mathematics 0n1 answer: fa b ng · 3.let a = fa,b,c,d,eg, b = fc,e,fgand c = fa,d,eg. which of the...

Mathematics 0N1

Solutions 11

1. Write the following sets in list form.1(i) The set of letters in the word “banana”.

Answer: {a, b, n}. �

1(ii) {x : x2 + 3x − 10 = 0}.Answer: {−5, 2} �

1(iii) {x : x an integer and 3 < x < 10}.Answer: {4, 5, 6, 7, 8, 9} �

2. Write the following sets in predicate form.2(i) {a, e, i , o, u }.

Answer:

{x : x is a vowel in English alphabet}�

2(ii) {√

2, −√

2 }.Answer: {x ∈ R : x2 − 2 = 0} �

2(iii) {5, 7, 9, 11, 13, 15, 17 }.Answer:

{x ∈ Z : 5 ≤ x ≤ 17 and x is odd}�

3. Let A = {a, b, c, d , e}, B = {c, e, f} and C = {a, d , e}. Whichof the following statements are true?

3(i) c ∈ A. Answer: True �

3(ii) f ∈ A. Answer: False �

1Homework 1 is due in Week 2.More generally, Homework N is due in Week N + 1.

1

2

3(iii) B ⊆ A. Answer: False �

3(iv) C ⊆ A. Answer: True �

3(v) B = {e, e, f , c}. Answer: True �

3(vi) ∅ ∈ B. Answer: False �

3(vii) ∅ ⊆ B. Answer: True �

4. Find all relations =, ⊆ or ∈ between pairs of

{r , t , s}, {s, t , r , s}, {t , s, t , r}, {s}, s.

Answer: s ∈ {r , t , s}, s ∈ {s, t , r , s},s ∈ {t , s, t , r}, s ∈ {s};{s} ⊆ {r , t , s}, {s} ⊆ {s, t , r , s},{s} ⊆ {t , s, t , r}, {s} ⊆ {s};{r , t , s} = {s, t , r , s} = {t , s, t , r}.Notice that if two sets A and B are equal then A ⊆ B and

B ⊆ A. This means that{r , t , s} ⊆ {s, t , r , s},{s, t , r , s} ⊆ {r , t , s},{r , t , s} ⊆ {t , s, t , r},{t , s, t , r} ⊆ {r , t , s},{s, t , r , s} ⊆ {t , s, t , r},{t , s, t , r} ⊆ {s, t , r , s}. �

5. Find all subsets of the set {w , x , y , z}.Answer: ∅, {w}, {x}, {y}, {z}, {w , x}, {w , y}, {w , z}, {x , y},{x , z}, {y , z}, {w , x , y}, {w , x , z}, {w , y , z}, {y , x , z}, {w , x , y , z}.

This makes 16 = 24 subsets altogether. �

0N1 MATHEMATICS • SOLUTIONS 2 • 04DEC18 3

0N1 MathematicsSolutions 2

1.1(i) If A ⊆ B is it necessarily true that B ⊆ A? Answer: NO �

1(ii) If A ⊆ B and A ⊆ C is it necessarily true that B = C? Answer:

NO �

1(iii) If A ⊆ B and B ⊆ C is it necessarily true that A ⊆ C?Answer: YES �

1(iv) If A ⊂ B is it necessarily true that A ⊆ B? Answer: YES �

1(v) If A ⊆ B is it necessarily true that A ⊂ B? Answer: NO �

1(vi) If A ⊂ B and B ⊆ C is it necessarily true that A ⊂ C?Answer: YES �

1(vii) If x ∈ B and A ⊆ B is it necessarily true that x ∈ A? Answer:

NO �

1(viii) If x ∈ A and A ⊆ B is it necessarily true that x ∈ B? Answer:

YES �

2. Which of the following sets are finite, and which infinite?

2(i) The set of even negative integers. Answer: INFINITE �

2(ii) {x : x ∈ Z and x2 < 9}. Answer: FINITE �

4

2(iii) {x : x ∈ R and x2 < 9}. Answer: INFINITE �

(i(v)[

12 , 1]. Answer: INFINITE �

2(v){

12 , 1}

. Answer: FINITE �

3. Let the universal set be the English alphabet (all 26 letters),i.e.

U = {a, b, c, . . . , z}.Let

A = {a, b, c, d , e},

B = {b, d , f , h, j},

C = {j , k , l , m, n}.Find

3(i) A ∩ B Answer: {b, d} �

3(ii) A ∩ C Answer: ∅ �

3(iii) B ∩ C Answer: {j} �

3(iv) A ∪ B

Answer:{a, b, c, d , e, f , h, j}

�

3(v) B ∪ C

Answer: {b, d , f , h, j , k , l , m, n} �


3(vi) (A ∪ B ∪ C)′

Answer:

{g, i , o, p, q, r , s, t , u, v , w , x , y , z}�

4. Let the universal set be the set of all integers, i.e. U = Z. Let

A = {x ∈ U : x is odd },B = {x ∈ U : x2 > 25},C = {x ∈ U : x is negative }.

Find

4(i) A ∩ B ∩ C

Answer:

A ∩ B ∩ C = {. . . ,−13,−11,−9,−7}.It can also be described as

A ∩ B ∩ C= {x ∈ Z : x is odd and x < −5}.

�

4(ii) A ∩ B′

Answer: A ∩ B′ = {−5,−3,−1, 1, 3, 5} �

4(iii) A ∩ C ′

Answer:

A ∩ C ′ = {1, 3, 5, 7, 9, 11, . . . }�

4(iv) B ∪ C

Answer:{. . . ,−7,−6,−5,−4,−3,−2,−1, 6, 7, . . . } �

5. Let A be any subset of a universal set U. Find

6

5(i) U ∩ A

Answer: A �

5(ii) A ∪ A

Answer: A �

5(iii) ∅′

Answer: U �

5(iv) ∅ ∪ A

Answer: A �

5(v) A′ ∩ A

Answer: ∅ �

5(vi) U ′

Answer: ∅ �

5(vii) U ∪ A

Answer: U �

5(viii) A′ ∪ A Answer: U �

5(ix) A ∩ A Answer: A �

5(x) ∅ ∩ A Answer: ∅ �

6. By drawing Venn diagrams, decide whether or not the follow-ing statements are always true when A, B and C are subsetsof a universal set U.


(i) (A ∩ B)′ = A′ ∩ B′

Answer: FALSE. �

(ii) (A ∩ B ∩ C)′ = A′ ∪ B′ ∪ C ′

Answer: TRUE. �

(iii) (A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A) = (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A)

Answer: TRUE. See Figure 1 on page 8. �

(iv) A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C

Answer: TRUE. See Figure 2 on page 9. �

Much harder (non-compulsory) questions

7. If elements of an infinite set A can be numbered by naturalnumbers 1, 2, 3, . . . , it is called countable. Examples:

– The set N of all natural numbers – its elements are theirown numbers.

– The set of even natural numbers E = {2, 4, 6, 8, . . . }: if

e ∈ E , its number ise2

.Prove that the following sets are countable:

(i) The set of all odd natural numbers.(ii) The set Z of integers.(iii) The set of all odd integers { · · · − 3,−1, 1, 3, 5, . . . }.(iv) The set of all possible sentences in English language.

[Answer to this question is likely to involve a very practi-cally useful concept: lexicographical order.]

8

FIGURE 1. Venn diagrams for Question 2.6(iii). Top row, fromleft to right:A ∪ B, B ∪ C, C ∪ A, (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A)Bottom row, from left to right:A ∩ B, B ∩C, C ∩ A, (A ∩ B) ∪ (B ∩C) ∪ (C ∩ A) As you cansee, the last two diagrams are the same. This proves that

(A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A) = (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A)

is true. �


FIGURE 2. Venn diagrams for Question 2.6(iv). Top row, fromleft to right:A, B ∪ C, A ∩ (B ∪ C)Bottom row, from left to right:A ∩ B, C, (A ∩ B) ∪ CAs you can see, the last set in the top row is part of the last set inthe bottom row. This proves that

A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C

is true. �

0N1 Mathematics

Solutions 3

1. Using the laws of Boolean algebra simplify the following ex-pressions.

(i) (A′ ∩ B′)′

Answer:

(A′ ∩ B′)′by (8)

= (A′)′ ∪ (B′)′

by (7)= A ∪ B

�

(ii) (A ∪ B) ∪ A′

10

Answer:

(A ∪ B) ∪ A′by (1)

= (B ∪ A) ∪ A′

by (3)= B ∪ (A ∪ A′)

by (7)= B ∪ U

by (6)= U

�

(iii) A ∪ (A′ ∩ B)

Answer:

A ∪ (A′ ∩ B)by (4)

= (A ∪ A′) ∩ (A ∪ B)by (7)

= U ∩ (A ∪ B)by (1)

= (A ∪ B) ∩ Uby (6)

= A ∪ B

�

(iv) A′ ∩ (A′ ∪ B)′

Answer:

A′ ∩ (A′ ∪ B)′by (8)

= A′ ∩ ((A′)′ ∩ B′)by (7)

= A′ ∩ (A ∩ B′)by (3)

= (A′ ∩ A) ∩ B′

by (1)= (A ∩ A′) ∩ B′

by (7)= ∅ ∩ B′

by (1)= B′ ∩ ∅

by (6)= ∅

�

2. Let A = {x ∈ R : x − 1 > 2 and x < 4} and let

B = {x ∈ R : 5 ≤ x2 ≤ 20}.Prove that A ⊆ B.


Answer: Let x ∈ A. Then x − 1 > 2 and x < 4. Hence x > 3and x < 4. Hence x2 > 9 and x2 < 16. Hence x2 ≥ 5 andx2 ≤ 20. Hence x ∈ B.

So every element of A belongs to B. Therefore A ⊆ B. �

3. Prove that {x ∈ R : x2 − 3x + 2 < 0} = (1, 2). (Here (1, 2) isthe open interval from 1 to 2.)

Answer: Write A = {x ∈ R : x2− 3x + 2 < 0} and B =]1, 2[.Let x ∈ A. Then

x2 − 3x + 2 < 0,

so

(x − 2)(x − 1) < 0.

Hence EITHER x − 2 > 0 and x − 1 < 0 OR x − 2 < 0 andx − 1 > 0. The first is impossible. Therefore x − 2 < 0 andx − 1 > 0. hence 1 < x < 2. Hence x ∈ B. Therefore A ⊆ B.�

4. A survey of 25 people was made to ask if they liked brandsA, B and C.

(1) 6 people liked A and B.

(2) 7 people liked B but not A.

(3) 2 people liked all three brands.

(4) 7 people liked B and C.

(5) 3 people liked A and C but not B.

(6) 5 people liked A only.

(7) 15 people disliked C.How many people liked B only?How many people liked C but not A?

Answer: Denote by a, b, c, d , e, f , g, h the number of peoplein each region of the Venn diagram as shown.

12

ab

c

f

d

eg

h

A B

C

(0) a + b + c + d + e + f + g + h = 25

(1) d + g = 6

(2) b + e = 7

(3) g = 2

(4) e + g = 7

(5) f = 3

(6) a = 5

(7) a + b + d + h = 15.From (3), (5), (6), g = 2, f = 3, a = 5. Therefore, from (1),d = 4. From (4), e = 5. From (3), b = 2. From (7), h = 4. From(0), c = 0.

Hence number of people who like B only is b = 2. Numberof people who like C but not A is c + e = 5. �

5. Construct truth tables for each of the following expressions.

(i) p →∼ p

Answer:p ∼p p →∼pT F FF T T

�


(ii) ∼ (∼ p ∧ q)

Answer:

p q ∼p ∼p ∧ q ∼ (∼p ∧ q)T T F F TT F F F TF T T T FF F T F T

�

(iii) p∨ ∼q

Answer:

p q ∼q p∨ ∼qT T F TT F T TF T F FF F T T

�

(iv) (p ↔ q) ∧ p

Answer:

p q p ↔ q (p ↔ q) ∧ pT T T TT F F FF T F FF F T F

�

(v) (p → (q∨ ∼ r )) ∧ (p → (q∨ ∼ r ))

Answer:

p q r ∼ r q∨ ∼ r p → (q∨ ∼ r ) (p → (q∨ ∼ r )) ∧ (p → (q∨ ∼ r ))T T T F T T TT T F T T T TT F T F F F FT F F T T T TF T T F T T TF T F T T T TF F T F F T TF F F T T T T

�

(vi) ((p → q)→ p)→ q

Answer:

p q p → q (p → q)→ p ((p → q)→ p)→ qT T T T TT F F T FF T T F TF F T F T

�

14

Much harder (non-compulsory) questions

5. It frequently happens that connectives can be expressed interms of other connectives – we already know, for example,that p → q ≡∼ p ∨ q, that is, conditional can be expressed interms of disjunction and negation.(i) Express conjunction in terms of disjunction and negation.(ii Express disjunction in terms of conditional and negation.

(iii) Hence express conjunction in terms of conditional andnegation.

(iv) Try to rewrite Boolean Laws in terms of conditional andnegation. You will see that this is possible, but the log-ical identities that you get that you get are ugly and notvery intuitive. The Boolean Laws have been chosen inthe way they are for good reason: they are simple andtransparent.

6. Recall that a statement is called a tautology if takes truthvalue T regardless of truth values of simple component state-ment involved in it. For example,

p∨∼ p, p → p, and p → (q → p)

are tautologies, but p → (p → q) is not.For statements built from component statements using only

conditional – one example is

(p → q)→ ((q → r )→ (o → (r → q))),

there is quick way of easily checking whether it is a tautologyor not. Find it.

Answer: Hint: p → q is F if and only if p ≡ T and q ≡ F .Given a statement built conditional only, try to analyse how itcould happen to be F . Start with a relatively short statementwith just two “variables”, that is, simple statements involved.�

7. Try to figure out how to produce a statement with a giventruth table. For example, can you find a formula of BooleanAlgebra with the following truth table?

p q ?T T FT F FF T TF F F


(This problem lies at the heart of the digital circuits theory inelectronic engineering.)

16

0N1 Mathematics

Solutions 4

1. Determine the truth value of each of the compound state-ments (a), (b) and (c) given the following information:

(i) The statement “Mr Black is taller than Mr Blue” is true.

(ii) The statement “Mr Green is shorter than Mr White” isfalse.

(iii) The statement “Mr Blue is of average height” is true.

(iv) The statement “Mr Brown has the same height as MrBlack” is false.

(a) If Mr Black is taller than Mr Blue, then Mr Green is notshorter than Mr White.

Answer: T .Let p be “Mr Black is taller than Mr Blue” TLet q be “Mr Green is shorter than Mr White” FLet r be “Mr Blue is of average height” TLet s be “Mr Brown has same height as Mr Black” Fp →∼q.p q ∼q p →∼qT F T T �

(b) If either Mr Brown does not have the same height as MrBlack or Mr Green is shorter than Mr White, then Mr Blueis of average height

Answer: T .(∼s ∨ q)→ r .

s q r ∼s ∼s ∨ q (∼s ∨ q)→ rF F T T T T �

(c) If Mr Brown does not have the same height as Mr Black,then either Mr Green is shorter than Mr White or Mr Blueis not of average height.

Answer: F .∼s → (q∨ ∼ r ).

s q r ∼s ∼ r q∨ ∼ r ∼s → (q∨ ∼ r )F F T T F F F �


2. Which of the following statements are tautologies?

(i) (p ∨ q)→ (q ∨ p)

Answer: Tautology.p q p ∨ q q ∨ p (p ∨ q)→ (q ∨ p)T T T T TT F T T TF T T T TF F F F T

�

(ii) p → ((p ∨ q) ∨ r )

Answer: Tautology. Always T when p is F . So we onlyneed to consider the case where p is T . But then (p∨q)∨ris T . Hence p → ((p ∨ q) ∨ r ) is T . �

(iii) p → (q → (q → p))

Answer: Tautology.p q q → p q → (q → p) p → (q → (q → p))T T T T TT F T T TF T F F TF F T T T

�

(iv) ((p → q)↔ q)→ p

Answer: Not a tautology.p q p → q (p → q)↔ q ((p → q)↔ q)→ pT T T T TT F F T TF T T T FF F T F T

�

(v) (p ∧ q)→ (p ∨ r )

Answer: Tautology. If p is F then p∧q is F so (p∧q)→(p ∨ r ) is T . So we only need to consider the case wherep is T .p q r p ∧ q p ∨ r (p ∧ q)→ (p ∨ r )T T T T T TT T F T T TT F T F T TT F F F T T

�

18

(vi) (p → q)↔ (q → p)

Answer: Not a tautology.p q p → q q → p (p → q)↔ (q → p)T T T T TT F F T FF T T F FF F T T T

�

3. Which of the following statements are logically equivalent tothe statement “Either it is raining or it is snowing”. [USETRUTH TABLES OR THE FUNDAMENTAL LOGICAL EQUIVALENCES,WHICHEVER IS EASIER.]

(i) It is not the case that it is not raining and not snowing.

Answer: Let p be “it is raining”; let q be “it is snowing”.The given statement is p ∨ q.

∼ (∼p∧ ∼q) ≡∼∼p∨ ∼∼q ≡ p ∨ q.

�

(ii) Either it is raining and snowing or it is snowing.

Answer: (p ∧ q) ∨ q.

p q p ∧ q (p ∧ q) ∨ q p ∨ qT T T T TT F F F TF T F T TF F F F F

Last 2 columns are different. So

(p ∧ q) ∨ q 6≡ p ∨ q.

�

(iii) Either it is raining or it is snowing and not raining.

Answer:

p ∨ (q∧ ∼p) ≡ (p ∨ q) ∧ (p∨ ∼p) ≡ (p ∨ q) ∧ T ≡ p ∨ q.

�


(iv) It is not raining and not snowing.

Answer:

∼p∧ ∼q ≡∼ (p ∨ q).

This has opposite truth values to p ∨ q. So

∼p∧ ∼q 6≡ p ∨ q.

�

(v) It is raining or it is not snowing.

Answer: p∧ ∼q.

p q ∼q p∧ ∼q p ∨ qT T F F TT F T T TF T F F TF F T F F

Last 2 columns are different. p∧ ∼q 6≡ p ∨ q. �

4. Using Boolean Laws, simplify the following statements.

(i) ∼p ∨ (p ∧ q).

Answer:

∼p ∨ (p ∧ q) ≡ (∼p ∨ p) ∧ (∼p ∨ q)≡ T ∧ (∼p ∨ q) ≡∼p ∨ q.

�

(ii) p → (q → p)

Answer:

p → (q → p) ≡ ∼p ∨ (q → p)≡ ∼p ∨ (∼q ∨ p)≡ ∼p ∨ (p∨ ∼q)≡ (∼p ∨ p)∨ ∼q≡ T∨ ∼q≡ T

�

20

(iii) (∼p ∧ q)∨ ∼ (p ∨ q)

Answer:

(∼p ∧ q)∨ ∼ (p ∨ q) ≡ (∼p ∧ q) ∨ (∼p∧ ∼q)≡ ∼p ∧ (q∨ ∼q)≡ ∼p ∧ T≡ ∼p.

[Using distributive law backwards!] �

(iv) ∼p ∧ ((p ∧ q) ∨ (p ∧ r ))

Answer:

∼p ∧ ((p ∧ q) ∨ (p ∧ r )) ≡ ∼p ∧ (p ∧ (q ∨ r ))≡ (∼p ∧ p) ∧ (q ∨ r )≡ F ∧ (q ∨ r )≡ F .

[Using distributive law backwards!] �

Somewhat harder (and non-compulsory) questions

5. Prove, without constructing full truth tables, that the followingstatements are tautologies.

(i) ∼ (p → q)→ p

(ii) p → (q → p)

(iii) (p ∧ (p → q))→ q

(iv) p → ((p → q)→ q)

(v) (p → q) ∨ (q → p)

(vi) ((p → r ) ∧ (q → r ))→ ((p ∨ q)→ r )

(vii) ((p → r ) ∨ (q → r ))→ ((p ∧ q)→ r )

6. Prove (or at least explain) that two statements p and q arelogically equivalent if and only if p ↔ q is a tautology.


7. Many of tautologies involving the conditional → are expres-sions of valid arguments. For example, 5(ii),

(p ∧ (p → q))→ q,

means that“if p is true and, as soon as p is true, q is also true,then q is true”,

which can be illustrated by a real life example:“If it is Friday night, then, since we know that on Fri-day nights John is in a pub, we have no choice butconclude that John is in a pub.”

For every statement in Question 5, find similar real life exam-ples of logically valid argumentation.

22

0N1 Mathematics

Solutions 5

1. In each of the following find values of x and y which makep(x , y ) true and find values which make p(x , y ) false. (Onlyone example of each is required.)

(i) p(x , y ) denotes x2 + y2 = 2.

Answer: p(1, 1) is true and p(0, 0) is false. �

(ii) p(x , y ) denotes x > y − 3.


(iii) p(x , y ) denotes x + y 6= xy .


2. Let p(x), q(x , y ), r (x , y ) and s(x , y ) denote the predicates x >0, x > y , x = y and x + y = 2, respectively. Find whether thefollowing statements are true or false.

(i) p(2) ∧ (q(1, 1)→ r (2, 0)).

Answer: True, because p(2) is true and q(1, 1) is false. �

(ii) p(−1)∨ ∼p(1).

Answer: False, because p(−1) is false and p(1) is true.�

(iii) ∼ (q(2, 1) ∧ s(1, 1)).

Answer: False, because q(2, 1) and s(1, 1) are both true.�

3. Write the following statements using quantifiers (e.g. (∀x)x2 >0):

(i) For all A and B, A ∩ B = B ∩ A.

Answer: (∀A)(∀B)A ∩ B = B ∩ A. �

0N1 MATHEMATICS • SOLUTIONS 5 • PREDICATE LOGIC • 04DEC18 23

(ii) x + y = y + x for all x and y .

Answer: (∀x)(∀y )x + y = y + x . �

(iii) 2x > 50 for some x .

Answer: (∃x)2x > 50. �

(iv) There exist A and B such that A 6= B.

Answer: (∃A)(∃B)A 6= B. �

(v) Given any x there exists y such that y < x .

Answer: (∀x)(∃y )y < x . �

(vi) There exists x such that x 6 y2 for all y .

Answer: (∃x)(∀y )x 6 y2. �

4. Let p(x , y ) denote the predicate “Person x answered questiony ”. Write the following statements using predicate notation(e.g. (∀x)(∃y )p(x , y )):

(i) There was one question which was answered by every-one.

Answer: (∃y )(∀x)p(x , y ). �

(ii) Everyone answered at least one question.

Answer: (∀x)(∃y )p(x , y ). �

(iii) All questions were answered by everyone.

Answer: (∀y )(∀x)p(x , y ). �

(iv) There was somebody there who answered all the ques-tions.

Answer: (∃x)(∀y )p(x , y ). �

24

5. Let p(x , y ) be as in Exercise 4. Suppose Ann, Bill, Carol andDick answered from a set of questions numbered 1, 2, 3, 4 asshown.

1 2 3 4Ann

√ √ √

Bill√

CarolDick

√ √ √ √

Which of the following statements are true?

(i) (∀x)(∀y )p(x , y ) Answer: False �

(ii) (∀x)(∃y )p(x , y )

Answer: False; Carol answered no questions. �

(iii) (∃y )(∀x)p(x , y ) Answer: False �

(iv) (∃x)(∀y )p(x , y )

Answer: True; this person is Dick. �

(v) (∀y )(∃x)p(x , y )

Answer: True; every question is answered by someone.�

(vi) (∃x)(∃y )p(x , y ) Answer: True �


0N1 Mathematics

Solutions 6

1. Find whether the following statements are true or false wherethe universal set is

Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . . }with the usual operations and inequality relations.

(i) (∀x)(∀y )(∃z)z < x + y

Answer: True. For every given x and y you may takez = x + y − 1. �

(ii) (∃z)(∀x)z < x4

Answer: True; for example, z = −1. Then

z = −1 < 0 6 (x2)2 = x4

(since squares are non-negative – see the Lecture Notes).Hence z = −1 < x4 for all integers x . �

(iii) (∀x)(∃y )y2 = x

Answer: False; x = 2 is a counterexample; number 2 isnot a square of any integer. �

(iv) (∀y )(∃x)y2 = x

Answer: True. It simply says that a square of an integeris an integer. �

(v) (∃x)(∀y )x + y = 2

Answer: False. The statement claims that there is an in-teger x such that x + y = for all integers y ; in particular, itmeans that it is true for y = 0 and y = 1, that is, x + 0 = 2and x + 1 = 2. But then 0 = 1 – a contradiction. Hencesuch x does not exist. �

(vi) (∃z)(∃y )yz = 6.

Answer: True; for example, z = 1 and y = 6. �

26

2. Let U be a universal set which contains at least 2 differentelements. Let x stands for any elements of U and let A and Bstand for any subsets of U. Which of the following are true?

(i) (∀A)(∀x)(x ∈ A ∨ x ∈ A′)

Answer: True. In plain English, the statement isFor all sets A and all elements x, x belongs toA or x does not belong to A.

This is obviously always true because P∨ ∼ P is a tau-tology, every statement of that form is always true. Forexample, the statement

I live on the Moon or I do not live on the Moonis true.Our particular statement

(∀A)(∀x)(x ∈ A ∨ x ∈ A′)

is logically equivalent to

(∀A)(∀x)((x ∈ A)∨ ∼ (x ∈ A))

(because x ∈ A′ is the same as ∼ (x ∈ A)). But, as wejust discussed,

(x ∈ A)∨ ∼ (x ∈ A)

is always true, and therefore

(∀A)(∀x)((x ∈ A)∨ ∼ (x ∈ A))

is true. �

(ii) (∀A)(∃x)x ∈ A

Answer: False; A = ∅ is a counterexample. �

(iii) (∀A)(∀B)(A ⊆ B ∨ B ⊆ A)

Answer: False. However, the assumption that U con-tains at least 2 elements is essential: the statement isT if U is empty or contains only one element. �

(iv) (∀A)(∃B)A ⊆ B.

Answer: True; take B = U (or B = A). �


(v) (∀A)(∀B)(A ⊂ B → (∃x)(x ∈ B∧ ∼ (x ∈ A)))

Answer: True �

3. Simplify the following statement.

(i) ∼ ((∀x)(∃y )(p(x , y ) ∧ (∃z) ∼q(z)))

Answer:

∼ ((∀x)(∃y )(p(x , y ) ∧ (∃z) ∼q(z))) ≡ (∃x) ∼ (∃y )(p(x , y ) ∧ (∃z) ∼q(z))≡ (∃x)(∀y ) ∼ (p(x , y ) ∧ (∃z) ∼q(z))≡ (∃x)(∀y )(∼p(x , y )∨ ∼ (∃z) ∼q(z))≡ (∃x)(∀y )(∼p(x , y ) ∨ (∀z) ∼∼q(z))≡ (∃x)(∀y )(∼p(x , y ) ∨ (∀z)q(z))≡ (∃x)(∀y )(p(x , y )→ (∀z)q(z))

�

(ii) (∀x) ∼ (∃y )(p(x , y )→ q(y ))

Answer:

(∀x) ∼ (∃y )(p(x , y )→ q(y )) ≡ (∀x)(∀y ) ∼ (p(x , y )→ q(y ))≡ (∀x)(∀y ) ∼ (∼p(x , y ) ∨ q(y ))≡ (∀x)(∀y )(∼∼p(x , y )∧ ∼q(y ))≡ (∀x)(∀y )(p(x , y )∧ ∼q(y ))

�

(iii) ∼ (∃x)(∀y )(∃z)p(x , y , z)

Answer: (∀x)(∃y )(∀z) ∼p(x , y , z) �

(iv) ((∃x)(∃y ) ∼p(x , y ))→ (∃x)q(x)

28

Answer:

((∃x)(∃y ) ∼p(x , y ))→ (∃x)q(x) ≡ ∼ ((∃x)(∃y ) ∼p(x , y )) ∨ (∃x)q(x)≡ (∼ (∃x)(∃y ) ∼p(x , y )) ∨ (∃x)q(x)≡ ((∀x) ∼ (∃y ) ∼p(x , y )) ∨ (∃x)q(x)≡ ((∀x)(∀y ) ∼∼p(x , y )) ∨ (∃x)q(x)≡ ((∀x)(∀y )p(x , y )) ∨ (∃x)q(x)

�

4. Find a counterexample to each of the following statements.

(i) Every even integer is divisible by 4.

Answer: For example, 2 is even but is not divisible by 4.�

(ii) If A and B are sets such that A 6⊆ B then B ⊆ A.

Answer: For example, A = {1}, B = {2}. �

(iii) If x and y are real numbers such that x2 = y2 then x3 =y3.

Answer: For example, x = 1 and y = −1. �

(iv) For all real numbers x , y and z, if x > y then xy > yz.

Answer: For example, x = 2, y = −1, z = 1. �

5. A contrapositive to the statement p → q is the statement∼q →∼p.

(i) Use truth tables to prove that the conditional statementp → q and its contrapositive ∼ q →∼ p are logicallyequivalent.

Answer: At this stage, it should be a routine truth table.�


A contrapositive to the statement

(∀x)(p(x)→ q(x))

is the statement

(∀x)(∼q(x)→∼p(x).

It follows from (i) that

(∀x)(p(x)→ q(x))

and(∀x)(∼q(x)→∼p(x)

are logically equivalent.

Write the contrapositive form of each of the following state-ments. The universal set is Z.

(ii) If x is odd then x2 is odd.

Answer: If x2 is even then x is even. �

(iii) If x2 = 4 then x < 3.

Answer: If x > 3 then x2 6= 4. �

(iv) If there exists y such that y2 = x then x > 0.

Answer: If x < 0 then for y2 6= x for every y . �

(iv) If x 6= 1 and x 6= 2 then

x2 − 3x + 2 6= 0.

Answer: If x2 − 3x + 2 = 0 then x = 1 or x = 2. �

Discuss why the contrapositive statements in (ii)–(iv) areequivalent to the original statements (that is, the both are trueor false simultaneously).

6. A converse of a conditional statement p → q is q → p.

30

(i) Show by real life examples, and/or by constructing truthtables that p → q and q → p are not logically equivalent.

Answer:

p: I am at home.q: The light is on.p → q: If I am at home then the light is on.q → p: If the light is on then I am at home.

p → q and q → p are not the same: I can leave homebut forget to switch off the light. �

(ii) Prove that is we start with a statement p → q and form itscontrapositive, and then the converse of the contraposi-tive, then the resulting statement is logically equivalent tothe contrapositive of the converse.

Answer: The contrapositive to p → q is ∼ q →∼ p, andthe converse of contrapositive is ∼p →∼q.If we now take the converse of p → q, we get q → p,and then take contrapositive of the converse, we againget the same ∼p →∼q. �


0N1 Mathematics

Solutions 7

1. Express in interval / segment notation

(i) { x : 1 ≤ x ≤ 4 }.Answer: [1, 4] �

(ii) { x : 1 < x ≤ 4 }Answer: ]1, 4] �

(iii) { x : x > 4 }Answer: ]4, +∞[ �

2. Given sets

A = {−2, 0, 1, 2, 3, 4},B = [−1, 3],C = ]− 2, 1[ ,

find(1) A ∪ B

Answer:

A ∪ B = [−1, 3] ∪ {−2, 4}�

(2) A ∩ B

Answer: A ∩ B = {0, 1, 2, 3} �

(3) A ∪ C

Answer:

A ∪ C = [−2, 1] ∪ {2, 3, 4}�

(4) A ∩ C

Answer: A ∩ C = {0} �

32

(5) B ∪ C

Answer: B ∪ C = ]− 2, 3] �

(6) B ∩ C

Answer: B ∩ C = [−1, 1[ �

3. Given intervals / segments

A = ]0, 3[ , B = ]2, 4], and C = [−1, 1],

find(1) A ∪ B

Answer: A ∪ B =]0, 4] �

(2) A ∩ B

Answer: A ∩ B =]2, 3[ �

(3) A ∪ C

Answer: A ∪ C = [−1, 3[ �

(4) A ∩ C

Answer: A ∩ C =]0, 1] �

(5) B ∪ C

Answer: B ∪ C = [−1, 1]∪ ]2, 4] �

(6) B ∩ C

Answer: B ∩ C = ∅ �

4. Find

(i) [0, 1]∪ ]1, 2]

Answer: [0, 2] �

(ii) ]−∞, 1] ∩ [0, +∞[.

Answer: [0, 1] �

5. We take R for the universal set U. Compute


(i) [1, 2] ∩ ( ]−∞, 1] ∪ [2, +∞[ ).

Answer: {1, 2 } – just two points! �

(ii) ( ]−∞, 1] ∪ [2, +∞[ )′.

Answer: The interval ]1, 2[ . �

6. [A much harder problem – not compulsory!] Suppose x isa positive real number. Prove, by contradiction, that

x +1x≥ 2.

Answer: Assume, by the way of contradiction, that

x +1x< 2.

Since x is positive, we can multiply the both sides of this in-equality by x and get

x2 + 1 < 2x ,

which can be rearranged as

x2 − 2x + 1 < 0

and then as(x − 1)2 < 0.

But squares cannot be negative – a contradiction. Hence ourassumption that

x +1x< 2

was false, which means that

x +1x≥ 2

for all positive real numbers x . �

0N1 Mathematics

Solutions 8

1. Solve the inequalities expressing the answers as segments,intervals, rays, half-lines, the whole line – or as the empty set.

(i) 2x + 1 ≥ x + 2

Answer: x ≥ 1, or x ∈ [1, +∞[. �

(ii) x + 1 < x + 2

Answer: x ∈ R.Indeed, since 1 < 2, adding an arbitrary x to the bothsides, we see that x + 1 < x + 2 is true for all x ∈ R. �

(iii) x + 1 > x + 2Answer: x ∈ ∅ (that is, no solution).

Indeed, if the inequality holds for some real number xthen, subtracting x from the both sides of the inequality,we get 1 > 2, an obvious contradiction. �

(iv) 2x < xAnswer: x ∈ ]−∞, 0[ . �

(v) x < 2xAnswer: x ∈ ]0, +∞[ .

Indeed, the equation can be rearranged as

0 < 3x ,

and, after dividing the both sides by 3, as

0 < x .

�

(vi) −2x < x + 1

Answer: −1 < 3x or

x ∈ ]− 13

, +∞[ .


�

(vii) −x > x

Answer: x ∈ ]−∞, 0[ . �

2. Solve the inequalities expressing the answers as Booleancombinations (that is unions, intersections, etc.) of segments,intervals, rays, half-lines.

(i) x < x2

Answer: x ∈ ]−∞, 0[ ∪ ]1, +∞[ . �

(ii) x ≤ x3

Answer: x ∈ [−1, 0] ∪ [1, +∞[ .

There are at least two possible approaches to this in-equality and at least two ways to solve it.

Solution 1. The inequality could be much simplified ifwe divide both its sides by x . This operation has to bedone with care.∗ We have to consider as a separate case the possi-

bility when x = 0 and we cannot divide by x .∗ We have treat separately the cases x > 0 and x < 0

when we can divide by x , but remember that· division by x > 0 does not change the direction

of inequality, but· division by x < 0 changes the direction of in-

equality.

So we have to consider three cases: x = 0, x < 0 andx > 0.

CASE 0. x = 0

Substituting x = 0 into the inequality x ≤ x3, we see that0 ≤ 03 = 0 is true, hence 0 belongs to the solution set.

CASE 1. x < 0

When dividing the both sides of the inequality x ≤ x3 byx < 0 we have to change the direction of the inequality,so we get

1 ≥ x2

36

which has the solution set [−1, 1] (or 1 ≤ x ≤ 1). Butwe have to remember our assumption x < 0 (or x ∈]−∞, 0[ ), so in this case the solutions form the set

]−∞, 0[∩[−1, 1] =]− 1, 0[ .

CASE 2. x > 0

Division by x does not change the direction of inequality,so we get

1 ≤ x2

which has the solution set

]−∞,−1] ∪ [1, +∞[ .

But we have to remember our assumption x > 0 (or x ∈]0 +∞[ ), so in this case the solutions form the set

(]−∞,−1] ∪ [1, +∞[ )∩ ]0 +∞[ = [1, +∞[ .

Now we have to assemble the three Cases 0, 1, and 2together: the solution set of

x ≤ x3

is

{0 }∪ ]− 1, 0[∪[1, +∞[= ]− 1, 0] ∪ [1, +∞[ .

Solution 2. This approach is more general and worksfor many inequalities between polynomial expressions.

We rearrangex ≤ x3

as0 ≤ x3 − x

and factorise the RHS of the inequality:

0 ≤ x3 − x = x(x2 − 1)= x(x + 1)(x − 1).

The expression x(x+1)(x−1) is zero, positive, or negativedepending of whether the individual factor x , x + 1, andx − 1 are zero, positive or negative. They turn into zerowhen x = 0, x = −1, or x = +1; it is easy to check that, intheses cases, x ≤ x3, so the points x = 0, x = −1, x = +1belong to the solution set.


The three points x = −1, 0, 1 divide the real line R into 4intervals

]−∞,−1[ , ]− 1, 0[ , ]0, 1[ , ]1, +∞[ ;

at each of these intervals the quantities x + 1, x , andx − 1 are either positive or negative but have their signsunchanged.we can easily make a list.∗ If x ∈ ]−∞,−1[ , all three of x + 1, x , and x − 1 are

negative, so their product is negative:

(x + 1)x(x − 1) < 0.

(just take x = −2).If we move the point x further in the positive directionand go over the boundary point x = −1 , x + 1 changessign, but x and x − 1 do not, so the sign of the product(x + 1)x(x − 1) changes:∗ if x ∈ ]− 1, 0[ , we have

(x + 1)x(x − 1) > 0.

Similarly, at x = 0 the quantity x changes the sign but x+1and x−1 do not, and the sign of the product (x+1)x(x−1)changes again:∗ if x ∈ ]0, 1[ , we have

(x + 1)x(x − 1) < 0.

Similarly, after another change of signs at x = 1 we have∗ if x ∈ ]1, +∞[ , we have

(x + 1)x(x − 1) > 0.

Now the solution set of the inequality

x3 − x = (x + 1)x(x − 1) ≥ 0

is the union

{−1 }∪ ]− 1, 0[∪{0 } ∪ { 1 }∪ ]1, +∞[= [−1, 0] ∪ [1, +∞[ .

�

(iii) −x ≥ x2

Answer: x ∈ [−1, 0]. �

3. Write the negations of the following inequalities.

38

(i) 2x + 1 ≥ x + 2

Answer: 2x + 1 < x + 2. �

(ii) 2x + 1 > x + 2

Answer: 2x + 1 ≤ x + 2 �

(iii) −x > x + 1

Answer: −x ≤ x + 1 �

4. Find solution sets for the negations of the following inequali-ties.

(i) 2x + 1 ≥ x + 2

Answer: The negation is

2x + 1 < x + 2,

it is equivalent to

x < 1

and has the solution set ]−∞, 1[ . �

(ii) 2x + 1 > x + 2


2x + 1 ≤ x + 2,

it is equivalent to

x ≤ 1

and has the solution set ]−∞, 1]. �

(iii) −x > x + 1


−x ≤ x + 1

and has the solution set

[−12

, +∞[ .

�

5. Solve the systems of simultaneous inequalities:


(i)

x + 1 ≥ 0x − 1 ≤ 0

Answer: x ∈ [−1, 1].Indeed, the system can be rearranged as

x ≥ −1x ≤ 1,

therefore the solution set is

{ x : −1 ≤ x ≤ 1 } = [−1, 1].

�

(ii)

x + 1 ≤ 0x − 1 ≥ 0

Answer: The solution set is empty, since the two in-equalities can be rearranged as

x ≤ −11 ≤ x

and contradict each other since they imply

1 ≤ x ≤ −1

and1 ≤ −1,

an obvious absurdity. �

(iii)

x + 1 ≤ x + 2x ≥ 0

Answer: The first inequality,

x + 1 ≤ x + 2,

holds for all x ∈ R, therefore it is only the second in-equality that matters. Hence the solution is x ≥ 0, andthe solution set is [0, +∞[ . �

40

(iv)

x + 2 ≤ x + 1x ≥ 0

Answer: The first inequality,

x + 2 ≤ x + 1,

has no solution, therefore the system of simultaneous in-equalities which includes it also has no solution. �

(iv)

2x + 2 ≤ x + 1x ≥ −1

Answer: The system is equivalent to

x ≤ −1x ≥ −1

and therefore has solution x = −1. �

The following three problems are much harder andnot compulsory.

They are loosely related to the harmonic mean andgeometric mean discussed in Lectures 15–16, andwell be discussed later. But if the rest of homeworkis too easy for you, you may wish to have a try now.No solution are provided for time being,

6. In the Manchester Airport, connections between terminalshave segments where passengers have to walk on their own,and segments where a “travelator’ ’, or a moving walkway, isprovided. A passenger is in a hurry, but needs to tie his shoe-strings. What is speedier: to stop on the usual pedestrianwalkway and tie the shoe-strings, or tie the shoe-strings whilestanding, instead of walking, on a moving walkway?

HINT: Buy a stop-watch and take bus 43 to Manchester Air-port. In a more serious way, introduce some notation: let T


be time, in seconds, needed for tieing shoe-strings, u speed,in m/sec, of the passenger walking, v speed, in m/sec, of thetravelator, and L the length of the travelator. Express the timelost in these two scenarios in terms of T , u, v , and L, andcompare the two outcomes.

7. A paddle-steamer takes five days to travel from St Louis toNew Orleans, and seven days for the return journey. Assum-ing that the rate of flow of the current is constant, calculatehow long it takes for a raft to drift from St Louis to New Or-leans.

HINT: The answer involves essentially the same expres-sion as the formula for geometric mean—but with a twist.

Answer: We need to somehow calculate the speed of thesteamer downstream and upstream, and for let us introducesome fantasy unit of length, say, lieue, and chose it such away that the distance between St Louis and New Orleans is35 lieue; then the speed of the steamer is 35÷5 = 7 lieue/daydownstream and 35 ÷ 7 = 5 lieue/day upstream. The differ-ence of upstream and downstream is twice the speed of thecurrent, hence the speed of the current is (7 − 5) ÷ 2 − 1lieue/day, and a raft will drift 35 days. �

8. Two hikers set out at sunrise and each walked with a constantspeed. One went from A to B, and the other went from B to A.They met at noon, and continuing without a stop, they arrivedrespectively at B at 4pm and at A at 9pm. At what time wassunrise on that day?

HINT: Can you imagine that the geometric mean appearsin the answer?

Answer: Assume that the two hikers walked from A to Band from B to A, respectively, and that they met at point M.Then the first hiker covered the distance from A to M in fromsunrise to noon and then distance from M to B in 4 hours.Since he walked at constant speed,

distance from A to Mdistance from M to B

=time from sunrise to noon

4 hours.

42

Similarly, for the second hikerdistance from M to Adistance from B to M

=9 hours

time from sunrise to noon.

Since it does not matter in which direction we measure dis-tance, from A to M or from M to A, etc.,

distance from A to Mdistance from M to B

=distance from M to Adistance from B to M

and consequently we get a proportiontime from sunrise to noon

4 hours=

9 hourstime from sunrise to noon

.

Solving it, we have

time from sunrise to noon =√

4× 9 =√

36 = 6 hours.

Therefore the sunrise was 6 hours before the noon, that it, at12− 6 = 6am hours. �


1. (Based on Exercise 4.3 of Schaum’s Intermediate Algebra)Graph the following lines using the intercept method.

(a) x − y = 2

(b) x + y = 2

(c) 3x + 4y = 12

(d) 2x + 4y = 8

(e) 2y − 3x = 6

(f) 2x + 3y = 4

Answer: See Figure 3 at page 44. �

2. Given a line 2x + 3y = 5, separate the points

A(−2, 1), B(1, 2), C(1,−2), D(1, 0), E(100,−1001)

in two groups that lie on opposite sides of the line.

Answer: B is one side, A, C, D, E are on another.Relative position of a point x , y with respect to a line 2x +

3y = 5 depends on the sign +, 0− of the value of the linearfunction 2x + 3y − 5 at this point. A simple calculation showsthat f (x , y ) is positive at B and negative at A, C, D, E . �

3. Graph the following linear inequalities in two variables x andy .

(a) x − 1 ≥ 3

(b) 2x + y < 2

(c) x − 3y > 3

(d) y ≥ −2


4. Graph the following systems of simultaneous linear inequali-ties in variables x and y .

(a) x ≥ 1, y ≥ −1

(b) x ≥ 1, y ≥ 1, x + y ≤ 3

44

FIGURE 3. Graphical solutions to Question 9.1.


(c) x ≥ 1, y ≥ 1, x + y ≤ 2

(d) x ≥ 1, y ≥ 1, x + y < 2


5. Solve inequalities involving the absolute value

|x | ={

x if x ≥ 0−x if x < 0

and express the answers in the interval form.

(a) |2x | < 8

Answer: x ∈ ]− 4, 4[ �

(b) | − 2x | < 8

Answer: x ∈ ]− 4, 4[ �

(c) |2x | > −8

Answer: x ∈ R �

(d) |x + 1| > 1

Answer: x ∈ ]−∞,−2[∪ ]0, +∞[ �

(e) |x + 1| > x

Answer: x ∈ R �

(f) |x − 1| ≥ x

Answer: x ∈ ]−∞, 12 ] �

Harder non-compulsory problems:

6. Sketch graphs of the functions:(a) y = |x |(b) y = x + |x |(c) y = x · |x |

46

(d)

y ={ x|x | if x 6= 00 if x = 0

7. Prove an inequality for all x , y ∈ R:

|x + y | 6 |x | + |y |.

8. Sketch graphs of the functions(a) y = |x |

-

6

��

@@@

x0

y y = |x |

(b) y = ||x | − 1|

-

6

��

@@@@

�� x0−1 1

yy = ||x | − 1|

(c) y = |||x | − 1| − 1|

-

6

�@ ��

@@@� x0−2 2

yy = |||x | − 1| − 1|



1. Solve the following quadratic inequalities

(a) x2 + 6x + 9 < 0

Answer: No solution. �

(b) x2 + 4x + 4 ≤ 0

Answer: x = −2. �

(c) x2 − 4x + 3 < 0

Answer: x ∈ ]1, 3[ . �

(d) x2 + 3x + 2 ≥ 0

Answer: x ∈ ]−∞,−2] ∪ [−1, +∞[ . �

2. The following inequalities can be rearranged into quadratic orlinear inequalities. Solve them. Warning: if you multiply /divide both part of an inequality by a number, take intoaccount the sign of this number!

(a) x(x + 2) > 3

Answer:

x ∈ ]−∞,−3[∪ ]1, +∞[ .

After opening the brackets and rearrangement, we havean equivalent inequality

x2 + 2x − 3 > 0.

After completing the square, we have

x2 + 2x − 3 = x2 + 2x + 1− 1− 3= (x + 1)2 − 4= (x + 1)2 − 22

= [(x + 1) + 2] · [(x + 1)− 2]= (x + 3)(x − 1).

48

Hence our inequality is equivalent to

(x + 3)(x − 1) > 0.

which has solution

x ∈ ]−∞,−3[∪ ]1, +∞[ .

�

(b) x + 4x ≥ 4

Answer: x > 0, or x ∈ ]0, +∞[ .We need to multiply the both part of the inequality by x .When x < 0, this will lead to change in the direction ofthe inequality. But for x < 0 4

x < 0, too, so in that case

x +4x< 0

and x < cannot be a solution of the original inequality.We also have to exclude the case x = 0 since division byzero is not permitted.So if x is a solution then x > 0 and we can multiply theboth parts of the inequality by x without changing thedirection of the inequality, obtaining

x2 + 4 ≥ 4x2 − 4x + 4 ≥ 0

(x − 2)2 ≥ 0

But the last inequality holds for all real x . Hence our so-lutions are bound only by restriction made in the processof rearrangements, x > 0, and therefore the solution setis ]0, +∞[ �

(c) 1x > x

Answer: Since the expression on LHS involves divisionby x , the value x = 0 should be excluded from the solu-tion set.CASE 1: x < 0. Multiplying by x , we change the directionof the inequality and get

1 < x2

which is equivalent to

x2 − 1 > 0


or, which is the same

(x + 1)(x − 1) > 0

Therefore x + 1 and x − 1 have to be of the same sign.Since we assume that x < 0, x − 1 < 0 and x + 1 < 0.But x − 1 < x + 1, therefore

x − 1 < 0 ∧ x + 1 < 0

is the same as x + 1 < 0 which is the same as x < −1.Hence in Case 1 the solution is

x ∈ ]−∞,−1[ .

CASE 2: x > 0. Multiplying by x , we get

1 > x2

which is equivalent to

x2 − 1 < 0

or, which is the same

(x + 1)(x − 1) < 0

Therefore x + 1 and x − 1 have to be of different signs,which is possible only if −1 < x < 1. Since we assumethat x > 0, this means that 0 < x < 1

x ∈ ]0, 1[ .

Combining these two cases, we see that the total of thesolution set is

x ∈ ]−∞,−1[∪ ]0, 1[ .

�

(d) x2 6 x3

Answer: x ∈ {0} ∪ [1, +∞[ . �

(e) x2 > x4

Answer: x ∈ [−1, 1]. Indeed, x = 0 belongs to the so-lution set. Let us look for other solutions: if x 6= 0, wecan divide the both parts of the inequality by the positivenumber x2 without changing the direction of the inequal-ity, and get x2 ≤ 1. We solved this inequality before, itssolution set is [−1, 1]; it contains 0, so we have not lost

50

solutions when divided by x2. �

(f)x + 1

x≤ 1

Answer: x ∈ ]−∞, 0[ . �

3. Determine which of the following points

A(3, 2), B(1, 2), C(1, 1)

lie inside of the triangle formed by the lines

−2x + y = 1, 2y − x = 2, 3x + y = 6.

Answer: B(1, 2). First of all, rewrite the equations of the linesas

−2x + y − 1 = 0−x + 2y − 2 = 0

3x + y − 6 = 0

and draw the following picture (purely symbolical, without anycoordinates or caring about realistic proportions

Solving system of linear equations,

−2x + y − 1 = 0−x + 2y − 2 = 0

we find that the point of intersection of these two lines is (0, 1),similarly the point of intersection of the lines

−2x + y − 1 = 03x + y − 6 = 0


is (1, 3), and the point of intersection of

−2x + y − 1 = 0−x + 2y − 2 = 0

is approximately (1.2, 1, 7). We represented this informationabout the vertexes of the triangle formed by the three lines inthe diagram

Now, for each vertex we compute the sign taken by the linearfunction representing the opposite side of the triangle; for ex-ample, the function 3x + y − 6 = 0 takes at the vertex (0, 1)the value 3 · 0 + 1 · 1− 6 = −2 thus is negative; we assign tothe side 3x +y−6 = 0 the sign −; we do the same for the restof the sides and add this information to the diagram:

Observe that the point (x , y ) of the plane liez inside of thetriangle if and only i if it lies on the half plane with respect toevery side as the vertex opposite sides, hence, if and only ifthe functions

−2x + y − 1, −x + 2y − 2, 3x + y − 6

takes at the point (x , y ) values of signs −, +,−, respectively.But it is ease to computes, that at the given points these signsare

A(3, 2) : −,−, +, B(1, 2) : −, +,−, C(1, 1) : −,−,−, .

52

Hence the point B lies inside of the triangle, while points Aand C are outside of the triangle. �

4. Sketch the solution sets of the following systems of simulta-neous inequalities in variables x and y .

(a) 1 ≤ x ≤ 2, −1 ≤ y ≤ 1

(b) 1 ≤ x + y ≤ 2

(c) x2 ≤ y ≤ 1

(d) 0 ≤ y ≤ x2

Answer: See Figure 6. �

5. [Examination January 2016] Solve the system of simultane-ous inequalities

−x2 + y + 1 ≥ 0x − y − 2 ≥ 0

Answer: The system has no solutions. That could be seenfrom sketching the solution sets of y ≥ x2−1 (the set of pointsin the plane over the parabola) and x ≥ y + 2 (the half-planeunder the line).

Alternatively, substituting x from the second inequality intothe first one, one quickly gets the inequality

y2 − 3y + 3 ≤ 0,

which has no solution because the discriminant of the qua-dratic function involved is negative. �


0N1 Mathematics

Solutions 11

Some of these problems were discussed in the lectures, but stillworth looking.

1. Prove, by induction on n, that

n < 2n

for every positive integer n.

Answer:BASIS OF INDUCTION.

1 < 2 = 21,therefore the basis of induction holds.

INDUCTIVE STEP. Assume that the statement is true for n = k ,

k < 2k .

Adding 1 to the both sides of this inequality, we get

k + 1 < 2k + 1< 2k + 2k

= 2k+1.

�


1 + 3 + 5 + · · · + (2n − 1) = n2


Answer: Was explained in the lecture notes. �


1 + 2 + 3 + · · · + n =12

n(n + 1)


Answer: Let pn be the statement

“1 + 2 + · · · + n =12

n(n + 1)”.

54

Then p1 is the statement

“1 =12× 1× 2”.

This is clearly true.Suppose pn is true for n = k , i.e.

1 + 2 + · · · + k =12

k (k + 1).

Then

1 + 2 + · · · + k + (k + 1) = (1 + 2 + · · · + k ) + (k + 1)

=12

k (k + 1) + (k + 1)

=12

k (k + 1) +12

2(k + 1)

=12

(k + 1)(k + 2).

Therefore

1 + 2 + · · · + k + (k + 1) =12

(k + 1)((k + 1) + 1).

�

4.. Prove, by induction on n, that

12

+122 + · · · + 1

2n = 1− 12n


Answer: Done in the lecture. �

5. Prove, by induction on n, that if q 6= 1 then

1 + q + q2 + · · · + qn−1 + qn =1− qn+1

1− q


Answer: BASIS OF INDUCTION. When n = 1,

1 + q =1− q1+1

1− q=

1− q2

1− q= 1 + q

is true.


INDUCTIVE STEP. Assume that the statement

P(k ) ≡ “1 + q + q2 + · · · + qk−1 + qk =1− qk+1

1− q”

is true. We need to prove

P(k + 1) ≡ “1 + q + q2 + · · · + qk−1 + qk + qk+1︸︷︷︸ =1 − q(k+1)+1

1 − q︸︷︷︸ ”

Here, the brace ︸︷︷︸ under the line shows parts of the formulaP(k +1) different from that of P(k ). So let us try to get P(k +1)by adding qk+1 to the both sides of P(k ):

1 + q + q2 + · · · + qk−1 + qk + qk+1︸︷︷︸ =1− qk+1

1− q+ qk+1︸︷︷︸

=1− qk+1 + qk+1(1− q)

1− q

=1− qk+1 + qk+1 − qk+2

1− q

=1− q(k+1)+1

1− q,

thus proving P(k + 1). �

6. Let x be any real number ≥ −1. Prove by induction that

(1 + x)n ≥ 1 + nx , for all n ≥ 1.

Answer: BASIS OF INDUCTION. When n = 1,

(1 + x)1 = 1 + x ≥ 1 + 1 · xis true.

INDUCTIVE STEP. Assume that the statement is true forn = k ,

(1 + x)k ≥ 1 + kx .Then

(1 + x)k+1 = (1 + x)k · (1 + x)≥ (1 + kx) · (1 + x)= 1 + x + kx + kx2

= 1 = (k + 1)x + kx2

≥ 1 + (k + 1)x .

56

�

7. Recall that, for a positive integer n,

n! = 1 · 2 · 3 · 4 · · · (n − 1) · nis the product of all integers from 1 to n, so that

1! = 1, 2! = 1 · 2 = 2, , 3! = ·2 · 3 = 6, etc.

Prove by induction that, for all integers n ≥ 4,

n! > 2n.

Answer:

BASIS OF INDUCTION. When k = 4,

4! = 1 · 2 · 3 · 4 = 24 > 16 = 24.

INDUCTIVE STEP. If

k ! > 2k ,

then multiplying both sides of inequality by k1, we get

k ! · (k + 1) > 2k · (k + 1) > 2k · 2 = 2k+1.

�

8. Prove by induction on n that for each n ≥ 3, the angles of anyn-gon in the plane have the sum equal to (n − 2)π radians.

HINT. Check that for a triangle (n = 3), then cut from an n-gone atriangle.

The following problems are harder and not compulsory.

9. Prove that, for all integers n > 10,

n3 < 2n.

Answer: Let pk be the statement

k3 < 2k

BASIS OF INDUCTION. Please notice that the inequality isfalse if n ≤ 10, so the induction starts at the first true state-ment p11, for k = 11,

113 = 1331 < 2048 = 211;

that is true, and therefore p11, the basis of induction, is true.


INDUCTIVE STEP. Assume that the k ≥ 11 and that thestatement pk ,

k3 < 2k ,is true. we want to prove

pk → pk+1,

that is, that(k + 1)3 < 2k+1

is a consequence of

k3 < 2k ,

for every k > 10.Now that pk is true, that k > 10, and compute:

(k + 1)3 = k3 · (k + 1)3

k3

= k3 ·(

k + 1k

)3

= k3 ·(

1 +1k

)3

< k3 ·(

1 +1

10

)3

(we used here that k > 10)= k3 · (1.1)3

= k3 · 1.331< k3 · 2< 2k · 2 (by pk k3 < 2k )= 2k+1.

This proves pk+1, hence proves the inductive step

pk → pk+1 for k > 10.

Hence all statements pk for k > 10 are true. �

10. Let x be a real number such that

x +1x

is an integer. Prove by induction that then

xn +1xn

58

is an integer for all positive integers n.

Answer: This problem is much harder than the previous prob-lems.

Denote

zn = xn +1xn

BASIS OF INDUCTION. The case n = 1 is given as the as-sumption of the problem. It is also useful to check the state-ment in the case n = 0:

z0 +1z0 = 1 +

11

= 2

is an integer.INDUCTIVE STEP. Assume that

zl = x l +1x l

is an integer for all l ≤ k . We want to prove that zk+1 is alsoan integer.

Multiplying zk by z1, we see that

zk · z1 =(

xk +1xk

)·(

x +1x

)is an integer. We can rearrange the expression on the RHSfurther:

zkz1 =(

xk +1xk

)·(

x +1x

)= xk · x + xk · 1

x+

1xk · x +

1xk ·

1x

= xk+1 + xk−1 +1

xk−1 +1

xk+1

=(

xk+1 +1

xk+1

)+(

xk−1 +1

xk−1

)= zk+1 + zk−1.

So we getzkz1 = zk+1 + zk−1,

which means that

zk+1 = zkz1 − zk−1.

Since zk−1 is an integer by the inductive assumption, we seethat zk+1 is an integer.


�

11. Prove that the sum of angles of a convex n-gon is (n − 2)π.

The principle of mathematical induction as we know it now was firstpublished by the British mathematician Augustus De Morgan (theone of the De Morgan laws in Boolean Algebra) in 1838 in The PennyCyclopedia of the Society for the Diffusion of Useful Knowledge. Thetitle of this encyclopedia clearly said that it was produced for thegeneral public. You can find the text of De Morgan’s original articleat http://bit.ly/2iD9jsi.

12. Fill in details in the last paragraph of De Morgan’s article:

There are cases in which the successive inductiononly brings any term within the general rule, whentwo, three, or more of the terms immediately preced-ing are brought within it. Thus, in the application ofthis method to the deduction of the well-known con-sequence of

x +1x

= 2 cos θ, namely, xn +1xn = 2 cos nθ,

it can only be shown that any one case of this theo-rem is true, by showing that the preceding two casesare true: thus its truth, when n = 5 and n = 6, makesit necessarily follow when n = 7. In this case the twofirst instances must be established (when n = 1 byhypothesis, and when n = 2 by independent demon-stration), which two establish the third, the secondand third establish the fourth, and so on.

13. [The “sudden examination” paradox]2 Friday afternoon, justbefore school lets out, a teacher promises his class that theywill have a quiz on one of the five days of the coming week.Moreover, he guarantees the students that the quiz will be asurprise in that they won’t be able to predict the night beforethat it will happen the next day. The class is dismayed until

2Adam Bjorndahl, Puzzles and Paradoxes in Mathematical Induction, http://www.math.cornell.edu/~mec/2008-2009/ABjorndahl/ppmi.pdf.

http://bit.ly/2iD9jsi

http://www.math.cornell.edu/~mec/2008-2009/ABjorndahl/ppmi.pdf

http://www.math.cornell.edu/~mec/2008-2009/ABjorndahl/ppmi.pdf

60

one of the students realizes that something fishy is going on.She reasons:

“The quiz can’t be given on Friday, for sure, because that’sthe last possible day, so we would be able to predict it Thurs-day night. So Friday is out. That makes Thursday the lastpossible day the quiz can be given. But then the quiz alsocan’t be given on Thursday, because Wednesday night wewould know it was coming the next day! And in the sameway we can eliminate Wednesday, Tuesday, and even Mon-day from the list of possible days for the quiz.”

This argument is good enough to convince the rest of theclass, who gleefully go about their business, content in thecertainty that there can be no surprise quiz. Tuesday morningcomes, however, and the teacher hands out a quiz sheet toeach student. There are, of course, objections: “You can’tgive this quiz! We already figured out that you couldn’t makeit a surprise no matter what day you gave it on!”

But the teacher is unperturbed. “You figured that out, didyou? Well, here’s the quiz. Aren’t you surprised?” The stu-dents reluctantly agreed that they were. But where did theirlogic go awry?

Answer: Students’ argument is self-referential, it refers totheir way of arguing. [Self-referential statements and para-doxes generated by them had been discussed in the lec-tures.] �

62


mathematics 0n1 answer: fa b ng · 3.let a = fa,b,c,d,eg, b = fc,e,fgand c = fa,d,eg. which of the...

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