elementary number theory and methods of proof

Chapter 4: Elementary Number Theory and

Methods of Proof

Section 4.7 Two Classical Theorems

MH1300 Lecture handout 5 (NTU) §4.7 Two Classical Theorems 38 / 50

Synopsis

Objective: Proofs of two famous theorems in mathematics:√

2 isirrational, and there are infinitely many prime numbers. Bothproofs use indirect arguments (i.e. proofs by contradiction) andwere well known more than 2000 yrs ago.

The irrationality of√

The infinitude of the set of prime numbers.

When to use indirect proof.

Open questions in Number Theory.

Theorem 4.7.1 Irrationality of√

2 is irrational.

Proof:

Suppose not. Then√

2 is rational.

Then there exists integers m and n

such that √2 =

(We may assume

m, n have no common factors other than 1 or −1

else we may divide them by their common factors to get a new pair m1

and n1 that have no common factors.)

Squaring both sides of equation gives

2 is irrational.

Proof:

2 is rational. Then there exists integers m and n

such that √2 =

(We may assume

m, n have no common factors other than 1 or −1

2 is irrational.

Proof:

such that √2 =

(We may assume m, n have no common factors other than 1 or −1,

2 is irrational.

Proof:

such that √2 =

2 is irrational.

Proof:

such that √2 =

Equivalently,2n2 = m2.

Note that left side of equation is even. Therefore, right side is even.

Therefore

m is even

(by Prop. 4.6.4).

Thereforem = 2k for some integer k .

Therefore2n2 = m2 = (2k)2 = 4k2.

Thereforen2 = 2k2.

Consequently n2 is even, and so by Prop. 4.6.4 again,

n is even

Hence both m and n have a common factor of 2.

This contradicts “

m, n have no common factors other than 1, −1

Note that left side of equation is even.

Therefore, right side is even.

Therefore

m is even

(by Prop. 4.6.4).

Thereforen2 = 2k2.

n is even

This contradicts “ m, n have no common factors other than 1, −1”.

Therefore

m is even

(by Prop. 4.6.4).

Thereforen2 = 2k2.

n is even

Therefore m is even (by Prop. 4.6.4).

Thereforen2 = 2k2.

n is even

Thereforen2 = 2k2.

n is even

Thereforen2 = 2k2.

n is even

Thereforen2 = 2k2.

n is even

Thereforen2 = 2k2.

Consequently n2 is even, and so by Prop. 4.6.4 again, n is even.

Thereforen2 = 2k2.

Example (Proposition 4.7.2)

1 + 3√

2 is irrational.

Proof: Suppose not.

Then 1 + 3√

2 is rational. Then there exists

integers a and b with b 6= 0 such that

1 + 3√

Rearranging, we find that3√

b− 1 =

a− b

Dividing both sides by 3 gives √2 =

a− b

Clearly a− b and 3b are integers. Also, 3b 6= 0 by the zero product

property. Therefore

√2 is a rational number

This is a contradiction to√

2 is irrational. Hence 1 + 3√

2 is irrational.

1 + 3√

2 is irrational.

Proof: Suppose not. Then 1 + 3√

1 + 3√

b− 1 =

a− b

property. Therefore

2 is irrational.

1 + 3√

2 is irrational.

1 + 3√

b− 1 =

a− b

property. Therefore

2 is irrational.

1 + 3√

2 is irrational.

1 + 3√

b− 1 =

a− b

property. Therefore

2 is irrational.

1 + 3√

2 is irrational.

1 + 3√

b− 1 =

a− b

property.

Therefore

2 is irrational.

1 + 3√

2 is irrational.

1 + 3√

b− 1 =

a− b

property. Therefore√

2 is a rational number.

2 is irrational.

1 + 3√

2 is irrational.

1 + 3√

b− 1 =

a− b

property. Therefore√

2 is a rational number.

2 is irrational.

The infinitude of the set of prime numbers

Proposition 4.7.3

For any integer a and any prime number p, if p|a then p - (a + 1).

Proof:

Suppose not.

Then there exists an integer a and a prime number p such

that p|a and p|(a + 1).

Since p|a, there exists k ∈ Z such that

a = p · k . (1)

Since p|(a + 1), there exists l ∈ Z such that

a + 1 = p · l . (2)

Proposition 4.7.3

Proof:

Suppose not. Then there exists an integer a and a prime number p such

a = p · k . (1)

a + 1 = p · l . (2)

Proposition 4.7.3

Proof:

a = p · k . (1)

a + 1 = p · l . (2)

Proposition 4.7.3

Proof:

a = p · k . (1)

a + 1 = p · l . (2)

(Proof of Prop. 4.7.3 continues)

Subtracting equation (1) from equation (2), we arrive at

1 = p · l − p · k = p · (l − k).

Clearly l − k is an integer. This shows that p is a divisor of 1.

Since p is prime, p is positive.

Therefore, by Thm. 4.3.1,

p ≤ 1. (If a, b ∈ Z+ and a|b, then a ≤ b.)

This is a contradiction to p being prime is greater than 1.

Therefore the supposition is false.

1 = p · l − p · k = p · (l − k).

Clearly l − k is an integer.

This shows that p is a divisor of 1.

1 = p · l − p · k = p · (l − k).

Theorem 4.7.4 Infinitude of the Primes

The set of prime numbers is infinite.

Proof: Suppose not.

Then the set of prime numbers is finite, say there

are only n many prime numbers.

Then we may list all the prime numbers in ascending order,

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

Consider the integer N :=

p1p2p3 · · · pn + 1.

Then N > 1, and so by Theorem 4.3.4 (Any integer > 1 is divisible by a

prime), N is divisible by some prime number p, i.e. p|N.

Also, since p is prime, it must be equal to one of p1, p2, · · · pn.

Clearly p|p1p2p3 · · · pn, hence, by P. 4.7.3, p - (

p1p2p3 · · · pn + 1

Hence p|N and p - N, which is a contradiction.

Proof: Suppose not. Then the set of prime numbers is finite, say there

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

p1p2p3 · · · pn + 1.

p1p2p3 · · · pn + 1

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

p1p2p3 · · · pn + 1.

p1p2p3 · · · pn + 1

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

Consider the integer N := p1p2p3 · · · pn + 1.

p1p2p3 · · · pn + 1

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

p1p2p3 · · · pn + 1

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

p1p2p3 · · · pn + 1

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

Clearly p|p1p2p3 · · · pn, hence, by P. 4.7.3, p - (p1p2p3 · · · pn + 1) = N.

p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . , pn.

Clearly p|p1p2p3 · · · pn, hence, by P. 4.7.3, p - (p1p2p3 · · · pn + 1) = N.

When to use indirect proof

When do we use indirect proofs instead?

No definitive answer.

Many theorems can be proved either way, and in such situations,usually indirect proofs are clumsier.

If no clues suggests an indirect proof, try direct proof first.

If it doesn’t work, try finding a counterexample.

If counterexample isn’t found, try look for a proof bycontradiction/contraposition.

If THAT doesn’t work, try to have some coffee or tea, or take a walk,and then try again ...

How to find clues?

Perhaps examine examples for small values of n.

Try to construct counterexample. If that fails, try to see whatproperties denied your counterexamples.

How to find clues?

Open questions in Number Theory

A ‘nice’ or efficient formula for obtaining primes?

Mersenne primes, i.e. primes of the form 2p − 1, p is prime.– Are there infinitely many Mersenne primes?

Fermat primes, i.e. primes of the form 22n

+ 1.– Are there infinitely many Fermat primes?

Are there infinitely many primes of the form n2 + 1?

Is there always a prime between integers n2 and (n + 1)2?

More open questions in Number Theory

Are there infinitely many primes of the form N = p1p2p3 · · · pk + 1?(Given that the pj ’s are all the initial primes)

Are there infinitely many twin primes, i.e. satisfying the property thatboth p and p + 2 are both primes?

For more information, see:

1 Richard K. Guy, Unsolved Problems in Number Theory.

2 P. Ribenboim, The new book of prime number records, 3rd edition,Springer-Verlag, New York, NY, 1995.

One way to prove that√

2 is irrational is to

1 assume√

2 = a/b for some integers a and b with

2 square both sides and multiply both sides by b2 to get ,

3 show that a and b have a .

One way to prove that there are infinitely many prime numbers is to

1 assume there is a largest prime number p,

2 construct the number ,

3 show that this number has to be divisible by a prime number that is

Question: Are the following logically equivalent?

∀x ∈ D, ∀y ∈ E , (P(x)∨Q(y)), and (∀x ∈ D,P(x))∨(∀x ∈ E ,Q(x)).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Part 1. If first statement form is true, then second statement form is true.

Suppose ∀x ∈ D,P(x) is true, then we are done.

Else, ∃c ∈ D such that ¬P(c).

Therefore, ∀x ∈ E ,Q(x) is true.

Hence Part 1 is true.

MH1300 Lecture handout 3 (NTU) §3.4 Arguments with Quantified Statements 68 / 69

Question con’t

Part 2. If second statement form is true, then first statement form is true.

Since second statement form is true, either ∀x ∈ D,P(x) is true

or ∀x ∈ E ,Q(x) is true.

Question con’t

Chapter 5: Sequences, Mathematical Induction,

and Recursion

Section 5.1 Sequences

MH1300 Lecture handout 6 (NTU) §5.1 Sequences 1 / 66

Synopsis

Objective: A review of sequences and sums.This is mostly a review of past knowledge of that you alreadylearnt from H2 Mathematics at A’levels or equivalent Mathsubject taken at high school.

Sequences.

Examples.

Summation notation, telescoping sum.

Product notation, computing products.

Factorial notation.

Properties of summation and products.

Change of variable.

Sequences

Definition

A sequence is a function whose domain is either

all the integers between two given integers, or

all the integers greater than or equal to a given integer

Sequences

Definition

A sequence is a function whose domain is either

all the integers between two given integers, or

all the integers greater than or equal to a given integer.

Sequences

Typically denoted asam, am+1, am+2, . . . , an,

each individual element ak (read as “a sub k”) is called a term.

The k in ak is called a subscript or index.

The integer m is the subscript of the initial term.

The integer n is the subscript of the final term.

The notationam, am+1, am+2, . . .

denotes an infinite sequence.

An explicit formula or general formula for a sequence is a rule thatshows how the values of ak depend on k .

Sequences

Example 5.1.3

Find an explicit formula for a sequence that has the following initial terms:

1, −1

9, − 1

25, − 1

36, . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

Denote by ak the general term of the sequence and suppose the first termis a1.

12, − 1

32, − 1

52, − 1

62, . . .

l l l l l la1 a2 a3 a4 a5 a6 . . .

(−1)k+1 1

for all integers k ≥ 1.

Example 5.1.3

1, −1

9, − 1

25, − 1

36, . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

12, − 1

32, − 1

52, − 1

62, . . .

l l l l l la1 a2 a3 a4 a5 a6 . . .

(−1)k+1 1

Example 5.1.3

1, −1

9, − 1

25, − 1

36, . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

12, − 1

32, − 1

52, − 1

62, . . .

l l l l l la1 a2 a3 a4 a5 a6 . . .

(−1)k+1 1

Example 5.1.3

1, −1

9, − 1

25, − 1

36, . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

12, − 1

32, − 1

52, − 1

62, . . .

l l l l l la1 a2 a3 a4 a5 a6 . . .

ak = (−1)k+1 1

k2for all integers k ≥ 1.

Summation notation

Definition

If m and n are integers and m ≤ n, then the symboln∑

ak , read

summation from k equals m to n of a sub k , is the sum of all terms

am, am+1, am+2, . . . , an.

We writen∑

ak = am + am+1 + am+2 + · · ·+ an.

k is the index of the summation,

m is the lower limit of the summation,

n is the upper limit of the summation.

Summation notation

Definition

ak , read

am, am+1, am+2, . . . , an.

We writen∑

ak = am + am+1 + am+2 + · · ·+ an.

Summation notation

Definition

ak , read

am, am+1, am+2, . . . , an.

We writen∑

ak = am + am+1 + am+2 + · · ·+ an.

Summation notation

Definition

ak , read

am, am+1, am+2, . . . , an.

We writen∑

ak = am + am+1 + am+2 + · · ·+ an.

Summation notation

Definition

ak , read

am, am+1, am+2, . . . , an.

We writen∑

ak = am + am+1 + am+2 + · · ·+ an.

Example 5.1.7

Express the following using summation notation:

n + 1+

n + 2+ · · ·+ n + 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

The general term is k+1n+k for integers k from 0 to n.

Therefore,

n + 1+

n + 2+ · · ·+ n + 1

n∑k=0

n + k.

Example 5.1.7

n + 1+

n + 2+ · · ·+ n + 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

Therefore,

n + 1+

n + 2+ · · ·+ n + 1

n∑k=0

n + k.

Example 5.1.7

n + 1+

n + 2+ · · ·+ n + 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

Therefore,

n + 1+

n + 2+ · · ·+ n + 1

n∑k=0

n + k.

Example 5.1.7

n + 1+

n + 2+ · · ·+ n + 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

Therefore,

n + 1+

n + 2+ · · ·+ n + 1

n∑k=0

n + k.

Example 5.1.8

What is the value of the expression

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

For n = 1, the sum is

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Although the sum seems to suggest that n must be at least 3 (since thefirst 3 terms are given), but there is no such suggestion implied.

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

For n = 1, the sum is1

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Example 5.1.8

1 · 2+

2 · 3+

3 · 4+ · · ·+ 1

n · (n + 1)

when n = 1? n = 2? n = 3?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 · 2=

1 · 2+

2 · 3=

1 · 2+

2 · 3+

3 · 4=

Caution

Telescoping sum

Example 5.1.10. Use the identity

k− 1

k + 1=

(k + 1)− k

k(k + 1)=

k(k + 1)

to find a simple expression for∑n

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

Solution:n∑

k(k + 1)

k− 1

1− 1

)+ · · ·+

n − 1

)= 1− 1

n + 1.

Telescoping sum

k− 1

k + 1=

(k + 1)− k

k(k + 1)=

k(k + 1)

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

Solution:n∑

k(k + 1)

k− 1

1− 1

)+ · · ·+

n − 1

)= 1− 1

n + 1.

Telescoping sum

k− 1

k + 1=

(k + 1)− k

k(k + 1)=

k(k + 1)

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

Solution:n∑

k(k + 1)

k− 1

1− 1

)+ · · ·+

n − 1

= 1− 1

n + 1.

Telescoping sum

k− 1

k + 1=

(k + 1)− k

k(k + 1)=

k(k + 1)

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

Solution:n∑

k(k + 1)

k− 1

1− 1

/− 1

/)+ · · ·+

n − 1

/− 1

= 1− 1

n + 1.

Telescoping sum

k− 1

k + 1=

(k + 1)− k

k(k + 1)=

k(k + 1)

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

Solution:n∑

k(k + 1)

k− 1

1− 1

/− 1

/)+ · · ·+

n − 1

/− 1

)= 1− 1

n + 1.

Product notation

Definition

For integers m and n with m ≤ n, the symboln∏

ak , read the

product from k equals m to n of a sub k , is the product of all theterms

am, am+1, am+2, . . . , an.

We writen∏

ak = am · am+1 · am+2 · · · an.

Product notation

Definition

For integers m and n with m ≤ n, the symboln∏

ak , read the

product from k equals m to n of a sub k , is the product of all theterms

am, am+1, am+2, . . . , an.

We writen∏

ak = am · am+1 · am+2 · · · an.

Factorial notation

Definition

For each positive integer n, the quantity n factorial denoted n!, is definedto be the product of all the integers from 1 to n:

n! := n · (n − 1) · (n − 2) · · · 3 · 2 · 1.

Zero factorial, denoted by 0!, is defined to be 1:

0! := 1.

Factorial notation

Definition

For each positive integer n, the quantity n factorial denoted n!, is definedto be the product of all the integers from 1 to n:

n! := n · (n − 1) · (n − 2) · · · 3 · 2 · 1.

Zero factorial, denoted by 0!, is defined to be 1:

0! := 1.

Example.

a. Compute∏6

(k−2)2−3 .

b. Simplify 3!·3!7! .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Example.

a. Compute∏6

(k−2)2−3 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:a.

6∏k=4

(k − 2)2 − 3=

22 − 3· 5

32 − 3· 6

42 − 3

b.3! · 3!

3! · 3!

7 · 6 · 5 · 4 · 3!=

7 · 6 · 5 · 4=

Properties of summation and products

Theorem 5.1.1

If am, am+1, am+2, . . . , and bm, bm+1, bm+2, . . . , are sequences of realnumbers and c is any real number, then the following equations hold forany integer n ≥ m:

n∑k=m

ak +n∑

bk =n∑

(ak + bk).

2 c ·n∑

ak =n∑

(c · ak) (generalized distribution law.)

n∏k=m

(ak · bk).

Proof: Simple application of mathematical induction in Section 5.2 andproperties of addition and multiplication. Left as exercise.

Theorem 5.1.1

n∑k=m

ak +n∑

bk =n∑

(ak + bk).

2 c ·n∑

ak =n∑

n∏k=m

(ak · bk).

Theorem 5.1.1

n∑k=m

ak +n∑

bk =n∑

(ak + bk).

2 c ·n∑

ak =n∑

n∏k=m

(ak · bk).

Theorem 5.1.1

n∑k=m

ak +n∑

bk =n∑

(ak + bk).

2 c ·n∑

ak =n∑

n∏k=m

(ak · bk).

Theorem 5.1.1

n∑k=m

ak +n∑

bk =n∑

(ak + bk).

2 c ·n∑

ak =n∑

n∏k=m

(ak · bk).

Ex. 5.1.60

Express the following as a single summation.

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1)

(6k2 + 8) +n∑

(10k2 − 5)

(16k2 + 3)

Ex. 5.1.60

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1)

(6k2 + 8) +n∑

(10k2 − 5)

(16k2 + 3)

Ex. 5.1.60

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1)

(6k2 + 8) +n∑

(10k2 − 5)

(16k2 + 3)

Ex. 5.1.60

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solution:

2 ·n∑

(3k2 + 4) + 5 ·n∑

(2k2 − 1)

(6k2 + 8) +n∑

(10k2 − 5)

(16k2 + 3)

Change of variable – Example 4.1.16a

Transform the following summation by making the specified change ofvariable.

summation:n+1∑k=1

)change of variable: j = k − 1.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:

When making the change of variable j = k − 1,

the lower limit, k = 1, is now j =

k − 1 = 1− 1 = 0

the upper limit, k = n + 1, is now

j = k − 1 = (n + 1)− 1 = n

since j = k − 1, the variable k is replaced by

the general term, kn+k , is now

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Solution:When making the change of variable j = k − 1,

k − 1 = 1− 1 = 0

j = k − 1 = (n + 1)− 1 = n

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

k − 1 = 1− 1 = 0

j = k − 1 = (n + 1)− 1 = n

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

the lower limit, k = 1, is now j = k − 1 = 1− 1 = 0.

j = k − 1 = (n + 1)− 1 = n

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

j = k − 1 = (n + 1)− 1 = n

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

the upper limit, k = n + 1, is now j = k − 1 = (n + 1)− 1 = n.

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

since j = k − 1, the variable k is replaced by j + 1.

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(j+1)n+(j+1) = j+1

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

the general term, kn+k , is now (j+1)

n+(j+1) = j+1n+j+1 .

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

n+(j+1) = j+1n+j+1 .

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

summation:n+1∑k=1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

n+(j+1) = j+1n+j+1 .

Therefore,n+1∑k=1

n∑j=0

(j + 1

n + j + 1

The expanded form ofn∑

ak is .

The value of a1 + a2 + a3 + · · ·+ an when n = 2 is “ ”.

If n is a positive integer, then n! = .

n∑k=m

ak + cn∑

)(n∏

Chapter 5: Sequences, Mathematical Induction,

and Recursion

Sections 5.2, 5.3 Mathematical Induction I & II

MH1300 Lecture handout 6 (NTU) §5.2, 5.3 Mathematical Induction I & II 17 / 66

Synopsis

Objective: A review of Mathematical Induction.This is mostly a review of past knowledge of that you alreadylearnt from H2 Mathematics at A’levels or equivalent Mathsubject taken at high school.

Principle of Mathematical Induction.

Sum of the first n integers.

Sum of geometric sequence.

Application.

Principle of Mathematical Induction

Let a be a fixed integer and let P(n) be a property that is defined forintegers n ≥ a.

Suppose the following two statements are true:

P(a) is true.

For all integers k ≥ a, if P(k) is true then P(k + 1) is true.

Then the statement

for all integers n ≥ a,P(n)

is true.

P(a) is true.

Then the statement

is true.

P(a) is true.

Then the statement

is true.

Method of Proof by Mathematical Induction

Consider a statement of the form“For all integers n ≥ a, a property P(n) is true”.

To prove such a statement, perform the following two steps:

Step 1. (basis step): Show that the property is true for n = a.

Step 2. (inductive step): Show that for all integers k ≥ a, if the property istrue for n = k then it is true for n = k + 1. To perform this step,

suppose that the property is true for n = k, where k is anyparticular but arbitrarily chosen integer with k ≥ a.[This supposition is called the inductive hypothesis]

Then show that the property is true for n = k + 1.

suppose that the property is true for n = k , where k is anyparticular but arbitrarily chosen integer with k ≥ a.[This supposition is called the inductive hypothesis]

Sum of the first n integers

Theorem 5.2.2 Sum of the first n integers

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

Proof (by mathematical induction):

n ≥ 1

, let the property P(n) be the equation

1 + 2 + 3 + · · ·+ n = n(n+1)2

Basis step: Want to show:

1 =1 · (1 + 1)

Simplifying the right side, we see that

1·(1+1)2 = 1·2

. Therefore P(1) istrue.

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

n ≥ 1

, let the property P(n) be the equation

1 + 2 + 3 + · · ·+ n = n(n+1)2

1 =1 · (1 + 1)

1·(1+1)2 = 1·2

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

For n ≥ 1, let the property P(n) be the equation

1 + 2 + 3 + · · ·+ n = n(n+1)2

1 =1 · (1 + 1)

1·(1+1)2 = 1·2

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

1 + 2 + 3 + · · ·+ n = n(n+1)2 .

1 =1 · (1 + 1)

1·(1+1)2 = 1·2

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

1 + 2 + 3 + · · ·+ n = n(n+1)2 .

1 =1 · (1 + 1)

1·(1+1)2 = 1·2

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

1 + 2 + 3 + · · ·+ n = n(n+1)2 .

1 =1 · (1 + 1)

1·(1+1)2 = 1·2

Therefore P(1) istrue.

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

1 + 2 + 3 + · · ·+ n = n(n+1)2 .

1 =1 · (1 + 1)

Simplifying the right side, we see that 1·(1+1)2 = 1·2

2 = 1.

Therefore P(1) istrue.

For all n ≥ 1,

1 + 2 + · · ·+ n =n(n + 1)

1 + 2 + 3 + · · ·+ n = n(n+1)2 .

1 =1 · (1 + 1)

Simplifying the right side, we see that 1·(1+1)2 = 1·2

2 = 1. Therefore P(1) istrue.

Inductive step: Let k ∈ Z with k ≥ 1 and suppose that P(k) is true, i.e.,

1 + 2 + · · ·+ k =k(k + 1)

Want to show: P(k + 1) is true, i.e.,

1 + 2 + · · ·+ (k + 1) =

(k + 1)(k + 2)

Note that the left side of equation (1) is

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

k(k + 1)

2+ (k + 1)

(by the inductive hypothesis)

(k + 1)

(k + 1)(k + 2)

which is the right side of equation (1).

Therefore P(k + 1) is true.

Therefore Theorem is true by mathematical induction.

1 + 2 + · · ·+ k =k(k + 1)

1 + 2 + · · ·+ (k + 1) =

(k + 1)(k + 2)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

k(k + 1)

2+ (k + 1)

(k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

k(k + 1)

2+ (k + 1)

(k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

k(k + 1)

2+ (k + 1)

(k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

=k(k + 1)

2+ (k + 1) (by the inductive hypothesis)

(k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

=k(k + 1)

= (k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

=k(k + 1)

= (k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

=k(k + 1)

= (k + 1)

(k + 1)(k + 2)

1 + 2 + · · ·+ k =k(k + 1)

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

2. (1)

1 + 2 + · · ·+ (k + 1)

= (1 + 2 + · · ·+ k) + (k + 1)

=k(k + 1)

= (k + 1)

(k + 1)(k + 2)

Sum of geometric sequence

Theorem 5.2.3 Sum of a geometric sequence

For any real number x except 1, and any integer n ≥ 0,

n∑i=0

x i =xn+1 − 1

x − 1.

For n ≥

, let the property P(n) be the equation∑n

i=0 xi = xn+1−1

x−1 .

x0+1 − 1

x − 1

Simplifying, the left side equals x0 = 1, and the right side equals x−1x−1 = 1.

Therefore, both sides of the equation agree, and so P(0) is true.

n∑i=0

x i =xn+1 − 1

x − 1.

For n ≥

, let the property P(n) be the equation∑n

i=0 xi = xn+1−1

x−1 .

x0+1 − 1

x − 1

n∑i=0

x i =xn+1 − 1

x − 1.

For n ≥ 0, let the property P(n) be the equation∑n

i=0 xi = xn+1−1

x−1 .

x0+1 − 1

x − 1

n∑i=0

x i =xn+1 − 1

x − 1.

i=0 xi = xn+1−1

x−1 .

x0+1 − 1

x − 1

n∑i=0

x i =xn+1 − 1

x − 1.

i=0 xi = xn+1−1

x−1 .

x0 =x0+1 − 1

x − 1.

n∑i=0

x i =xn+1 − 1

x − 1.

i=0 xi = xn+1−1

x−1 .

x0 =x0+1 − 1

x − 1.

n∑i=0

x i =xn+1 − 1

x − 1.

i=0 xi = xn+1−1

x−1 .

x0 =x0+1 − 1

x − 1.

Inductive step: Let k ∈ Z with k ≥ 0 and suppose that P(k) is true, i.e.,k∑

x i =xk+1 − 1

x − 1.

Want to show: P(k + 1) is true, i.e.,k+1∑i=0

xk+2 − 1

x − 1

k+1∑i=0

xk+1 +k∑

xk+1 +xk+1 − 1

x − 1(by the inductive hypothesis)

xk+1 · (x − 1) + xk+1 − 1

x − 1=

xk+2 − 1

x − 1,

which is the right side of equation (2).Therefore P(k + 1) is true.Therefore Theorem is true by mathematical induction.

x i =xk+1 − 1

x − 1.

xk+2 − 1

x − 1

k+1∑i=0

xk+1 +k∑

xk+1 +xk+1 − 1

xk+1 · (x − 1) + xk+1 − 1

x − 1=

xk+2 − 1

x − 1,

which is the right side of equation (2).Therefore P(k + 1) is true.Therefore Theorem is true by mathematical induction.MH1300 Lecture handout 6 (NTU) §5.2, 5.3 Mathematical Induction I & II 24 / 66

x i =xk+1 − 1

x − 1.

x i =xk+2 − 1

x − 1. (2)

k+1∑i=0

x i = xk+1 +k∑

= xk+1 +xk+1 − 1

=xk+1 · (x − 1) + xk+1 − 1

x − 1=

xk+2 − 1

x − 1,

which is the right side of equation (2).Therefore P(k + 1) is true.Therefore Theorem is true by mathematical induction.MH1300 Lecture handout 6 (NTU) §5.2, 5.3 Mathematical Induction I & II 24 / 66

elementary number theory and methods of proof

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