discrete mathematics, 1st edition kevin ferland chapter 4 indexed by integers 1

Discrete Mathematics, 1st EditionKevin Ferland

Chapter 4

Indexed by Integers

1

Sequences

A sequence is simply an ordered list of real

numbers. All sequences have an initial term,

but only finite sequences have a final term.

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4.2 Sigma Notation

Given a sequence {sn }, one may be interested

in a sum of several of its terms:

S = sa + sa+1 + sa+2 + ・・・ +sb−1 + sb , (4.10)

where a, b Z. The notation used to represent ∈the sum in equation (4.10) is

(4.11)

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THEOREM 4.1

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THEOREM 4.2

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THEOREM 4.3

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Product Notation

The product

P = sa ・ sa+1 ・ sa+2 ・・・・・ sb

is represented by

When b < a, it is assigned the value 1.

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EXAMPLE 4.16

Let n ≥ 1. We represent a factorization of

x2n − y2n in product notation.

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EXAMPLE 4.16 (Cont.)

Solution.

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4.3 Mathematical Induction, an Introduction

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Well-Ordering PrincipleEach nonempty subset of the nonnegative integers has a smallest element.

THEOREM 4.6

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THEOREM 4.6

Proof. Assume conditions (i) and (ii) in the hypotheses of the theorem.

Suppose it is not true that P(n) holds ∀ n ≥ a.

Let S be the set of those integers n ≥ a for which P(n) does not hold.

By our assumptions, S is nonempty.

Hence, by the Generalized Well-Ordering Principle, S has a smallest element, say s.

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THEOREM 4.6 (Cont.)

Since P(a), P(a + 1), . . . , P(b) all hold, it must be that s > b.

Therefore, s − 1 ≥ b. Since s − 1 S, it follows that P(s − 1) holds. However, for k = s − 1, by condition (ii), since

P(k) holds, P(k + 1) must also hold. That is, P(s) holds.

This contradicts the fact that s ∈ S.

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OUTLINE 4.1

(Proof by Mathematical Induction).

To show: ∀ n ≥ a, P(n).

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OUTLINE 4.1

Proof by induction.

1. Base cases:

Show: P(a), . . . , P(b) are true.

2. Inductive step:

Show: ∀ k ≥ b, if P(k) is true, then P(k + 1) is true.

That is,

(a) Suppose k ≥ b and that P(k) is true.

(b) Show: P(k + 1) is true.

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EXAMPLE 4.18

Show: ∀ n ≥ 0, 2n ≥ n + 1.

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EXAMPLE 4.18

Proof.

(By Induction)

Base case: (n = 0)

It is straightforward to see that 20 = 1 ≥ 0 + 1.

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Inductive step:

Suppose k ≥ 0 and that 2k ≥ k + 1.

(Goal: 2k+1 ≥ (k + 1) + 1.)

Observe that

2k+1 = 2(2k )

≥ 2(k + 1) ← By the inductive hypothesis

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

≥ (k + 1) + 1.

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4.4 Induction and Summations

EXAMPLE 4.26

Show: ∀ n ≥ 0,

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EXAMPLE 4.26

Proof. (By Induction)

Base case: (n = 0)

It is straightforward to see that

Inductive step:

Suppose k ≥ 0 and that

(Goal: )

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Observe that

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4.5 Strong Induction

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Theorem 3.7

Proof.Assume conditions (i) and (ii) in the hypotheses

of the theorem.For each n ≥ a, let Q(n) be the statement that

P(i) holds for each a ≤ i ≤ n.Since Q(n) implies P(n), it is sufficient to show

that Q(n) holds ∀ n ≥ a.We accomplish this with a proof by regular

induction.24Ch4-p203

Theorem 3.7 (Cont.)

Base cases: (n = a, . . . , b)

The truth of Q(a), . . . , Q(b) follows from the truth of

P(a), . . . , P(b).

Inductive step:

Suppose k ≥ a and that Q(k) holds.

That is, P(i) holds for each a ≤ i ≤ k.

From (ii), it follows that P(k + 1) holds.

By definition, Q(k + 1) also holds.

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Strong Induction

OUTLINE 4.2

To show: ∀ n ≥ a, P(n).

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OUTLINE 4.2

Proof by strong induction.

1. Base cases:

Show: P(a), . . . , P(b) are true.

2. Inductive step:

Show: ∀ k ≥ b, if P(a), . . . , P(k) are true, then P(k + 1) is true.

That is,

(a) Suppose k ≥ b and that P(i) is true for all a ≤ i ≤ k.

(b) Show: P(k + 1) is true.27Ch4-p202

Theorem 3.5

Every integer greater than 1 has a prime

divisor.

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Theorem 3.5

Proof.

Base case: (n = 2)

Certainly, 2 is already prime and divides itself.

That is, 2 | 2.

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Theorem 3.5 (Cont.)

Inductive step:

Suppose k ≥ 2 and that each integer i with

2 ≤ i ≤ k has a prime divisor.

Case 1: k + 1 is prime.

Obviously, k + 1 divides itself.

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Theorem 3.5 (Cont.)

Case 2: k + 1 is composite.

We can express k + 1 as a product k + 1 = rs, where 2 ≤ r ≤ k and 2 ≤ s ≤ k.

Consequently, there exists a prime p that divides r.

That is, r = pt for some integer t.

It follows that k + 1 = rs = pts.

Therefore, k + 1 is divisible by the prime p.

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Standard Factorization

DEFINITION 4.1

The expression of an integer n > 1 as a product of the

form

where m is a positive integer, p1 < p2 < ・・・ < pm

are primes, and e1, e2, . . . , em are positive integers, is

referred to as the standard factorization of n.

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mem

ee pppn 2121

THEOREM 4.8

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THEOREM 4.8

Proof.

Existence and uniqueness are proved

separately, but each by strong induction.

Existence is handled first.

Base case: (n = 2)

Certainly, 2 = 21 is a standard factorization.

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THEOREM 4.8 (Cont.)

Inductive step:Suppose k ≥ 2 and that each integer i with 2 ≤ i ≤ k has a standard factorization.(Goal: k + 1 has a standard factorization.)

Our proof naturally breaks into two cases.Case 1: k + 1 is prime.Here, k + 1 = (k + 1)1 is already a standard

factorization.

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THEOREM 4.8 (Cont.)

Case 2: k + 1 is composite.

We can express k + 1 as a product k + 1 = rs, where 2 ≤ r ≤ k and 2 ≤ s ≤ k.

By the inductive hypothesis, r and s have standard factorizations.

By appropriately grouping the primes in the product rs, we obtain a standard factorization for k + 1.

Now we handle uniqueness.

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THEOREM 4.8 (Cont.)

Base case: (n = 2)Since 2 is prime, 2 has the unique standard

factorization 2 = 21. (Note that any other product of powers of primes is greater than 2.)

Inductive step:Suppose k ≥ 2 and that each integer i with 2 ≤ i ≤ k has

a unique standard factorization.(Goal: k + 1 has a unique standard factorization.)

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THEOREM 4.8 (Cont.)

Suppose k + 1 has two standard factorizations

Since p1 | (k + 1), Corollary 3.18 tells us that for some i.

Moreover, Corollary 3.19 tells us that p1 | qi .

Since p1 and qi are prime, it must be that p1 = qi .

Since the primes in a standard factorization are listed in increasing order, we have q1 ≤ qi = p1.

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THEOREM 4.8 (Cont.)

By a symmetric argument (reversing the roles of the p’s

and the q’s), we see that p1 ≤ q1. Thus, p1 = q1.

The integer now has the two standard factorizations

By the inductive hypothesis, those standard factorizations must be the same.

Therefore, the two standard factorizations for k + 1 must be the same.

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The Fibonacci Numbers

The sequence of numbers

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .

is known as the Fibonacci sequence.

If we denote the Fibonacci sequence by {Fn}n≥0, then

F0 = 1, F1 = 1, and ∀ n ≥ 2, Fn = Fn−2 + Fn−1.

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EXAMPLE 4.27

Show that the Fibonacci sequence can be

expressed by the formula

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EXAMPLE 4.27

Proof.

Base cases: (n = 0, 1)

It is straightforward to check that

and

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Inductive step:

Suppose k ≥ 1 and that

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Observe that

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4.6 The Binomial Theorem

The Binomial Theorem is a result that tells us

how to expand an expression of the form

(a + b)n for some integer n ≥ 0.

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(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(4.19)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

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… …

This infinite array of integers is known as Pascal’s

triangle. To express this, let cn,k denote the kth entry in

the nth row. Here, both n and k start counting from 0.

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The edges of the triangle display the identity

∀ n ≥ 0, cn,0 = cn,n = 1.

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The internal entries of each row are determined from the previous row by Pascal’s identity

∀ n ≥ 2 and 1 ≤ k ≤ n − 1, cn,k = cn−1,k−1 + cn−1,k . (4.20)

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THEOREM 4.9

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EXAMPLE 4.29

Use the Binomial Theorem to expand each of

the following.

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(a) Expand (x + y)6. Solution. We use a = x, b = y, and n = 6 in Theorem 4.9.

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(b) Expand (2x + 3y)5.

Solution.

We use a = 2x, b = 3y, and n = 5 in Theorem 4.9.

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(d) Expand (x − y)n. Solution.

We use a = x and b = −y in Theorem 4.9.

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EXAMPLE 4.30

Find the coefficient of x40 in (1 + 2x)50.

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EXAMPLE 4.30

Solution.

By the Binomial Theorem,

Consequently, the coefficient of x40 (that is, when i = 40) is

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EXAMPLE 4.31

Verify each of the following identities.

(a)

Proof

By Theorem 4.9,

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(b)

Proof

By Theorem 4.9,

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(c)

Proof

By Theorem 4.9,

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discrete mathematics, 1st edition kevin ferland chapter 4 indexed by integers 1

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