=a + 2d and y = xr, z = xr3.


T rT _ h c a afrf . ,a+2d , 3a+3d 3a+2d L. H. S . =x -y • z = x • (xr) • (xrs) =* •r . ( 1 )

TD IT o c a b a + 2d , ,a . . . o+d 3a+3rf 3a + 2</ . . . . R . H . S. = * -j> = * • (xr, (xr3) =x r . ( i t )

.'. L. H . S. = R. H . S. and hence the result

E x . 5 . If the 5th term of a G. P. is 2 and the 8th term is 61, find the sum cf first six terms.

Here, we have, f 5 = ar* = 2. . . . . . . ( i )

27 as = ar7 = — , ( i i )

27 3 Dividing ( i i ) by ( i ), we have r3 = - r - • r = — • o 2

Substituting this value of r in (i), a ^ j = 2

„ ( r6 _ j ) The sum S6 of the first six terms of the series is — -

32 ST

r - 1

S e ~ 8 1

36 _ 2« _ 729 - 64 _ 665 = T l 81 81

Ex. 6 . Find the sum to 2» terms.

1 + -L + -2 + X + -A + . L + . . . 7 + 3* 3 s 3' 3* 3"

( i ) S2„ = = [ - § - + | r + jr + t 0 « tern>s J

+ ̂ -̂ r + yr + IT + t0M terms J


Ex. 7. Sum the following series

3 + 33 + 333 + 3333 to n terms.

We have S„ = 3 + 33 + 333 + 3333 + to n terms

= 3 [ 1 + 11 + 111 + 1111 + to n terms ]

= ~ ~ [ 9 + 99 + 999 + 9999 + to n terms ]

= [ ( 10 - 1) + ( 10* - 1 ) + ( lC - 1 )

+ ( 104 - 1 ) + to n brackets ]

= [ (10 + 10» + 10s + + 10")

- ( 1 + 1 + 1 + 1 + to » terms ) ]

= ^ [ 1 0 n , 1 - 9 „ - 1 0 ] .

Exercise 9 ( b ) 1. Find the nth term and also the 10th term of each of the

following sequences.

( i ) 5 ^ - , 3 ^ - , 1^ - ( i i ) 162, 270, 450, 750, 1250

W ( i V > 2 4 3 ' 3 2 4 ' 4 3 2 <" •

2. Find the nth term and the sum to n terms of the G. P. if

1 3 ( i ) the 2nd term is 4—, and the 5th term is 15—, 2 16

( i i ) the 5th term is 1 ^ - and the 9th term is • oU 5

3. If x, y, z are the nth, (2n )th and ( 3n )th terms of any G. P., show that y2 = xz.

4. If x, y, z be the pth, gth and rth terms of any G. P., show that x"'r y-fi' zi>~<> = 1.


5. Find three numbers in G. P., such that

( i ) their sum is 21, and the sum of their squares is 189;

( i i ) their sum is 38, and their product is 1728.

6. Find five numbers in G. P., such that

( i ) their product is 32, and the product of the last two is 108;

3 ( i i ) their sum is 7 - j - and their product is unity.

7. How many terms of the G. P. 1, 4, 16, must be taken to have their sum equal to 341 ?

8. Sum the following series : ( i ) 9 + 99 + 999 + to n terms. ( i i ) 1 + 11 + 111 + to n terms. (iii) 7 + 77 + 777 + to n terms. (iv) 1 + 11 + 101 + 1001 + to n terms. ( v ) -5 + -55 + -555 + to n terms. (vi) 1-4 + 3-04 + 5-004 + 7-0004 + ton terms.

[Hint : Split the series into two series one in A. P. and the other in G. P. ]

9. Sum the following series to n terms

( i ) l + ( l + x ) + ( l + x + x 2 ) + ( l + x + jc2 + x 3 ) + . . . ( i i ) 1 + * ( 1 + x ) + x2 ( 1 + X + X2)

+ x 3 ( 1 + x + x2 + x 3 ) +

10. Sum the following series :

( i ) - | - + 3 4 - + J - + ! - + J - + ~ + . . . to2«terms;

_4_ _ 5 -L 4 _ A 4 , v 1 1 ) 7 7 2 7 3 7 4 "^75 76

to ( 2« + 1 ) terms. 11. Show that the sum of the geometric progression

, . Zfr-a2

a, b, / i s —t—— • b — a


12. If a, b, c are in A. P. and a, b, ( c +1) are in G. P., show that a = ( a — b ) .

13. The »th terms of two progressions are denoted by an and bn. It is given that

i?rs±L _ -£«±! and Z>n+1 - bn = bn+2 - 6„+1 an an+X

State which terms are in A. P. and which are in G. P.

Further if at = 2 and -3s±L = - L find 2 ar. an i r=l

14- r f + yTJ = s b o w t h a t * » y ' 2 a r e i n G - p -

15. If a, b, c, d are in G. P., prove that

( i ) ab — cd a + c b*-c2 ~ b '

( i i ) ( a b + bc + cdy = ( a 2 + 62 + c 2 ) (b^ + c^+d2). 16. If a, b, c be the pth, qth and rth terms both of an A. P .

and of a G. P., then show that 6 - c c-a a~b

a • b • C =1.

17. If a, b, c, d are in G. P. show that ( i ) ( a - b y , ( b - c y , (c — d y are in G. P. ( i i ) (at + b2), (ab+bc), (b 2 + <?) are in G. P.

18. If S„ represents the sum of n terms of a G. P. whose first terms and common ratio are a and r respectively, then prove that o I c i c i , c na ar (1 — r" > S! + S2 + S3 + - - . + SB = J — ( 1 - r ) » •

19. If tr is the rth term of a G. P. and tt = a and f j = b, prove that

r = 1 b ~

a

20. If p is the product of n terms in G. P., s their sum, and s

the sum of their reciprocals, then prove that p2 =


21. If a, b, c, d are in G. P. and a > 0, r + 1 0, then prove that ( i ) a + d > b + c; ( i i ) a2 + d2 > b2 + c2.

23. The sum of the first three terms of a G. P. is to the sum of the first six terms as 125: 152. Find the common ratio of the G. P.

24. Prove that

25. Answer the following as required :

( i ) Determine k if 2, k, 5 are in (1) A. P. (2) G. P

( i i ) The sum of the first n terms is 2n2 — n. Find the nth term. Write down the first three terms and state whether the terms are in A. P. or G. P.

( i i i) The third term of a G. P. is 0-08 and the 7th term is 0-000128. Find the first term and the common ratio.

( i v ) If the pth, qt h, and rth terms of a G. P. are them-selves in G. P., show that p, q, r are in A. P.

( v ) If three positive numbers are in G. P., show that their logarithms are in A. P.

x y z 22. If a is any positive number and a , a , a are in G. P.

then show that

(x - y)2 + (y - z)2= y ( z - x ) 2

Answers. Exercise 9 (b)

(iii) -b , ; 0 r - „ + l ar-9

248 : COLLEGE ALGEBRA /

2 2 according as r = H or — — • 3 3

5 - - U ) 3 , 6 , 1 2 : ( i i ) 8 ,12 ,18 .

6- ( i ) f . f . 2 . 6 , 1 8 ; ( U ) 1 , i , 2 | 4. 7. 5.

8. ( i ) -i- ( 1 0 n + 1 — 9« — 10) ; ( i i ) 1 ( 1 0 " + 1 - 9 » - 1 0 ) ; ' ol

(iii) ~ ( lo"+ 1 - 9n - 10) ; (iv) n + (lo" - 10) ; 81

q , n _»» * ( ! - * " ) (1 - * " ) (1 - *"+ 1)

' ' 1 - * (1 - * ) ' 1 U l ) U - * ) (1 - * * ) '

io. (i) 2 f i _ J _ V ( i i ) 2 2 _ i . _ J _ + i . . _ L . 8 V 9" / ' 48 12 72«+l T 48 72„

13. a , , . are in G. P. ; are in A. P. ; 2 a , = 3 • 1 3 "

23. 25. ( i ) ft = 7/2 ; k = VlO.

( i i ) = 4« - 3 ; a, = 1, a , = 5, = 9 Terms are in A. P.

(ii i) a = 2, r- = 1/5.

Harmonic Progression

14. Definition of Harmonic Progression :

If the reciprocals of the terms of any sequence form an arith-metic progression, then the sequence is said to be an Harmonic Progression.

An Harmonic Progression is briefly referred to as H. P. T k r

1 1 1 1 1

Thus ( i ) T , _ > T , T , T ,


(••_L _L _L ± ( 11 ; 7 ' 11 ' 15 ' 19

are examples of harmonic progression.

Since the general form of an A. P. is taken as a, a + d, a + 2d, a + 3d, ...

the general form of an H. P. can be taken as

J _ 1 1 _ 1 a ' a + d ' a + 2d' a + 3d

15. The nth term of an H P. whose first two terms are « and ft.

Let a, a + d, a + 2d, be the corresponding A. P. so that the given H. P. must be

_L 1 1 a 'a+d *a + 2d'

Now, we are given that

— = « . . . ( i ) and — i - j = .... ( i i ) a a+d

:. a = — and a+d = • « P

1 1 ' the common difference d = -5-/3 «

nth term of the corresponding A. P. = a + (n — \ ) d

- £ + ( » - n ( « - ) 8 )

the required nth term of the given H. P.

y8 + ( n - l ) ( « - / 8 ) '

250 : COLLEGE ALGEBRA /

Note—No formula can be obtained to give the sum of an number of terms jn an H. P. as we have obtained in the case of the two progressions A. P. and G. P.

16. To show that if a,b,c be three consecutive terms of an H. P. then

a — b __ a_ b — c ~~ c

v a, b, c are H. P.

I l l • A « 1 1 1 1 •• — — are in A. P. — - — • a b c b a c b

a — b _ b — c . a — b ab . _a_ ab ~~ be b — c be ' ' c

This property is sometimes taken as the definition of an H. P. and accordingly a, b, c are said to be in H. P. if

a_ _ a—b c ~ b—c

Starting from this definition we, now, prove that

1 1 1 A T , — , - r - , — are m A. P. a b e

We have by definition. — = ' c b — c

ab — ac = ca — cb. Dividing by abc throughout, we get

1 1 1 1 . I l l — = — .. — _ — are in A. P. c b b a a b c

17. Illustrative examples. Ex. 1. Find the nth term and 9th term of the H. P.


• A- L V • 1 5 1 1 7 3 . . the corresponding A. P . is ^ . ^ • & • ^

the »th term of this = ( » - 1 ) ^ ^ ^

_ 1 5 ~ 4 n + 4 = 19 ~ 29 29

29 the »th term of the given H P. = ^ __ ^ •

29 12 the 9th term of the given H. P = i. e - 1 — •

Ex. 2 . The 7th term of an Harmonical Progression is

— and the 12th term is ~ • Find the 20th term. 10 25

Let a be the first term and 4 the common difference of the corresponding A. P Hence the 7th and 12th terms of this corresponding A. P. are 10 and 25 respectively.

a + dd ~ 10... ( i ) and a + \ld = 25 ( i i ) .

Solving these two equations for a and d, we get, a = - 8 and d = 3. /. the 20th term of the corresponding A. P. = - 8 + 19 ( 3 ) i. e . 49.

the 20th term of the given H. P. = ~ •

Thus, it can be seen that the H. P. can be completely determined if any two terms are known.

Ex. 3 . If the mth term of an H. P. be n , and the nth term be m, prove

mn that the Zth term «s —— •

Let a be the first term and d be the common difference of the correspond, i n g A. P.

o + ( m - l ) d = — . . . . . . ... ( i ) tt

and a + ( » - l ) i = — • .. . — ••• ( i i ) tn

Subtracting ( i i ) from ( i ) , we get

/ , . 1 1 tn - n . . 1 ( m — n ) d — = •• a = — • 1 tt m mn mn

Substituting this value of d in ( i ) , we have a + ( m - 1) • ~ ~ = — •


, 1 1 1 • „ - 1 a -i 5 —- — • - a — n mn n mn

/th term of cor esponding A, P.

— + ( / - 1) - — i e. — • mn mn mn

mn

:. required Jth term of H. P. = — •

Ex. 4 . Find the 9th ter n of the progression 6 6 , 4 , 3 Ans - y

Ex. 5 . Find the nth term of ;

1 2 1 Ans. 21 ' 41 ' 20 ' "' " 43 - n

E x . 6 . The 6th term of an H P. is and the 17 th term is Show

that the 10th term is ^ r • 87

1 4 Ex. 7 If the 5th term of an H . P. is l-r- and the 8th term is —, find

3 5

12 4 the nth and the 20th term of the H. P. Ans ; — •

2n — 1 13

Means 18. Definitions :

If a, A, b are in A. P., A is called the Arithmetic Mean

between a and b. If a, G, b are in G. P., G is called the Geometric Mean

between a and b. If a, H, b are in H. P., H is called the Harmonic Mean

between a and b.

19. To find the values of A, G and H—the arithmetic, geometric and harmonic means-between two numbers a and b.

( i ) v a, A, b are in A. P. A - a ~ b - A, .\ 2A = a + b,

. * a + b A = —s— '


( ii ) V a, G, b, are in G. P.

; G* = ab; G = Jab. G b

It is customary to take the positive value of the square-root for the Geometric Mean and a > 0, b > 0.

( i i i ) v a, H, b are in H. P.

1 1 1 • A D T ' h ' T a r e i n A - p -

j i_ _ J i_ H a b H '

2 _ 1 J _ . . 2_ _ a+ b . H ~ a b ' " H ab '

• H - 2 a b •

We notice tha t : H = 1 ~ l 1 — p —

L t [ - 7 + T )

20. Theorems on Means.

Theorem I. If A, G, H be the A. M., G. M. and H. M, between two positive numbers a and b, then A • H = G2.

We have A = q ~ - , G = ^ f a b and H = •

... A . H = « ± £ X = = = 2 a + b

:. A • H = G2.

Theorem II. If A, G, H be the A. M., the G. M. and the H. M. between two unequal positive numbers a and b, then

A > G > H.


We have A — G = - \ f a t>

_ a —2sfab + b _ (\[a - \ f b y ~ 2 ~ 2 y

since a and b are unequal positive quantities.

.'. A > G. ( i )

Since AH = G2 , A_ = G G H

But as A > G, > 1. A ^ > 1, and G > H. u r l

. . A > G > H. Note—We may, now, define the means of n quantities.

Definition : If a,, a2,..., an are n quantities, then

a, + aj + • • • + a„ „ / —1 , Vtfi a2 - • • a„

and 1

± ( - L + J L + . . + ± ) n \ a\ a2 a„ j

are called the Arithmetic, Geometric, and Harmonic Means respectively of the given n quantities.

21. To insert n Means between two numbers. ( i ) To insert n A. M.'s between a and b. If a, A t , A2, A 3 , . . . A„, b are in A, P., then Ai, A2, A3 ... A,,

are called n A. M.'s inserted between a and b.

Let d denote the common difference of this A. P. Including the given terms a and b, there are ( n + 2 ) terms in this A. P.

b = ( n + 2 )th term of this A. P. = a + 0+ 1 ) d.


/ . Ai = a + d = a + b-a « + l '

A7 = a + 2d = a +2 ^ ^ > ,

A r = a + rd = a + r(b - a) « + 1 '

a i j , n(b — a) A , = a + nd = a + —JT+~1 '

( i i ) To insert n G. M.'s between a and 6 ( a > 0 , 6 > 0 ) . If a, G„ G2, Gj, ... GN, 6 are in G. P., then GT, G2( G3, .. G„

are called n G. M.'s inserted between a and b. Let r denote the common ratio of this G. P. Including the

given terms a and b, there are ( n + 2 ) terms in this G. P.

b = (n + 2 )th term of the G. P. =

Gi, G2 , G3 , • • • • G n , are equal to • ( ^ • ' ( ^ • • ( t ) * - °(T f' respectively,

( iii ) To insert n H. M.'s between a and b. If a, Hj , H2, H 3 • • • H„, b are in H. P., then H t , H 2 H3 • • H„

are called n H. M.'s inserted between a and b. a, Ht, H2, H 3 . . . H„, b are in H. P.


/. including the terms — and - j - , there are ( n + 2 ) terms a b in this A. P.

1 , = ( n + 2 )th term in this A. P. b

= — + ( n + 1 ) d where d denotes the common a '

difference of this A. P.

n + l 1_ b a J ( n + l

1 = J- + d = ~ +

-b ) ab

Ha a a (n + \)ab'

H 2 ~ « + a + (n + l)ab '

J J_ , _ 1 n ( a - b ) H „ ~ a + fl + ( „ + l ) a b '

, 1 1 , r(a~b) ( n + l ) b + r ( a - b ) In general, ^ = - + ^ y ^ = ^ (n + Tfab

- (» + l - r ) b+ra (n+\)ab

( n + 1 ) ab : . rth harmonic mean, H , = 7—-r-. , , • (n + l — r ) b + ra

Putting r = 1, 2, 3,...n, the required n harmonic means H l f H2 , H 3 . . .H„ are

( n + 1 ) ab ( n + l )ab (n+l) ab (n + l)ab nb + a ' ( n - l ) 6 + 2 c j ' ' " ' 2 6 + ( n - l ) « ' b + na

The formulae for inserting n means between two given numbers need not be remembered. In any particular problem, it is convenient to obtain the required number of means between two numbers ab initio.


22. Illustrative examples. Fx. 1. Insert ( i ) 7 A. M ' s . (ii) 7 G. M.'s, (iii) 7 H. M.'s between 4 andl6.

( i ) Let 4, A., Aj, A3, . . . A ; , 16 be in A, P.

There are in all 9 terms in Ibis A. P, 3

9th term. 16 = 4 + ( 9 - l ) r t ; d = — •

the required seven arithmetic means are

4 + l , 4 + 2 . 1 . 4 + 4 . 1 , 4 + 4 - 1 4 + 7 - f

11 , 17 I n 29 1 e- 7- i ' 10 y

( i i ) Let 4, Gi, G3, G, G,, 16 be in G. P. There are in all 9 terms in this G. P.

.*. 9th term. 16 = 4 >"« ^ = 4 ; r = (4 = 21'4.

.". the required seven geometric means are

, „1/4 1/2 3/1 . j i* 4-2 4-2 4-2 ... 4'2 . fiii) Let 4, H ; , H2 , Hs , . . . H;, 16 be in H. P.

There are In all 9 terms in this H. P.

The corresponding A P. is

_L 1 - 1 1 i " 4 ' H,' Hi' II.' "'H,' 16'

= gth term of the A. P. = 1 + 8d.

-3 12S

= ( r + 1) th term of the corresponding A. P. H r

1 1 ( - 3 \ 1 3r 32-3r + - - - - - - - lag- "

Hr = 1 2 8 32 —3r

Putting r = 1, 2, 3, 7 we get the required harmonic means

128 128 128 128 29 ' 26 ' 23 U

C. A . — 17

258 : (,oi 11 (ii. AIA.I BRA

Ex 2 Prove that the sum of the n arithmetic means inserted between any two numbers is n times the single A M. between the two numbers.

Let a, b be any two given numbers. Let the n arithmetic means between ft and b be

a + d, a -r Zrf, a + id a + nd.

The value of d is given from the fact that b is the ( n + 2 ) th term of the A. P. a, a + d, a + 2d, , a + nd, b.

.". b = a + ( n + 1 ) d. /. d = — • « + 1 .'. the sum of these n A. M.'s

= £ (a + d) -r (a + nd ) J , [ formula (iii 1 of article 4. ]

- f [ 2 f l + („ + 1 w ] = f + ]

~ j^a -r b J - u • +2

b J = n ( single A. M. ).

Ex. 3. The arithmetic mean of two numbers exceeds their geometric mean by3/2, and the geometric mean exceeds the harmonic mean by 6 5 . Find the numbers.

Let a and 6 denote the two numbers. Let A, G and H be their arithmetic, geometric, and harmonic means respectively.

.'. from the given conditions of the problem, we have

A = G + - i - ( i ) and H = G - y ...( ii )

Now, we know that AH = G2.

, f o + 4 ) ( c - - § - ) - * ;

G2 + ~ G - | • - G'; .-. G = 6.

We also know that G2 = ab, .. ab = 6 ( i i i )

( i ) reduces to -= 6 + i e . a + b = 15 ... ( iv )

To solve equations ( i i i ) and ( i v ) for a and b, we use the identity ( a - b y i = ( « + &y*-4<»6.

(a - b)' = 225 - 144 ; ( a - 6 ) s = 8 1 ; a - b = ± 9. ( v )

Solving ( i v ) aod ( v ) , we bave a = } 2 or 3 and ft = 3 or 12. Hence the required numbers are 12 and 3.


Ex. 4 . If a, b, c.d.e be five quantities such that a, b, c are in A, P , b, c, d are in G. P., and c, d. eare in H. P., then prove that a, c, e axe in G.P.

a + c , . \

V a, b, c are in A. P . , b = —. — — ( 1 )

V 6, c, d are in G. P. . c' = bd• ... . . . ( i i )

c, d, e are in H. P . , .'. d = . . . . ... ( i i i ) c + e Now a, c, e will b j in G. P , if c ' = ae. . . . . . . ( i v )

Obviously, in order to obtain relation ( i v ) , we have to eliminate b, d from ( i ), ( i i ) and ( i i i ).

Thus. M . c » « « ± J L . 2 c + e

( a + c ) a , , , „ .. c — — — - — • • c 1 + ce = ae + ce ; c m ae. c + e

Hence a, c, e are in G. P. E x . 5 , if a. b, c are in H. P., b,c, d are in G P. and c, d,e, are in A.P.,

ab' show that e = (2a - b)1

tac a, b, c are in H. P. .". b = ( i )

( i i ) . ( m )

a + c

V b, c, d are in G P. cs — bd V c, d e are in A. P. 2d = c + c.

We have to obtain a relation between a, b and e which is" therefore, to be obtained by eliminating c and d from ( i ), ( i i ) and ( i i i )

From ( i ) we have 6 ( a + c ) = 2ac.

c ( Ii - 2a) = - nb ; .'. c = • ia — b Substituting this value of c in ( i i ) , we get

= _ _L = d b ' b (2it — b)1 (2a-b)!

We ha"", tuns, found the values of c and d in terms of a, b which can be substituted in v i i i) for their elimination.

„, . . . . . . a" b a b Thus, ( HI ) reduces to 2 ^ ^ _ " _ b + <"•

a*b - (2a - b) ab = ab [2a - 2a + b J _ ab1

•'• e~ (2a-b)3 (2 a - b p [2a - b)1

ab'


Ex. 6. ]f a* = 6V = o" = dv = . . . and <r, 6, c. d,... are in G. P. then show that x, y, z, w ... are in H. P.

r i * i? * j"' . . Let a = 6 = c = d = = k. so that

„ - i,Vx 3. - b11' >.1/z ,) .i/w a = k , 6 = fe , c = fe , d ~ k V i , c, rf, ... are in G. P.

_&_ c_ __ a 6 a

e- =

_ i - A_ J . J_ i - JL k ' * =* 3 v = *

. j L = J i = i y x z y w z 1 1 1 1 • . ^ . . — , — , — , — . . . . are in A P. x y s w

x,y,z,w. .. . . are in H. P. Exercies 9 ( c )

1. Find the nth term and also the 10th term of each of the following sequences. . . 10 10 10 10 ( 1 } 67 • 60' 53' 46 * '

1 4 1 ( 11 ) 3, 2 —, 1 -y , 1 -y ,

J L _ L _ __L_ 1-4' 3-7' 5-10' 7-13'

. . . 1 1 1 1 ( iv ) 3-3 ' 6 5 ' 12-7 ' 24-9

Find the /ith term of the H. P. *s if the 12 ( i ) first and second terms are 2 and yy

1 3

( ii ) 6th and 12th terms are 3 y j and 1 —

(iii ) 7th and 12th terms are 7 and 12.


3. If the /th term of an H. P. is /, and mth term is m, find the nth term. Deduce that the ( / + m - 1 ) th term is Im.

4. If the /th term of an H. P. is m and the mth term is I.

show that the ( / + m )th term is ~ — l + m

5. Three numbers form of an H. P. The product of the first two numbers is equal to the third, and the sum of the three numbers is 23. Find the numbers.

6. If the pth, tfth, rth terms of an H. P. be a, b, c, respectively, prove that

ab ( p — q) + be ( g - r ) + c a ( r — />) = 0. 7. Find the values of x for which the three numbers

2+x, 3 + *, 9+jrare ( i ) in G. P. ( i i ) in H. P. 8. Find the two numbers, if

( i ) their G. M. and H. M. are 5 and 3 ; ( ii ) their A. M. and H. M. are 2 and 7/8.

9. If the H. M. between two numbers is to their G. M. as 12:13, prove that the numbers are in the ratio 4 :9.

10. The A. M. of two numbers exceeds their G. M. by 15 and their H. M. by 27. Find the numbers.

11. If a, b, c are in A. P. then show that ( i ) a2 (b + c), 6 3 ( c + a ) , c2(a + b) are in A. P. ( i i ) ( 6 + c ) J - a * , ( c + a )2 — 62, ( a + f t ^ - c 2

are in A. P.

( i i i ) 1 , f - 1 , • r - are in A.P. \}b+\c V c + v a V a + V b

12. If a, b, c are in H. P., then show that

( i ) T T - . —T"—» A — are in H. P. v b + c' c + a a + b

. .. > a b c • t t r« ( ii ) , —— -. —r— are m H. P. v b+c — a c + a — b a + 6 — c


13. Prove that J - + — — + - - L . js e q ual to — or — o r — b b—a b—c n b b b

according to as a. b, c are in A. P. or G. P. or H. P. 14. If a, b, c are in A, P., b, c, d are in G. P., and r , d, c are

in H. P., show that e = . a

15. If a, b, c are in A. P., b, c, d in G. P., and c, d, e in H. P., show that c2 = ae.

16. If a, b, c are in A. P., b, c, din H. P., and c, d, e in G. P., show that a2e = b2 ( 2b — a ).

17. If a, b, c, d are in G. P., prove that ( i ) a2-b2,b2- c2, c2-<P are in G. P., ( i i ) a2 + b2 + c2 = (a + b + c) (a-b + c), ( i i i ) (6 + c ) ( 6 + rf) = (c + a ) ( c + rf).

18. Insert means as given below :— ( i ) 10 A. M.'s between 7 and - 15 :

( ii ) 17 „

(iii) 6 G. M.'s

( i v ) 5 „

( v ) 5 H. M.'s

( vi) 5

19. Prove that the product of n geometric means inserted between a and b is ( ab )*'2.

20. Two arithmetic means At, A2, two geometric means Gi, G2) and two harmonic means Hj, H2 are inserted bet-ween any two numbers. Prove that „ 2 = ^ • Hi + H 2 Hi H2

- 4 1 2 '

28 _7_ 32 '

3 4 „ " 4 0 - 1 - ;

11 „' - 3 ;

2 1 1 4 2 " T » " 1 T T '


21. If Ai, A2, • • • Ar • • • A„ be n arithmetic means between b and a, and Hi, H2, • • • Hr, • • • H„ be n harmonic means between a and b, then show that A, • Hr = ab.

22. If a, b, c are in H. P., show that

. . . 2 1 . 1 b+a h+c ( l ) —r- — r + r ( ii ) + y = 2 . v ' b b—a b — c b — a b — c

23. If a, b, c, d are in H. P., show that ab f be 4- cd = 3ad. 24. If a, b, c are in H. P., show that be, ca, ab are in A. P. 25. If a is the A. M. between b and c, and b is the G. M.

between c and a, then show that c is the H. M. between a and b.

26. If ( b + c ), ( c + a ), ( a + b N are in H. P., then show that a2, b2, c2 are in A. P.

27. If a, b, c are in H. P., show that ( i ) a ( b + c ), b ( c + a ), c ( a + b ) are in A. P.

( " ) I. / I \ » — / • ! I I ui L IN areinH.P. be ( a + 1 ) ca ( b + 1 } ab (c+1)

28. If a, b, c are in A. P., p, q, r are in H. P., ap, bq, cr are

in G. P., then show that — + — =•£- + — . c a r ^p

29. The arithmetic mean between two numbers exceeds their geometric mean by 10 and harmonic mean by 16; find the numbers.

30. A. M. between two numbers is to their G. M. as 5 :4 and the difference of their G. M. and H. M. is 16/5. Find the numbers.

Answers. Exercise 9 ( c )

. , . , 10 5 . . . . Q *

(Hi)

( i v )

74 - 7» ' 2 n + 2 ' 4

i ( 2 # » — 1 ) ( 3 » + 1) ' 19-31 '

1 1 1 3 - 2 - 1 ( 2 n + l ) ' 6 3 2 9


2. " . <„ , - « L (iii) 3 . *» "n — \ 5n-4 ' 19 —n 19-n 1+ m -n

5. 3, 5, 15 or - 4. - 9 . 3 6 . 7. ( i j - M i ) - - | -

2 7 1 8- ( i > I — . 1 5 . ( i i ) _ , _ _ _ . 1 0 1 2 0 , 3 0

18. ( i ) 7 - 2r, O = l , 2 1 0 ]

( " J 4 - ( 7 - 5 r ) , ( r = l . 2 , 3 ,17); 2

(iii) 2S

(iv)

T ) • = 1 , 2 6 ] ;

32 / 3 j l ± y ) 1.2 5] ;

(v ) j ^ r . O = 1 , 2 , - . . , 6 ] ;

15 ( v | ) i > = i . 2 , . . . . , 5 ] ;

2 9 . 4 5 , 5 . 3 0 . 32 ,8 .

Chapter 10

Summation of Series 1. Introduction. 2. Sigma notation. 3 The three sums i

2 r, and 2 c \ 4. The sum of the series whose rth term is ar1+br'' + cr+d. l l 1 5. Some miscellaneous methods: The method of differences, the method of partial fractions and the arithmetico geometric series 6. The Infinite Geometric Series. 7. Recurring Decimals. Exercise 10.

1. Introduction. In the last chapter we have already considered the sum of

Arithmetic and Geometric series. In the chapter 8 we have proved by mathematical induction that

( i ) the sum of the first n natural numbers is

\ n ( « + 1),

( i i ) the sum of the squares of the first n natural numbers is - j - n ( n + 1 ) ( 2n + 1 ) o and

( i i i ) the sum of the cubes of the first n natural numbers is

~ n J ( « + I.)2-

In this chapter, we shall consider methods for finding t h e sum of some other simple types of series.

In the next paragraph we shall explain how the " sigma " notation can be conveniently used in examples on the summation of series. In the following paragraphs we will explain how the above sums ( i ), ( i i ) and (iii ) can be obtained by the method of differences.

265

266 : COLLEGE A.LGERKA

2. Sigma notation. The letter " 2 " of the greek alphabet (pronounced as

" sigma " ) is used to denote the sum of a given series. The letter 2 is placed before the > th term, say, ar. We, thus, wr te 2 a,, to denote the sum of terms of the type ar. If we want to sum up terms ar for values of r corresponding to r = 1,2, 3, • • • • n, we denote the sum by

r=n « 2 ar. or by 2 ar.

r = l 1

Thus, we write m

at + a2 + a3 + + ar + + •• + <?„ = 2 ar. I

The results for the sums of the series mentioned inparagraph 1 above can be conveniently written as

" n (n + 1 ^ 1 + 2 + 3 + •.. + r • • • + » = 2 r = l l J . ( i )

l 2

12 + 22 + 32 + • • • + J"2 + • • • + n2 = 2 r2

= - g - n ( n + 1 ) (2n + 1) ... (2)

13 + 23 + 33 + • • • + r3 + • • • + rf = 2 r3 l

= - J - n 2 ( n + l ) 2 ... (3)

The sum of the Arithmetic series whose first term is a and the common difference is d can be written thus :

a +(a + d) + {a + 2d) +•••• + (« + V^l d)

= 2 ( a + ^~ld)= -%-[2a + ( n - 1 )d]. ... (4) l L

Similarly the sum of the Geometric series whose first term is a and the common ratio is x ( # 1 ) can be written thus :

a + ax + ax2 + +

= < 5 >

SUMMATION OF SERIES : 2 6 7

3. The three sums.

In this paragraph we shall prove the formulae CI J, ( 2 ) and ( 3 ) of the article 2 by means of identities based on differences of squares, cubes and fourth powers of consecutive integers ( x + 1 ) and x.

( i ) To prove that

We have the indentify ( x + 1 )2 - x2 = 2x +1 .

Put x = 1, 2, 3, • • • , ( n — 1 ), n, successively in this identity.

We, then, get 22 - I2 = 2-1 + 1, 32 - 22 = 2-2 + 1, 42 — 32 = 2-3 + 1,

n2 - O - l ) 2 = 2 - ( « - 1 ) + 1, (n + 1 )2 - «2 = 2-n + 1.

Adding these n equalities columnwise, we have ( « + l ) 2 - l2 = 2 ( 1 + 2 + 3 + ... + » )

+ ( 1 + 1 + 1 + times).

.*. n2 + In = 2Si + n, where S t = 1 + 2 + 3 + + ».

S, = 1 + 2 + 3 + - . . . + » = 2 r = It n ( n + 1 )

2

n 1 + n = 2Si ; 5 n ( n + 1 ) 2

Si i. e. 5 r = 1 + 2 + + n = n (n + 1 )

( ii ) To prove that

S2 = I2 + 22 + • • • ' + n2 = 2 r2 n ( » + 1 ) ( In + 1 ) . 6


In the last case we used an identity based on the difference of squares of two consecutive numbers x + 1 and x. We will use here, similarly, an identity formed by taking the difference of the cubes of consecutive numbers x + 1 and x.

We have ( * + 1 )3 - x3 = 3x2 + 3x + 1. Put x = 1, 2, 3, • • • • , ( n — 1 ), it successively in this

identity.

We, then, get 23 - l 3 = 3-12 -t- 3-1 + 1, 33 - 23 = 3-22 + 3-2 + 1, 43 — 33 = 3-32 + 3-3 + 1,

n3 - (n - l ) 3 = 3 - ( n - l ) 2 + 3 . ( n - 1) + 1 ( n + 1 )3 - n3 = 3-n2 + 3-n + 1.

Adding these n equalities columnwise, we have ( n + l )3 —13 = 3 ( l 2 + 22 + 32 + • • • + « 2 )

+ 3 ( l + 2 + 3 + - - - + n ) + ( l + l + l + - . . n times).

( n + l J ' - P s s S S a + aSt + fl, where S2 = l 2 + 2 2 + 3 2 + . . • + n 2 and Si = 1 + 2 + 3 + . • . + n .

n3 + 3n2 + 3n = 3S2 + 3 • + «.

3S2 = n3 + 3n2 + 3n - -J- n (n + 1 ) - n

= ~ { 2 n 2 + 3n + 1] = - y (n + 1 ) (2n + 1 ). n

n ( n + l ) ( 2 n + l ) 6


Thus, we have C - ,1 , ,1 , . 2 - n ( n + 1 ) ( 2n + 1 ) S2 1. e. 2 r2 = l2 + 22 + • • • • + nJ = ^ —

1

( iii ) To prove that n2(n + 1 )2

S3 = 2 r3 = l3 + 23 + • • • + h3 = 1 4

As in the two previous cases we will now use here, the identity based on the difference of the fourth powers of the consecutive numbers x + 1 and x. We have

( x + 1 ) 4 - * 4 = 4x3 + 6;c2 + 4;t + 1 . Put x = 1, 2, 3, • • , ( n - 1 ), n successively in this identity.

We, then get 2 1 - 1 4 = 4 . l 3 + 6-l2 + 4.1 + l , 34 — 24 = 4-23 + 6-22 + 4-2 + 1, 44 — 34 = 4-33 + 6-32 + 4-3 + 1,

u4 - ( n - 1 )4 = 4 ( n - 1 )3 -(- 6 ( n - 1 )2 + 4 ( n - 1) + 1, ( n + 1 y - n* = 4n3 + 6n2 4- 4» + 1.

Adding these » equalities columnwise, we have ( n + 1 ) 4 - l 4 = 4 [ 13 + 2 3 + 33 + - + h3 ] + 6 [ l 2 + 22 + - + « 2 ]

+ 4 [1 + 2 + 3 + • • • + » ] + [ 1 + 1 + 1 +•••» times].

.'. ( » + l ) 4 - l = 4 S J + 6 S 2 + 4 S 1 + « , where S3 = l3 + 23 + 33—I- «3 and Sa and S2 have the meanings as in the last two cases. / . 4 S 3 = ( » + l ) 4 - l - 6 S 2 - 4 S 1 - » «

- t 4 -1V , . M 6 » ( " + 0 ( 2 , 1 + 1) , , ( » + ! ) = ( » + 1 )4 - ( » T-1 ) - O g 4

= ( « + 1 ) [ ( » + I )3 ~ I - « ( 2« + 1 ) ~2n ] = « + l ) [ ( » + l ) 3 - ( 2 n + l ) - n ( 2 « + 1 ) ] = ( » + l ) [ ( « + ! ) 3 - ( 2 » + l ) ( « + 1 ) ] = ( « + 1 /2[ ( ;» + I ) 2 — ( 2;» + 1 ) ] = „ 2 ( « + l )2.

. <j n\n+ I ) 2

• • *3 = A •


Thus, we have

S3 i. e. 2 r3 = l 3 + 23 + • • • + n3 = ( " 4

+ ' ...

4. The sum of series whose r th terms is car3 + br2 + cr + d.

The method to find the sum of any series, the rth term of which can be expressed in the form ar3 + br2 + cr + d, where a, b, c, d are some constants is given below.

Let the rth term of a series, a, = ar3 + br2 + cr + d. Puting r = 1, 2, 3, • • •, « successively, we have

a , = a - l 3 + M 2 + c - l +d, a2=a-23 + b-22+c-2 + d, a3 = a-23 + b-32 + c-3+d,

q„ = q-n3 + b-n2 + c-ra + d.

Adding columnwise, we get n n

2 o r = a ( 2 r 3 ) + b( 2 r 2 ) + c ( 2 r ) + ^ d • i i I i i

n n it Substituting the values of 2r3 , 2r2 , 2 r and observing that

i I I

n

2d=d + d + d+ --ton terms i. e nd, the required sum can

be obtained. This method is illustrated below by solving the following

examples Ex. 1. Find the sum

1-2 + 2 - 3 + 3 - 4 + + « ( n + 1) . Here ar = r (r + 1).

•-• 2 « , = f r( r+1) = 2(r» + r) = 2 + 2 r t l t 1 \

» ( » + 1) (2n + 1) n (n + 1) 6 + 2

^ > L l ^ L i j [ 2 o + 1 + 3] = n(n + l ) ( » + 2 ) .


Ew 2 . Find the sum

l a + 32 + 5a • • • + 29'2. It can be easily seen that the general term is a, = [1 + ( r - 1) -2 ] 2 --= (2c - 1 f = 4r" - 4r + 1. Thus, a r is expressed in the required form. To find the number of terms

in the given series, we put 2r — 1 = 29 r = 15, and hence there are 15 terms in the yiven series.

V a,- = 4r- - 4r + 1

IS 1 a t i [?-]-4 Pf']•[?•]

= 4 £ 15 ( 15 + 1 ) (2• '5 + 1 ) J _ 4 15(15 + 1) j + ^

= 160 X 31 - 30 X 16 + 15 = 4495.

the required sum = 4495.

Ex. 3. Find the sum 1 1 - 3 + 2 3 5 + 3 5-7 + 4-7-9 + • • • • to » t e r m s .

The first factors of the terms form the sequence 1, 2, 3, 4, • • • whererth term is r. Similarly, the second factors of the terms form the sequence 1 . 3 , 5 , 7 , • where rth term is ( 2r - 1 . Similarly, the third factors of the terms form the sequence 3, 5, 7, 9, • • • where rth term is ( 2r + 1

.'. the rth term of the given series, ar - r (2r - 1 ) ( 2r + 1 ) = 4r'i - r.

4 2 2

. . , n (n + 1) (2«' + 2n - 1 ) . - the required sum = — — - •

Ex. 4 . Find the sum (2* - l 3 ) + ( 4» - 3») + ( 63 - 5 : i) + • • • • to « brackets.

We can write the bracket as [ ( 2r ) 3 - (2r - 1 )3 ] Thus, a-t = ( 2 r )5 - ( 2r - 1 )3 = Sr1 - (?r' - 1 2 ^ + 6r - 1 ).

a r = 12r* — 6r + 1.

••• 2 a , = 12 2 r> - 6 2 r + 2 1 1 i l l

, „ » < » + ! ) 2n + l ) _ 6 . n . ( . n _ + l ) + w

6 2 = « [ 2 ( n + l ) ( 2 n + l ) - 3 ( « + 1 ) + 1 ] = « [ 4 n s + 6n + 2 - 3» - 3 + 1 ] = n* [ 4 » + 3 ] .

the required SU ED = h"1 ( 4 « + 3 ).

, 2 1 2 : COLLEGE ALGEBRA

5. Some miscellaneous methods.

We, now, proceed to i l lus t ra te few o the r me thods for s u m m i n g u p some t y p e s of simple seHes.

1, The Method of differences.

I n t he case of some seris in wh ich the d i f ferences of successive te rms

f o r m an A. P . or G . P . . t he fol lowing method can be employed to find the n t h

t e rm . T h e sum of euch a series to n t e rms may then be o b t a i n e d .

Ex. 1. F ind the « t h t e r m a n d the sum to n t e rms of t he ser ies

1 + 7 + 17 + 31 + 49 +

We have a% — a\ =6.

«3 - a-i = 10. a, - a., = 14.

« - a = (n - 1 ) th te rm of t he sequence 6, 10. 14

A d d i n g co lumnwise , we get

a n - a i = 6 + 10 + 14 + 18 + to ( « - 1 ) te rms.

«,, - a, = [ 2-6 + ( » — 2 ) • 4 ].

= 1 [ 12 + 4„ - 8 ]

[4m+ 4] = (n — 1) ( 2 « + 2 ) .

an = 1 + 2 (n - 1) (n -f 1). [ V « , = l ] ,

an = 2mj - 1. a , — 2r1 - 1,

I l l 6

= " ( " + 1 ) <2n + 1) - 3 « = n (« + 2 ) ( 2« - 1 ) 3 - 3


Ex. 2 . F i n d the n t h t e r m a n d t h e sum to » t e r m s of t h e series 3 + 5 + 9 + 17 + 3 1 +

W e have — a t = 2 .

«s _ «> = — = 8.

a * - a » - l = ( n - 1 ) t h t e rm of t he sequence 2, 4, 8, wh ich i s a G, P .

A d d i n g co lumnwise , we get a„- a> = 2 + 4 + 8 + t o ( n - l ) t e r m s

2 ( 2 - 1 - l ) - — a i r [ 2 2-

.'. = 2 " - 2 + 3. [ V a i = 3 . ]

a„ = 2 + 1. •'• a,= 2r + 1.

| « , = 2 + 2a + 23 + 2< + + 2" J + 2 1

= n f = l l + „ = 2«+1 + „ - 2. 2 - 1

.'. t he requi red sum = 2" + 1 + « — 2.

I I . Method of Pa r t i a l fractions, If t he general te rm of a ser ies cons is ts of the p r o d u c t s of t h e reciprocals

of two consecut ive t e rms of an A P . , then the term c a n be split u p i n t o part ial f r a c t i o n s and t h e series c an be summed . T h e m e t h o d is i l lust ra ted in t h e fol lowing example

Ex. 3 . S u m the series

+ ihr + Tr-y + whs +""10 M terms-H e r e , the f ac to r s in t h e d e n o m i n a t o r s a re the p r o d u c t s of two successive

t e rms of an A. P 3 , 7, 11, >5, 19,

.'. rth t e rm of t h e given series, ar (4 r-l)(ir+3)

Express ing a w a s t h e d i f fe rence of i t s part ial f r a c t i o n s , we h a v e

= i r _ i i _ n . a r 4 | _ 4 r - 1 Ar + 3 J

C. A.—18


B y p u t t i n g r = 1, 2, 3 ( n — 1 ) , n in success ion, we get

* - t [ T " n ]

= T [ iT " B ]

j. r _ i L _ 1 " n 1 = 4 | _4« - 5 4 « - 1 J

= a r_j i_l " " 4 | _4« "

1 + 3 J A d d i n g columnwise , we get

^ = 7 [ T " W ' V s ] = 3 ( 4 « + 3 )

•'• t he required sum = r - r - " • 3 ( 4 « + 3 )

^ Note . If t he general t e rm of a series consis ts of t he p roduc t s of t he rec iprocals of two consecu t ive t e rms of a n A P . t he formula for t h e sum of t he ser ies c a n b e established by t h e method of i nduc t ion also.

I I I . Arithmetico-Geometric Serie .

A ser ies wh ich is fo rmed by mult iplying the co r r e spond ing t e rms of an A. P . and a G. P . is called an ar i thmet ico-geometr ic ser ies . T h u s in t h e series

a + (a+d)r + ( a + 2d) r3 + ( a+3d ) r 1 •••• + f " 1 ... ( i ) t he t e r m s are formed by tak ing t he p roduc t s of cor responding t e rms of the

A. P . a+ (a + d ) + (a + 2d) +••••+ (a + n - id), and t he

G . P , 1 + r + + r> + + r n ' i .

H e n c e the series ( i ) is an ar i thmet ico-geometr ic series T h e me thod of summing u p such series is explained in the following

example.

Ex. 4 . S u m the series

2-1 + 4-3 + 6-9 + 8-27 + 10-81 + to n t e rms .

Le t S = 2 1 + 4-3 + 6-9 + + ( 2n - 2 ) • 3" + 2 » - 3 " ( i )

Mul t ip ly ing by 3, the common ra t io of the Geometr ic series, we get

3 -S = 2-3 + 4-9 + 6 - 2 7 + • • • + ( 2 « - 2 1 a " " 1 + 2 n - 3 " . . . . ( i i )


S u b t r a c t i n g ( i i ) f r o m ( i ) , w e get

( 1 - 3 ) S = 2-1 + $3 ( 4 - 2 ) + 9 ( 6 - 4 ) + 2 7 { 8 - 6 ) +

... + 3 " " 1 ( - 2M-3".

( —2 )»S = 2-1 + \2 ( 3 + 9 + 2 7 + ... to ( n - 1 ) t e r m s ) \ -Zn-l".

= 2 + 2 £ 3 • 3 i — U j - 2 n - 3 n

= 2 + 3 n - 3 - 2 « - 3 m = - 1 - 3 " ( 2 » - l ).

S = j [ 1 + 3 " ( 2 « - 1 >]

6. AD infinite geometric series.

A series in which every te rm is followed by a n o t h e r is called an infinite series. I f , in an in f in i t e series, t he te rms a re i n G- P . , t h e series is called an infinite geometric series.

If the sum S „ of » t e r m s of an inf in i te series approaches a def in i te n u m b e r S . a s n is made larger a n d larger , then S i s defined as t h e sum of t h e inf ini te series.

Consider t he inf in i te geometr ic series

i + l + T + 8 +

The sum S n of n t e rms of this series is

— i e 2

2

An >t is made larger and larger H i " becomes smaller a n d smaller a n d ult imately would approach t he n u m b e r zero. H e n c e S « approaches t h e n u m b e r 2 as n t e n d s to inf in i ty . The re fo re , t he sum of t h e above inf in i te geometr ic series is 2,

Le t u s now c o n s i d e r t he inf ini te geometr ic series

a + ar + at* + ar" +

a l l - r " ) The sum Su of n t e rms of th i s G. P . — • 1 - r


W h e n r < 1 [ a p roper f r a c t i o n pos i t ive o r n e g a t i v e ] , t h e numer i ca l « n

va lue of r decreases a s « i nc rea se s . T h e r e f o r e , r a p p r o a c h e s the a a m b e r zexo a s n i s m a d e larger a n d l a rge r .

—-— • rn t e n d s to zero a s n t e n d s to in f in i ty , 1 — r

S H t e n d s to a s n t e n Q S t 0 in f in i ty .

^ a - i s t h e s u m of t h e given in f in i te geomet r ic ser ies by

def in i t ion ,

a t h e sura S of t h e in f in i t e geomet r ic series is ^ - ^ - ^ i f — 1 < r < + l .

If r i s n o t a posi t ive or nega t ive p rope r f r ac t ion i. e . r 1 numer ica l ly ,

t h e n u m e r i c a l va lue of r" wou ld n o t decrease a s n i nc reases ; a n d h e n c e t h e in f in i t e geometr ic series h a s no sum f o r such values of r.

7. Recurring Decimal. R e c u r r i n g dec imals f u r n i s h a good i l lus t ra t ion of t h e in f in i t e geometr ic

ser ies .

C o n s i d e r «3333 T h i s i s briefly wr i t t en as -3*, a n d i s called a re-c u r r i n g dec imal . T o eva lua te 3 , w e h a v e

•3 = -3333 = 3 + 03 + -003 + -0003 +

10 ^ 10" 10' + 10< +

= _ 3 A° _ JL = 1

= i -1,10 9 r • 3 1

s ince in t h i s inf in i te geomet r ic ser ies a " — r = — a n d t h e sum of t h e

in f in i t e geomet r ic ser ies i s - ( r < 1 ) , 1 — f"

Exercise 10

Find the sum of the following series : 2 2 2 1

1. 1 + 2 + 3 + - - + 30 3 3 3 3

2. 1 + 2 + 3 + + 2 0 ,

3. 21 2 + 222 + 233 + +50 2

4. 11 3 + 12 3+ 132 + +253.


5. 502 - 492 + 482 - 472 + + 2 2 - l 2 .

6. 4 8 3 - 4 6 3 + 4 4 3 - 4 2 3 + + 43 - 23.

7. l 2 - 52 + 92 - 1 3 2 + - 9 3 2 + 972

Write down the r th terra, and find the sum to n terms of each of the following series :

8. l 3 + 33 + 5 3 +

9. l 2 + 52 + 9 2 + 10. 1-2 + 2 - 3 + 3 - 4 + 11. 1-2-3 +2-3-4 + 3-4-5+

12. l-22 + 2-32 + 3 -4 2 + 13. 2-5 + 5-8 + 8 - 1 1 + 14. 1-2-4+ 2-3-5+ 3-4-6+ 15. 3 - 1 + 5 - 8 + 7 - 1 5 + 9 - 2 2 + 16. l + ( l + 2 ) + ( l + 2 + 3 ) + ( l + 2 + 3 + 4 ) + »»

1 7 . + L ± 2 ± _ 3 + i ± 2 + 3 ± l + . . .

i , I ' a- l 2 + 2 2 + 32 , l W + 4 J ,

19. ( 53 - 3 3 ) + ( 9 3 - 7 3 ) + ( 1 3 3 - l l 3 ) +

20. l-« + 2 ( « - l ) + 3 ( n - 2 ) + ••••

21. l 2 . « + 2 2 - ( » - l ) + 3 2 - ( n - 2 ) +

22. 32 + 72 + l l 2 + 1 5 2 + Find the nth term and the sum to n terms of each of the

following series : 23 . 4 + 13 + 28 + 49 + 7 6 + 24. 4 + 1 4 + 30 + 52 + 80 + 114+ -25. 4 + 10 + 1 8 + 2 8 + 4 0 + 26. 3 + 5 + 11 + 2 9 + 83 + 2 4 5 + 27. 3 + 9 + 2 1 + 4 5 + 93 + 189+ 28. 28 + 32 + 52 + 152 + 652+

2 9 , T 5 + 2-3 + 3-4 + 4-5 +


3 0 ' 2 T + 5 T + 8 T l + i m +

31. - J - + - J L + _ L _ + _ ! _ + 4-13 T 13-22 ^ 22-31 + 31-40 +

,2 3 4. 5 , 7 , ~2 2~ + 2 2 + " T T + 1 - 2 2 - 3 3 -4

33. 3 + 7 - 2 + ll-22 + 15-23 + 19-2* +

34. 3 + 3 - 9 + 5 - 2 7 + 7-81+

35. 5 + 1 2 * + \9x + 26*3 + 36. Express as fractions

( i ) 1-6, ( i i ) S23, ( i i i ) -234, ( iv ) -2103.

Answers—Exercise 10

1. 9,455. 2 . 44,100. 3 . 40,055. 4 . 1,02,600.

5 . 1,274. 6. 58,752. 7 . 4 ,897.

8 . ( 2 r - l ) s ; « s ( 2 « a - 1 ) . 9 . ( 4 r - 3 ) 2 ; ( 1 6 » a - l 2 « - 1 )

1 0 . + - y - » ( » + l ) ( » + 2 ) .

11. r(r + 1) (r + 2 ) : « (» + 1 ) { » + 2) (» + 3 >•

12. + i r t ( t t + i ) (» + 2) (3» + 5).

13. ( 3 r - 1 ) ( 3 r + 2 J ; n (3»a + 6n + 1 ) .

1 4 . r ( r + l ) ( r + 3 ) ; ( » + l ) ( » + 2 ) ( 3 » + l 3 ) .

1 5 . ( 2 r + l ) ( — 6 ) ; - g - • » (28m® + 27» — 3 7 ) .

16. - i - n ( » + l ) { » + 2).

17. - | - ( r + l ) ; - i - » ( n + 3 ) .


1 8 . ~ ( 2 r + l ) ; - » ( » + i ) ( 4 * + 3 ) .

19. (4r + l ) » - ( « » • - 1 ) » ; 2»(16h«+ 24« + 9 ) .

20. r ( » - f + l ) ; — « (» + 1 ) (» + 2).

21. + +

22. (4r - 1)» : 16«s + 12» - 1).

23. 3»a + 1: • »{2ns + 3n + 3). 24. 3«s + n ; « ( » + l)a.

25. » ' + 3 » ; - y - - « U + l ) U+5). 26. 3B _ 1+2; { 3" - 1) + 2n

1 27. 3 ( 2* — 1); 3 ( 2m+1— n — 2). 28. e " " ^ ; — ^ + i7».

OO 1 _ . on I " , » ( » + 1 ) ' n+1 ( 3 » - l ) (3« + 2 ) ' 2 (3»+2)

„ 1 » 2n+l , w(n+2) 3 l ' (9» - 5 ) (9n + 4 ) ' 4 {9rt+2) n ^ f t + l ) 1 ' ( » + V

33. (4»—1) 2*-1; 2* ( 4n — 5 ) + 5.

34. (2« — 1) • 3 n ; 3*+1 (tt — 1) + 3 .

35. (7n — 2}x . l _ x + ( i - * ) » 1 - *

, , . . . 5 ,.. , 518 , . . . ,232 . 2101 36. (i ) - J - , («) gg » (»i) ^ j , (iv) -3950-.

Chapter 11

Permutations and Combinations

1. I n t r o d u c t i o n . 2. Def in i t ions . 3. No ta t ion . 4. F u n d a m e n t a l P r inc ip l e . 5. N u m b e r of P e r m u l a t i o n s of « d i f ferent th ings t aken r a t a t i m e . 6 . Factor ia l no ta t ion . 7 P e r m u t a t i o n s with res t r ic t ions . I l lus t ra t ive examples* Exerc ise 11 ( a ) . 8. C o m b i n a t i o n s of n d i f ferent t h i n g s taken r a t a t ime . 9. Complementa ry c o m b i n a t i o n s . 10, Combina t ions wi th res t r i c t ions . 11. I m p o r t a n t iden t i ty : n + 1 C , = " C , + "cr-i. 12. I l lus t ra t ive examples. Exerc i se 11 ( b ) . 13 P e m u t a t i o n s of t h i n g s n o t all d i f f e r e n t 14. Pe rmuta -t i o n s wi th repe t i t ions 15. C o m b i n a t i o n s of ti d i f ferent t h i n g s taken any n u m b e r a t a t ime. 16. Kemarks r ega rd ing solut ion of problems. Exerc ise 11 (c).

1. Introduction.

In this chapter we shall study the number of different possi-ble ways of grouping or arranging some things taken at a time out of a given number of things.

Let, for instance, a, b, c, d denote four different given things Consider the different groups which can be formed out of these four things when we take two, three or four things at a time.

( i ) Groups of these 4 different things, two at a time, are ab, ac, ad, be, bd, cd i. e. 6 in number.

( i i ) Groups of these 4 different things, three at a time, are abc, abd, acd, bed i. e. 4 in number.

( i i i ) Group of these 4 different things, four at a time is abed i. e. only one.

If , however, the order of things in any group is considered, we get different arrangements from one and the same group.. For instance, from a group of two things, ab, we' get two different arrangements ab and ba.

280

PERMUTATIONS AN]D COMBINATIONS : 2 8 1

The different arrangements of the above groups ( i ) , ( i i ) , ( iii ) , therefore, can be given as follows :

( i ) Arrangements of the four different things, two at a time, are ab ac ad be bd cd

ba ca da cb db dc and are 12 in number.

( i i ) Arrangements of the four different things, three at a abc abd acd bed acb adb ode bde bac bad cad cbd bca bda cda cdb cab dab dac dbc cba dba dca deb

and are 24 in number.

(iii) Arrangements of the four different things, four at a abed bacd cabd dabc abdc bade cadb dacb acbd bead aedb beda adbc bdac adeb bdea

and are 24 in number.

Counting and writing the groups or the arrangements in cases when there are more given things will, obviously, be a tedious job. Many a time, however, we require only the number of the groups or the number of arrangements. Hence, in the following articles, methods for finding the number of of groups and the number of arrangements shall be considered,

2. Definitions.

Combinations. The different groups C or selections ) that can be formed out of a given set of things by taking some or all at a time, without any considerations for the order of things, are called the combinations.


Thus, we have 4 different combinations abc, abd, acd, bed taken three at a time out of four given things, a, b, c, d.

Permutations. The different arrangements in a line of the things in a group, taken out of a given set of things, are called permutations.

Thu , we have 12 different permutations of two things taken at a time out of four given things. 3. Notation.

( i ) The total number of combinations of n different given things, taken r at a time, is denoted by the symbol

)' Thus, C2 i. e. the number of combinations of 4 things

taken 2 at a time = 6. Similarly 4C3 = 4 and iCi = 1. ( i i ) Similarly, the total number of permutations of n

different given things taken r at a time is denoted by the symbol "P, or „Pr or P ( n, r).

Thus, 4 Pj i. e. the number of permutations of 4 things taken 2 at a time = 12. Similarly, 4P3 = 24 and 4P4 = 24.

4. Fundamental Principle.

Before considering the general formulae for the nu mber of permutations and the number of combinations of n things taken r at a time, we wilt explain, here, by an example, the general principle which is of great use in this theory.

Example. There are four different routes for going from the place P to the place Q. Find the number of ways in which a man can go from P to Q and return back by a different route.

Let A, B, C, D denote the four different routes between P and Q, A man can go by any one of the four routes from P to Q and having gone there, he can return by any one of the three remaining routes. Thus, for each one of the four ways of going, theree are three ways of coming back. Hence the total number of ways of going and coming back is 3 + 3 + 3 + 3 i. e. 4 x 3 or 12.

Cr or nC, or C (n, r ) or ( C


The 12 different ways of going and coming back can be indicated in terms of the routes as follows :—

A B B A C A

A C B C C B

A D, B D, C D,

The method, used in the above example, leads us to the following Theorem.

Fundamental Theorem —

If a certain thing can be done in m ways, and after having done it in any one of these m ways, a second thing can be done in n ways, then both the [things can be done together in mx n different ways.

Let au fl2. «3, ••• am be m different ways of doing the first thing, and bu b2, b3> - bn be n different ways of doing the second thing.

Suppose the first thing is done in the way a,. Now the second thing can be done in any one of the n ways bt, ba, 63,— bn. Thus, corresponding to the way a\, the two things can be done in the n ways,

namely, aj feJt axb2, atb3, — at bn. Similarly for a2: a2 bu a2 b2, a2 fa, ••• a2 bn.

And for am : ambu amb2, a„,b3 •» aj>„.

Thus, the total number of different ways is n+n+n-h — — to in terms i. e. mn.

5. To find the number of permutations of n different things taken r at a time.

The number of permutations of n different things, taken r at a time, is the same as the number of ways, in which the following r blank spaces, arranged in a line, can be filled with


r things taken out of n given things, each blank space being filled with one thing only.

1 2 3 4 r-1 r

Now, the first blank space can be filled with any one of the n things. Hence, there are n ways of filling it.

Now, there remain (n — 1 ) things and the second space can be filled with any one of these ( n — 1 ) things. Thus, for each one of the n ways of filling the first space, there are ( n — 1 ) ways of filling the second. Therefore, by the Funda-mental Theorem, the first two spaces can be filled in n ( n — 1) ways.

After having filled the first two spaces in any one of the « ( tt — 1) ways, the third space can be filled with any one of the remaining ( « — 2 ) things. Hence, by the Fundamental Theorem, the first three spaces can be filled inn (n — 1 ) (n — 2 ) different ways. Proceeding in this way, it can be seen that r blank spaces can be filled in

[n(n — 1 ) ( « — 2 ) ( » — 3) t o r factors ] ways,

i . e . n ( n — l ) ( n —2) ( w — 3 ) ( n - r - 1 ) ways, i . e . « ( n — 1 ) ( n — 2 ) ( « — 3 ) . . - ( n - r + 1 ) ways. Hence, we have

J>r = n ( n - 1 ) ( n - 2 ) ( n - r + 1 ) - I

6. Factorial Notation.

In order to express the product of consecutive integers, it is convenient to use the following notation.

The product 1.2.3--« is expressed as n ! or j_n ( pronounced as n factorial or factorial n ).

If there be a product of the type 11-12-13 25, it can be expressed as

( 1 .2-3-10) • 11-12-13—25 . 25J ( 1 -2 -3 -10) 1 - e < 10!*

The advantage of the factorial notation can be seen, in particular, in the case of the formula for "P r of the last article.


For, «Pr = n(n - l)(n - 2 ) - - - ( « - r + 1 ) _ n ( » - l ) ( « - 2 ) - - - ( n - r + l ) x ( w - r ) ( n - r - l ) - - - 3 - 2 - l

(n-r)(n-r- 1> - - -3-2-1

« ! = ! '

Thus, we have V , = ^ " ^ • II

Corollary. The number of permutations of » different things, taken all at a time, i . e . np„ can be obtained by substituting r = » i n I.

Thus, "P„ = n ( « - 1 ) ( n - 2 ) ( « - « + ! ) = n O - 1 ) ( n - 2 ) 2 1 .

npn = n ! . HI Definition of 0 !

n i If we substitute r = n in II, we get "Pn = -^-j •

Now, from the definition of n !, we can not interpret 0 ! and hence the symbol 0 ! has no meaning until and unless we assign a meaning for it. In order to have the same value for MP„ as obtained from I and by substituting r = n, in II we shall have to define j 0_as 1.

Definition. 0 3 = 1 . IV Examples.

( i ) F i n d t h e va lues of 5 !. SP6 , 10P5 a n d SP 8 . Ans. 120,720,30243,40320.

( i i ) Show t h a t 9 ! = 72. [ 7 .

( i i i ) Show t h a t ( 1 ) » ! « = # » • ( « - 1 ) I (2) *! = «• (» - 1 ) < n - 2) ! ( 3 ) « ! = » • ( » - ! ) ( » — 2 ) . . ( » — r + l ) • ( n — r ) I

( i v ) P r o v e t h a t " p r = » x " 1 P r - i .

7. Permutations with restrictions.

The number of permutations will also depend upon the different conditions given in the problem. The following


illustrative examples will show how the restrictions may change the number of arrangements.

Illustrative examples. Ex. 1. H o w m a n y p e r m u t a t i o n s can be formed ou t of t h e le t ters of the

word ' triangle ' ? How m a n y of these will begin wi th t a n d e n d w i th e ?

T h e r e a re 8 le t te rs in t h e word ' triangle ' P e r m u t i n g t h e s e 8 l e t t e r s in all possible ways, t he n u m b e r of p e r m u t a t i o n s is 8 P 8 i. e. 8 !.

I f , however , t occupies t h e f i rs t place and e t h e last place, there remain 6 let ters . If t he se 6 le t te rs a r e p e r m u t e d in t he r e m a i n i n g 6 places in all possi-ble ways , t he requi red n u m b e r of pe rmuta t ions will be 6 P 6 i. e. 720.

Ex. 2 . H o w m a n y n u m b e r s of 4 digi ts can be formed wi th the digits 3 , 5 , 7, 8, 9, n o digi t b e i n g repeated ? H o w m a n y of these will be greater t h a n ' 0 0 0 ?

T h e r e are 5 given digi ts 3, 5, 7, 8 and 9. Of these any four will form a f o u r digi t n u m b e r . H e n c e t he total n u m b e r of four d ig i t -numbers will be 'P -i . e. 123.

If the n u m b e r is to b e greater t han 7 M t \ the lef t h a n d digi t c a n ei ther be 7, 8 or 9 b u t no t 3 OT 5. T h e r e f o r e , t h e f irst lef t h a n d dig i t can b e chosen in 3 ways. T h e r ema in ing t h r e e d ig i t s can be any th ree o u t of t he remain ing four digi ts a n d h e n c e t he n u m b e r of their a r r a n g e m e n t s is ' P . .

H e n c e by the F u n d a m e n t a l Theorem, a fou r digi t n u m b e r greater than 7000 can b e formed wi th t he he lp of the given f ive digi ts 3 x 'P., d i f ferent ways i . e. 72 ways.

Ex. 3 . In how m a n y ways can 5 Engl ish books. 3 M a r a t h i books and 3 Guja ra th i books be a r ranged on a shelf, if all the books in tbe same language a r e to b e together ?

T h e five Engl ish books c a n be a r ranged amongs t themselves in ' P 5 i . e. 5 ! ways. Similarly, the th ree Mara th i a n d three Gu ja r a th i books can be ar ranged a m o n g s t themselves in SP3 a n d SP3 i. e. 3 ! a n d 3 ! ways respect ive ly .

A r r a n g e m e n t s of these g roups can be made in 3P3 i e- 3 ! ways.

H e n c e , by t he F u n d a m e n t a l T h e o r e m , t he requ i red n u m b e r of ar range-m e n t s is 5 1 x 3 1 x 3 1 x 3 ! i. e. 25, 920

Ex. 4. Six s t u d e n t s i n c l u d i n g two bro the rs are to be seated on six chairs in a l ine for a pho tog raph F i n d t he n u m b e r of dift r e n t a r r a n g e m e n t in which

( i ) the b r o t h e r s d o n o t occupy the e n d sets , ( i i ) t he b r o t h e r s occupy seats ad jacen t to each o the r .

( i ) Exc lud ing t he e n d seats t he two b ro the r s can occupy any two out of t b e 4 middle seats. T h i s can b e d o n e in *P3 ways.

N o w , t h e r e m a i n i n g f o u r s t u d e n t s can occupy t he r e m a i n i n g fou r seats in any o n e of the 4 P, ways.


H e n c e by the F u n d a m e n t a l Theorem, t he requ i red n u m b e r of a r range-m e n t s for t he pho tog raph i s

' P i X «P4 i . e . 12 x 4 ! i, e , 288. { i i ) Imag ine t h e two a d j a c e n t sea ts t o b e occupied by t h e two b r o t h e r s

fo rming one combined seat . T h u s , t h e r e are S d i f fe ren t sea ts to b e occupied by t h e s tuden t s . T h i s c a n b e d o n e in SP5 ways. T h e two a d j a c e n t sea t s fo rming t he combined seat c a n b e occupied by t h e two b r o t h e r s in ' P j ways.

Hence , by tbe F u n d a m e n t a l T h e o r e m , the requ i red n u m b e r of a r r a n g e ' m e a t s is 6PS X "P., i , e. 5 ! x 2 i . e 240.

Ex 5 . F ind t h e n u m b e r of a r r a n g e m e n t s in which 7 b o y s a n d 4 girls can b e a r ranged in a row so t h a t n o two girls may be toge the r ,

Cor respond ing to a n y o n e a r r angemen t of 7 boys , we h a v e 8 possible pos i t ions for the girls, namely t h e two posi t ions a t t he two e n d s a n d si* in between the b o y s . F o u r gir ls can occupy any f o u r o u t of these eight pos i t ions in e P 4 ways.

B u t the seven b o y s can be a r ranged amongs t themse lves in 7P7 d i f fe rent ways.

H e n c e by the F u n d a m e n t a l T h e o r e m , t he requ i red n u m b e r of a r range-men t s is »P4 X vP f i. e. 16b0 X 71 i . e. 84,67,200.

t x . 6 . Show tha t t he n u m b e r of pe rmu ta t i ons of n d i f fe ren t th ings all a t a t ime in which p given t h i n g s shall always c o m e toge ther is

I "ZLP +J1 * ]JL

Since p given t h i n g s a r e always to come toge ther , w e may cons ider t h e m as f o r m i n g o n e e l emen t . T h u s , t he given th ings {at, at ... a / , ) , « „ + i aji+2, an will f o r m 1 + ( « — p ) i. e. n - p + 1, d i f f e ren t e lements , which can be pe rmuted a m o n g themselves in ) n - p + 1 ways . B u t in each of these ways, t h e p th ings can b e pe rmuted a m o n g themse lves wi thout touch-ing t he o the r s in ) p ways . H e n c e , t he required n u m b e r of p e r m u t a t i o n s i s

| w - P + 1 X | pj_

Note. If we sub t r ac t t h i s n u m b e r f rom the t he total n u m b e r of pe rmuta -t ions of n th ings , we get the n u m b e r of pe rmu ta t i ons in which p pa r t i cu la r t h i n g s are never f o u n d all together as | n — \ n - p + I x ( P .

Exercisc 11 (a)

1. There are five different routes to go from Dadar to Churchgate. Find the number of ways in which a traveller can go from Dadar to Churcheate and come back. Find also the number of ways in which be can go and come back by a different route.

2. There are five candidates for the office of the president, 7 candidates for that of the secretary and 8 candidates for that of


the treasurer. Find the number of different ways in which the elec-tion of 3 Officers caa take place.

3. In how many ways can a consonant and a vowel be chosen out of the letters of each of the words ( i ) Logarithm, ( i i ) Equation ?

4. In how many ways can, three different prizes be given to 10 boys, (i) if a boy is eligible to get only one prize, and (ii) if a boy is eligible to get any number of the 3 prizes ?

5. How many numbers of three digits can be formed by using the figures 1,2, 3 9,

( i ) if no digit is to be repreated in any number and (ii) if a digit can be repented any number of times in a nnmber ?

6. How many of the numbers in example 5 ( i ) and 5 ( i i ) will be greater than 400 ?

7. Find n, if

( I ) " P 7 = 2 l O . "P5, ( i i ) 2 " P3=100.mP2 , ( i i i ^ P a =2- "P4.

8 Show that

( i ) (2«)! ==l-3-5...(2« — l)-2".

(ii) (n + 1 ) (« + 2). . . (In -1) • In = 2-6-10-14... to n factors;

... 1 2 3 n , 1 (m) 2-f + 3 1 + 4 ~ j + ... + - J j r + l j r = 1 - ( F + l ) l •

9. How many different arrangements can be made by using all the letters of the word ( i ) Monday ?, (ii) Oriental ? How many of these arrangements begin with a and end with n ?

10. How many different arrangements can be made jvjth all the letters of the word " Equation " ? How many of these begin and end with a vowel ?

11. A number of four different digits is formed by using the digits 1, 2, 3, 4, 5, 6, 7 in all possible ways. Find (i) how many such numbers can be formed, and ( ii ) how many of them are greater than 3400.

12. A numbers of five different digits is formed by using the digits 1, 2, 3, 4, 5, 6, 7 in all possible ways. Find how many of these numbers are exactly divisible by (i) 2, ( i i ) 25 and (iii) 4.


13. Eight boys including two brothers are to be seated on chairs in a row for a photograph. Find the number of different arrangements in which the photograph can be taken

( i ) if the two brothers are to occupy the chairs at the end; (ii) if none of the two brothers are to occupy tbe chairs

at the ends.

14. In how many different ways can the books, 6 on Physics, 7 on Chemistry and 8 on Mathematics be arranged on a shelf so as to keep all the books on the same subject together ?

15. How many different arrangements can be made by using all the letters of the word courage if all the vowels are to be toge-ther and all the consonants are to be together ?

16. Eight papers are to be set at an examination, two of which are on Mathematics. In how many different ways can the exami-nation be arranged so that the two papers on Mathematics are not consecutive ?

17. If all the permutations of the letters of the word "chalk" be written down as in a dictionary, what is the rank of this word ?

18. In how many different ways can a party of 4 girls and 8 boys seat themselves in a railway compartment with two benches of six seats each, so that the girls occupy the end "seats ?

19. Eight men and four women are to be seated on chairs in a row for a photograph. Find the number of different ways in which the group can be arranged so that no two women occupy the chairs adjacent to each other.

20. In how many ways can 5 women, 5 girls and 5 boys be seated in a row so that all the women are together and so are all the girls and all the boys ?

21. A number is formed by using the digits 0, 1, 2, 3,4,5,6,7, no digit being repeated. How many of such numbers are greater than 100 and less than 1,00,000 ?

22. Show that the number of different permutations of n things in a line, so that two particular things are not together, is (m — 2 ) | < i - l .

23. There are 6 students of whom 2 are Indians, 2 Americans and the remaining 2 are Russians. They have to stand in a line

C. A. —19


so that the two Indians are together, the two Americans are toge-ther and so also the two Russians. Show that there are 48 different ways of arranging the students.

24. Without using the formula for "P„ prove that

"Pr = nX ""Vi-

Hence deduce the formula for "Pr.

25. The number of permutations of n things taken r at a time in which k particular things always occur is

P r-j X rPA.

Answers—Exercise 11 (a)

1. 25, 20. 2 . 280. 3 . ( i ) 18, ( i i ) 15. 4 . ( i ) 720, ( i ) 1000. 5 . ( i ) 504, ( i i ) 729. 6 . ( i ) 336, ( i i ) 486. 7 ( i ) 20, ( i i ) 13, ( i i i ) 8. 9. ( i ) 720, ( i i ) 40, 320 a n d ( i ) 24, ( i i ) 720.

1 0 . ( i 4 0 , 3 2 0 . i i ) 1 4 , 4 0 0 . 1 1 . < i ) 840, ( i i ) 560. 12. ( i ) 1080 ( i i ) 120, ( i i i ) 600. 1 3 . ( i ) 1440, ( i i ) 21, 600. 1 4 . 3 ! x 6 ! x 7 ! x 8 ! . 15. 2 ! X 3 ! X 4 ! i . e . 288. 16. 7 ! x o i. e. 30, 240. 1 7 . ( 4 ! + 3 ! + 1 ! + 1 ) t h i . e . 32nd . 1 8 . 4 ! X 8 !. 19- 8 ! x 9-3-7«6.

2 0 . 3 ! ( 5 ! ) 8 - 2 1 . 7 x 7 ! x £ + - L J .

8. Combinations of n different things taken r at a time.

We now proceed to illustrate, by a numerical example, how

the formula for "Cr can be derived from that of "Pr. Let us consider the number of combinations of 5 different

things taken 3 at a time. Let the number of such combinations be x. We have to find the value of x.

Each combination will consist of three things taken out of five. If the three things in any such group are arranged in a line and permuted in all possible ways, we get 3 ! arrangements ( permutations ). Thus, each combination gives rise to 3 ! per-mutations and there are x such combinations. Hence, the total number of such permutations is x x 3!. But these per-


mutations will be exactly the same as the permutations of 5 different things, taken 3 at a time i. e. they are SP3 in number,

sp,

x X 3 ! = sp3 . ... X = •

5 x 4 x 3 ••• * = - T 2 T - = 1 0 ' But x denotes, in this case, the number of combinations of

5 different things taken 3 at time i. e. SC3. 5Pa SC3 = y3 , = 10.

We will, now, make use of this method to find the formula

for "Cr. To find the number of combinations of n different things

taken r at a time. Let the n given different things be denoted by

a2,a3 a„-u a„ . Let us consider the combination of r things taken out of

these n things, say, ( a-y, a2, a3, • • - • ar ). If the r things in this group are arranged in a line and permuted in all possible ways, we will get r ! different arrangements ( i . e. permutations ) .

Every such combination of r things, taken out of n given things, will give rise to r ! permutations. But there are "Cr

such combinations. Therefore, the total number of such per-mutations will be "Cr X r!. But these permutations will be exactly the same as the permutations of n different things taken r at a time i, e. they are "P, in number.

» « n P C r X r ! = P r, -

. V _ n ( n - l ) U - 2 ) - - - ( n - r + l ) T

or C. = T~, r-. • U r ! ( n - r ) !

Corollary. "C„ = "C0 = 1.


Putting r =n in I, we have "C„ = " O L - J J L ^ l H . - 1.

Substituting r = n in If , and remembering the definition tha tO! = 1, we have the same result, namely, "C„ e 1. The result is also, otherwise, obvious since "C„ denotes the number of combinations of n things taken all at a time which must be equal to 1.

Again, putting r = 0 in II, we get

"C0 = -Q-7^-7 = 1 • The symbol "C0 has no meaning as we

cannot select zero things out of n. But owing to the result just

obtained, we assign the meaning of unity to the symbol "C0.

9. Complementary Combinations.

To prove that nCr = "Cn.r. Let the n given things be denoted by

ax, a 2 , a 3 — an — a„-1, a„. Consider a group of r things, namely, a\, a2, a3, ••• ar.\, ar.

When we select this group, we leave behind another group of ( n — r) things, namely, ar+1, ar^2, a„-i, a„.

In general, whenever we select a group of c things we automatically leave behind another corresponding group of [n — r) things. Thus, corresponding to every group of r things, there is a group of remaining (n — r ) things.

V — "r1 ^r — ^ n - f

Alternatively, we may arrive at this result by using the

formula of "Cr which is valid for values of r < n.

We have "Cr = . . n '—r-r • r r ! ( n — r ) ! Changing r to n — r, we get

»r w ! ~ ! [ n - ( . n - r ) ] !

= — ~ — r which is the same as nC r . ( n - r ) !r !


Hence, the number of different selections of r things out of n is equal to the number of selections of ( n — r ) things out of n.

This is called the Principle of complementary combina-tions.

Note 1 . T h i s P r inc ip le o f t en he lps u s to s impl i fy numer ica l calculat ions

part icular ly in cases w h e r e r > • T h u s , 13CI2 = ' 'C3 = = 455. 2 I 'A 'J

. . . . . • , x , i x i.i. 15 14-13.. . . to 12 fac to r s which is m u c h more s imple to ca lcula te t han Ci2 = — — X 'A

Note 2 . If i n f o r m u l a " C , = "Cn-r, we p u t r - n. we get " c „ = "C0.

W e know tha t n C „ = 1. H e n c e , fo r t he a b o v e fo rmula to b e t rue for r = n

we have to give to " C 0 a m e a n i n g cons i s ten t wi th t he above result a n d

h e n c e " c 0 is to b e t aken as equal to 1.

Note. 3 . If "Cx =» " C y , t h e n e i ther t h e two c o m b i n a t i o n s conta in t h e same n u m b e r of t h i n g s in which case x — y or they a r e compl imenta ry , in which case , n — x = y. Examples .

1. F i n d the va lues of «C8, SSC24, 80C2 8 . Ans . 20, 25, 435.

2 . F i n d n, if " C 3 = 30 X " c „ Ans . 28.

3 . If "cr= , 0 C , + 2 ' find rCe. Ans . 84. IX 18 r

4 . If Cr= C r + 2 » find C 4 . An%. 70.

5 . F i n d n , if " C , - 5 x WP3. Ans . 123. 6 . If 3 P S + 5 x ' P i = " P , , find r. Ans . 5.

7 . If 1 0 P r = 6.04,800 a n d wCr =• 120, show tha t r = 7.

10. Combinations with restrictions.

In permutations the nature of the restriction is with regard to the order or position of certain things; whereas in combinations it involves the inclusion or exclusion of certain things in a given selection.

I. The number of combinations of n different things taken r at a tim e when p particular things out of n given things are always to be included is n i 'Cr^ i ).

As every combination is to include p particular things, we have only to select ( r - p ) things out of ( n - p ) remaining


things to form a group of r things taken out of n . This can be done in " t 'C r .p ways. Therefore, the required number of com-binations is " PC r .p .

II. The number of combinations of n different things taken r at a time, when p particular things are to be excluded from every selection is " ^C,.

As every combination is to exclude p particular things we have only ( n — p ) things remaining from which r things are to be

selected. This can be done in " pCr ways. Therefore, the requi-red number of combinations i s " ~*Cr.

11. Important identity.

To prove that n + 1 C r = "Cr+ nCr-i. Let a1% a3,...an, b be ( « + l ) different things. The

number of combinations of these ( n + 1 ) things taken r at a

time is C r . Now, these combinations can be divided into two classes.

One class will consist of all those combinations in which a particular thing, say, b is always to be excluded. The number of these combinations is the same as the number of combina-tions of the n things a\, a2, a3,...an taken r at a time which is nCr. The second class will consist of all those combinations in which the same particular thing, say, b is always to be included. As every combination has to include b, we have only to select ( r — 1 ) things out of the n things au a2,...,a„. This can be done

in "C r- i ways. Hence, there are "Cr_i such combinations.

Alternatively the identity can also be obtained by using the

formula for nCr. » yi 1

Thus, we have C r = ) f

Changing r o x r - l , "c r_ a = ^ r _ 1 } , r + l )


n! , n\

= - +

! ( n —r ) ! ( r - 1 ) ! ( n - r + 1 )!

n ! , « ! r-'(r — I)! (n — r)\ + ( r - 1 ) ! ( « - r + l > (n-r) !

r _ L + L — i ~ ( r - l ) ! ( « - r ) ) ! L r » — J

n ! n + 1 ~ ( r - 1 ) ! ( n - r ) ! r ( « - r + 1 )

_ ( " + ! ) ' • _ * + V

" r ! ( « - r + 1 ) ! " " r"

/ . we have n + 1 C r = "C, + "C r_ t .

12. Illustrative examples. Ex. 1 . F i n d « , if " C 6 : " * 3 C a = 9 1 : 4.

I » - s I n — 3 ,,. , «_ I— , " i f , ' W e have C 3 = l T T ^ 6 a n d C s = | j ' [ « - 6

n [3 |» - 6 " tt-3c. " I l l ^ i s * !•» - 3

1 _ w ( w - 1 ) ( n - _ H _

~ 3 * o -5-4 6 -5 -4

. » (w - 1 ) ( n - 2 ) = 9 1 _ 6 - 5 - 4 4

» ( n - 1 ) ( n - 2 ) = 6 -5 -91 = 5 6 - 7 - 1 3 = 15-14-13, expressing t h e p roduc t on R . H . S . a s a p roduc t of th ree c o n s e c u t i v e in tegers in de scend ing o rde r as on L . H . S .

n = 15.

Ex. 2 . T h e r e a r e 24 d i f fe rent books, i n c l u d i n g 9 novels , 8 d r a m a s a n d 7 essay-books- F i n d t h e n u m b e r of d i f fe ren t w a y s in w h i c h ( i ) a selection of six books can b e m a d e so a s to have 2 books f r o m each g roup ; ( i i ) a selec-t ion of / books c a n b e m a d e so as to have at least two books f r o m each group .

( i ) 2 novels , 2 d r a m a s a n d 2 essay-books can be selected f r o m 9 novels , 8 d r a m a s a n d essay-books in 9 C 3 , 8 C j a n d r C 3 d i f fe ren t ways .

b y t h e F u n d a m e n t a l T h e o r e m , t h e requ i red n u m b e r of select ions of six bocks c o n t a n i n g two books of each g roup

= 'C, x . C . x ' C - ^ f x ^ x ^ ! - * ! , 168


( i i ) A selection of 7 books c o n t a i n i n g at least 2 books f r o m each g r o u p can be made ei ther by se lect ing 3 books f r o m the novels , or f r o m t h e d r amas , or f r o m t h e essay-books a n d the r e m a i n i n g fou r books be ing selected two f rom each of t he r ema in ing two groups .

T h e d iSe ren t se lect ions c a n , there fore , be m a d e in

(1) 8Cj x 6C2 x ' C 2 . ( 2 ) SC, x SC 3 x 'C, and (3 ) 9C,. x 8Qs x "'C3 ways.

.'. the requi red tota l n u m b e r of select ions.

= " C V C r ' Q + »C1-8CJ-7CJ + 'Ca-'Ca-'C, 9 8*7 8 -7 7 6 { 9 - 8 ^ 8 - 7 - 6 J7_-6 9 -8 8-7 7 - 6 - 5 3 - 2 - 1 2 - 1 2 - 1 2 -1 3 - 2 - i 2-1 2 1 2 - 1 3 - 2 - 1

__ 9 - 8 8 -7 7 -6 I 7 6 5 | „ , 18

= 21,168 x 6 = 1,27.008.

E x . 3 . T h e r e a r e 7 male s t u d e n t s a n d 3 lady s tuden t s . I n how m a n y ways can a commi t t ee of six b e fo rmed ( i ) if t h e r e is n o r e s t r i c t ion , i i ) if t he three lady s t u d e n t s a r e always to be inc luded a n d ( i i i ) if a t least two lady s tuden t s are to be inc luded ?

( i ) I n this case, we have to fo rm a commit tee of 6 ou t of 10 persons . Th i s can be d o n e in J<,C6 i . e . 210 d i f fe ren t ways-

( i i ) If in t he commi t t ee of six, three lady s t u d e n t s a r e a lways to be inc luded , then we have really only to select 3 male s t u d e n t s out of 7 which can b e d o n e in "Cn i. e. 35 ways.

( i i i ) If at least two lady s t u d e n t s are to be inc luded , t h e n there m a y b e e i ther 2 ladies or 3 ladies in t h e commit tee . W e have , t he re fo re , to cons ider two cases. If two ladies a r e to be inc luded , then they can be selected f rom the given 3 ladies in ; |C2 ways and the r ema in ing 4 m e m b e r s of t he commi t t ee c a n be selected f rom 7 male s t u d e n t s in 7C ( ways. T h u s , t h e n u m b e r of ways of fo rming the commi t t ee where only two ladies are inc luded is 3C a x 7C4

ways.

As in case ( i i ) above , t he commi t t ee which inc ludes all the th ree ladies can be formed in ' C 3 x 7CS ways . There fo re , t h e tota l n u m b e r of ways requi red is

sCa• 'Ci + 3C., • 'C3 i . e. 3 X 35 + 1 x 35 o r 140 ways.

Ex. 4 I n how m a n y d i f ferent ways can a cricket team of eleven be chosen f r o m 15 players ? If out of t he 15 players 9 are ba t smen a n d 6 a re bowlers , find also the n u m b e r of d i f fe ren t ways in which the team can be fo rmed so as to i nc lude exactly 7 ba t smen a n d 4 bowlers.

Ou t of 15 players a t eam of 11 can be selected in I _ ,Cu ways . B u t , by the useful ident i ty about t he complementa ry combina t ions , we have

is,- 15 X 14 X 13 X 12 •'CM = loCi = — ; ; — R — = 1365 ways. 1 4 x 3 x 2 x 1 J


In the second case 7 ba t smen can b e selected o u t of 9 ba t smen °C: ways , a n d 4 bowlers can b e selected ou t of 6 bowlers in 8 C , ways H e n c e , by t h e Fundamen ta l Theorem, the team cons is t ing of 7 ba t smen a n d 4 bowlers can be selected in 6C7 x 6 C ( .

i e. 9 C j x 6C2 t "Cr = "Cn-rJ

9 x 8 6-5 . i. e. — x ~ r > 540 ways. 2 x 1 2 1

Ex. 5 . A game of mixed doub les t enn i s is to b e a r ranged f r o m a g roup of 12 players inc lud ing 4 ladies a n d 8 gent lemen, so t ha t each s ide of the g a m e consis ts of one lady a n d o n e gen t leman. F i n d t h e n u m b e r of d i f fe rent w a y s in which the game can be a r r a n g e d .

Fo r a r r ang ing a game of mixed doubles t e n n i s two lad ies a n d two gen t l e -men mus t be selected first. T h e two lad ies can b e selected o u t of 4 ladies in "Ci i. e. 6 ways. T h e 2 gen t l emen can be chosen o u t of 8 gen t lemen in *C4

i. e. 2f ways. H e n c e , t he fou r players—two ladies , say L< a n d L*, a n d two gent lemen, say Gi a n d G 2 , can be selected in 6 x 28 = 168 ways. Af t e r selecting the fou r players i n c l u d i n g two ladies a n d two g e n t l e m e n , t he g a m e can be arranged in 2 ways accord ing as t he lady L i h a s a p a r t n e r Gi o r G a . T h u s corresponding to each of t h e 168 ways of se lect ing fou r players , i nc lud ing two ladies a n d two gen t l emen , the re are two ways of a r r a n g i n g t he game . Hence , the game can b e a r ranged in 168 x 2 i. e 336 ways.

Ex. 6 . T h e r e a r e 10 po in t s in a p lane of wh ich 4 a r e coll inear . F i n d ( i ) the n u m b e r of t ra ingles t h a t can be formed wi th ve r t i ce s a t these p o i n t s and ( i i ) the n u m b e r of s t ra ight l ines ob ta ined b y j o i n i n g these p o i n t s in pairs .

W e can select 3 po in t s o u t of 10 in , 0 C 3 ways. B u t if these 3 po in t s a re selected out of 4 col l inear po in t s , a t r iangle can n o t b e f o r m e d . T h e n u m b e r of such selections will b e 4 C , a n d they do n o t give u s a n y tr iangle. H e n c e t he required n u m b e r of t r i ang les tha t can b e fo rmed is 1 0C3 — *C3 i. e. 120 — 4 which is 116.

T o fo rm s t ra ight l ines w e have to select 2 p o i n t s o u t of 10. T h i s can b e done in I 0C a ways . Of these , ' C 3 are t he se lec t ions in which bo th the p o i n t s lie on the given i ine con t a in ing the 4 coll inear po in t s . H e n c e , we ge t only one straight l ine a n d n o t *C2, H e n c e , t he n u m b e r of s t ra ight l ines which can be formed is 10Ca - 4C2 + 1, i. e. 45 - 6 + 1 wh ich is 40.

Exercise 1 1 ( b )

1. If "C* = "Cy, show that x = y or x + y = n.

2. Find n, if ( i ) "C* = 495; ( i i ) n C 5 = " C 6 ; ( ii ) l s C n = 1SCn+ 2 ; (iv ) 2"C3: "C, = 44 :3 ; ( v ) 3 x " Q = 5 x " 1 C S .


3. If 2 0 C „ = 2 0 C 6 , f i n d ( i ) " C 3 a n d ( i i ) " c , 0 .

4. Find r , if ( i ) " c , , : 22C2r_2 = 138 : 95; 24

( i i ) C j r : C2r-i =225 :11.

5. A committee of four is to be formed from 16 members. Find the number of different ways in which the committee can be formed.

If 16 members include three particular members A, B and C, find further, in how many different ways can the committee be formed ( i ) if A is to be excluded from the committee; (ii) if A is to be included but B and C are to be excluded.

6. In how many ways can a committee of 4 gents and 3 ladies be formed out of a club consisting of 10 gents and 6 ladies ?

In how many of these selections of the committee will a parti-cular lady be included but two particular gents be excluded ?

7. There are 12 points in a plane of which 5 are collinear.

Find ( i ) the number of triangles that can be formed with vertices at these points, and

( i i ) the number of straight lines obtained by joining these points in pairs.

8. How many ( i ) triangles ( i i ) diagonals can be formed by joining the angular points of a ten sided plane polygon ?

9. If 15 parallel straight lines are intersected by 13 other parallel straight lines, find total number of different parallelo-grams thus formed.

10. There are 4 Professors and 16 students. Find in how many different ways can a committee of 5 be formed so as to include ( i ) exactly 3 Professors and ( i i ) at least 3 Professors.

11. From 8 gentlemen and 4 ladies a committee of 5 is to be fo rmed . In how many ways can the committee be formed so as to include at least 1 lady ?

12. There are 5 English, 7 Marathi and 9 Gujarathi recom-mended books. In how many different ways can a selection of 9 prize-books including 2 English, 3 Marathi and 4 Gujarathi books be made ?


13. From a class of 15 students, how many different parties can be formed so as to include at least 3 and not more than 5 students in any party ?

14. A cricket team of 11 players is to be formed from 16 players including 4 bowlers and 2 wicket-keepers. In how many different ways can a team be formed so that the team contains ( i ) exactly 3 bowlers and 1 wicket-keeper: ( i i ) at least 3 bowlers and at least 1 wicket-keeper ?

15. In how many ways can 9 different prizes be awarded to A, B, C so that A gets 2, B gets 3, and C gets 4 prizes ?

Find also the number of ways, if in the above distribution, B is not eligible for a particular prize.

16. For an examination a candidate has to select 7 subjects from three different groups A, B and C. The three groups A, B, C contain 4, 5, 6 subjects respectively, In how many different ways can a candidate make his selection if he has to select al least 2 subjects from each group ?

17. In how many ways can a committee of 5 be chosen from 10 candidates ( i ) so as to include both the youngest and oldest, ( i i ) so as to exclude the youngest if it includes the oldest ?

18. In how many different ways can a committee of 5 be formed from 6 Congressmen and 4 Socialists so as to include at least two Socialists ?

19. A person has 10 acquaintances of whom 3 are relatives. In how many different ways can he invite 7 of his acquaintances so as to include at least 2 of his relatives ?

20. A party of 6 is to be formed from 10 boys and 7 girls so as to include 3 boys and 3 girls. In how many different ways can the party be formed if two particular girls refuse to join the same party ?

21. If p parallel straight lines are intersected by q other parallel straight lines, find the total number of different parallelo-grams which are formed.

22. There are p points in a plane no three of which are in a straight line except q which are all collinear. Find ( i ) the number of different triangles with their vertices at these points and


( i i ) the number of different straight lines formed by joining these points.

23. Find the number of diagonals that can be drawn by joining the vertices of a plane polygon of n sides.

24. Without using the formula for "C r prove that

r X C, = n X C r . i .

Hence, deduce the formula for "C r.

25. Show that the number of different ways of distributing {p + q + r ) different things to three persons A, B, C so that A gets p, B gets q, and C gets r things respectively, is

( p + q + rV- e p ! q ! r !

26. A game of tennis doubles is to be arranged from a group of 10 players including 6 men and 4 women, so that each side of the game consists of one man and one woman. Find the number of different ways in which the game can be arranged.

27. In how many ways can a mixed doubles tennis game be arranged from 7 married couples, if no husband and wife play in the same game ?

28. Show that 10 men, of whom 4 are brothers can be arranged in a row, so that no two brothers are together in 120 x 71 different ways.

29. In how many ways can a number of 5 digits be formed by using the digits from 1 to 9 so that even digits occupy even places, odd digits occupy odd places and no digit is repeated ?

30. Twelve persons get into a railway compartment with 6 seats on either side. Two of the persons refuse to face the engine and one of them dislikes travelling backwards. In how many ways can they be seated ?


2 . ( i ) 12. ( i i ) 11. ( i i i ) 8. ( i v ) 6 . ( v ) 10.

3 . ( i ) 20 o r 364. ( i i ) 1001 only . 4 . ( i ) 10. ( i i ) 7.

.5. 1820. ( i ) 1365. « ii ) 286. 6 . 4200, 700.

7 . ( i ) 210. ( i i ) 57. 8 . ( i ) 120. ( i i ) 35. 9 . 8190.

0 . ( i ) 480. ( i i ) 196. 1 1 . 736. 1 2 . 44,100. 1 3 . 4823.


1 4 . ( i ) 960. ( i i ) 2472. 1 7 . ( i ) 56. ( i i ) 196.

21- -~tg{p- I) ( q - D •

1 5 . 1260 ; 840. 1 6 . 2700. 1 8 . 186. 1 9 . 98. 2 0 . 3600.

2 2 . ( i ) ( i i ) + 2 3 . ^ J ^ — 1 •

2 6 . 2 - 6 C 2 - 4 C 2 i.e. 180. 27. 2 - , C a - s C a i. e. 420. 2 9 . 720.

3 0 . ' P ^ ' P i - g ! i. e. 180 x 9 !.

13. Permutations of things not all different.

We have considered upto this time permutations of things which were all different. We will now consider in this article permutations of things, some of which are similar, taken all at a time. We will illustrate the method, first by the following example.

Example. Find the number of permutations of the letters of the word " Infinity " , taken all at time.

We have, here 8 letters of which 3 are i, 2 are n, and the three others, namely / , t, y, are different. If all the 8 letters would have been different, the number of permutations (taken all at a time ) would be 8 ! .

Let x denote the number of permutations required in the example. Take any such permutation, say, Infinity". If we replace the three latters i by the different letters z"i, i2, z'3

then this permutation w J1 give rise to j_3_ permutations by permuting 3 letters amongst themselves in all possible ways, and without altering the positions of other letters. Hence, if this change is made in each of the x-permutations we would obtain i x [ 3 permutations.

Similarly, if in each of x X |3 permutations, the two letters n were replaced by two different letters « t and n2, we would get a; x | 3 x |_2_ permutations

But now all the things have become different, and therefore, the total number of permutations must be the same as the number of permutations of eight different letters taken all at a time i. e. 8 ! . I8

x x 13 x 12 = 18. . ' . x = — — — ,


We shall now use this method to prove the following Theorem.

Theorem. The number of permutations of n things taken all at a time, when p of them are similar and of one type, q of them are similar and of another type, r of them are similar and of a third type, and the rest all different is

n ! p \ q\ r\

If all the n things had been different, the number of permutations, all at a time, would have been n !.

Let x be the required number of permutations in the proposition. Take any one of these permutations and replace in it, the p similar things by p things different from each other and also different from the rest. This change will give rise to p ! permutations by permuting these p different things amongst themselves in all possible ways and without altering the positions of other things. If this change is effected in each of the ^-permutations, we shall get x-p ! permutations.

Similarly, replace the q similar things of the second type by q things, different from each other and different from the rest. Then, by permuting these q different things, amongst themselves in all possible ways, in the above x-p ! permutations we shall obtain x-p I q I permutatations.

Repeating the process with the r similar things of the third type, we shall obtain x-p ! -q ! -r1 ! permutations.

But, now, all the n things have become different and hence the total number of their permutations must be n ! as already observed in the beginning.

n ! x-p ! • q ! • r ! = n !; .'. x = ; r • ^ p ! • q ! • r !

the required number of permutations is p ! - q I -r ! Ex. 1 . F i n d t he n u m b e r of pe rmuta t ions t ha t can b e m a d e ou t of the

: f i t ters of t he w o r d s ( i ) Mathematics ; ( i i ) Inedpendetice ; ( i i i ) Allahabad

, , 11! , .. . 12! , . . . , 9 ! Ans. ( i ) - 2 , , 2 ! , 2 ! ' • TTTiTT ' ( m ' 4T2T '


Ex. 2 . The re are four copies of a book on Algebra, t h r e e copies of a n o t h e r book on Tr igonometry and th ree copies of a th i rd book on Geomet ry . F i n d t h e different number of ways in which they can b e a r r anged in a l ine on a shelf . How many of these a r r a n g e m e n t s will have two books on Geomet ry at t h e two ends ?

r A n s . - J - i - . i L 4 ! 3 ! 4 ! 4 ! 3 ! J

14. Permutations with repetitions. To find the number of permutations of n different things taken

r at a time when each thing may be repeated any number of times in any arrangement.

The required number of permutations of n different things taken at a time is the same as the number of ways in which r blank spaces arranged in a line ( as in article 5 ) can be filled with r things out of n given things, each blank place being filled with any one of the n things.

Here, any one blank space can be filled by any one of the n things i. e. in n ways.

.". all the r blank spaces can be filled in [ n X n X n X to r factors ] i. e. n different ways.

E x . 1 . A boy has five pockets a n d th ree pieces of ten Pa i se . In h o w many ways can he p u t t h e m in to his pockets ?

T h e first piece of ten P . can b e put in 5 ways a s i t may b e " p u t in a n y of the 5 pockets. T h e second piece of ten P may also b e pu t in 5 ways fo r t h e boy is not p reven ted f r o m using t he same pocket aga in . Similar ly , t he thi rd piece of ten P . can be pu t in 5 ways. H e n c e , t h e r equ i r ed n u m b e r of ways = 5 x 5 X 5 = 5 ' i . e. 125.

E x . 2 . H o w many n u m b e r s of 4 digi ts can be fo rmed wi th t he d ig i t s 1 to 9 ? [ Ans. 9* ]

Ex 3 . I n how many ways can five pr izes b e given away to fou r b o y s when each boy is eligible for all t he prizes ? [ Ans , 4* ]

15. Combinations of n different things taken any number at a time.

To find the number of combinations of n different things taken any number at a time.

In making a selection of n different things any number at a time, each thing may be dealt with in 2 ways, viz. it may either be selected or rejected. Since either Way of dealing with any one thing may be associated with each way of dealing


with any one of the n others, the number of selections is 2 X 2 X 2 X — to n factors i e. 2" ways. But this includes the case where all the n things are rejected, which is inadmissible since no selection is formed.

required number of combinations of n different things, taken any number at a time, is 2" — 1.

Corollary. The total number of combinations of n different things taken any number at a time is clearly made up of the combinations of the n things taken one at a time, two at a time, • • • • up to all at a time.

we have " Q + "C2 + nC3 +••••+ "C„ = 2" - 1.

or "C0 + " Q + "C2 + "C3 + • • • • + "C„ = 2".

[ "c0 = 1, may be taken to represent the case where all the n things are rejected. ]

E*. 1. A salesman places three dif ferent t ies before a cus tomer . I n how m a n y ways can t he c u s t o m e r make a purchase cf o n e or more of them ?

T h e cus tomer has 2 ways of disposing the first tie ( e i ther to purchase it or n o t t o purchase i t ) . Similarly, here are two ways of deal ing wi th second a n d two again of dea l ing with t he th i rd . 3 t ies may be deal t with in 2-2 2 i. e. 8 ways. B u t th i s i nc ludes t he case where n o tie is purchased .

excluding t he inadmiss ib le case, the cus tomer has ( 2s - 1 ) i . e . 7 ways = J C i + 3 C a + 3 C 3 = 3 + 3 + 1 = 7.

Ex. 2 . In how m a n y ways can a man choose o n e o r more chai rs o u t of 6 cha i r s? [ A n s i 6 — 1 i. e. 63 ]

Ex . 3 , T h e r e are 3 Socialists, 4 Congressmen a n d 2 Communis t s . In how m a n y ways can a selection be made so as to inc lude at least o n e of each par ty ?

[ A n s . ( 2 3 - 1 ) (2« - 1) ( 2 s - 1 ) i. e. 315 ]

16. Remarks regarding solution of problems.

In solving problems on permutations and combinations the students are advised to pay attention to the following points.

( i ) Whether the problem relates to the arrangements or the selections of the given things. In a selection or a group, we pay no attention to the order of the things taken. When the order of the things is relevant in the problem, it becomes a problem in permutations.


( ii ) Whether any conditions are imposed on the manner of arranging or grouping of things.

( iii ) Whether the Fundamental Theorem is applicable or not.

( i v ) Whether the given things are all different or not . If some of the things are alike, the effect on the problem of arrangements or of selections should be carefully considered.

( v ) Whether the repetitions of the things in the problem are admissible or not.

Exercise 11 ( c )

1. Find the number of permutations that can be made out of the letters of the words

( i ) Calcutta, ( i i ) Institutions. 2. How many of the permutations in Example 1, begin and

end with t ? 3. How many different numbers can be formed by permuting

the digits, 2, 2, 2, 3, 3, 4 in all possible ways ? How many of these numbers are greater than 3,00,000 ?

4. Find Ihe number of arrangements that can be made from the letters of the word Algebra, without altering the relative positions of the vowels and consonants.

5. In how many ways can the letters of the word College be arranged so that ( i ) the two /'s are never together, ( i i ) the two

s are together but not two e's, ( iii ) neither the I's nor the two e's are together ?

6. In how many ways can a man distribute his 3 votes among 5 candidates, it being permissible to give more than one vote to the same candidate?

7. In how many ways can 5 scholarships be awarded to 4 boys, when each boy is eligible for all the scholarships ?

8. How many different numbers, each consisting of 4 digits, can be formed with the figures 1, 2, 3, 4, 5 if repetitions are allowed ?

9. At a certain examination a candidate has to secure a certain minimum percentage of marks in each of the six different subjects. In how many different ways can he fail ?

C. A.—20


10. In how many ways can n things be given to p persons, when there is no restriction as to the number of things one may receive ?

11. A candidate has to answer one or more questions out of the 8 questions in a paper, each question being given an alter-native. The candidate may answer a question, or its alternative or may not answer it at-all. In how many different ways can the candidate attempt the paper ?

12. In how many ways can a selection be made from four black, three white and five red balls so as to have at least one of each colour ?

13. There are 4 different books on Physics, 5 different books on Mathematics and 3 different books on Chemistry. In how many ways can a selection of books be made so as to include at least one book of each ?

Answers—Exercise 11 ( c )

( 2 ! ) ( 3 1 , - 21) ( 2 1 ) ' ( 3 1 ) . ( 2 ! )

_ . . . 6 1 , . . . 5 1 , 51 . . . 3 ! . 3- ( l ) 3-T2T- ( u ) r r + - n T 2 r - 4• 4 ! T T 1 « - 7 2 -

5. (i) —If »•••*». (") Vr - 51 1 e- 24°-(21) 2 1

( i i i ) —"—r — 2 X 240 — 5 1 i e . 660. ( 2 1 )

6 . 5 s i . e . 125. 7 . 45 i . e . 1024. 8 . 54 i . e. 625.

9 . 2* — 1 i . e . 63. 1 0 . p". 1 1 . 3« - 1 i . e . 6560. 1 2 . ( 2 4 - l ) ( 2 5 - l ) ( 2 4 - l ) i . e 3255.

1 3 . ( 2 « - l ) - ( 2 » - 1 ) ( 2 3 - X) i . e . 3255.

Chapter 12

Binomial Theorem

1. I n t r o d u c t i o n . 2. C o n t i n u e d P r o d u c t of B i n o m i a l F a c t o r s . 3 . B i n o -mial t h e o r e m f o r a pos i t ive i n t eg ra l i n d e x . 4. Proof of t h e B i n o m i a l T h e o r e m by I n d u c t i o . 5. P a r t i c u l a r cases . 6 . T h e g e n e r a l t e r m . 7. Midd le t e r m s . Exerc ise 12 ( a ) . 8. B i n o m i a l coeff ic ients . 9 . R e l a t i o n s b e t w e e n B i n o m i a l coefficients. 10. I l l u s t r a t ive examples . Exerc i se 12 ( b ) .

1. Introduction.

Any algebraic expression consisting of two terms is called a Binomial Expression. Thus x -f- a, 3x — 5, 2a + 2b.... etc. are examples of binomial expressions, (x + a)" is the nth power of the binomial expression ( x + a ) The Binomial Theorem deals with the expansion of a power of a binomal expression. We will discuss in this chapter, a particular case of the binomial theorem giving the expansion of a power of a binomial expression, when the index is a positive integer. To "begin with, we may infer the result from some of the known few powers of the binomial.

By actual multiplication, we have ( x + a)2 = x2 + 2xa + a2 ; ( x + a y = x* + 3x2a + 3xa2 + a3; ( x + a y = x4 + 4x3a + 6 x2a2 + 4*a3 + a4 ;

and so on. Thus, if n is considered as the positive integral index, in the

expression of C x + a )", we should have, ( i ) the first teim as xn, ( i i ) the second term as n x" 1 a, (iii ) the indices of x steadily decreasing by one and those

of a steadily increasing by one from term to term. 307


( i v ) We also obsreve thai the coefficients in expansions given above, are the values of 2C0, 2Ci, 2C2, in the first, 3Co, 2Ct, 3C2, 3C3 in the second and 4C0, 4Clt *C2, *C3, 4C4 in the third. Hence, in the expansion of ( x + a ) " the various coefficients in order, can be expected a s :

"n "n "n " r nC ' ' ' » W ' " ' » These observations lead us to infer the general result

/ i \n n n n-1 . k_ " ' 2 „ 2 i ( x + a ) = C 0 x + Q x a + C2 x a +

, n n-r r , n

— + Crx a +•••+ c„ a . It may be noted that ( i ) the number of terms in the expansion is ( n + 1 ) i. e.

one more than the index n of the binomial, ( ii ) the suffix of C is the same as the index of a, (iii) the sum of indices of x and a, in any term in the

expansion, is n, i. e. the index of the binomial. Example Expand (x + a) •

6 W e have ( * + a) *

= "Co*8 + 6Ci x*a + "C zx'a* + 6 C ; , * V +tClx1a* + 6C,,*a5 + «C6a°. = x' + 6 x ' a + 15* V + 20x*a' + 1 5 * V + 6xa 5 + a".

2. Continued product of binomial factors.

In obtaining the expansion of ( x + a )", we have to multiply the binomal expression ( x + a ) by itself n times. The formation of the product of such factors can be considered as follows.

We have ( x + ^ i ) ( x + o 2 ) = x2 + xat + xa2 + a2. It is convenient to group the terms in the product having

the same index o f ' x ' together. Thus, ( x + fli) ( x + a2 ) = x2 + x ( ai + a2 ) •+ a\ a2. Similarly, ( x + a i ) ( x + a 2 ) ( x + »3 )

= x3 + x2 (tfi + a2 + a3) + x ( ata2 + + o2a3 ) + ata2a3. Similarly ( x + a 1 ) ( x + a 2 ) ( x + a 3 ) ( x + ai)

= x4 + x3(at-{-a2-ha3+a4) + x2 { ata2 + a\a3 + a^a^ + a2a3 + a2ak + a3a.) + x ( aia2a3 + aia2cn + a\a3a± + a2a3<n ) + axa2a3ai.

BINOMIAL THEOREM : 3 0 9

It will be seen that each term in the product is obtained by choosing one term from each factor, x or a, and multiplying them together. All the terms in the product are obtained by doing this in all possible ways. The expansion, thus, is the sum of all such partial products.

It may be convenient to write

(X + flj ) 0 + <J2) 0 + fl3) 0 + fl4> = X 4 + J i X 3 + J2X2 + +

where = the sum of the terms a2, a3, s2 = the sum of their products, two at a time, 53 = the sum of their products three at a time, 54 = the product, four at a time.

This process of writing the product of continued binomial factors can be generalized to any number of factors.

Thus,

(* + 0 + a2) (* + <*$) (x + an) h n - 1 n - 2 n - 3 n-r . .

= X + S l X + S 2 X + S3X + . . . + S,X + . . . + S „

where si contains "C t terms, (a-i+a2 + az + + a „ ) ,

s2 contains "Cj terms, ( ata2 + flifl3 + ),

and in general, sr contains "Cr terms because it consists of terms of selections of n things taken r at a time.

If in this result, we put at = a2 = a3 = ... = an = a,

we get Si = "Ci-a, s2 = "C2-a2, s3 ="C3-a3,..., sr — "Crar,

... and sn = nCn-a" i. e. a". Also the L. H. S. becomes (x + a ) "

Hence, we have

r i " i " - 1 i "/-> n ~ 2 1 i (x +a ) = x + C t x a + C2 x a2 + . »_ n-r r n — + C,x a + a .

which is the Binomial Theorem,

By following the procedure of continued multiplication of n binomial factors, explained in this article we give in the next article a shorter proof of the Binomial Theorem by the use of combinations.


3. Binomial Theorem for a positive integral index.

If x and a are any two numbers and n a positive integer, then / i \n n i nn T "/-i n ' 2 2 i (* + a ) =s x + C i x a + C 2 x a +

— + C,x a + +a . We have ( x + a ) " = ( x + a ) ( x + a") ( x + a ) - to n factors.

We shall get a term in the expansion by selecting a term from each of the n factors and multiplying these terms together. All the terms in the expansion will be obtained by doing this in all possible ways.

Select x from each of the n factors and multiply. We get

a term in the expansion, namely, xn. This can be done only

in one way and hence the term x* occurs only once in the expansion.

Select a from any one of the n factors and x from the remaining ( n — 1 ) factors and multiply. We get a term

x a. The selection of the factor for a f rom the n factors can

be done in " Q ways. Hence x a occurs "Cj times in the

product. Therefore, " Q X" 1 a is the term in the expansion. Similarly, select a from any two of the n factors and x

from the remaining ( n — 2 ) factors and multiply. We get a

term x" 2 a . The selection of the two factors for a from the

n factors can be done in "C2 ways. Hence x" 2 a occurs " Q

times in the product. Therefore, "Qj x" a2 is the term in the expansion.

In general, select a from any r of the n factors and x from the remaining ( n — r ) factors and multiply. We get a term

x"" a. The selection of the r factors for a from the n factors

can be done in "C r ways. Hence, x" ' a occurs "C r times in

the product. Therefore, "Cr x" r a is the term in the expansion.

Finally, select a from each of the n factors and multiply.

We get a term a". This can be done only in one way and

hence the term a occurs only once in the expansion.

BINOMIAL THEOREM : 311

Adding all the terms, we have the expansion

(x + a) =x + Cxx a+ C2 X a + 4- «_, n - r r . n ••• T C r * a + . . . a .

4. Proof of the Binomial Theorem by Induction.

We will prove in this article the binomial theorem by the method of mathematical induction.

We have to prove that

( x + a )" = x" + "Cix^ 1 a + "C2x"~2 a + ... + ...

...+ nCrx'-ra +•••+ a ( i ) where n is any positive integer, and x and a are any two numbers.

Assume that the theorem is true for a particular value of n, say n = k.

Therefore, by the assumption, we have , .ft fe . fc„ fc-1 . k-2 2

( x a ) = x + Cj x a + C2 x a + k-r+1 r-1 k_ k-r r , ft / • •

+ c r-! x a + cr x a + • • • + a . ... ( n ) Multiply both sides by x + a.

; , ( x + a ) ( x + a ) *

= xfe+1 + fcC1xfe.a+ k Cj xft 1 a + . . . . + *C r x*" 4 1 * ' + . . . . + * • « *

+ xka + kC1xk-1a2+ + , h . fe+i

(* + fl) = + ( *Cl + 1 ) x a + ( *C, + *Ca ) x""1 a2 +

+ ( * C r + * C r - i ) •<»'+... + ...

+ ( I + kCk~! ) x-ak+ a f t+1 . Jj+l , ft . fi+V fc-l 2 ,

= x + Q x a + C2 x -a +

+* + 1 C r x h ^ r a + + ft+1C,.xafe + a fe+1 ( i i i )


and &Ci + hC0 i. e. ^C^ + 1 = ' i+1Ci and so on. )

The expansion ( iii ) is of the same form as the expansion ( i ) and can be obtained from it by putting n = k + 1 just in the same way as ( ii ) has been obtained by putting n = k in (i).

Thus, if the theorem is assumed to be true for n = k, it is also true for n = the next integral value k + 1.

Now if n = 2, we have ( x + a )2 = x2 + 2C txa + a2

= x2 + 2 xa +• a2

and if n = 3, we have ( x + a )3 = x3 + 3C tx2a + 3C2xa2 + a3, = x3 + 3 x2a + 3 xa2 + a3.

Therefore, the theorem is verified for n — 2 and n — 3 ; and hence it must be true for n = 4 and hence for n — 5; and so on.

Thus the theorem is true for all positive integral values of n.

Ex. \ . Find the expansion of (2c + 1 ) ' . Here, we have 2x and J for x and a in the above standard expansion.

( 2 * + - ! - ) - ( a * ) ' + »Ci(ar)» ( 4 " ) +'C2(2*)«

+ ' C 3 ( 2 * ) ' ( \ y + 'C,(2*)» ( 4 - ) '

+ ' C 5 ( 2 * ) > ( - L ) V C 3 ( 2 * ) ( - L ) 6 + ( ± } j

= 2'-*' + • 25-*3 +~62-2ix + 2x*

7-6-5-4 , 7J5 X J _ J _ + "1-2-3-4" ' 2 1 f 12 * 1 ' 25 * + 2'

= 128s7 + 224*' + 168*5 + 70*4 + x* + x1 + * + ^ •

Ei. 2 Expand

A 34-A J L j . A 1 4.A J 2 X 2 x + 4 " * + 16 " x3- 32 ' *s

Ex, 3. Expand ( + ~ j • Ans. x3 + 4 6X' + 4 • ~ + ~


The following points about the above standard expansion in the Binomial Theorem should be noted.

( i ) The number of terms in the expansion is ( n + 1 ) i. e. one more than the index of the binomial.

( i i ) The sum of the indices of powers of x and a in any term is always n.

(iii ) The index of the power of a, in any term, is the same as the suffix of C.

( iv) Since, nCr = "C„.r, the coefficients of the terms from

the two ends are equal, and the coefficients of x"and a" are both equal to 1.

( v ) The rank of the term in the expansion is one more

than the suffix of C. Thus the rank of the term "Crx" *a is ( r + 1). 5. Particular cases.

We have (x-ra) =x + Q x a + C2x a + ... i "/-i n-r r , , n /• • \ ... + Cr x a +... + a. ... ( i )

This is an identity which is true for any value of x and a, rt being a positive integer. Following convenient expansions can be obtained as particular cases by giving suitable values to x and a.

In ( i ) , writing — a for a, we get , t t n «_ «-l , N . n-2 . -2 ( x - a ) —x + Qi ( - f l ) + C2x ( - a )

, «™ n-r . . . , x" + ...+ Cf x ( -a) +... + ( - a) .

n « n-1 , «_ n-2 2 . . ( x - a ) —x — Qx a+ C-x a +

. . . + ( - 1 ) . C ,x a + . . . + ( - 1 ) a . ... (11) In this case, the terms are alternately positive and nega-

tive and the last term is positive or negative according as n is even or odd.

Again in ( i ) . writing 1 for x and x for a, we get

( 1 + x ) " = 1 + "C tx + "C2x2 + "C3*3 + ...

+ . . . + " C r * r + . . . + x " . . . . ( i i i )


This result gives the simplest form of the Binomial Theorem.

Again in ( i ) , writing 1 for x and - x for a, we get

(1 - *)" = 1 + " Q ( - x) + "C2( - J C ) 2 + . . . + " C r ( - * ) ' + . . . + ( - * ) " .

( 1 — x ) " = l— "Cj x + "C2 x2 —"C^x3 +

+ ( - l ) ' V + . " + ... ( i v ) Here, as in ( i ), the terms are alternately positive and nega-

tive, and the last term is positive or negative according as n is even or odd.

E*. 1. Expand ^ x - ^ • Ans. * 5 - 5 * 3 + 1 0 * - I 0 - -

E x . 2 . Expand ( 1 -2x)'.

Ans. 1 - 12* + 60*a - 160*s + 240*4 - 192*® + 64*e.

E* . 3. F ind the value of { 1 + ^2) - ( 1 - ^2)"- Ans. 5^2.

<S. The General term.

In the expansion {x + a)n, the ( r + 1 ) th term is given by

T - r n - » U - l ) Q - 2 ) . ••(»-»•+ 1) n-r j. r + l — A U — , A *u • r!

All the terms in the expansion can be obtained by giving to r, positive integral values from 0 to n ; and hence this term is called the general term in the expansion. It should be noted that the rank of this term is ( r -M )th in the expansion. This is an important result and is used to find any particular term of the given expansio n.

/ 2

Ex. 1. F ind the 10 th term in the expansion of ( 3x — — 1

We have T r + l = nC, xn ' a, _ 2

Here, in this example, 3x is the first term, and — is the second term of

the given binomial. Also, » - 12 and r + 1 = 10 i . e. r = 9.


T , + i = » C 9 ( 3 * ) 1 2 " 9 ^ y = »C8 ( 3 „ S ( - I , S ( 1 ) 9

=» - "C3 3 3 - 2 9 ' = - 1 2 ' " ' 1 0 - y ^ x - i -1 • Z • J

required JOtb term «• - 2u -33 -55 x -« .

Ex. 2. F i n d the coefficient of Xs in the expansion of ^ 2x - J

We have, w i th the usual notation, T r + i = "Cr r a".

Here, i n the given example, first and second terms of the binomial are

2* and - y • Also n = 7. We have to find r such that T r + 1 contains

T r + 1 - 7C r ( 2 x f r ( - } ) ' = »Cr ( 2 ) 7 - ' ( - D r ( a

The value of »• is given by 7 — r = 3 i . e, r = 4.

X 4 + 1 = 'C4 ( - 1 >«.* ( { ) ' • * > = + »C. ( I ) = { I f • \ •

35 .*. the required coefficient of * s is — •

ion of ^ x - ~ j Ex. 3. F ind the term independent of x i n expansion

W i t h the usual notation, we h a v c T r + i = * C r x" r a".

Here, i n the given example, first and second terms of the binomial are x

— 2 and —> • Also « = 8.

We have to find r such that T r + i is independent of x and thus wi l l not involve a power i n x.

T r + l = "Cr ( * ) * " '

= « C r ( - l ) r - ( 2 ) r - * 8 " r . ~ = ( - l ) ' » C , ( 2 ) ' - * 8 " 4 r -

the value of r is given by 8 - 4r « 0 i . e. r = 2.

T2 + 1 = ( — 1 ) ' • eCa (2)2 • x° => = U ,

•*. the required term independent of %, is 112,


I / 1 V Ex. 4 . Find the coefficient of in the expansion of I * - J

Find also the rank of this term in the expansion.

With the usual notation, we have T r + i = nCr x" ' a .

Here, in the given example, the first and second terms of the binomial

are x and - respectively. Also « = 10.

W e have to find r such that T r + i contains -^r • xm

- 1 - 1)' • i • •"C, ." - • - b - I - ' > " ~r »c,«™-3'• A X A

•'• T r + t c o n t a i n s - ^ j - i . e. * 5 if 10 — 3r = — 5, .'. r = 5.

••• T s + i =s ( — 1 )5 2 "S 10C6x~S

1 10-9 -8 -7 -6 - s 63 1 r» X • .. T« . - • 32 1 -2 -3 -4 -5 8 8 * 5

.'. the required coefficient of is - ~ jc5 8

It can be seen that the rank of this term 6 .

Ex. 5 . Find { i ) the 7th term, ( i i ) the coefficient of in the expansion of

( } * - 2 ) 9 . Ans ( i ) ; ( i i )

7. Middle terms. To find the middle term or the middle terms in the expansion of ( x + a )".

Case ( i ) : Let n be even and be put in the form 2p. The number of terms in the expansion is ( n + 1 ) i. e. 2p + 1 . Therefore, Ti+1 is such that there are p terms preceding it and p terms following it, and hence it is the middle term.

T r + 1 will be the required middle term if r = p i. g. n/2 where n is even.


Case ( i i ) : Let ti be odd and be put in the form 2p + 1. The number of terms in the expansion is n -f 1 i. e. 2p + 2. Here, it can be easily seen that there is no one term equidistant from the two ends in the expansion.

However, the term Ip+i is preceded by p terms and the next term Ty,+2 is followed by p terms in the expansion; and hence both the terms and T M 2 are called the middle terms

in the expansion. Since n = 2p + 1 we have p = n • •-- •

/ . T r + 1 will give the two middle terms if we put r = 1

and r = where n is odd.

Ex. 1 . Find the middle term in the expansion

Here, the total number of terms in the expansion is 7. Hence, 4th term is the middle term.

We have, with the usual notation. T r + i "Cr x" r a .

Here, in this case, we have T ? + i «• 6CS ( )6 3 ^ ^ •

••• T, = H 4 • ( ~ 1 )s. - s = ~ 20A;3. 1 1 2 3 Xs

.'. the required middle term is - 20*®.

/ 1 V Fx. 2 . Find the middle terms in the expansion of f 2x — — I •

Here, the total number of terms in the expansion is 10. Hence, the 5th and 6th terms n the expansion are the middle terms.

W e have, with this usual notation, T r + i = " C x" r a .

Here, in this case weget the middle terms by putting r = 4 and 5.

f 1 V / 2" 9 8 7 6


1 1 63 1

t h e r equ i r ed m i d d l e t e rms a r e — x a n d — • — • 4 32 x

/ i \1 0 Ex 3 . F i n d t h e m i d d l e te rm in t h e expans ion of I * — J .

Ans . - 252. / 2 * \ "

Ex . 4 . F i n d t h e m i d d l e t e r m s in t h e expans ion of I — + — - J •

u 2 1 ii * A n s . C 5 — i, e . 924 a n d C 0 • i . e. 231 * . x x 2

Exercise 12 ( a )

i . Expand the following:

( i ) ( 2 x - f l ) ' ; ( i i ) ^ - A j ; (iii) ( l + y ) 7

( i v ) ( ^ - ^ ( v ) ( * + t ) '

2. Write down the expansions o f :

( i ) (x + V2)4 + (x - \J2)*:

( ii) (x +yfa)6 + (x - \ f a ) 6 ;

(iii) (x + 1 ) (x• + 2)(x + 3)(x + 4 ) .

3. Find the values o f :

( i ) (\[2 + 1)5 - ( V I - I ) 5 ;

( H ) ( 2 + \ / 3 ) 7 + ( 2 - \ { 3 y -

(Hi j ( yfs + 1 )6 + ( \fS - 1 )6.

10

Find : ( i ) 7th term

( i i ) 5th term

in ( l ;

in - ;


10

10 / v n 1

( iv) 4th term

Find the coefficient of : ( i ) x5 in 0 - 2 ) 9 ; ( i i ) in ^ x - ;

( i i i ) x18 in ( a x - bx f ; ( iv ) -i? i n ^ ** ~ ~T j *

Find the term independent of x in : / 2 \ 9 / i \ 10

( i ) ( $ - £ ) ; < « > ( > R - 4 ) =

o u ,

Show that the term independent of x in the expansion of :

Find the middle term or middle terms in the expansions o f :

( i i i ) U v ) ( > - 4 - ) •

Simplify :

( i ) ( x - l f + 5 ( x - l ) V l O ( x - 1 ) 3

+ 10 ( x - 1 ) 2 + 5 ( x - l ) + l .


( i i ) ( x - 1 )4 + 4(x— 1 ) 3 ( x + 1) + 6(x - l ) 2 ( x + l ) 2

+ 4 ( x - l ) (x + l f + Cx + l f .

10. Use the Binomial Theorem to find ( i ) 954 ; ( ii ) l i s .

Answers—Exercise 1 2 ( a )

1. { i ) 128x7 - 448x<> a + 672xs a3 - 560x* a8 + 280*3 a4

~ £4x> a + 14xa* - a\ + I 11 >

, 7*8 , x i + -rr +

( i i ) * 8 64

3*3 15 _ _ + _ _ 20 Xs + " x"

96 x'

( U i ) 1 + 7x , 21*» ,

T 4 35*3

8 + 35x4

16 , 21a;5

+ 32

( i v ) 64*® - 96*' + 60*' - 20 + 15 4x' ' 8x4 +

64 T 128

64 a;6

( v ) a1 0 + 10x* + 45*8 + 120** + 210*2 + 252

210 120 45 10 1 a:1 + ** + x f *8 + '

2 . ( i ) 2xl + 24x* + 8. ( i i ) 2x« + 30x'a + 30x'a> + 2a3.

I iii) x* + 10x3 + 35*» + 50A; + 24.

3 . { i ) 82. ( i i ) 10084. ( i i i ) 1152.

^ , . , 105 , , .. . 1120 . . , . x 1

4- ( i ) 32 *6- ( u ) —gj— x* a4. ( i i i ) - 120-^j- •

5 . ( i ) 2016. ( i i ) - 5 6 . ( i i i ) 84a86«. ( i v ) - 1 3 6 5 .

6 . ( i ) X . ( U ) 405. ( i i i ) .

8. ( i ) - 2 5 2 . ( i i ) ~ ~

- 4 2 9 „ 2n! (m) *14. (iv) ( - 1 ) 16 ' v ' v ' « < • m I 9 . ( i ) Xs. iii) I6x4.

10, ( i ) 9 , 6 0 . 5 9 . 6 0 1 , ( i i ) 1 , 6 1 , 0 5 1 .

8. Binomial coefficients. The binomial expansion is given by

n » n n-\ . n n-2 ! ( x + a ) = x + Cix a + C2x a + • . n-r r . . « • • • + C-x a -f • • • + a .


Here, the coefficients of x* and a" are both equal to 1.

Since, MC0 = "Cn = 1, we may for the sake of uniformity write "C0

as the coefficient of x" and "CB as the coefficient of a . Then, the binomial expansion may be written as

, . .» « . n-1 . n„ »-2 -( x + a) = C 0 x + C i * a + C^x a2

i nr~i n~ r r i i "/-< "

... + Crx a + + Cna ,

The ( n + 1 ) coefficients in the expansion of (x + a)",

VIZ. , L-2 . . .

are called the Binomial coefficients of the nth order. When we are concerned with the binomial coefficients of the same order n, we may drop the prefix n for the sake of brevity and write these coefficients simply as COJ CJ, C2 , Cf, Cn . 9. Relations between Binomial coefficients:

I. In the expansion of (1 -fx)" the sum of the coefficients

is 2*.

We have the binomial expansion,

( l + x ) " = C0 + C 1 x + C 2 ^ + ... +Crxr + ...Cnx", ... ( i )

where C, i. e. "c, denotes the binomial coefficient ctf order n. In the above identity, put x = 1, we, then, get

2" = C0 + C1 + C 2 + . . . + C, + . . . + C n . the sum of the binomial coefficients of order n is 2".

Note —We have C, + C2 + C 3 + + C„ = 2 " - 1 ; that is the total number of selections from n different things, tak-ing any number of things, at a time is 2" — 1.

II. In the expansion of (I +x )" the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms, each being equal to 2" \

In the above identity ( i ) , putting x = — 1, we get

( l - l ) n = C0 + C 1 ( - l ) + C 2 ( - l ) 2 + ... + C r ( - l ) r +

+ c„(- n". C. A —21


0 = C 0 - C 1 + C 2 - C 3 + + ( - l ) " C r + . . . ( - l ) "C„ . C0 + C2 + C4 + C 6 + . . . = C1 + C3 + C5 + C 7 +

and each of them = + C1 + C2 + C3 + . . . + C „ j

= - | - f 2 " ] = 2 - 1 ( by I )

the sum of even coefficients = the sum of odd coefficients = 2" .

10. Illustrative examples :

We shall now work out a few miscellaneous examples to illustrate some applications of the Binomial Theorem.

Ex. 1. Simpl i fy

( * — l ) * + 4 ( * . — 1)" + 6 ( * — 1)* + 4 ( * — 1) + 1.

The coefficients 1, 4, 6, 4 , 1 can be easily recognized as the binomial coeffi-cients of the fourth order. Hence the substitution x - 1 = y, will reduce the given expression to a simple binomial expansion in y. Thus, the given expres-sion

- y4 + 'Ci • y3 + 'Ciy a + lCsy + 1

= (y + I)' = (* - 1 + 1 ) = x*.

/— 8 s Ex. 2 . Show that ( V 3 + 1 ) — ( V 3 — 1 ) is an integer and hence

find the integral part of ( V 3 + 1 ) •

Let V 3 = * .

We have ( * + 1 )e = * s + iCi* 4 + >C2x3 +>* C3,*J + 5C4* + 1 ; and (x - i y = x6 - eC,*4 + »Cj*» - "Cj** + lC,x - 1.

(x + 1)" - (x - 1 ) ' = 2 [ e C , * 4 + s C f *«+ 1 ] .

Substituting the value of x, we get

(VT +1) 5 - ( J T - i)5 = 2 [ 5 - ( V l V + io (J3 )' + 1] = 2 [ 4 5 + 30 + 1 ] = 152, which is an integer.

.*. ( V f + 1 )5 = 152 + ( V f - 1 )5. ( 1 »

Now. we have 1 < 3 < 4, 1 < « / 3 < 2 .


0 < >/ 3 - 1 < 1, subtracting 1 from each side.

.'. ( .J 3 - 1 ) i s a positive proper fraction i less than 1 ) ,

( \ f 3 — 1 j5 is also a proper fraction, say/, so that 0 < / < 1.

from ( i ) , the integral part of ( V 3 + l ) 5 i s 152. 7

Ex. 3 , Evaluate (1*02) correct to 4 decimal places.

W e have' ( 1 -02 )7 = ( 1 + -02 )7

= 1 + 'Q ( -02 ) + 7Ca (-02)2 + 'Cs (-02 )8

+ 7C, (. 02 )' + ... + = 1 + 7 ( -02) + 21 (-0004) + 35 ( -000008) + = 1 + -14 + -0084 + -000280 + = 1-14868 = 1-1487, correct to 4 decimal places.

Note—We have considered only the first four terms of the expansion s ince the remaining terms do not affect the digits in the first four decimal places.

Ex. 4 . Find the coefficient of xl in the expansion of

(1 - * + *a - Xs )"•

W e have l - * + * > - * 3 = = ( l - * ) + * , U - * ) = ( l - * ) > < ( l + * a )• 19 19 19

(1 - * + * » - *3) = ( 1 - * ) ( 1 + * ' )

= [ 1 - MCi * + »Cj *> - llC3 + »C4 *«...] x [1 +»C, (* ') + l,C2 ( x l ) + ... + ...].

As we want the coefficient of we have not written down' termscontain-ing higher powers of x.

Multiplying together, suitable terms from the two brackets, the co -efficient of A4

= l . " C a + » C a - l s C i + " C 1 - l = 66 + 66-12 + 495 = 1353.

Exercise 12 (b)

1. Expand and simplify ( x + a )" — (x — a)". What will be the last term in the above simplification if n is ( i ) even, ( ii ) odd ?

2. If ( y/5 - )5 = A\f5 + B-v/3, show that A - B = 504.

3. Find the coefficient of x4 in the expansion of

( i ) ( 1 - x ) 2 ( 1 + JC)5; ( i i ) ( 1 - x ) 5 ( l + 2x) 4 .


4. Find the coefficient of :—

( i ) xMn ( l + x + x^ + x 3 ) 1 1 ; 7

( ii ) x3 in (1 — x — x2 + x3) ;

( i i i ) z7 in ( l + x + ^ + x5)6 . 5. Evaluate, using the Binomial expansion,

( i ) ( -99)5 to 4 decimal places ;

( ii ) ( 1-0025 )10 to 4 decimal places ;

( i i i ) ( 102)4 ; ( i v ) (49) 4 . 6. Simplify:—

( i ) ( a + 6)3 - 3b(a + bf + 3b2(a + b) - b3\ ( i i ) (2x+1)4—4 (2x+lf + 6 (2x+l)2—4 ( 2 x + l ) + 1 . •

7. By considering the values of ( \ [5 + 2 ) — ( \ / 5 - 2 )5 ,

show that the integral part of (V5 + 2 ) 5 is 1364.

8. Show that ( ^ 2 + 1 )6 + ( V ? - 1 ) ' = 198.

Hence show that the integral part of ( \ f 2 + 1 ) 6 is 197. 9. Prove that

"P "P, "P " P i + 2 T + 3 ! ! + = 2 » - l .

j^Hint: Use "Cr = j '

10. + ^ + 3 + + 1 „(„+!). Co m y-2 z

[ Hint: Use the formula for "Cr. ]

Answers—Exercise 12 (b)

1. ( i ) 2a\ ( i i ) 0. 3. ( i ) - 5 , (iii) 21. 4. ( i ) 990, (ii) -344, (iii) 120. 5. ( I ) 9510, ( i i ) 10253, (iii) 10, 82, 43, 216. (iv) 57, 64, 801. 6. ( i ) a3. (ii) 16*'.

Chapter 13 Determinants

1. Determinant of second order. 2. Consistency of two linear equations in one unknown. 3 Two linear simultaneous equations in two unknowns. 4. Determinant of third order. 5. Consistency of three simultaneous linear equations in two unknowns. 6 Three linear equations in three unknowns. Exercise 13 ( a ) . 7. Properties of determinants. 8. Illustrative examples. Exercise 13 (6 ).

1. Determinant of second order:

Definition 1—A square array ( or arrangement ) of 4 i. e. 2 numbers in two horizontal rows and two vertical columns enclosed between two vertical bars is called a determinant of the second order. For example, if we have 4 numbers alt bu a2, b2

we can arrange them as at bi

a2 b2

The four number s ax, b\, a2, b2 are called the elements of the determinant. The elements of a horizontal line form a row ; those of a vertical line, a column. The determinant is said to be of the second order, since there are two rows and two columns.

The diagonal, from upper left to lower right, is called the ' Principal Diagonal \

The value of the symbol of a second order determinant is defined as the product of the elements of the principal diagonal minus the product of the elements of the other diagonal. Thus, we have the notation

ax b t = a1 b2 — a2

a2 b2 a \ t>2 ~~ a2 is called the expansion of the determinant. In every term in the expansion, there is one and only one element from

325


each row and one and only one element from each column crt b3

is called the leading term in the expansion. The determinant is sometimes denoted by bracketing the leading term as ( a\ b2 )•

We illustrate the evaluation of second order determinants by the following examples.

I l lustrations:—

4 ( i )

(ID

(Hi)

5 7

1 0

5/7 - 3

ax + b

ay + b

4 x 7 - 5 x 3 = 23 - 15 = 13.

e I X ( - 3 ) - { X ( 0 ) = — 3 — 0 = — 3,

= ( a * + 6 ) x 6 — (ay + 6 ) x b

= abx + 6" - aby — 6* = ab (* - y).

tan 9 sec 0 sec 6 tan 9

= tan* Q - sec* 9 = - 1.

E*ampV»— Evaluate the following determinants:—

( i v ) tan 0 x tan Q — sec 9 x sec Q

3 7 - 1 7 - 1 a h a )

2 9 . ( I I )

5 0 . ( i i i )

h b

( I v ) sin A

— cos A

cos A

sin A . ( v )

* + y x - y

X - y x + y . ( v i )

2/3 7/3

- 1 / 5 4/5

Ans.—( i ) 13. ( H ) 5. ( i i i ) ( i v ) l . ( v ) Axy. ( v i ) l .

2. Consistency of two linear equations in one unknown:

Let the two linear equations in the unknown viz. x be a\X + bi = 0 • • • • ( i ) and a2x + b2 = 0. ... ( i i )

These two equations are said to be consistent if they are satisfied simultaneously by the same value of the unknown x. Thus, the equations ( i ) and ( i i ) are consistent

i f _ h. = -a\ 02

i .e. if aib2 — a 2 bx=0. ... (iii)

DETERMINANTS : 3 2 7

If the condition (iii) is statified by the constant numbers in equations ( i ) and ( ii ), the equations ( i ) and ( ii ) will be consistent. Two equations, in general, may not always be consis-tent. For example, we can not find a value of x satisfying both the equations

3x + 5 = 0 and 2x 4- 7 = 0 simultaneously. The condition ( i i i ) of consistency of equations ( i ) and ( ii ) can be very conveniently put in the form of the determinant notation as

h

a2 = 0. ( i v )

Result (iii) or (iv) is also called the eliminant of equations ( i ) and ( i i ) . The eliminant can be conveniently written in the determinant form by suppressing the variable and enclosing the constant coefficients and the constant terms with their proper signs on the 1. h. s. ( r. h. s. being zero ) by two vertical bars.

For example, the eliminant or the condition of consistency of the two equations

px — q — 0 and rx + 5 = 0

is - 1

s = 0 i. e. ps + qr = 0.

Remark 1. Students who are familiar with Co-ordinate Geometry can easi ly see that when the condition of consistency (jv > is satisfied, the equa-t ions ( i ) and { i i i represent one and the same straight line.

Remark 2 , If the equations are a^x + 6iy = 0 and a2x + bty = 0, then dividing by y and treating xjy as one unknown the condition of consistency for these equations can be written as

a-i bt

a2 b2

3. Use of second order determinants to solve two linear simulianeous equations in two unknowns :

1. Cramer's Rule : We shall, now, show in this article how the second order determinants can be used to give the solution of two simultaneous linear equations in a convenient form. Students are

0.


already familiar with the method of solving two simultaneous linear equations in two unknowns.

Let the given equations be written in the form aix + b±y = cx ... ( i ) a2 x -f b2y = c2. ... ( ii )

Note that the constant terms in both the equations have been arranged to be on the right hand side. To find the value of x we eliminate y by multiplying ( i ) by b2 and ( ii ) by b\, and then subtract the latter from the former; we then get

(axb2 ~ a2bx~) x = cxb2 - c2bx. ... ( i i i ) Similarly to find the value of y we eliminate x by multiplying

( i ) by a2 and ( ii ) by ax and then subtract the latter from the former ; we then get

( & i a 2 - axb2) y = cxa2 - c2ax. ... ( i v ) The values of x and y as given by ( i i i ) and ( i v ) can be

written as jc _ y 1

cx b2 - f2 bx ~ a^c2-a2cx ~ axb2~a2bx

This solution can be conveniently written in the determinant form as follows :—

x _ y _ 1 C\ h «1 Cl a \ bt 1

C2 b2 a2 a2 a2 b2

ci bx

Cl b2

a\ bi D

«2 b2


We observe that the denominator for each unknown is the determinant in which the elements are the coefficients of x and y arranged as in the two given equations. This determinant, we shall call, as the determinant of coefficients and will denote it by D. Observe further that the numerator for any unknown is the same as D except that the column of coefficients of that unknown is replaced by the column of constant terms. Let us call for convenience the determinant in the numerator for x by D t and the determinant in the numerator for y by D2. The rule embodied in the solution given in terms of determinants as described above is known as Cramers Rule.

Note—Students who are familiar with Co-ordinate Geomatry know that equations ( i ) and { i i ) being linear in x y represent two straight l ines. T h e values of x. y given in solution ( v ) give the co-ordinates of the point of intersection of l ines ( i ) and ( i i ) . If «i bt - «3&i = 0, the equations ( i ) and ( i i ) are not satisfied by finite values of x and y and the l ines become parallel. However , if <ix b% — a,i b, 4=0 , the l ines ( i ) and ( i i ) intersect in a finite point whose co-ordinates are given by ( v ).

We, now, illustrate the use of Cramer's Rule for the solution of the s imultaneous linear equations in two unknowns .

Illustrations—

( i ) Solve : 3 * - 3y - 4 = 0, * + 4y - 2 = 0.

Arrange the constant terms on the right hand side. The equations t h e n become

Sx - 3y = 4 . . . ( i ) and x + 4y = 2. ... ( H )

W e have by Cramer's Rule

4 - 3

Z>i 2 4 1 6 - ( - 6 ) 22 D ~ 3 - 3 = 12 - ( - 3 ) _ 15

1 4

3

D and y = " o 3 - 3

1 2 ~ 12 - ( - 3 )

6 - 4 2 " 15

1 4

X V X V ( l i ) Solve— + f = 1. J + f = 1.


Let us rewri te the equations as

3 * + 4 y = 12, 4x + 5y = 2C.

W e have by Cramer's Rule

12 4 I

20 5 1 6 0 - 8 0 - 2 0 A tO ~ 80 15 - 16 - 1 = 20

D > and y =

12

20 60 - 48 15 - 16

+ 12 - 1 - 12.

II Rule of cross-multiplication : Let the equations be in the form

a{x + bty + C\ — 0, ( i ) a2x + b2y + c2 = 0. ( i i )

By taking the constant terms on the r. h. s. we can apply the Cramer's rule to solve them. However we give here an alternative process to solve the equations as they stand. The values of x and y on elimination can be found as

x ~ &1C2 - b2CX _ alC2 - .

axb2 — a2b\' a2bt — axb2

These values of x and y can be expressed as x - v _ 1

b\c2 — b2cx ~ atc2 — a2Ci ~ atb2 — a2bx

This solution can be put in the determinant form as

x - y _ 1 bx Cl Cl bx

b2 C2 ^2 a2 b2

or as x y 1 bx Cx Cx ax ax bx

b2 C2 C2 a2 a2 b2 ... ( i v )


'' e" biC2 — b2Ci cta2 — c20j a%b2 — <22^1

Note 1. In the determinant form of the solution ( i i i ) , we observe that the numerators have alternately positive and negative signs and the denominators of x, — y and 1 are obtained by forming second order determinants with coefficients in equations ( i ) and ( i i ) by, omitting the coefficients of x, by omitting the coefficients of y and by omitting the absolute terms respectively.

Note 2. In the solution ( iv) , the numerators have the same positive sign and the denominators are obtained by taking the values of the determinants formed by the coefficients in the same cyclic order beginning with the second coefficients. The method of writing the solution as given by ( i v ) is called the Rule of Cross-Mnltiplication.

Examples—Solve the following equations;—

( i ) 3 * + 7y - 27 = 0: 5* - y - 7 = 0. Ans. * = 2, y = 3.

2 3 ( U ) 11* - 4y-2 => 0 ; 35* - 75y + 31 = 0. Ans. * = — ; y = _ .

2 3

( i i i ) 5* + 2y = 45 ; + y j = 5. Ans. * =J; y = 5.

( i v ) J L + = 10; - * - - Z f - + 5 = 0. Ans. * = 25; y = 35.

( v ) x + y = 1 ; a * + 6 y = l . Ans. ^ = * 4. Determinant of third order :

Definition—Any nine numbers, ar, b„ cr, [ r = 1, 2, 3] arranged in a square form of three horizontal rows and three vertical columns enclosed between two vertical bars in the form

ax h Cl

«2 b2 C2

«3 b3 C3


are said to form a determinant of the third order. The terms element, row, column and principal diagonal are defined as in the case of the second order determinants. The determinant is said to be of the third order as there are three rows and three columns. Since we have many occasions to refer to this determinant, we will use hereafter the letter D to denote this determinant, unless otherwise mentioned,

The value of this determinant D is defined as

b2 C2 C2 b2

t) III - h + C! h C3 a3 c3 «3 b3

The expression corresponding to the determinant D is obtained by taking the sum of products of each element of its first row and a determinant obtained by suppressing the row and column in which it occurs, positive or negative sign being placed before each product according as the element in the first row occupies an odd or even rank.

We now illustrate the evaluation of the third order deter-minant by the following examples.

Illustrations—

( i ) The determinant

1 4 2 4 2 1 = 1 - 2 + 3

4 1 3 1 3 4

= 1 [1-1 - 4-4] - 2 [2-1 - 3 - 4 ] + 3 [2-4 - 3-1] = 1 [1 — 16] — 2 [ 2 — 12] + 3 [ 8 — 3 ] = - 15 + 20 + 15 « 20.

i i ) The determinant

1

- 2

3

= 3 3 - 2 4 - 2 4 3

- 2 + 1 - 4 3 5 3 5 - 4


= 3 (9 - 8 ) - 2 (12 + 10) + 1 ( - 1 6 - 15) = 3 - 44 - 31 = - 72.

Examples. Evaluate the fol lowing determinants: —

1 1 1 0 1 2 a b o

( i ) 2 3 4 • (H) 1 0 2 • ( i i i ) b o a

4 5 6 2 1 0 o a b

1 X -y 7 3 - 5

( i v ) — * I z • ( v ) 2 7 3

y — z 1 - 3 - 5 1

a h S ( v i ) h b f •

8 f c

Ans. ( i ) o. ( i i ) 6. ( i i i ) 3 abo - a8 - b* + z5 + 1, ( v ) 66. ( v i ) abc + 2 \fgh - af* - bg1 - ch*.

5. Consistency of three simultaneous linear equations in two unknowns :

Let the equations be

at x + bty + Cj = 0, ... ( i )

a2 x + b2 y + c2 = 0, ... ( ii )

a3x + b3y + c3 = 0. ... ( i i i )

Equations ( i ) , ( i i ) and ( i - i ) are said to be consistent if the same values of x and y satisfy all the three equations simultan-eously. We may solve any two of the three equations for x and y and substitute these values in the remaining equation in order to get the condition for consistency r f the three given equations, Solving equations ( ii ) and (iii), we get

x _ —y 1 c2 «2 C2 a2 h

h C3 c3 a3 b3


Substituting these values in equation ( i ) and simplifying, we get

1. e.

bi C2 C2 a 2 b2 — bx + Cx

b3 c3 1 «3 f3 h

bx cx

a2 b2 C2 = 0. •

«3 h 3

= 0,

( i v )

Thus, equations ( i ), ( i i ) and ( iii ) are consistent if condi-tion ( iv ) is satisfied. We have, here assumed that a2 b3 - 0362 =£0.

We can easily see that the result ( iv ) is the eliminant of equations ( i ) , ( i i ) and ( i i i ) and can be written by suppressing x and y in the given equations and writing in a determinant form the coefficients and the constant terms arranged as in the given equations. Students who are familiar with Co-ordinate Geometry know that equations ( i ), ( ii ) and ( i i i ) being of the first degree represent three straight lines and ( iv ) is the condition for these straight lines to be concurrent.

Illustrations—

( i ) Show that the following three equations are consistent. x + y - 3 = 0, 2* + y — 4 = 0, * - y + 1 = 0.

The given equations are consistent

1 1 - 3

2 1 - 4

1 - 1 1 = 0

i . e . if 1 ( 1 - 4 ) - J ( 2 + 4 ) - 3 ( - 2 - 1 ) - 0

i. e. if - 3 - 6 + 9 « 0, which is true.

Hence the given equations are consistent.

( i i ) Find a if the following three equations are consistent, ( a - 2 ) x + (a - ) )y - 17 = 0. ( o - l ) * + ( a - I ) y - 18 = 0,

x + y - 5 = 0.


The three equations will be consistent if

a-2 a-1 - 1 7

a —I a —2 -18 =0, 1 1 - 5

1, e. if ( a - 2 ) [ - 5a + 10 + 1 8 ] - ( a - 1 ) [ - 5a + 5 + 1 8 ] - 17 [a - 1 - a+ 2 ] = 0 ,

i. e. if [ a - 2 ) ( - 5a + 28) - (a - 1) ( - 5a + 23 ) - 17 ( 1 ) = 0.

i. e. if ( - 5a5 + 38a - 56) - ( - 5a» + 28a - 23) - 17 = 0.

i. e. if 10a - 50 = 0, f- e. if <7 = 5. a = 5

( i i i ) Find the condition for the following three equations to be consistent.

OiX + b,y + Ciz = 0, a*x + b2y + c.2z = 0, a<,x + b y + c^z = 0.

Assuming s * 0 , and dividing by z and treating */z and ylz as unknowns , the condition of consistency can be written as

«i bi ci

O.i

CS

c8

0.

6. Three Linear equations in three unknowns: Cramer's rule or the rule of cross multiplication stated for

the solution of two linear simultaneous equation can be extended to the solution of three linear equations involving three unknowns. We extend the Cramer's rule.

Let the equations be written in the form arx + bry + crz = dr, r = 1, 2, 3.

The determinant of the coefficient will be fli h cx

D = «2 b2 c2

a3 b3 c3

The solution is given by Cramer's rule as D i „ - 5 Ds

x = ^ ' y ~ D • z ~ D whcre Dl' D* Dz

are obtained from D by replacing the columns of coefficients of .v, y, z respectively by the constant terms dlt d2 and d3.


Exercise 13 (a)

1. Evaluate the following determinants:-

( i )

2 3 4

3 2 1

4 1 2

1 a -b

( i i i ) —a 1 c

b -c 1

2. Solve the following equations:—

( i i )

; (iv)

5

- 3

- 4

— a

c

- 1

3

5

- 2

b

1

c

4

2

5

1

-b

— a

2 1 1 X - 1 2

( i ) - 1 X 3 = 0 ; 2 1 2

- 1 2 2 - 3 1 X

X 2 3 x-1 1 3

( i i i ) x + l 3 4 = x-2 3 4

X — 1 5 6 x-3 5 6

= 0;

3. If a , b, c, d are real, prove that a —ib c — id

= 0 — c —id a-h ib

if, and only if, o, b, c and d are all zero.

4. Find the missing terms shown by dots in the following:— 8 5 1

( i )

( i i )

5 ... 1

3 3 1 a* + 2a 2a + 1

2a + 1

3 3

= 1:

1

1

1

= ( f l - 1 ) 3 .


= (x + 2 y ) ( x - y y .

•5. Show that

x y y

y x y

y y x

Hence or otherwise show that

a+x a-x a—x

a—x a + x a—x

a—x a—x a+x

Show that

bc(b + c) c a ( c + a) ab (a + b)

be ca ab

1 1 1

, 7 . If p + q + r = 0, show that

pa

= 4 x 2 ( 3 a-x).

= 0.

qc

rb

qb

ra

pc

rc

pb

qa

— PQr

a

c

b

b c

a

18. Find the value of A such that the determinant

a2 3 5

a 2 1

1 - I ( X - l )

is exactly by ( a + 3 ).

-9. Find the value of x if the determinant

1 5 5 — x 4

3 2 1

a 1 1

is exactly divisible by ( a — 1).

C. A. 22


10. Solve the following equations using determinants:— ( i ) 3 x - 7 j > = 2 3 ; 2x + 5^ + 4 = 0.

( i i ) T + T = 1 ; T + i = L

( i i i ) ax + hy + g=0; hx + by + f==0.

( i v ) x+2y + z = 4, 2 x - y + z = - 1 , x+y-z = 4.

11. Find the value of fc if y = 1 satisfies the simultaneous equations

2x + ky = k + 2, 3 x - ( k - l ) y = k-2. /

Find also the value of x which satisfies the above equations.

12. Find the value of A if the three straight lines represented by the equations

4x + 7y — 9 = 0 , 5.*+X}> + 1 5 = 0 , 9 x - y + 6=0 are concurrent.

13. Show that the following sets of three equations are consistent ( i . e., simultaneously true for the same values of x and y ) and find, in eadh case, their common solution:—

( i ) x + y = 3, 2x + 3y = %, 5x + 6y = \7 ;

( i i ) ( i + c ) x + (c + a)y = ( a + 6 ) , ax + by = c, x+y = 1.

14. Find the value of k if the following equations are consistent—

( i ) \4x — 3j> — 7 = 0, foe + 1 l j — 81 = 0, lx=2y ;

( i i ) 2x-y + 3=0,kx-y+l=0,5x-y-3=0.

15. Show that the equations x + 2y — z =0, 3x-y + 4z = 0,4x + y + 3z = 0

have a common solution other than x = y = z = 0. x v

[ Hint. Since z # 0, divide by z and regard — and as

unknowns ].


^ y — 1 L \ — x -16. If a = , b = , c = x — y, x y show that a + b + c+abc= 0. [ Hint. Eliminate x and y in the given equations ].

17. If atx + bx = a2x + b2 = a3x + b3,

at bt 1

a2 b2 1 show that

03 1

= 0.

18.

[ Hint. Put each expression = — X and eliminate x and X],

If ax + by = bx + cy = cx + ay, prove that ab + be + ca = a2 -(- b2 + c2.

>19. If ( J i ) , ( x2, y2 ), ( *3> y3 ) satisfy the equation ax + by + c = 0,

1

show that

y\

y*x y%

= 0.

20. If the equations ax2 + bx + c= 0 and Ix + m = 0 have a common root, then show that

a b c

I m o = 0.

0 I m

Answers—Exercise 13 ( a )

1. ( i ) - 20, (ii ) 270. (iii) and I Iv) each = 1 + a' + 6' + c\ 2. ( i ) * = 3, (ii) * = ± 4 i. (iii) * = 5/2. 4. ( i ) 4, (ii) a + 2. 8 X = 10/27. 9. * = 1/2.

10. ( i ) * = 3, y = - 2, (ii ) * = - 12, y = 20,


12. A = - 8.

c - 6 a — b

(U) fe = 3.

a - c a-b'

II. k=3,x = l.

13. ( i ) * = 1. y = 2,

14. ( I ) k = 2,

7. Properties of Determinants : Elementary Transformations. We prove in this article some properties of determinants.

These properties are found very useful in finding the values of determinants. It will be seen that the actual expansions of determinants can be avoided in many cases with the help of these properties. Although the proofs given below are in respect of determinants of the third order, they are valid in respect of determinants of any order. The students can very easily verify these properties in case of determinants of second order.

Property I. The value of a determinant remains unaltered if the rows and columns are interchanged.

Let the given determinant D — Oi

a2

a 3

h

b2

b3

C2

C3 Interchange the rows and columns in D. Let the resulting

determinant be denoted by D\ so that

ax «2 «3

D' = bx b2 h

C2

We have to prove that D' = D. By definition

D' = a1ib2c3-cj>i)-a2(bici-c1h^+az(b1c2-c1b3>)

- <*i (b2 c3 b3 c2) - br (oj c3 - a3 c2) + ct (a2 b3 a3 b2) = D.

[This theorem is extremely useful and proves that any property which is true in respect of rows is also true in respect of columns and vice versa. As in article 4, the determinant, therefore, can also be expressed in terms of the elements of the first column].


Property n . If two rows (or columns) of a determinant be interchanged, the determinant is unaltered in numerical value but is changed in sign only.

Let, with the usual notation, the given determinant be D. Interchange the first two rows. Let the resulting determinant be denoted by D', so that

a2 b2 c2

D' - «i h Cl

a3 b3

We have to prove that D' = - D. By definition

D = a2(bic3-b3c1)-b2(a1e3-a3 c t ) + c2 ( fli 63 - a3&i) =S - [ at ( b2c3 - hc2 ) - h i a2<- 3 ~ a3c2 ) + c i ( a2b3 - < J 3 b j ) ]

at h Cl

= — <h h c2

% h c3 For example, we have

1 4 17 4 1 17 5 2 18

2 5 18 =s — 5 2 18 = 4~ 1 17

3 6 19 6 3 19 6 3 19

(interchanging the first two columns in the first step and interchanging the first two rows in the second step).

Property III. If a determinant has two rows (or columns ) identical, its value is zero.

Let D' denote a determinant having first two rows identical, so that

h Cl

D = at h

03 C3

We have to prove that D' = 0.


Interchanging the first two rows D' changes into — D' by Property II. But here, in the case of this determinant, the actual interchange of the first two rows keeps D' unaltered.

.*. we have D' = - D'. .\ 2D' = 0, D' = 0 .

This can also be verified by actually expanding the deter-minant D .

For example, we have 1 1 x

( i ) 2 3

2

3

= 0 , ( i i )

13 17

a b

19

13 17 19

= 0.

Property IV. If all the elements of any row (or column) be multiplied by the same factor, then the determinant will be multiplied by that factor.

Let, with the usual notation, Z> be the given determinant. Multiply all elements of the first row by a factor k. Let the resulting determinant be denoted by D \ so that

ka-i kb-i key

D' = «2

<•h

c-t

We have to prove that D' = kD. By definition,

Z>' =kax (b2 c3 - b3 c2) - kbx (a2 c3 - a3 c2) + kct (a2 b3 - a3 b2) = k [at(b2c3 - b3c2) - bt (a2e3 - a3c2) + ca (a2 b3 - a3 b2% = kD.

It can be seen from this property that if all the elements of any row ( or column ) have a common factor k, then the factor k can be taken outside, as a common factor of the determinant.

For example, we have 6 9 15 2 3 5

( i ) 23 29 31 = 3 23 29 31

a b c a b c


1 2 3

7 11 13

17 19 23

1 2 3

21 33 39

17 19 23

( i i ) I f D = 7 11 13 then 5D =

also D =

5 10 15

7 11 13

17 19 23

Property V. If each element of any row ( or column ) can be expressed as a sum of two terms, then the determinant can be expressed as the sum of two determinants.

Let D' be a determinant in which each element of the first row is expressed as a sum of two terms, so that

+ xi h + }'i c \ + z \

D' = a2 b2

h

C 2

a3 b* c,

We have by definition, D' = (at + xt) (b2c3 - b3c2) - (bx + yx) ( a 2 c 3 - a 3 c 2 ) + ( c j + ^ j )

(a2b3-a3b2)

= [ fi (b2c3-b3 c2)~ bx ( a 2 c3 - a 3 c2 ) + c ^ f o f>3 - a3 b2~) ] + [xt(b2c3-b3 c2) - (a2 c 3 - a 3 c2) + zi ( a 2 - a 3 6 2 ) ]

J i

a2 b2 c2

Ol

a2

a3

h

b2

b3

c 1

C,

+ yi

b2

a3 h c3

Conversely, it can be easily seen that the sum of two deter-minants differing only in one row or column can be expressed as one determinant of the same order.

For example, we have x y z

I m

w

+ X V z

I m' n

U V W

x y + y z

I m+m' n

u v + v' w


I t can be seen that if every element of two rows or of three rows is expressed as a sum of two terms, then the deter-minant can be expressed as a sum of 22 i. e. 4 determinants or of 23 i. e. 8 determinants.

Property VI. The value of the determinant is not altered by adding to the elements of any row (or column) the same multiples of the corresponding elements of any other row (or column ).

Let, with the usual notation, D be the given determinant. Multiply all the elements of the second row by m and add them to the corresponding elements or the first row. Let the resulting determinant be denoted by D\ so that

at + ma2 + mb2 + mc2

D = a2

We have to prove that D' = D. We have, by Property V,-

Ol h ma2 mb3 mc2

D' = a2 b2 C2 + a2 b2 C2

b3 Ci h C 3

D + m

u2

b2

h

C-i [ by Property IV ]

[ by Property III ] = D + m (0) = D.

By a simple extension of the above property, we can see that at + ma2 bt + mb2 ct + mc2

D which = D\ namely,

«3

is also =

a j + ma2 + na3 bt + mb2 + nb3 ct + mc2 + nc3

a2 b2 c2

a3 b3 c3


We may, similarly, show that + ma2 bt + mb2

D =

[ + m<\

a2

a3 + not b3 + nbi c3 + nc t

We note that in any one application of this property at least one row ( or co lumn) must remain unchanged and further if the multiples of the elements of any column ( or row ) are added to the elements of any other column ( or row) , then the multiples of this second column ( or row ) should not be added to the first at the same time.

For example, we have

t. i ) 1964

1965

1966

1967

1964

1

1966

1 = - 2 .

1 1 1 0 1 1 0 0 1

( i i ) 11 10 9 - 1 10 9 1 1 9

101 100 99 1 100 99 1 1 99

= 0 ;

[ In (ii) we subtract the elements of the second column from the corresponding elements of the first column in the first step and subtract the elements of the third column from the corres-ponding elements of the second column in the second-step ). 8. Illustrative Examples:

W e now illustrate application of the above properties, in evaluating determinants, by the following examples.

20 27 36

Ex. 1. F ind the value of | 2 3 4

1 2 3 Multiply the elements of the second row by 9 and subtract them from t h e

corresponding elements of the first row. W e may briefly write this as i?i - 9

2 0 - 1 8 2 7 - 2 7 3 6 - 3 6

the given determinant

= 2 ( 9 - 8 ) - 0 ( 6 - 4 ) + 0 ( 4 - 3 ) = 2.


Ex. 2, Find the value of

11

12

1 3

12

1 3

1 4

13

1 4

15

Multiplying all the elements of the first row by - 1 and adding them to the corresponding elements of the second and third rows, ( w e may briefly indicate this as Ra - Ri and R s — R l respectively.)

= 2 x 0 = 0,

= 0.

Let a denote the determinant on the left hand side of the equation. W e will use the symbol A , in this chapter, for a determinant. Multiplying all the elements of the third column by - 1 and adding them to the corresponding elements of the second column ( say C.3 — C 3 ) , we have

1 1 1 2 1 3 1 1 1 2 1 3

Determinant = 1 1 1 = 2 1 1 1

2 2 2 1 1 1

1 4 - 2 * 1 0 9

Ex. 3 . Solve the equation 1 0 - 3 * 1 6 1 5

6 -x 1 3 1 2

1 4 - 2 *

A = | 1 0 - 3 *

0-x

9

1 5

12

( by Property V I )

Multiplying all the elements of the first row by - 1 and adding them to the corresponding elements of the second and third rows, we get

1 4 — 2 * 1 9

A = — 4 — * 0 6

- 8 + * 0 3

Interchanging the first two columns, we get

1 1 4 - 2 * 9

A = — 0 -4 — x 6

0 — 8 + * 3

( b y Property VI )

( by Property I I )

Since the interchange of rows into columns and vice versa does not alter the value of the determinant, we can as well expand the determinant in terms of the elements of the first column.


4 = - [ l | ( - 4 - * ) ' 3 - ( - S + * ) ' 6 j ]

= - [ - 12 - 3x + 48 - 6* ]

= - [ - 9* + 3 6 ] .

.'. the given equation A = 0 reduces to 9x - 36 = 0.

.*. x = 4 is the required solution.

Ex. 4 . Show that

= 3abc - a ' - b3 - o

The L. H . S. determinant say, A , can be expressed as a sum of four determinants.

b — 6 a

b + c a — b a

c + a 6 - c b

a+b c — a c

A =

a

b

o

.". A =

b a a

a b b

a c c

The I _ s first three determinants on the R . H . S. vanish as they have two of

their columns identical,

A = - } o ( c a , c a - ab ) - b (aa - 6 a ) + a (a1 - be) \ = - ( a» + 6" + c" - 3a6o) .

/ . A = 3abo - a* - b*- c 8 .

Ex. 5 . Show that a + b + 2c a b

a b + c + 2a b = 2 ( ® + & + c ) 8

c a c + a + 2b

Add the second and third columns to the first; then

2 ( « + & + c ) a b

b + o + 26 6

a c + a + 26

a 6 1 6 + c + 2a 6

1 a c + a + 2 6

2 (a + b + c)

2 (a + 6 + o) 1

=2 (a + b + c)


Subtract the first row from the second and third ; then l a b

0 a + b + c 0

0 0 a + b + o

A = 2 (a + b + o)

A = 2 ( a + 6 + c )*, on expanding the determinant in terms of the elements of the first colnmn.

Ex. 6. Factorize a b + c a8

b a + a

c a+b c'

Adding the elements of the first colnmn to the corresponding elements of the second colnmn, we get

a a + b + a

A = {a + b + o)

b a + b-t-c

a a + b + c

a 1 a*

b 1 6*

a• 6* r«

C 1 C1

A =• ( a + b + c ) A ' , say.

I f a «« b, b => o or c — a be substituted in A ' , then two rows become identical and hence A ' vanishes,

/ . (a — b), (b — c), (c — a) are factors of A .

Now A ' is of the third degree in a, b, c and hence i t cannot have a fourth factor except a constant, say k.

A ' = fe(o-b) ( J - c l ( c - 4 ... ( 1 )

By comparing the coefficient of any term, say «ca. from both sides w« have 1 = — ft.

Otherwise, to find the valne of k give convenient values to a, b, a la ( i ) , say, a - 0, 6 = 1 and c = - 1.

.'. relation ( i ) reduces to 0 1 0

1 1 1 - * (0 - 1) (1 + 1) ( - 1 - 0)

- 1 1 1 i. e - l ( l + l ) = * ( 2 ) l . e . - 2 - 2 * * = - 1.

A = - (« + * + c ) (« - 6 ) (b-a ) (o - a).

DETERMINANTS : 349

Exercise 13 ( b )

1. In the following determinants, express two of the elements in the first row as zero and hence evaluate them.

( i )

1 1 1

10 11 12

1000 1001 1002

1 1 1

x + 2 * + 3 JC+4

y+5 y+6 y + 7

2. In the following determinants, express any two rows or columns so as to have the same elements and hence find their values.

10 13 16

( i )

1 1001 17

3 3G03 19

5 5005 23

30 33 36

47 50 53

ap + bg ar + bs

cp + dq cr + ds

as the sum of four determinants and hence prove that it equals

3. Write

4. If D =

in terms of D.

a b p 9 X •

c d r s

1 X X2 x2 y> z2

1 y y* , express X y z

1 z z2 2 2 2

5. Without expanding the determinant, show that 1 1 x

1 x'

x3

= 0.

Verify the result by actually expanding the determinant


6. Without expanding or evaluating any determinant prove that

1 2 3

2

3

1

2

2

1

- 2

3

- 2

1

Further express the right hand side as a determinant in which two elements of the first column become zero and hence show that the value of the determinant is equal to — 18.

7. Verify, without expanding any determinant, that 7 2 5

- 9

15

-3 1

4 8

- 1

5

- 3

2 5

•3 1

4 8

= 2

3 2 5

- 2 - 3 1

6 4 8

State the properties about determinants used in the above proof.

8. Without direct expansion, prove that 0 b c

( i ) -b 0 a = 0

— c — a 0

0 a-b a - c

( i i ) b-a 0 b — c

c — a c-b 0

= 0.

Verify the results by actually expanding the determinants. [ H i n t ; Take ( — 1 ) common from Ci, C2, C 3 so that

D' = ( — 1 )3 £>.' Interchange columns into rows in D' so that D' = D].

Without expanding any of the determinants in examples 9 to 18, show that :

a b c X y z y b q

9. X y z = P q r = X a P

P <7 r a b c z c r

DETERMINANTS : 351

ax by cz a b c

10. x» y2 z2 = X y z

1 I 1 yz zx xy

be a a2 1 a* a3

11. ca b b2 1 62 63

ab c c2 1 c2 c3

b + c c+a a+b a

12. a + b b + c c + a = 2 c

c + a a+b b+c b

j y-z y + z y

13. z—x z + x z

j

1 = 0.

x-y x+y X

a + 1 a + 2 a + 3

14. a + 4 a + 5 a+ 6 = 0.

a + 1 a + f> a + 9

p + a P+b p + c

15. q + a q + b q + c = 0.

r + a r + b r + c

1 1 - x 11 10

16. 1 7 - x 17 16 = 0.

1 4 - x 14 13

j 4 11 10 5 11 10

17. I 7 2 6 = 3 3 2 6 = 0.

1 1 5 4 2 5 4


a7 + 2ab a 1

18. 3b* b 1

c2 + 2bc c 1

a2

b2

a

b

c

19, Solve the following equations:—

3 5 - 7

( i )

(H)

(Hi)

1 9 31

9 15 2x + l

x + 2 x + 6

x+6 x—l

x-1 x+2

a — x c b

c b — x a

b a c—x

= 0,

x - 1

x + 2

x + 6

= 0.

= 0,

Find the factors of the following determinants.

1 a a 2 1 a be

20. 1 b b* 21. 1 b ca

1 c c2 1 c ab

1 1 1 a b c

22. a b c • 23. a2 b2 c2

a 3 b3 c3 be ca ab

Without developing the determinants, as far as possible, prove that

a-b-c 2a 2a

24. 2b b — c-a 2Jb

2c 2c c-a-b

( a + fc + c ) 3 .

26.

27.

28.

29.

30.

31.

32.

C. A. 23


a a — c a-b

b c + a b-a

c c — a a+b

1 X X2 -yz

1 y y2 — zx

1 z z2 -xy

= 4abc.

= 0.

y + z z y

Z Z + X X

y x x + j>

1 + a 1 1

1 1 + 6 1

1 1 1 + c

a — b b—c

= 4 xyz.

— abc

c — a

a-b

b-c

r i + _ l + i + ± i 1 a b c J

= 0. b—c c—a

c—a a—b

1 a b + c

1 b c + a = 0.

1 c a + b

O - l ) 2 x2+l x

( . y - i y y2 +1 y = 0 .

( z _ 1 )2 22 + 1 z

a 2 + 2a 2 a + 1 1

2a + 1 a + 2 1 = ( a - l ) . 3

3 3 1


1 + a2 — b2 2ab

33. 2 ab 1 -a2 + b2

2b -2a

x2 x

34. If J>2 y —

Z3 + l z2 z

-2b

2 a

l-a2-b2

-=(1 + a2+b2)\

= 0 and x, y, z are all different.

prove that xyz = - 1.

35. Solve the following equations using determinants.

( i ) 2x + y-z = 2, x-2y + z = 5, x+y + 2z = 3: ( i i ) 2x + 3y-z-2 = 0,x + 2y + z + 3 = 0 .

3x + y-2z-l = 0.


4 -2D. 1. ( i ) 0. ( u ) 0. 2 . ( i ) 0. ( i i ) 0.

19. ( i ) * - 10. ( i i ) y = - 7/3.

( i i i ) x - a + b + o, * « ± V a '+f t '+o" — ab- bc — ca.

2 0 . + (a-b) (b-c) (c-a). 2 1 . ( a - b ) (b-c ) {o-a).

22. {« - i ) ( 6 - c ) (c - a) (« + 6 + o). 2 3 . (a-b) (b-c) (c-a) (ab+bc+ca).

35. ( i ) * = + 2 , y = - 1 , * - 1; ( i i ) * = - 2 , y = 1, * = - 3

Appendix 1 LOGARITHMIC TABLES

0 1 2 3 4 S 6 7 8 » Mean Differences

0 1 2 3 4 S 6 7 8 » 1 2 3 | 4 3 6 7 8 9

10 0000 0043 0086 0128 0170 0212

0645

0294

0719

0374 5 9 13 4 8 12

17 21 26 16 20 24

30 34 38 28 32 36

11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755

4 8 12 4 7 11

15 19 23 15 18 22

27 31 35 26 29 33

12 0792 0828 0864 0899 0934 0969 1004

1335

1038 1072 110$ 4 7 11 3 7 10

14 18 21 U 17 20

25 28 32 24 27 31

U

1 *

1139 1173 1206 1239

1553

1271

ISM 1303

1004

1335 13SL

« 7 3

1399

1703

1430 3 6 10 3 6 10

13 16 19 13 16 19

23 26 30 22 25 29

U

1 * 1461 1492 1523

1239

1553

1271

ISM 1614 1644

1931

2203

13SL

« 7 3

1399

1703 1732 3 6 9 3 6 9

12 15 19 12 15 17

22 25 28 20 23 26

IS 1761 1790 1818 1847 1875 1903

1644

1931

2203

1959 1987 2314

2279

3 6 9 3 6 8

11 14 17 11 14 17

20 23 26 19 22 25

16 2041 2068 2095 2122 2148

2405 m

2430

2672

1644

1931

2203 2227 2253

2314

2279 3 6 8 3 5 8

11 14 16 10 13 16

19 22 24 18 21 23

17

I s

T f

2301 2330

2577

2355 2380

2148

2405 m

2430

2672

2455

2695

2430 2504 2529 3 5 8 3 5 8

10 13 15 10 12 15

18 20 23 17 20 22

17

I s

T f

2553

2330

2577 2601 2625 '2648

m

2430

2672

2455

2695 2718 2742 2765 2 5 7 2 4 7

9 12 14 9 11 14

16 19 21 16 18 21

17

I s

T f 2788 2810 2833 2856 2878 2900 3 9 p 2745 2967 J989

2 4 7 2 4 6

9 11 13 8 11 13

16 IS 20 15 17 19

26 21 22 t s 24

25 26:

28

30 81 32 33 34

35 36 37 38 39

40 41 42 43 44

43 46 47 48 49

3C10 3222 3424 .3617 3802

3979 4150 43M 4472 4 6 *

4771 4914; 5051! 5185 5315

5441 5563 5682 5798 5911

6021 612S 6232 6335 6435

6532 6628 6721 6812 6902

3032 3213 3444 3636 3820

3997 4166 4330 4487 4639

4786 4928 5065 '5198 <5328

5453 5375 5694 58(9 5922

60S 6138 6243 6315 6444

6542 6637 6730 6S21 6911

3054 3263 3464 3656 <W38

4014 4183 4346 4582 4654

4806 4942 5079 5211 5340

5465 5537 5705 5821 5933

6012 6149 6253 6355 6454

6551 6646 6739 6839 6920

3.175 3231 3483 3674 3856

4031 4200 4362 4518 4669

4814 4955 5092 5234 5353

5478 5599; 5717 5832 5944

6053 616.1 6263 6365 6464

6561 6656 6749 6839 6928

30% 3394 3502 3692 3874

4348 4216 437S 4533 4683

4829 4969 5105 5237 5366

5490 5611 S729 i843 S955

6064 61% 6271 6375 6471

6571 6665 6758 6843 6937

3118 3324 3522 371? 3892

4065 4232 4393 4548 4698

4843 4983 5119 5250 5378

5502 5623 5740 5855 5966

6075 61 SO 6284 6335 64a4

6580 6675 6767 6857 6946

3139 3315 3511 3729 3909

4082 4249 4409 4564 4713

4857 4997 5132 5263 53)1

5514 5635 575* 5366 5977

6085 6191 6294 63 5 6193

6590 6684 6776 6866 6955

3160 J36S 3560 3747 3927

4,99 4265 4425 4579 472J

4871 5011 5145 5276 5403

5527 5647 5763 5877 5988

60% 6201 6304 6405 6503

6599 6693 6785 6S75 6961

3181 33S5 3579 3766 3945

4116 4281 4440 »594 4742

4886 5024 5159 5289 5416

5539 5658 5775 5838 5999;

6107 6212 6314; 6415 : 6513 :

6609 6702 6794 6884 6972

3201 3404 3598 3784 3%2

4133 4298 445$ 4609 4757

190.) 5038 5172 5302 5428

5551 5670 5786 589; 6010

6117 6222 6323 6425 6522

6618 6712 6803 6893 6981

2 4 6 2 4 6 2 4 6 2 4 6 2 4 5

2 3 5 2 3 5 2 3 5 2 3 5 1 3 4

1 3 4 1 3 4 1 3 4 1 3 4 1 3 4

1 2 4 1 2 4 1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

8 11 13 8 10 12 8 10 12 7 9 11 7 9 11

7 9 10 7 8 10 6 8 9 6 8 9 6 7 9

6 7 9 6 7 8 5 7 8 5 6 8 5 6 8

5 6 7 5 6 7 5 6 7 5 6 7 4 S 7

4 5 6 4 5 6 4 5 6 4 5 6 4 5 6

4 5 6 4 5 6 4 5 5 4 4 5 4 4 S

15 1719 -14 16 18 14 15 17 13 15 17 12 14 16

12 14 IS 11 13 15 11 13 14 11 12 14 10 12 13

10 11 13 10 11 12 9 11 12 9 10 12 9 10 11

9 10 11 8 10 11 8 9 10 8 9 10 8 9 10

8 9 10 7 8 9 7 8 9 7 8 9 7 8 9

7 8 9 7 7 8 6 7 8 6 7 8 6 7 8

355


0 1 2 3 4 6 6 7 8 9 1 2 3 4 S 6 7 8 9

50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 3 3 4 5 6 7 g 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 3 4 5 6 7 8 S2 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 4 5 6 7 7 S3 7243 7251' 7259 7267 7275 7284 7292 7300 7303 7316 1 2 2 3 4 5 6 6 7 54 7324 7332 7340 7348 7356 7364 7372 7380 7383 7396 1 2 2 3 4 5 6 6 7

SI 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 3 4 5 5 6 7 56 7482 7490 7497 7505 7513 752P 7528 7536 7543 7551 1 2 2 3 4 5 5 6 7 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 3 4 5 5 6 7 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 3 4 4 5 6 7 59 7709 7716 7723 7731 7733 7745 7752 7760 7767 7774 1 1 2 3 4 4 5 6 7

60 61

7782 T789 77% 7833 7810 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 b £ 60 61 7853 7860 7868 7875 7883 7889 7896 7903 7910 7917 1 1 2 3 4 4 5 6 0 < 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7937 1 1 2 3 3 4 5 6 o A 63 7993 8000 8007 8014 8021 8028 8035 8041 8348 8055 1 1 2 3 3 4 5 5 o 64 8062 8069 8075 8082 8089 8096 8102 8109 3116 8122 1 1 2 3 3 4 5 5 D

6

« 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 1 2 3 3 4 5 5 66 8195 8202 8209 8215 8222 8228 8235 8241 8748 8254 •i 1 2 3 3 4 5 5 o •7 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 1 2 3 3 4 5 5 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 3 3 4 4 5 6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 2 3 4 4 5 6

6

70 8451 8457 8463 8470 8476 8482 8488 8494 8501 8506 1 1 2 2 3 4 4 5 ^ 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 2 3 4 4 5 72 8573 8579 8545 8591 8597 8603 8609 8615 8621 8627 1 1 2 2 3 4 4 5 ^ 71 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 2 3 4 4 5 5 74 8692 869" 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 2 3 4 4 5 5

75 8751 8756 8262 8768 8774 8779 8785 8791 8797 8802 1 1 2 2 3 3 4 5 e 76 8808 8814 8821' 8825 8831 8837 8842 *848 8854 8859 1 1 2 2 3 3 4 5 77 8865 8871 8876 8*82 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 5 e 78 8921 8927 8932 8938 8943 8949 8>54 8960 8965 8971 1 1 2 2 3 3 4 4 9 e 79 8976 8982 8981 8993 8998 9004 9009 5015 9020 9025 1 1 2 2 3 3 4 4 9 5

<0 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 1 2 2 3 3 4 4 c 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 1 2 2 3 3 4 4 J 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 1 2 2 3 3 4 4 83 9191 9196 92 1 9206 9212 9217 9222 9227 9232 9238 1 1 2 2 3 3 4 4 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 1 2 2 3 3 4 4 5

5

85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 3 3 4 4 5 86 9345 9350 9355 9360 9365 9370 9375 9330 9385 9390 1 1 2 2 3 3 4 4 5 87 9395 9400 94U5 9410 9415 9420 942S 9430 9435 9440 0 1 1 2 2 3 3 4 4 88 9445 9450 9155 9460 9465 9469 9474 9479 9434 9489 0 1 1 2 2 3 3 4 4 8* 9494 9499 9504 9509 9513 951.5 9523 9528 9533 9538 0 1 1 2 2 3 3 4 4

to 9542 9547 9S52 9557 9562 9566 9571 9576 9531 9586 0 1 1 1 2 3 3 4 4 91 9591 9595 3601) 9605 9609 9614 9619 9624 %:s 9633 0 1 1 2 2 3 3 4 4 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 968!) 0 1 1 2 2 3 3 4 4 93 9685 9689 9694 9699 97)3 9708 9713 9717 9722 9727 0 1 1 2 2 3 3 4 4 •4 9721 9736 9741 9745 9750 975» 9759 9763 9768 9773 0 1 1 2 2 3 3 4 4

95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0 1 1 2 3 3 4 4 96 9823 9827 9832 9836 9841 9645 9850 9854 9859 9863 0 1 1 2 2 3 3 4 4 97 9868 9872 9877 9881 9686 9890 9894 9899 9903 9908 0 1 1 2 2 3 3 4 4 98 9912 9917 9921 9926 9930 99H 9939 9943 9948 9952 0 1 1 2 2 3 3 4 4 99 9956 9961 9965 9969 9*74 9978 9963 9987 9901 9996 « 1 1 2 2 3 3 3 4

ANTI-LOGARITHMIC TABLES

0 1 t 3 4 5 6 ) 8 • 1 2 3 4 5 6 7 8 1

•00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 0 0 1 1 1 1 2 2 2

•01 1023 1026 1028 1030 1033 1035 1038 1040 1012 1045 0 0 1 1 1 2 2 2 •02 1047 1050 1052 1054 1057 1059 1062 1061 1067 1069 0 0 1 1 1 2 2 2 •03 1072 1074 1076 1079 1081 1081 1086 1089 1091 1094 0 0 1 1 1 2 2 2 •04 10% 1099 1102 1101 1107 1109 1112 1111 1117 1119 0 1 1 1 2 2 2 2 •OS 1122 1125 1127 1130 1132 1135 1138 1140 1143 1116 0 1 1 1 2 2 2 2

•06 1148 1151 1153 1156 1159 1161 1164 1167 1169 1172 0 1 1 1 2 2 2 2 •07 1175 1178 1180 1183 1186 1189 1191 1194 1197 1199 0 1 1 1 2 2 2 2 •OR 1202 1205 1208 1211 1213 1216 1219 1222 1225 1227 0 1 1 1 2 2 2 3 •09 1230 1233 1236 1239 1212 1215 1247 1250 1253 1256 0 1 1 1 2 2 2 3 •10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 0 1 1 1 2 2 2 3

•11 1288 1291 1294 1297 1300 1303 1306 1309 1312 1315 0 1 1 2 2 2 2 3 •12 1318 1321 1324 1327 1330 1331 1337 1310 1313 1316 0 1 1 2 2 2 2 3 •IS 1349 1352 1355 1358 1361 1365 1368 1371 1374 1377 0 1 1 2 2 2 3 3 •14 1380 1384 1387 1390 1393 1396 1400 1103 1406 1409 0 1 1 2 2 2 3 3 • is 1413 1416 1419 1422 1126 1129 1432 1135 1439 1112 0 1 1 2 2 2 3 3

•1« IMS 1449 1452 1455 1159 1462 1466 1169 1472 1476 0 1 1 2 2 2 3 3 •17 1479 1483 1486 1489 1193 14% 1500 1503 1507 1510 0 1 1 2 2 2 3 3 •18 1514 1517 1521 1524 1528 1531 1535 1538 1512 1545 0 1 1 2 2 2 3 3 •19 1549 1552 1556 1560 1563 1567 1570 1574 1578 1581 0 1 1 2 2 3 3 3 •20 1585 1589 1592 15% 1600 1603 1607 1611 1614 1618 0 1 1 2 2 3 3 3

•21 1622 1626 1629 1633 1637 1641 1644 1618 1652 1656 0 1 2 2 2 3 3 3 •22 1660 1663 1667 1671 1675 1679 1683 1687 1690 1694 0 1 2 2 2 3 3 3 •23 1698 1702 1706 1710 1711 1718 1722 1726 1730 1734 0 1 2 2 2 3 3 1 •24 1738 1742 1746 1750 1751 1758 1762 1766 1770 1771 0 1 2 2 2 3 3 4 •25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 0 1 2 2 2 3 3 4

•26 1820 1821 1828 1832 1837 1841 1815 1819 1851 1858 0 1 2 2 3 3 3 4 •27 1862 1866 1871 1875 1879 1881 1888 1892 1897 1901 0 1 2 2 3 3 3 4 •28 1905 1910 1914 1919 1923 1928 1932 1936 1911 1915 0 1 r 2 3 3 4 4 •29 1950 1954 1959 1963 1968 1972 1977 1982 1986 1991 0 1 2 2 3 3 4 4 •30 1995 2000 2004 2009 2011 2018 2023 2028 2032 2037 0 1 2 2 3 3 4 4

•31 2042 2046 2051 2056 2061 2065 2070 2075 2080 2084 0 1 2 2 3 3 1 4 •32 2089 2094 2099 2101 2109 2113 2118 2123 2128 2133 0 1 2 2 3 3 4 4 '33 2138 2143 214$ 2153 2158 2163 2168 2173 2178 2183 0 1 2 2 3 3 1 4 •34 2188 2193 2198 2203 2208 2213 2218 vm 2228 2231 1 1 2 2 3 3 1 4 5 'S3 2239 2244 2249 2254 2259 2265 2270 zas 2280 2286 1 1 2 2 3 3 1 1 5

•36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 1 2 2 3 3 1 1 S •37 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 1 1 2 2 3 3 1 4 5 •38 2399 2404 2410 2415 2121 2427 2132 2438 2143 2119 1 1 2 2 3 3 4 4 5 •39 2455 2460 2466 2172 2477 2483 2489 2495 2500 2506 1 1 2 2 3 3 1 5 S •40 2512 2518 2523 2529 253S 2541 2547 2S53 2559 2561 1 1 2 2 3 1 4 5 5

•41 2570 2576 2582 2588 2S94 2600 2606 2612 2618 2624 1 1 2 2 3 1 4 5 5 •42 2630 2636 2642 2619 2655 2661 2667 2673 2679 2685 1 1 2 2 3 4 4 5 6 "43 2692 2698 2704 2710 2716 2723 2729 2735 2742 2748 1 1 2 3 3 1 1 5 6 •44 2754 2761 2767 2773 2780 2786 2793 2799 2805 2812 1 1 2 3 3 1 4 S 6 •45 2818 2825 2831 2838 2811 2851 2853 2861 2871 2877 1 1 2 3 3 1 5 5 6 •46 2884 2891 2897 2901 2911 2917 2921 2931 2938 2911 1 1 2 3 3 1 5 5 6 "47 2951 2958 2965 2972 2979 2985 2992 2999 3006 3013 1 1 2 3 3 4 S 5 6 •48 3020 3027 3034 3011 3048 3055 3062 3069 3076 3083 1 1 2 3 1 1 5 6 6 •49 3090 3097 3105 3112 3119 3126 3133 3141 3148 3155 1 1 2 3 4 4 5 6 6

357


0 1 X 3 4 5 6 7 8 9 I X 3 4 5 6 7 8 9

-to 3162 3170 3177 3184 3192 3199 3206 3211 3221 3228 1 1 2 3 4 4 5 6 7

•51 3236 3243 3251 32S8 3266 3273 3281 3289 3296 3301 1 2 2 3 4 5 5 6 7 •51 3311 3319 3327 3334 3342 3350 3357 3365 3373 3381 1 2 2 3 4 5 5 6 7 53 3388 3396 3104 3412 3120 3428 3136 3413 3151 3459 1 2 2 3 4 5 6 6 7 '54 3467 3475 3483 3191 3499 3508 3516 3521 3532 3510 1 2 2 3 4 5 6 6 7 '55 3548 3556 3565 3573 3581 3589 3597 3606 3611 3622 1 2 2 3 4 5 6 7 7

56 3631 3639 3648 3656 3664 3673 3681 3690 3698 37U7 1 2 3 3 4 5 6 7 8 '57 3715 3724 3733 3711 3750 3758 3767 3776 3784 3793 1 2 3 3 4 5 6 7 8 '58 3802 3811 3819 3828 3837 3816 3855 3861 3873 3882 1 2 3 4 4 5 6 7 8 '59 3890 3899 3908 3917 3926. 3936 3915 3954 3963 3972 1 2 3 4 5 5 6 7 8 to 3981 3990 399? 1009 4018 1027 1036 4046 1055 1061 1 2 3 4 5 6 6 7 8

•61 4074 4083 4093 1102 4111 1121 1130 4140 1150 4159 1 2 3 1 5 6 7 8 9 '62 4169 4178 4188 1198 4207 1217 4227 4236 1216 1256 1 2 3 1 5 6 7 8 9 '63 4266 4276 4285 4295 4305 1315 4325 4335 4345 4355 1 2 3 4 5 6 7 8 9 64 4365 4375 4385 4395 4406 1116 4426 4436 4446 4457 1 2 3 4 5 6 7 8 9 '65 4467 4477 4487 4498 4508 1519 4529 4539 4550 1560 1 2 3 4 5 6 7 8 9 '66 4571 4581 4592 4603 4613 1624 4631 4615 4656 4667 1 2 3 4 S 6 7 9 10 67 4677 4688 4699 4710 4721 4732 4742 4753 1761 1775 1 2 3 1 5 7 8 9 10 '68 4786 4797 4808 4819 4831 4812 4853 4864 1875 4887 1 2 3 4 6 7 8 9 10 *6* 4898 4909 4920 4932 4943 1955 1966 4977 1989 5000 1 2 3 5 6 7 8 9 11 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 4 5 6 7 8 9 1 1 •71 5129 5140 5152 5164 5176 5188 5200 5212 5221 5236 1 2 4 5 6 7 8 10 11 •72 5248 5260 5272 5281 5297 5309 5321 5333 5346 5358 1 2 4 5 6 7 9 10 11 •73 5370 5383 5395 5106 5420 5133 5445 5458 5470 5183 1 3 4 5 6 8 9 10 11 •74 5495 5508 5521 5531 5516 5559 5572 5585 5598 5610 1 3 4 5 6 8 9 10 12 '75 5623 5636 5619 5662 5675 5689 5702 5715 5728 5741 1 3 4 5 7 8 9 10 12 •76 5754 5768 5781 5794 5808 5821 5831 5848 5861 5875 1 3 4 5 7 8 9 11 12 •77 5888 5902 5916 5929 5943 5957 5970 5984 5998 6012 1 3 4 5 7 8 10 11 12 78 6026 6039 6053 6067 6081 6095 6109 6121 6138 6132 1 3 4 6 7 8 10 11 13 •79 6166 6180 6191 6209 6223 6237 6252 6266 6281 6295 1 3 4 6 7 9 10 11 13 "80 6310 6324 6339 6353 6368 6383 6397 6112 6427 6142 1 3 4 6 7 9 10 12 13 •81 6457 6471 6486 6501 6516 6531 6546 6561 6577 6592 2 3 5 6 8 9 11 12 11 82 6607 6622 6637 6653 6668 6683 6699 6711 6730 6715 2 3 5 6 8 9 11 12 U 'S3 6761 6776 6792 6808 6823 6839 6855 6871 6887 6902 2 3 5 6 8 9 11 13 14 "84 6918 6934 6950 6966 6982 6998 701S 7031 7047 7063 2 3 5 6 8 10 11 13 15 •85 7079 70% 7112 7129 7145 7161 7178 7194 7211 7228 2 3 5 7 8 10 12 13 IS •86 7244 7261 7278 7295 7311 7328 7345 7362 7379 73% 2 3 5 7 8 10 12 13 15 87 7413 7430 7447 7164 7482 7499 7516 7534 7551 7568 2 3 5 7 9 10 12 U 16

•88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 2 4 5 7 9 11 12 :1 16 •89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 2 4 5 7 9 11 13 11 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2 4 6 7 9 11 13 15 17 •91 8128 8147 8166 8185 8204 8222 8211 8260 8279 8299 2 4 6 8 9 11 13 15 17 •92 8318 8337 8356 8375 8395 8414 8433 8453 8472 8492 2 4 6 8 10 12 14 15 17 •93 8511 8531 8551 8570 8590 8610 8630 8650 8670 8690 2 4 6 8 10 12 11 16 18 •94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 2 4 6 8 10 12 14 16 18 •95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2 4 6 8 10 12 15 17 19

•96 9120 9141 9162 9183 9204 9226 9247 9268 9290 9311 2 4 6 8 11 13 15 17 19 97 9333 9354 9376 9397 9419 9441 9162 9481 9506 9528 2 4 7 9 11 13 15 17 20

'98 9550 9572 9594 %16 9638 9661 9683 9705 9727 9750 2 4 7 9 11 13 16 18 20 '99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 S 7 9 11 14 16 18 20

Appendix 2

Impor tant Formulae F r o m School Algebra.

1. ( a + b )2 = a2 + 2ab + b2

2. (a- by = a2 - lab + b2. 3. ( a + b ) ( a - b) = a2 - b2. 4. a2 + f>2 = ( a + b)2 - lab = ( a - b)2 + 2ab. 5. ( a + 6 ) 2 = ( a - 6 ) 2 + 4afc. 6. ( a - 2>)2 = (a + b)2 - 4ab. 7. (x + a) (x + b) = x2 + (a + b)x + ab. 8. ( a + fc)3 = a"* + 3a26 + lab2 + b* = a3 + b3 + 3ab (a + b).

9. ( a - ft)3 = a 3 - 3 a 2 b + 3ab 2 -b* = ( a - 6 ) . 10. a3 + £< = ( a + 6 ) (a2-ab + b2). 11. = ( a - 6 ) (a2 + ab + b2). 12. ( a + b + c )2 = a2 + 62 + c2 + lab + 2bc + lea 13. a 3 + i>3 + c3 — 3abc = ( a + 6 + c ) X

( a2 + i 2 + c2—ab — bc — ca) . 14. If a + b + c = 0, then a3 + Z>3 + c3 = 3abc. 15. ai + a2b2 + bi = (a2 + ab + bi) ( a 2 - a & + 62 ,). 16. ( a + 6 + c ) ( 6 + c - a ) ( c + a - 6 ) ( a + 6 - c )

=2a2b2 + 2b2c2 + lc2a2 - a4 - b* - c*.

17. If 4 " <= 4 r , then — = — • (Invertendo ). b a a c

18. = then -fL = A . (Alternendo ).

19. If 4 - = 4 - - t h e n = ' (Componendo ) . b a b a

20. If = , then = . ( Dividendo ). b a b a

359


21. I f A ' t h e n b a a — b c — d

( Componendo and Dividendo )• a c c

22. I f - ^ - = - j — - y — then each of the equal fractions

— fa + /«c + »e + ~ lb + md+nf +

where /, m, n , . . . are + ve or — ve numbers.

Important Formulae and Results from F. Y. Algebra. Chap. 1. A set is a well defined collection of objects.

The objects constituting a set are called its elements. Sets can be described either by tbe Roster or Rule method.

A set is called a finite set if it has a finite number of elements; a set that is not empty or finite is called an infinite set.

Two sets A and B are said to be equal if they have the same ( i . e. identical) elements and we write A = B.

The order in which the elements are listed is immeterial.

If an object occurs more than once, it is listed only once in the set.

A null or empty or void set is a set containing no element. It is denoted by the symbol <t>.

Diagrams representing sets are called Venn diagrams.

A set with one element is called a singleton or a unit set. A set A is a subset o f a set B if x e A x e B; and we write it

a s k c B . Every set is subset of itself. The null set $ is a subset of every set.

From a set containing n elements, 2" subsets can be formed and a set consisting of these 2" subsets is called a power set of a given set.

Two sets are called equivalent if the elements of one set can be put into one-to-one correspondence with the elements of the other set.

IMPORTANT FORMULAE FROM F. Y. ALGEBRA : 3 6 1

A set A is called a proper subset of a set B if every element of A is an element of B and there is at least one element in B which is not in A. We write this as A c.B. Thus A c B if a <? A, then a e B and 3 b e B s. t. b £ A.

Two sets are called disjoint if they do not have any common element.

Given sets can always be considered as subsets of some fixed set called the universal set, denoted by, say, X.

The compliment of any set A, which is a subset of some universal set X, is a set of those elements of X which do no t belong to A. We write it as A'.

Thus, A' = \x\xeX;x(£A\.

The union of two sets A and B ( written as A U B ) is the set of elements which belong to A or to B or to both A and B.

Thus, AU B = \ x\xeA and/or e 3 | .

The intersection of two sets A and B ( written as A fl B ) is the set of those elements which belong to both A and B.

Thus, AV\ B — \ x\xe A and x e B

Following results may be carefully noted.

If A = B and B = C, then A = C. A = B iff A c B and B C A. AUA = AIADA = A. AU<t> = A;Af)X = A. A U X - X; A D 0 = 0. A U A' = X; AH A' = 0 . (A')' =A;(X)' =0; 0' = X.

Commutative Laws : AU B = BU A; AC]B = BCl A. Associative Laws : ( ^ U B ) U C = , 4 U ( 5 U C ) ;

M n ^ n c = ^ n ( 5 n c ) .

Distributive Laws : AU(B[^C) = AV\{BUC) = ( ^ n J B > u ( ^ n c ) .

De Morgan's Laws : (A U B)' =A' fl B'; ( ^ f l B)' =A'U B'


Chap. 2 . Positive integers are called natural numbers A number which can be expressed as a ratio of two integers, say plq where q i * 0, is called a rational number.

A number which can not be expressed as a ratio of two integers, say p/q, qj= 0, but which corresponds to some point on the number axis is called an irrational number.

Rational and irrational numbers together form a set of real numbers.

The square of every real number is non-negative. Every real number can be represented by some point on the number axis and conversely every point on the number axis corresponds to some real number. The set of real numbers is ordered and dense. A set of real numbers satisfies the properties of ( i ) closure with respect to addition, subtraction, multiplication, division except by zero and involution, ( ii ) commutativity, (i i i ) associativity and ( iv ) distributivity.

Approximate rational value, of any irrational number can be calculated.

The properties of real numbers as given in articles 4 and 9 should be carefully remembered.

Chap. 3 . \j — 1, denoted by i, is the unit of imaginary numbers. We have i1 = — 1, i3 = — i, i* = 1, ... etc. A number of the type a + ib where a and b are real numbers is called a complex number. If a = 0, the number becomes purely imaginary and if b = 0 it reduces to a purely real number a.

Two complex numbers are said to be conjugate complex numbers if either of them can be obtained from the other by changing i to — i.

Two complex numbers a + ib and a' + ib' are equal if a = a' and b - b'. If a + ib = 0, then a = 0 and b = 0.

The sum and the product of two conjugate complex numbers are purely real while their difference is purely imaginary.

The complex numbers can be represented in the Argand's diagram on a plane with rectangular coordinate axes. Algebraic operations with complex numbers as defined in article 3 should be carefully noted.


Chap. 4 . If a and b are real numbers and m, n are positive integers, then

( i ) a™ x a" == a m + n ,

( i i ) am -s- a» = a " - " , ( a = £ 0 ) ,

( i i i ) ( a » ) » = a"*" = ( a » ) m ,

( i v ) ( a b ) m = ambm,

<*> (f)"-£ These laws are true for any rational values of m and n

(provided a > 0, b > 0 ) .

Definitions: ( i ) If a js any real number and m a positive integer, then

am = a x a x a x m times. ( ii ) If a is any non-zero real number, then a.0 — 1.

( i i i ) If a is any positive real number and pfq a positive rational number, then

i>h <i a = v a*.

( i v ) If a is any positive real number and r any positive rational number, then

1 a = — • ar

( v ) If a is any positive real number, x an irrational number and ( x„ ) a sequence of rational numbers such that

xn Lt x„ = x, then a* = Lt a .

n—±oo n —> oo

If a # 0, a 4= ± 1 and a* = a\ then x = v .

Chap . 5 . If a and N are positive real numbers and a # 1, then a* = N ^ x - loga N.

Ion N We have the identity : N — a *a . log„ 1 = 0 and loga a = 1, a e R and a > 0 and 1.


If a, TO, n e R and m > 0, / i > 0 , a > 0 and # I, we have

( i ) loga mn = loga m + loga n,

( ii ) loga (mjn ) = loga m - loga «,

( i i i ) logaCm") = nlo&m, ( i v ) Iog6m = log a w -7- log t t6, > 0 and =£ 1.

Chap . 6 . A surd is an irrational root of a rational number. If a and x are rational numbers and sfb, are real surds,

then

( i ) a +\[b = x + yf~y a = x and b = y; ( ii ) a + \fb = x + \[~y => a -\f~b =x - \ f y ;

( i i i ) \[7> = x + \[~y x = 0 a n d b = y;

( i v ) /Ja+\[T =\Tx +s[y =>Ja - \[~b =yf7-yfyi

( v ) ^Ja +y[~b =sf x + y f y if a > 0 and a2 — b is a

perfect square and x + y = a and 4xy = b.

Chap. 7 . ( a ) The roots of ax2 + bx + c = 0 are given by

— b ±\[ b2 — 4ac V = 5 • X 2a

(b) If b2 -4ac > 0, the roots are real and unequal;

= 0, the roots are real and equal ;

< 0, the roots are imaginary;

=s a perfect square, roots are rational

We have assumed that a, b, c are rational.

( c ) Sum of the roots = — — ; v ' a

Product of the roots = — • a


( d ) The equation whose roots are known can be written as x2 — ( sum of roots ) x + ( product of roots) = 0.

( e ) Conditions for ax2 + bx + c = 0, a'x2+b'x + c' = 0 to have

( i ) one common root is (ca'-c'a)2 = (ab'-a'b) (bc'-b'c),

( ii ) both the roots in common is — = = — • v ' a b c ( / ) Irrational and imaginary roots of a quadratic equation

always occur in pairs ( g ) If /3, Y are the roots of the cubic equation

ax* + bx2 + cx + d = 0,

then 2<* = — —, 2«/3 = — and °c/3y = — — • a a a

(h) If «, ft, y, $ are the roots of the biquadratic equation axi + foe3 + cx2 + dx + e = 0, then

a a ' a a Chap. 8. Method of Induction is a method used to prove the

results ( which are already given or known ) to be true for positive integral values. If P (ni) denotes a statement containing a property of the natural number n, the method consists of the following two steps:—

(a) P ( 1 ) or P ( 2 ) is verified. ( b ) Assuming P ( « ) to be true, say for n = k, we show

that P ( & ) = > P (fc + 1 )• With the help of step ( a ) , the repeated use of step

( b ) establishes the result y n.

Chap. 9. With usual notation, we have— ( i ) wth term of an A. P. = a + ( n - 1 ) d;

( ii ) Sum of n terms of an A. P. = 2a + ( n -1) d ]

= + n ;

( iii ) nth term of a G. P. = ar


( i v ) Sum of n terms of a G. P. = a ( r " } } i f r > l , r — 1

= i f r < 1 , 1 — r

= na if r = 1; ( v ) a, b., c are in A. P., G. P. or H. P. according as

a—b a a a . , o r ~b °T ~c r e sPec t , ve ly a n d conversely;

( vi ) Sum of infinite geometric series, if - 1 < r < 1, a is

1 -r '

(vii) A — arithmetic mean between a and b =

G = geometric mean between a and b= \fab.

H = . harmonic mean between a and b = - : a + b

(viii) A > G > H and AH = G2.

Chap. 10. Following formulae must be remembered.

( i ) 1 + 2 + 3 + ... + « i . e . 2 r - " f " 2+ 1 > .

( ii ) l2 + 22 + 32 + ... + n2

i .e. 2 r J = n ( n + I ) (2n + 1 ) _ l 6

( i i i ) l 3 + 23 + 33 + ... + »3

. n2 in + 1 l. e. 2 r3 = • l 4

( iv ) 2 ( a r 3 + br2 + cr + d) l = 2<zr3 + 2ir2 + 2cr + 2</

= a 2 r3 + 6 2 r» + c 2 r + nd. i t I

Chap. 11. ( i ) w! = n (n - 1 ) (n - 2 ) . . . 3-2-1.

nl ( ii ) "P, =n(n - 1 ) ( n - 2 ) ... (n-r + l ) =

( i i i ) C r = Pt n ! r\ r! (n — r ) !


( iv ) " C = " C „ - r . X = "C f l = 1, 0 ! = 1.

( v ) "Cr + "c,-t =

Chap. 12. ( i ) ( x + a)" = x» + "C1x"'1a+ "C^"'2 a 2 - f

+ nCrx"'r a 4 - . . . + a\

( ii ) Genera] term, T r + j = "C^x" r a .

_ n (« — 1 ) f » — 2 ) . . . ( » — r + 1 ) r ! x*'rar.

( i i i ) is the middle term if r — when n is even.

T r + 1 will be the middle term if

r = and r = 1 * when n is odd.

( i v ) - C o + Q + C2 + . . . . + C„ = 2 ;

C0 + C2 + C4 + = C1 + C3 + C 5 + ... = 2 n-1

Chap. 13.

( i ) a2

= CTifcj — "2

( i i ) Two equations A t x-f = 0 and a2x + b3=0 are

a2 b2

consistent if = 0.

( i i i ) The solution of equations written in the form a-ix + biy- and a2x + b2y = c2 is given by the Cramer"! rule

x v 1 ct bt a1 C\ a 1 bx

C? b2 «2 c2 <>2 bt


( i v ) The solutions of equations written in the form fljx + i 'iJ ' + ci «= 0 and a2x+b2y + c2 — 0is given by

x —v 1 bx Cl ax cx ax bx

b2 C2 a2 c2 a2 b2

or by

i. e.

y bx

b2

c\

c2

c-i

c2 a2

y

a\

a 2

1

bx cx

( v ) a2 b2 c2 = ax (b2c3

a3 b3 C3

bi c2 — b2 ci c-i a2 — c2 a\ b2 — a2 bt

( This is known as the Rule of Cross-Multiplication ) .

+ ?! ( a2b3 - a3b2).

( v i ) Three equations arx + bry + cr = 0, r = 1, 2, 3 are consistent if

0.

(vii) The solutions of three equations a,x + bry + crz = dr is given by Cramer's rule

ax bx cx

a2 b2 c2

a3 b3 c3

x dx bx cx

d2 b2 c2

d3 h c3

2

<t\ d\

a3 d3

1

c\

C2

Cl

ax

a2

a3

h

b2

h

dx

d 2

<*3

at bi c t

a2 b2 c2

a3 b3 c3

(viiij Various properties of determinants should be remem-bered.

Appendix 3 TEST PAPERS

Test Paper I

1. (a) Given two sets A and B, define AVB and Af\B. Give one illustration of each. Prove : ( i ) AUB = BUA, ( i i ) U D f l ) fl C = Af) (Bf)C). If A = ) 2, 4, 7, 9 B = } 3, 5, 7, 9 { and the universal set X = J x \x is a + ve intger < 11 }, find ( AUBy and ( A'?\B') and verify that these sets are equal.

(b) A is the set of students studying mathematics. B is the set of students studying sanskrit and C is the set of students studying Gujarati. Interpret the sets

( AU B) f l C and ( At\B ) U C.

Draw Venn diagrams to illustrate your answer .

2. (a) Explain the meaning of the symbol a* where a is + ve real number and x is irrational.

(b) Define a rational number and show t h a t ^ 2 can not be a rational number. Find three rational numbers in ascending order giving better and better approximate value of V 2.

(c) State whether the following statements are true or false. If a statement is true, prove i t ; if you consider a statement to be false, give an example in support of your answer. ( i ) The product of two rational numbers is rational.

( i i ) The sum of two irrational numbers is irrational.

(iii) The product of two odd integers is an odd integer.

(iv) If x is less than y, then x2 is less than y2. ( v ) Given any real number x, we can find a real num-

ber y such that xy = 1. C . A . 2 4 3 6 9


(vi) A = \ - 1,0, 1 ( ; then A is closed under multi-plication.

(d) If one root of the equation x2 + x + 1 = 0 is a then show that ( i ) the other is ca2 and ( i i ) w3 = 1.

3. (a) What do you understand by the statement " <x is a root of the equation ax2 + bx + c = 0 " ? Solve the equation ax2 + bx + c = 0 ( a 0). If * and ft are the roots of this equation, express * + ft, « ft and o<2 + jn terms of a, b, c.

(£>) If both roots of the equation ax2 + bx + c - 0, where at is a non-zero rational number, are rational, prove that ( i ) b and c are rational and ( i i ) ( b2 — 4ac ) is the square of a rational number,

(c) If the equation x3 - bx2 + cx — d = 0 has the sum of two of its roots equal to zero, show that be — d.

4. (a) If a is a real number and m and n are positive integers, prove that

(ttt v fl tttft a ) = a .

If m and n'are positive integers, does am — an imply that m = n for all real numbers a ? Give reasons for your answer.

If 3 f c + 1 - 3" = 3*+3 - 9, find x.

(b ) If a, b, c are rational numbers, but neither b nor c is the square of a rational number, prove that a + Vb = Vc implies a = 0, b — c.

If a + 2 ^ 5 - 156 = 7a - (b + 4)\[\25, where a and b are rational, find a and b.

5. (a) Define loga x, where a and x are positive real numbers and a =£1. Find the positive integer n such that log3 96 lies between n and n + 1. If A, b, c, d are positive real numbers such that

a = b2 = c3 = d* ( # 1), find log0 ( bed).

TEST PAPERS : 371

(6) Prove, making suitable assumptions (which must be stated), that

( i ) log„ ( m » ) = loga m + log0 n and

( u ) ( l o g 0 6 ) ( l o & f l ) = 1.

If loga m = x a n d 1 °Sa n — y> where a, m, n are positive, a # 1 and m n # 1, express log a in terms of x and y.

mn

6. (a) Assuming that nPr = » ( n - l ) ( n - 2 ) ( n - r + 1) , obtain a formula for the number of combinations of » different things taken r at a time.

Deduce that ( i ) "c r - i + "Cr = "+1Cr and

( i i ) Cr + 2 C r+i + C r+2 = Cr+J. (£>) A guard of 12 men is formed from a group of n soldiers

in all possible ways. Find

( i ) the number of times two particular soldiers A and B are together on guard, and

( i i ) the number of times three particular soldiers C, D and E are together on guard.

Also find n if it is found that A and B are three times as often together on guard as C, D and E are.

7. (a) Prove the binomial theorem for a positive integral index.

If n is a positive integer such that the coefficient of JC6

in the expansion of ( 1 + x ) " is 593775, find the ti—6

coefficient of x in the same expansion.

Use the expansion of ; 1 + x ) " to prove that "Co + n C 1 + " C 2 + + "C„ = 2".

(Z>) «, ft and y are the roots of the equation 3xS + 4x2+5x + 6= 0

Find the values of ( i ) 2<x2 and ( i i )


8. Attempt the following : ( i ) Find the sum of all odd positive integers from 1 to

149, excluding those divisible by 7.

( i i ) The sum of the first n terms of a series is

( 3 n —2" )/2" 2.

Find the n«» term of the series and prove that the series is a geometrical progression.

( i i i) Find the sum of the first n terms of the series

1 + 1 + 2 ^ 1 + 2 + 3

Answers.

1. (a) A LJB = } 2 , 3 . 4 . 5 , 7 . 9 / . ( A U B ) ' = ) 1, 6 , 8 10 \ . A' = $1.3 5 . 6 8. 0 and JB' - \ 1 . 2 . 4 . 6 . 8, 10 \ A ("|B' - $1 .6 ,8 . 10$. Hence etc. ( b ) {A LIB) f | c •= set of students who have taken ( i ) Gnjarathi, Mathematics but not Sanskrit ( i i ) Gnjarathi Sanskrit but not Mathematics and ( i i i ) Gujarathi, Mathematics, and Sanskrit See fig. 10 page 26 This set is shown by ( 4 ), 5 ) and ( 7 ) Similarly the set ( A f l B ) |J C is shown by ( 7 ) . ( 4 ) . ( 5 ) , ( 6 ) and ( 2 ) .

2 . ( a ) ( i ) True. Let pjq, rjs be two rational numbers so that p, q, r

and s are integers and q, s #= 0. The product = — is a rational number qs

since pr, qs are integers add qs 4= u.

( i i ) False, 3 + V5,3 - J 5 are irrational numbers but their sum is 6 which is rational.

( i i i ) True Let 2m + 1. 2» + 1 be two odd integers. Their product = 4mn + 2m + 2n + 1 which is an odd integer of the from 2k + 1 since 4mn, 2m, 2n are all even and their sum which is even is increased by 1.

( i v ) False. - 4 < 3 and 16 > 9.

( v ) False. If we take a real number zero, then we can not find any real number whose product with zero is equal to 1.

( v i ) True, Multipl cation of any elements gives an element of the set.

4 . i a ) a m =» a" does not Imply that m = t» for all real numbers a. In fact m =t= « when a 0 or ± 1 : * => - 1 or 2.

( 6 ) a = 11 and 6 = - 22/5.

TEST PAPERS : 3 7 3

5 . ( » ) 81 < 9 6 < 243. 3* < 96 < 3* . 4 < logs 96 < 5 .

n - 4 ; loga (bed) ~ jL •

log^ » n Iog^ m + Iog ( * + y

6. ( a ) ( i i ) nCf + 2*0. + 1 + * C r + 2

= [ " C r + " C r + 1 ] + [ " C r + 1 + " C I + J ]

n+l_ «+l n+2 W+l + 2 S+2 '

(6) ( n "~2c , ( « ) n " 3 c 9 . 10 * - 2 *r ~ 3

ri is given by the relation C^ = 3 C^; « =• 32.

7. ( a ) The coefficient of x* is " c The coefficient of 6 i s *C , 6- n

Bat we have the relation that nC = "c ^ rt r»

/ . "C( - " c „ _ 6 = 59377S as given.

It & fl ff Put * = 1 in the expansion of ( 1 + * ) to get 2 C — 2 • r=0

( 6 ) ( i ) - 14/9 ; ( ii ) - 23/36.

8 . ( i ) 5798. • 3" _ 2" 3 " - 1 - 2 " - 1 _ 4 / 3. \ r (.1) = - = -3 ^ 2 )

We have — — - 4 - for all « .\ au ait ch a„ a r e lnG .P ' V 1 2

fm\ 1* +2* +3' + +r* ( i U ) 1 + 2 + 3 + • ••• + r

„ *•(«•+!) (2r 1) x 2 _ . j 6 X r ( r + 1) 3


Test-Paper II. UNIVERSITY OF BOMBAY, April 1969.

1. (a) State when A C B ( A is a subset of B ). Explain the term " Empty S e t I f 0 is an empty set, state whether the following statements are true or false. ( Assume that A is not an empty set ) :— ( i ) 4>CA, ( i i ) ACA.

Write all the possible subsets of the set ) 1, 2, 3 (. Define the complement A' of a set A with respect to the universal set S and write down S'.

(b) Given two sets A and B, define A U Band AO B. If A = ) x\x*~ 3x + 2 = 0 (,

B = j + 4x — 12 = 0 write AD B and AO B. (c) State with reasons whether the following statements are

true or false :— ( i ) A CB and B CC=> ACC, ( i i ) AD B = A U C => B = C. ( Assume that A, B, C are three non-empty sets ) .

Or ( For students appearing under the old course only )

1. (a) Prove that 1 +a 1 1

1 1 + b 1

1 1 1 +c

= abc ( l + — + 4" + —V \ a b c J

([b) If the three equations in x and y x + y - 1=0, ax + (a + \ )y - ( a + 3 ) = 0,

a2x + ( a + l f y - (a + 2 ) 2 = 0 are simultaneously true, find a.

(c) The product of three numbers in a G. P. is 216 and their sum is 26. Find the three numbers.

2. ( a ) Attempt any two of the following :— ( i ) Give any two values of x, one rational and one

irrational, such that \f 5 < x < yf 7.

TEST PAPERS 3 7 5

( i i ) If a is any positive real number, show that :—

a + — > 2 . a

(iii) If n is a positive integer and « < 5 — -^37 < « + 1, find n.

(b) ( i ) Find four rational numbers /, m, x, y such that

I < / < x < \[2 < y<m< 2 and represent these numbers (roughly) on a straight line.

( i i ) If = ) 1 , 3 , 5 , 7 , . . . . 5 = ) 2 , 4 , 6 , 8 , . . . . examine whether the sets A and B are closed under the operations of addition and multiplication.

Or

For students appearing under the old course only ]

(6) ( i ) If x = » where / = \ j ~ T , obtain the

value of x3.

( i i ) I f x = « / 4 + \ f T 5 + — = ,

y 4 + \fl5~

show that x3 - 3x - 8 = 0.

X 3. (a) Define a when :—

( i ) x = m, ( i i ) x = , ( iii ) x = 0,

where m is a positive integer stating in each case the restric-tions, if any, on the value of a so that a* becomes a real number.

If a is a real number and m, n are positive integers; prove that—

m n m+H a x a = a


(b) Evaluate :— - - 2 - - 1 1 2 5/6 , - - 3 , - 1 / 2 2/3 * -1/6 , 2 - 3 • a + 2 -3 • a — 10 ( 2 7 a ) where a = 64.

(c) Arrange the following numbers in the ascending order of magnitude :—

lOO/TT J T \[27, , 3V 2 .

[ Hint : Express each number in the form a where x is expressed in decimal notation ].

4. (a) Define log^m, stating the usual restriction put on m and a. Assuming that the necessary restrictions are satisfied by a, m, n prove that

1°8a ( m") = n loga m. (b) Without using tables, find x if

- i - log10 (11 + 4 \ f 7 ~ ) = log i o (2 + x ).

(c) If ' ^ — ^ = a + ib, where a and b are real and

i — \[ — 1, find a and b and plot the points a + ib and a — ib in a plane mentioning the necessary set of axes.

5. (a) Define a surd. State one positive rational value of a

such that \f a is not a surd.

If a + \f b = x + \l y, then prove that a = x and b = y stating the necessary restrictions on the values of a, b, x and y.

TO If + 5 ^294 + - J l J j - J ' J J )

= find a.

(c) Find x and y if ( \ f x + \f~y)2 = 4 + •\f~lT. Deduce

Hence Snd k if ( 4 + yflS f + ( 4 - f" = k \ f 1 0 .

TEST PAPERS : 3 7 7

(fl) Obtain the roots of the quadratic equation

ax2 + bx + c = 0. State the restrictions on the values of a, b, c under which the roots are—

( i ) equal, ( i i ) real but unequal, ( i i i ) rational,

( i v ) purely imaginary.

(b) If P are the roots of x2 - 2x + 3 = 0, find the oc Q

quadratic equation whose roots are - 3 - , — • p <x

(c) If j3, y are the roots of the cubic equation 4x3 - 5jc 4 - 7 = 0 find the values of—

(i) + ft2 + y2. (ii) + - J - + \ •

Or

( For students appearing under the old course only ) (c) For what values of a, are the roots of the equation

x2 + a? = 8x + 6a real and unequal ? (А) Prove by Mathematical Induction or otherwise the

formula—

r a ( x" - 1) , , 2 ax = — r—, x=F 1. ,=0 x - I '

State also the sum when x = 1.

(б) I f S „ = fl(f x ± 1

prove that—

ax ( x" — 1) tta

( x - l ) 2 Sr= .

r=l ( ~ 1 \ 2 X - l

(c) For an examination, a candidate has to select 7 subjects from 3 different groups A, B and C which contain 4, 5, 6 subjects respectively. In how many different ways can a candidate make his selection if he has to select at least 2 subjects from each group ?


8. (a) Prove the Binomial Theorem : -n n « n-r r

(x + a) = 2 Cr x a, where n is a r=0

positive integer.

(ft) Using the Binomial Theorem, find the value of ( 2-02 )5

Correct to 5 places of decimals. 3 3

Find 2 a if a , = 1 + 2 + • • • • + .

r=1 r2

Answers. 1. ( 6 ) A = \ * ! ( * - 2) ( * - 1) - 0 j,

B = \ x\ ( * + 6) ( * - 2 ) = 0

A - \ 1 . 2 j . B = } 2, - 6

••• A U B = ) 1. 2, - 6 ( and A f | B = \ 2\.

( c ) A c B and B S C A £ C . This statement is correct.

x e A o * e £ , since A C B.

Similarly * e B * e C, since B C C.

Thus * e A * f C. Hence i C C , ( i i ) .A U B = A U c => B = C This statement is wrong. The

elements in C which are common to A need not be the same as the elements in B which are common to A. Consider A = \ 1 . 2 . 3 B «= ) 2. 3. 4 C = \ 1. 2 .3 . 4 ) We have A L)B = \ 1. 2, 3, 4 \ = A U c : but B 4= C.

Alternative Question : 1 ( 6 ) a = - i - . 1 ( c ) 2, 6, 8.

2 . ( a ) ( i ) Rational value : 2-5, irrational value : ^

( i i ) ^ - ^ > 0, since the square of a real numbti

is non-negative. ( i i i ) n = 2.

( 6 ) ( i ) The value of V 2 correct to three places of decimals is 1-414. Hence we may take J = 1 2 , * = 1-3. y = 1-5 and m = 1-6.

( i i ) The set A is closed for multiplication but not for addition. The set B is closed both for addition and multiplication,

Alternative : 2 (b) ( i ) *» = 1.

TEST PAPERS : 3 7 9

3. (6) 5/ V 3.

( c ) 3 M < 3 1 ' 4 1 < 3 ^ 2 < 3 1 ' 5 . since V"2 - 1-414, correct

three places of decimals.

. 5 / T 100 / l 4 1 V~2 •• V 3 < V 3 < 3 < V 2 7 .

5 . ( a ) Sard is an irrational root of a rational number.^ 4 is not a surd.

(6) a = 41. (c) ft-7. 6 . ( 6 ) 3 * s + 2 * + 3 = 0. ( c ) ( i ) 5/2, ( i i ) 5/7.

Alternative ( o ) — 2 < a < 8.

7. (o) «Ca - 'C, • *C, + 'C3 «C3 • eCj + 'Ca • 5Ca • «C2

•= 1200 + 900 + 600 - 2700.

8 . ( 6 ) 33-63232. ( c ) a r = .

... + 2 » ( n + l ) + w j .

Test-Paper ffl. UNIVERSITY OF BOMBAY, October 1969.

1. (a ) Explain the term ' Empty Set ' and give two examples of it. (6) Given sets A = [ Natural numbers 1 to 10 ],

B = [ n\n being a natural number and 2n < 20 ] and C = [ Even integrers between 1 and 9 ], write down the following sets : — ( i ) Bf\C. ( H ) ( ^ n B ) n ( A n c ) . ( i i i ) Complement of ( BUC ) with respect to A. ( i v ) All subsets of the set in ( i ).

(c) Given set X = [ 2, 22, 23, 2i ] and Y = [ AH positive even integers ] state with reason whether XC Y or Yci X. State also whether X is closed with respect to addition and multiplication.

(d) In a hostel 15 members take tea, 8 members take coffee and 6 members take milk. If 5 members take tea and


1, 3, 7 3, - 1 , 5

2, 6, 3x + 2 + - 2 , 5, 9

11, 2, 3 1, 4, 14

coffee both, 4 members take tea and milk both and if none of them takes coffee and milk both or all three (i. e. Tea, Coffee and Milk) , find the number of members in the hostel. [ Every member takes at least one of the three beverages ]. Hint:—Use Venn Diagrams.

Or

( For students appearing under old course )

1. ( a ) Obtain the condition in the form of a third order determinant, that the equations

ar x + bry + cr = 0. [r = 1 , 2 , 3 ] are simultaneously true.

(b) Solve —

= 0.

(c) If p \ q h and rih terms of a G. P. be themselves in G. P. prove that q is the arithmetic mean of p and r.

2. ( a ) For any real number a > 0, give meanings of

( i ) a and ( i i ) a*'® (p}q is a positive fraction.)

Prove cT -r- a = a" " for positive integers m and n.

(6) Simplify:—

M ^ r r [ w - r (c) Solve :—

3 , + 3 - 5 X 2 ' + 1 = 7 x 2* 1 - 3 , + 2 = 1.

i d ) I f x = ^ 3 2 - V 2 ?how that x2 (x* + 24x2 +144) = 900.

3. (a ) When do you say log„ m = x 1 State restrictions on m. a and x. Prove with usual restrictions on m, n and a

= log* m - log, n.

TEST PAPERS : 381

(ft) ( i ) Find the value of—

) a'2 l 0 K«3 { X ) log927 { X $ log27 (3n/3") f.

( i i ) If log2 ) log, (81) i = 2, find log, ( 3x2 ).

(c) Show without using tables of logarithms, that—

< logio 2 < - i - •

(a) State with reasons whether the following statements are true or false :—

( i ) If x is rational and y is irrational, then xy is irrational.

( i i ) If x is irrational and y is rational, then x y i s irrational.

( i i i ) Roots of { x — p ) ( x — q ) = k2 are always real for real values of p, q and k.

( iv ) The smallest real number greater than 4 is 4-00001.

( v ) The sum of squares of two conjugate complex numbers is always a positive real number.

( v i ) / = ( 2 3 ) 2 .

(ft) ( i ) Find the positive integer n such

n < (\j2 — 1 )'2 <n + 1. 7 — 3

( ii ) Find the greater of the numbers v5, / 2.

( i i i ) Show that 3 ^ 2 + 2 J i e s between\J2 and

Or

( For students appeoring under the old course ).

(ft) ( i ) I f * = Y ^ J i find t h e value of x3 + x2 - x + 22.

( ii ) Find a and b if

3 + ^ 6


5. (a) Solve the equation :— ax2 + bx + c = 0 where a > 0 and a, ft, c, are ra t iona l

If 2 (a + i\[3) be a root of x2 - 4x + k = 0

determine a and k. [ i = \f — 1 ]. (ft) Find conditions on a, ft, c if roots (3 of equat ion

given in ( a ) be such t h a t : — 2

( i ) (v + t ) =1' (ii) 0-£>2 = 5-(c) If «t, /8, y be the roots of the cubic equation

2x3 - 3x - 10 = 0, prove that :—

«

Py + A + y 3 «/3 5

Py + yL.—3

Or ( For students appearing under the old course ).

(c) Prove that the roots of equation : — (a1 + ft2) x* - 2b(a + c) x + ( f t 2 + c 2 ) = 0 ,

are equal if a, ft, c are in G. P. [ a, ft, c are real ] .

6. (a) If a + \f ft = \f c where a is rational a n d \ f b , \fc are surds, then show that a = 0 and ft = c. Prove tha t :—

( 2 V 5 + 3 V 2 ) ( 1 9 - 6 V I 0 ) ^

( 2 ^ 5 - 3 ^ 2 )

(ft) If ( h - t ) ( 1 + t ) + = a + ^

find a and ft and represent numbers ± a ± ib by points in a plane.

(<•) If the sum of the A. P. : — - 4 7 , - 4 2 248 be 6030, find the sum of all its positive terms.

TEST PAPERS : 3 8 3

7. (a) ( i ) Give expressions for "Pr and CT in factorial notation.

( i i ) Prove that nCr + "C r - i = "+ 1C r .

( i i i ) Find x if " c 5 + 2 1 2C4 + 12C3 = U C r

(ft) In how many ways a party of 7 can be selected from 5 boys and 6 girls so that boys are in majority in every selection ?

(<•) Use method of Mathematical Induction to prove that » ( » + ! ) ( / » + 2 ) is divisible by 6 for all positive integers n.

6

8. (a) ( i ) Write down expansion of

simplifying each term. M )

( ii ) Find the value of (1-02 ) using Binomial Theorem. 10 / !r 1

(ft) If the constant term in the expansion of j

be 405, find the value of k.

(c) Show by Mathematical Induction that

and hence find the sum l 2 + 32 + 52 + +(29)2-

Answers.

( « ) Two examples— ( i ) A = $ * | x - - ve integer > 3 („

( i i ) A = ) x'x-- natural number which is the root of " 4 * 1 - 9 « 0 "

( 6 ) A = 2 . 3 1 0 ( . B = p . 2 , 3 , 4 = \ 2 , 4 , 6 . 8

( i ) B O C - j 2 . 4 | , ( i i ) ( A f l B) D ( A f l C ) = ) 2. 4

( i i i ) ( B U C ) ' with respect to A = ) 5, 7. 9. 10

( i v ) ) 2 ) 4 (, j 2 . 4 <*>. I

( c ) X C y . X is not closed with regard to addition as 2 + 2 —6 which does not belong toX. X is closed with regard to multipli-cation; since 2 m x 2 B = 2 m + " eX(ormeN,ne N.


( d ) Numbers of members taking only tea, only coifee and only milk are 6, 3 and 2 respectively total number required = ( 6 + 3 + 2) + (5 + 4 ) = 20.

OR 1. ( 6 ) * = 4. 2 . ( 6 ) Exp. = ^ ) ' ( « ) * = 3, y 1.

3. (6) ( i ) -3J- ( i i )

4 . ( a ) ( i ) False; Take * = 0, y = V2". xy = 0 which is rational.

( i i ) False; * =J 2, y = 2; x = ( J z )2 = 2 a rational number^

(iii) True, a = - g )2 + > 0.

( i v ) False. The number J ^ 1 is real, greater than 4 and A smaller than 4-0001.

( v ) False, (a + ib ) 2 + ( a - ib)2 = 2o» - 26'= 2 ( « ' - & ' ) . It is negative if a s < 6'.

(vi) False. L. H. S. =2° and H. H. S. = 2 .

( 6 ) ( i ) n = 5. ( i i ) V ~

4 . Old course ( 6 ) ( i ) 7. ( i i ) a = 0, 6 = 1.

5. (a) a = 1, £ = 16. ( 6 ) ( i ) 6s = c\ ( i i ) 6" - 4ac = 5a1

6. (a) Exp. = (19)2 - (6^10 )? = 1. ( 6 ) « = — 3, b - - l . Points ( - 3 , - 1 ) , ( - 3, 1), (3, - 1),

( 3 , 1 ) . ( c ) 6275.

7 . ( a ) ( i i i ) x = 5- ( 6 ) 115.

8 . ( a ) ( i ) 729 *« - 729*5y + — — x* y"

135 . , , 135 , . 9 . , y«

( i i ) 1-08 24 32 16. ( 6 ) ft = 3. ( c ) 4495.

Test Paper IV. UNIVERSITY OF BOMBAY, 1970

1. (a) Give four sets E, F, G, H such that E C F(Z GC.H.

(b) If A and B are two sets, define A fl B and show that At\B = Bfl A.

(c) Draw Venn diagrams illustrating the relation between the sets mentioned below :—

( 0 A\JB =2 A, (ii) C O D = C, (iii) EOF = tf>.

( d ) If A C E, define A' the complement of A in E. If J is the set of all integers and N is the set of all natural numbers what is N' in J?

(e) If A is the set of all quadrilaterals, B is the set of all squares, C is the set of all parallelograms, and D is the set of all rectangles state with reasons which of the following are true or false :

( / ) S C f l , (/';) AC.B, (iii) B-D, (iv) D C C C A.

( for students appearing under the old course only ).

1. (a) Find the value of

Or

2 3 - 5

4 - 1 2

3 - 4 1

ib) If a + b + c = 0 show that

ax by cz x y z

bz cx ay = abc z x y

cy az bx y z x

(c) If ax + by = bx + cy = cx + ay, prove that a1 + b2 + c2 = ab + be + ca.

C AT 25. 335


2. ( a ) Define a quadratic surd. If a + \Jb = y[c where a is a

rational number and \[b andVc are surds, prove that a = 0 and b = c.

(b) If - p = a + bs/3 5\f3 + 4 \ f 2 --v/72 - -N/ 108 + 8 + 2

find a and ft. (c) If a and ft are real and if

( i 4 + 3/) a + ( / - 1) ft + 5/3 = 0 find a and ft. (d) State with reasons whether the following statement is

true or false : If a is real then a 2 — 8a + 16 is always -f ve.

3. (a) Solve the quadratic equation ax2 + bx + c = 0.

(ft) If one root of a quadratic equation is 3 + \ [ — 4 find the equation.

(c) Find the value of k if the roots of the equation x(x - 6 ) = 3 * ( 1 - x)

are equal in magnitude but opposite in sign.

(d ) Prove that the roots of the quadratic equation (x — a) (x-b) + (x-b) (x-c) + (x-c) (x-a) = 0 are equal if and only if a = ft = c.

4. (a) Obtain the value of the following :

( / ) ( 2 " 3 ) 2 , ( / / ) ( 125 ) 2 '3, (Jit) ( 2 3 > ) ,

f . . (4 + \ f 5)° ( / V ) ji-J- .

(ft) Show that

3 X 3" 9"+ 1 1

(3n)n~* (3M~y , + 1 9

(c) Prove that

1 + 1 | = 1 ' , . I T , J. , . A — 1 • 1+X"'m + xt~m 1 + Xm'" + XP-» l+X">-i> +x"-*

(d) Determine x if 3 2 l + 1 - 3* = 3*+3 - 9.

B. u . PAPER 1 9 7 0 : 3 8 7

5. (a) Prove the following results : ( i ) log aM + \ogaN = l o g a ( M N ) ;

( i i ) logax = Iog&* log6a

(b) Find x in the following cases :

( 0 logy-y x — - 4, (ii) log., 125 = 3, ( / / / ) X=221ob*5 •

(c) If y = log2 ( log 2 x) find y when x - 2 and find x when y = 2.

(d) Prove that

log\f 27 + logV8 - logy/"l25 _ 3_ _ log 6 - log 5 = 2

(а) Prove by the method of induction or otherwise

n i + i , 1 1 1-4 + 4-7 + 7-10 + " ' + ( 3 « - 2 ) ( 3 n + 1 )

_ n ~ ( 3/i + 1 ) ;

(ii) l 3 + 23 + 33 + + «3 = " 2 ( _ " 4+ 1 ) 2 .

(б) Find the sum to n terms of any two of the following series:—

( / ) 2 r ( 6 r - 2 ) , ( » ) 1 + 5 + 9 + 13 + 17 + • • • r= 1

(j/I) 2 + 22 + 222 -+ 2222 + , ( a ) Define nPr and nCr and derive the formula for nCr.

(b) If ^ = 5 find n. nr 3

(c) Find r if 21Cr = 2 1C3 r-8 . (d) The staff of a bank consists of the manager, the deputy

manager and 10 other officers. A committee of 4 is to be selected. Find the number of ways in which this can be done so as to always include ( i ) the manager, (ii) the manager but not the deputy manager, (iii) neither the manager nor the deputy manager.


8. (a) If n is a positive integer and x and a are any real numbers prove by the method of induction, ( x + a ) " = x» + nCxxn-xa + nC2 x»"2 a2

+ + n C,. xn'r ar + + a».

(ft) Obtain the coefficient of in the expansion of

(c) If a is a real number and if the middle term in the

expansion of + 2 j is 1120 find a,

Answers 1. ( a ) £ = 5 l | , F = J l , 2 ( , G = j l , 2 , 3 [ , tf- | 1 , 2 , 3 . 4

(<*) N' = J 0, - 1. - 2 - n , ... j .

( e ) ( i ) True; since every square is a rectangle. ( i i ) False; since every quadrilateral cannot be a square. (iii) False; since every rectangle cannot be a square. (iv) True; since every rectangle is a parallelogram and every

rectangle or a parallelogram is a quadrilateral.

Or

1. (a) 85, (b) L . H . S. = abc ( AT8 + y8 + z s ) — xyz ( a s + 6 s + c 3 ) = abc ( Xs + y* + e8 - 3*y2 ) = R. H. S. ( c ) Put each

expression equal to 0). Form three equations and eliminate'*, y, k.

2 . (6) a = 14, b = 9, (c) a = 5/4 = b. (d) (a - 4 )a > 0 statement i3 false since for a = A, the expression is equal to zero which is a signless integer.

3 . (b) - 6* + 13 = 0, (c) k = 2.

4 . (a) (i) 1/64. ( i i ) 1/25, (i i i ) 1/512. ( i v ) 1/4. (d) x = - 1 or 2.

5 . (6) ( i ) 1 /9 , ( i > ) 5 . ( i i i ) 25. (c) y = 0 , * = 16.

6. (6) (i) 2»ia (« + 1), (ii) n (2« — 1), (iii ) £ ML ( 10"-1)J 7 . (6) n = 123, (c)r = 4, W) ( i ) 165. ( i i ) 120, ( i i i ) 210.

8 . (b, 165, (c) a = 2

B. u . PAPER 1 9 7 1 s 3 8 9

Test Paper—V

University of Bombay, 1971

1. Attempt any three :—

( i ) Define :—

(a ) The complement A of a set A with respect to the universal set X.

(b) a subset A of a set B, ( c ) A O B* and A U B .

( ii ) If A and B are the subsets of the universal set X and

A C B, prove that B C A .

( i i i ) If X = J a, b, c, d,e\, state two non-empty subsets

A and B of X such that A fl B = 0 .

( i v ) If A = | 2, 4, 6 B = (4 , 5 , 6 , 7 ) and $ = ) n | n i s a + ve integer < 10 | is the universal set, then verify

the result that ( A U B ) ' = A ' n b'.

( v ) Draw a Venn digram indicating the three non-empty subsets A, B, C of the universal set X and satisfying the following requirements

A f l B # 0 , A f l C = 0 , B f l C i M .

2. Attempt any three:— ( i ) Define a rational number and show that there is no

rational number whose square is 2.

( i i ) Define a surd number and state two irrational numbers which are not surds. State a real value of a for which

•tya is not surd. State a rational between \ [ 5 and \f 6.

( i i i ) If x = ( 4 + y [ l 5 ) 1 / J + ( 4 + s [ l 5 ) _ 1 , \ prove that x3 — 3x — 8 = 0.

.. . T. \f~3 - V 2 , , \f2 ( i v ) If a= : and b=

\[3+\f2 N/3-V2 find the value of a3 + ft3.


4 — 2 / ( v ) Express the number p - ^ - r in the form a + ib where

a and b are real and hence represent this number in a plane mentioning the necessary set of axes.


( i ) Define loga m, stating the usual restrictions put on m and a. Prove with the usual notation that loga (m«) = n log a m.

( i i ) Simplify and show that

l o g i o ( 3 4 3 ) = 3 .

1 + i l o g 1 0 (49 /4 ) + | log 1 0 ( 1/125)

( i i i ) If (2-5)* = (-25 )* = 1C00, prove that - I - — = - | -

( i v ) Simplify and find the value of

(•008 ) " 2 / 3 + 1 0 (

( v ) I f 2 x = \[5/3 - y/S/5,

( • 0 0 8 ) " 2 / 3 + 1 0 ( 2 - 2 2 ) + ( 2 5 6 ) 2 " 3 .

snow that ^ 1 + A - — = 4 - -x+\! I + x2 5


( i ) State without proof the roots of the quadratic equation ax2 + bx + c = 0. If a, b, c are rational find the conditions under which the roots are ( i ) real, ( i i ) rational, ( i i i ) imaginary.

( ii ) Find the range of values of p for which the roots of the equation / > ( x + l ) ( x + 3 ) + 2 = 0 a r e not real.

3 3 ( i i i ) If <* — fi=z4 and * - / 3 =208, find a quadratic equation

whose roots are <* and ft. ( iv ) Solve the equation 9x3 - 36x2 + 23* + 12 = 0, it being

given that one of its roots is half the sum of the other two.

B. U . PAPER 1 9 72 : 3 9 1

( v ) Use the method of induction to prove the following

result : (xn) =n xH~\ where n is a positive integer,

(The results used in the proof should be clearly stated before using them )

Attempt any three :— ( i ) Prove with the usual notation the formula :

2 [ a + ( r - l ) r f ] = -^[2a + ( n - l ) d ] . r=1 *

( i i ) I f S„ = 2 tr, where tr = a + ( r - \ ) d , r=l

S'„= 2 t'r, where t \ = d + ( r - 1 ) d' r=1

, SH 3n + 5 . x t , and gr- = » prove that t* = t\.

( i i i ) Define "P r and "C r and state the relation between them.

Prove that 2 i ( V ) = 2n -1. r=i r !

( i v ) There are 25 seats vacant in a double decker bus, 18 on the lower deck and 7 on the upper deck. In how many ways can 25 persons be accommodated in the bus of 6 persons refuse to go up and" 3 insist on going up ?

Attempt any three:— ( i ) Prove the Binomial Theorem :

( x + a ) " = 2 "cr x" r a where n is a positive r=i)

integer.

( i i ) Find the coefficient of x 17 in the expansion of

(X - X ) .

( iii) Show that ( \ f 3 + 1 )5 - { s f l - 1 )S = 152.

Hence find the integral part of ( \ f J + 1 )5 .

( i v ) Find 2 ( 2r — 1) ( 2 r + l ) ( 2 r + 3 ) . r = l


Answers

1. ( i i i ) a ' - \ a, b j , B = j d, • A = ) e, d, e\, B » j a, b. o \ ( i v ) each set - $ 1, 3, 8 , 9 , 1 0 . 1 1 f . ( v ) Refer to page 40. Fig. 22.

2 . ( i l ) ir, « ; « = 27; 2-3. ( i v ) 970. ( v ) 1 - 3*.

3 . ( iv) . 23 + 1 + 2 = 28. 4 . ( i i ) 0 < £ < 2, (Hi) *» T 8* + 12 - 0, -<«*> =f' f 3 -

5 . ( i v ) "Cu = 1820, assuming that the order of occupying the seats is immaterial.

6 . ( i i ) - 1363. ( i i i ) Integral part = 132.

( i v ) 2»" ( » + 1 ) ' + Zn ( » + 1 ) (2» + 1 ) - « ( « + ! ) - 3.

Test Paper VI. University of Bombay, 1 9 7 2

1. Attempt any three :— (a) For the two given sets A and B, define:

CO AU B (ii) Aft B, (iii) A', (iv)AcB. (b) A and B are subsets of the universal set X where

X = j x | x is an integer and - 5 < x < 6 X X

A = 5 x | - y is an integer B = j x | y is a positive integer

Find ( i ) AU B, ( i i ) ^ f l 5 , ( i i i ) 5 ' , ( i v ) all the subsets of A'.

( c ) If X is the universal set, with A and B as its non-empty subsets, show that ( AU B)' = A' C\ B'.

( d ) 9C00 students appeared for two papers in Mathe-matics at the F. Y. Science Examination. Exactly 7,400 and 6,600 students passed in Papers I and II respectively. 6,400 students passed in both the papers. Draw a venn-diagram to indicate these results and hence or otherwise find the number of students who have failed in both the papers.

B. u . PAPER 1972 : 393


( a ) State any two values of x, one rational and one irrational such that V 2 < * < \ / T '

If x is a real number, find the least value of x2 — 4x — 10.

( b ) Examine which is greater 21/3 or 31/5.

If x > 0, solve the equation ( x )X"J* = ( x s f x )*. ( c ) Simplify and show that

- i i ~1/2

J 1 - (l-flsj^J I = a, ( a > 0 and a # 1 ).

(</) If Va =b+ \Jc where b is rational a n d \ f a , \Jc are quadratic surds, then prove that b = 0 and a = c.

(e) If 2x =\f3 + ~ , find the value of _\f3

sf x2- 1 X - \ F X 2 - 1

3. Attempt any three :— ( a ) Prove with the usual notation that

log6 m = (log« m ) -r ( log a b ) and deduce that logj a • log a b = 1.

( b ) Show that logy ( x 3 ) • log, ( j 3 ) < log* ( z 3 ) = 27.

( c ) Without using log-tables, show that

y < logio 6 < y •

(d) If x, y, z are all > 0 and # 1, and x=z, yz=x,

z* = y, then prove that xyz = 1.

(i -/3V ( e ) Represent the number I p p p 1 in a plane by

expressing it in the form a + ib where a and b are real and

i = \f - 1.


4. Attempt any three : — ( a ) If a # 0, solve the equation ax2 + bx + c = 0.

State the sum and product of its roots.

( b ) Solve the equation - 3 ( 2*+2 ) + 32 = 0. ( c ) If A: is real, find its possible values for which

x2 + kx + 4 = 0 has no real root. ( d ) Find the equation whose roots are the reciprocals of

the roots of x3 + px + q = 0. 5. Attempt any three:—

( a ) With the usual notation prove that

( b ) A number of 4 different digits is formed by using the digits 1, 2, 3 ,4, 5, 6, 7 in all possible ways. Find —

( i ) how many such numbers can be formed and ( i i ) how many of them are greater than 3,400. ( c ) Prove, with the usual notation, that

fe=i r ~ 1

Find the sum when r = 1.

( d ) If 2 a, = 4 - ( 3 " - 1 ) for all n, find an. r= 1 ^

Show that - f l"+ 1 is constant. Find the value of a7. aM

6. Attempt any three :— ( a ) State and prove the Binomial Theorem for a

positive integral index. (b) If 2nd, 3rd and 4th terms in the expansion of

( x + y)" are 240, 720 and 1080 respectively, find x, y and rt. ( c ) Prove by Induction that

(a" — b") is divisible by (a — b). ( d ) Find the sum to n terms of the series

i 3 . I ' + a ' i ' + a ' + a ' , 1 1 + 2 1 + 2 + 3

B. u . t>APER 1 9 7 2 •. 3 9 5

Answers

i . ( 6 ) X = ) - 4 . - 3, - 2. - 1 . 0 . 1 , 2 . 3 , 4 A = } - 4. - 2. 0. 2, 4 B = ) 2, 4 ( i ) A u B = ) - 4, - 2. 0. 2, 4 ( i i ) ADB = ) 2, 4 (iii) B ' = } - 4, - 3. - 2. - 1. 0. 1, 3. 5 f . (iv) )2,4\, \2\,)4{.<t>.

( c ) See the solution of Ex. 12 ( i ) on page 40.

( d ) In the follwing venn-diagram S denotes the universal set of

9000 students. A denotes the set of 7400 students who have passed in Paper 1. B denotes the set of 6,600 students who have passed in Paper I I . A fl B contains 6400 students who have passed in both the papers.

f rom the diagrom, the number of students who have passed in both or in either paper = 6400 + 1000 + 200 = 7600.

the number of students who have failed in both the papers =» 9000 - 7600 = 1400,

Alternative Method,

We want A' fl B ' . By the result in (c ) above, we have A' n B ' = (A u B )' = 1400, since A ( j B = 7600 and S = 9000.

2. ( a ) Rational number = V 2 •25 i .e . 1-5; irrational nnmber x1 - 4* - 10 = ( a - 2 )2 - 14. Least value = - 14 V ( x - 2 ) » ^ 0.

l/s is / 1/5 15/ 11> ( 6 ) 2 ' = V32 and 3 - / 2 7 . 2 > 3 ; * = 9/4. ( e ) Expression = 1.

3 . ( e ) We get a = 0 and b = - 2. .'. the point (0, — 2 ) represents the number in Argand's diagram.

4. (6 ) * = 2 or * = 3 ; (c ) - 4 < k < 4 ; (d) qx* + fix* + 1 = 0 .


5. { b) ( i ) 'p, i. e. 840. ( i i ) 560,

(d)an~ 3 " " 1 -2H±L = 3 . a, = 3« i, e. 729. OH

6. (b) x = 2, y — 3 and n = 5 (d) 1 > (w + 2> .

Test Paper VH-University of Bombay, 1 9 7 3

1. Attempt any three:—

(а) State with reasons whether the following statements are true or false. If they are false, correct them :— ( i H 1 , 2 , 3 , 4 , 5 ( Utf = JO, 1 ,2 ,3 ,4 , 5 ( , ( i i ) J.5 ( e } 1 ,5 (iii) There are 16 subsets of a set having 4 elements. (iv) A = } 1,2 \ and B = ) 1, 2, 1, 2, 1 ,2 {, the = 5 .

(б) ( i ) Define the complement A' of a set A with res-pect to the Universal set S.

( i i ) If A = J 1, 2, 3 f, B = $ 2, 3, 4 \ and the Universal set S = ) 1, 2, 3, 4, 5, 6, 7 {, then verify that

(A U B)' = A' fl B'. Draw a Venn-diagram to illustrate this result.

(c) Give an example of three sets A, B and C such that AnB <t>,B0C4= <f>&nd AO C<£(f> but Aft BftC=<l>.

(d) « a - 4 * + 3 = 0 | , B = \x | sc2 — 7x + 1 2 = 0 \ and C as j x | x is a positive even integer less than 6 j ,

verify that A U (B fl C) = (A U B)0(A U C).

2. Attempt any three :— (a) Define a rational number and show that V~2~ can-

not be a rational numbeh (b) A = } - 1 , 0 , 1{. B = \ 2, 22, 2\ 2*

examine whether the sets A and B are closed under the opera-tions of addition and multiplication.

(c) ( i ) If m and n are positive rational numbers,

show that is a rational number between 1 and 2. m+ n

6. PAPER 19?3 : 397

( i i ) !f 7m - 3 > ( m + 1 )2 > 5m - 1, find the integral value of m.

(d) Find real numbers x and y such that (2~i)x + (l+3i)y + 2 = 0.

(e) I f a = b+ \[ c where b is rational and \ja,

\J c are quadratic surds, then prove that b = 0 and a = c.


(a) Define log„ N, stating the usual restrictions put on N and a. Prove, with usual restrictions on N, a, b, that

tafctf-jSi^. a log5 a

(b) Without using logarithmic tables, prove that

~ < log10 5 < •

(c) If a2 + i 2 = lab, show that

2 I o g | y ( f l + i ) | = ( log « + log b).

(</) Show that ^ ( 3 7 - 2 0 ^ 1 i s r a t i o n a l .

5 —2 \f 3

(c) I f (2 -381 )* = ( -2381 )" =10", prove that

J L - i + i . x y z

4. Attempt any f/iree

(a) Solve the quadratic equation ax2 + bx + c= 0. If n and /S are the roots of this equation, express <x/3 and «2 4- in terms of a.ft.c. Hence find the equation whose roots are

« 0

(4) If 3 + 5/ is one root of the quadratic equation ax2-6x + c=s 0, find the values of a and c. a I


(c) If two roots of the equation x3 + px* + qx r = 6 are equal and of opposite sign, show that pq=r.

(d) If ( 1 + 2 / ) ( 2 - 0 ( 2 + i ) _ 1 —a + ib, where a and b are real and i = \f — 1, find a and b and plot the points a + ib and a — ib in a plane mentioning the necessary set of axes.

• ( e ) I f 3 2 * + 1 - 3 * = 3 * + 3 - 9 . f i n d * .


(a) Assuming the formula for nPr, obtain a formula for the number of combinations of n different things taken r at a time.

(b) How many of the numbers formed by using all of the digits 2, 5, 6, 8 ,9 are divisible by 2 or 5 ?

(c) From 7 teachers and 4 students a committee of 6 is to be formed; in how many ways can this be done, (1) when the committee contains exactly 2 students, (2) at lest 2 students ?

(d ) Prove by the method of induction or otherwise 4 + 1 4 + 30 + 5 2 + + n ( 3 « + l ) = « (n + 1)2.

6. Attempt any three :— (a) State and prove the Binomial Theorem for a positive

integral index. (b) Find the term independent of x in

( T x 2 - i (c) Use the method of induction to prove that n2(n + 1 ) 2

is divisible by 4 for all positive integral values of n. (d) Find the sum to n terms of the series :—

3 + 33 + 333 + 3333+

Answers 1. ( a ) ( i ) False. •: A A. If the set on L. H. S. is A, and the

set on R. H. S. is B, A 4= B as OeB but O g A .

Correct statement = j 1, 2, 3 , 4 , 5 j U 0 = i 1. 2, 3, 4, 5 S. ( i i ) False. \ 5 [ is a set; it cannot belony to J 1, 5

Correct statement : 5 e \ 1. 5 [ or ) 5 ( C \ 1, 5

( i i i ) True. If a set has n elements it has 2n subsets.

B. u . PAPER 1973 : 399

( iv) True. There are two elements 1, 2 in both the sets. They are repeated in B. Repetition does not affect equality.

(b) (H) A u B = 5 1,2,3.4 (A u B ) ' = j 5, 6, 7 }. A' = } 4, 5, 6, 7 B' = ) 1, 5, 6. 7 A ' O B ' = )5, 6, 7\.

(A u B ) ' = | 5, 6, 7 ( A ' U B ' = $ 5 , 6 . 7 }

( c ) A = $ 1 . 2 B = $ 2 , 3 C = $ 3 , 4

( i ) A = j l , 3 i , B = S 3 , 4 i , C = ! 2 1 4 ! , B n c = ) 4 j , A U ( B D C ) = | 1 , 3 , 4 } -

A u B = ) 1, 3, 4 A u C = \ 1, 2, 3, 4

( A ( j B ) f l ( A u C ) = } l , 3 , 4 j, Hence the result.

2 . (6 ( A = \ - 1, 0, 1 It can be easily verified that if x, y e A, (x + y) e A and xy e A. Hence A is closed under addition and multiplication.

B = )x]x = 2n , » e N 2* + 28 = 1 + 2 ) = 2a-3 £ B. -Hence

B is not closed under addition. If m, n € N, 2m- 2" = 2"*+" e B and 2"'eB If

2 e B. Hence B is closed under multiplication.

( c ) ( i ) m , n e Q ( w + 2 w ) e Q and (»» + » ) e Q. m + 2» V m + w =j= 0 as m, n > 0, m + n

m + 2n

€ Q.

Show that m + - 2 — - 1 > 0 and 2 - '" ^ — > 0 (m, n > 0) . m + ft m + n (ii) (m + 1 )s < 7m - 3 => m8 - 5m + 4 < 0

(w - 1) (m - 4) < 0=> 1 < m < 4. ( i )

4 0 0 : C O L L E G E ALGEBRA

Also ( m + 1 )a > 5m — 1 m1 — 3m + 2 > 0 (m - 1) (m - 2 ) > 0=> m < 1 or m > 2. . . . ( i i )

F rom ( i ) and ( i i ) m •= 3,

( d ) Equating real and imaginary parts we get

2x + y + 2 = 0 and - x + 3y - 0.

Solving ' * and y •

3 . ( b ) 10a < 5* 2 < 3 log,0 5;' 54 < 10s 4 log,,, 5 < 3. J

( c ) The result is true if | - j - ( a + b ) j = ab i. e. if a' + b' = lab.

Id) EXD , (5 + 2^3)° (37 - 20Vf) (d) Exp. fr

= - j y ( 3 7 + 20VF) . (37 - 20^3")

' - - j | ( 3 7 ) ' - (20 -JT) \ = a rational.

(«) x log (2-381) = y log (-2381 ) = z log 10 = z.

Use now the fact 2-381 = 10 x -2381.

4 . (a) Equation with roots <x//8 and /8/<x i s a c * ' - ( 6 ' - 2 a c ) * + a c = 0.

(6 ) 3 — 5 * is also a root, a = 1, c = 34.

( c ) If « , - « and /3 are the roots then /S = - p, - « a = q, — oca j5 ™ - r,

Hence pq = r.

(d) Exp, = , ••• « = - j - , 6 = • Plot the points

^ taking the 1 and y axes as the axes of reals and

imaginary numbers respectively.

x 3 -t * ( e ) Put 3 = y. Then 3y - y = 2 7 y - 9 ^ > y = 3 or 3 =i> * = - 1 or 2

5. ( 6 ) 3-4 ! + 1-4 ! = 96. ( c ) ( 1) 210, ( 2 ) 210 +.140 + 21 = 371.

6. (6) 7/18.

(c ) Hin t . If a H = « ' ( » » + 1 ) ' , a „ + 1 - a n = 4 ( n + 1)».

< < o £ ( 1 0 " - l

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