Discrete Mathematics, 1st Edition Kevin Ferland Chapter 4 Indexed by Integers

A sequence is simply an ordered list of real 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. 2 Ch4-p168

S = sa + sa+1 + sa+2 + ・・ ・+sb−1 + sb , (4.10) 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) 3 Ch4-p177

THEOREM 4.1 4 Ch4-p178

THEOREM 4.2 5 Ch4-p178

THEOREM 4.3 6 Ch4-p179

When b < a, it is assigned the value 1. Product Notation The product P = sa ・ sa+1 ・ sa+2 ・ ・・・ ・sb is represented by When b < a, it is assigned the value 1. 7 Ch4-p181

Let n ≥ 1. We represent a factorization of EXAMPLE 4.16 Let n ≥ 1. We represent a factorization of x2n − y2n in product notation. 8 Ch4-p181

EXAMPLE 4.16 (Cont.) Solution. 9 Ch4-p181

4.3 Mathematical Induction, an Introduction Well-Ordering Principle Each nonempty subset of the nonnegative integers has a smallest element. 10 Ch4-p185

THEOREM 4.6 11 Ch4-p188

Suppose it is not true that P(n) holds ∀ n ≥ a. 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. 12 Ch4-p188

Since P(a), P(a + 1), . . . , P(b) all hold, it must be that s > b. 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. 13 Ch4-p188

(Proof by Mathematical Induction). To show: ∀ n ≥ a, P(n). OUTLINE 4.1 (Proof by Mathematical Induction). To show: ∀ n ≥ a, P(n). 14 Ch4-p86

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. 15 Ch4-p186

EXAMPLE 4.18 Show: ∀ n ≥ 0, 2n ≥ n + 1. 16 Ch4-p186

It is straightforward to see that 20 = 1 ≥ 0 + 1. EXAMPLE 4.18 Proof. (By Induction) Base case: (n = 0) It is straightforward to see that 20 = 1 ≥ 0 + 1. 17 Ch4-p186

EXAMPLE 4.18 (Cont.) 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. 18 Ch4-p187

4.4 Induction and Summations EXAMPLE 4.26 Show: ∀ n ≥ 0, 19 Ch4-p195

EXAMPLE 4.26 Proof. (By Induction) Base case: (n = 0) It is straightforward to see that Inductive step: Suppose k ≥ 0 and that (Goal: ) 20 Ch4-p197

EXAMPLE 4.26 (Cont.) Observe that 21 Ch4-p197-198

EXAMPLE 4.26 (Cont.) 22 Ch4-p198

4.5 Strong Induction 23 Ch4-p203

P(i) holds for each a ≤ i ≤ n. 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. 24 Ch4-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. 25 Ch4-p203

Strong Induction OUTLINE 4.2 To show: ∀ n ≥ a, P(n). 26 Ch4-p202

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. 27 Ch4-p202

Every integer greater than 1 has a prime divisor. Theorem 3.5 Every integer greater than 1 has a prime divisor. 28 Ch4-p202

Certainly, 2 is already prime and divides itself. That is, 2 | 2. Theorem 3.5 Proof. Base case: (n = 2) Certainly, 2 is already prime and divides itself. That is, 2 | 2. 29 Ch4-p202

Suppose k ≥ 2 and that each integer i with 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. 30 Ch4-p202

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. 31 Ch4-p202

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. 32 Ch4-p203

THEOREM 4.8 33 Ch4-p204

Existence and uniqueness are proved 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. 34 Ch4-p204

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. 35 Ch4-p204

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. 36 Ch4-p204

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.) 37 Ch4-p204

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. 38 Ch4-p204

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. 39 Ch4-p204-205

F0 = 1, F1 = 1, and ∀ n ≥ 2, Fn = Fn−2 + Fn−1. 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. 40 Ch4-p205

Show that the Fibonacci sequence can be expressed by the formula EXAMPLE 4.27 Show that the Fibonacci sequence can be expressed by the formula 41 Ch4-p205

It is straightforward to check that and EXAMPLE 4.27 Proof. Base cases: (n = 0, 1) It is straightforward to check that and 42 Ch4-p205

EXAMPLE 4.27 (Cont.) Inductive step: Suppose k ≥ 1 and that 43 Ch4-p205

EXAMPLE 4.27 (Cont.) Observe that 44 Ch4-p205-206

EXAMPLE 4.27 (Cont.) 45 Ch4-p206

(a + b)n for some integer n ≥ 0. 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. 46 Ch4-p211

(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 … … 47 Ch4-p211

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. 48 Ch4-p211-212

The edges of the triangle display the identity ∀ n ≥ 0, cn,0 = cn,n = 1. 49 Ch4-p212

∀ n ≥ 2 and 1 ≤ k ≤ n − 1, cn,k = cn−1,k−1 + cn−1,k . (4.20) 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) 50 Ch4-p212

THEOREM 4.9 51 Ch4-p213

Use the Binomial Theorem to expand each of the following. EXAMPLE 4.29 Use the Binomial Theorem to expand each of the following. 52 Ch4-p213

EXAMPLE 4.29 (Cont.) (a) Expand (x + y)6. Solution. We use a = x, b = y, and n = 6 in Theorem 4.9. 53 Ch4-p213-214

EXAMPLE 4.29 (Cont.) (b) Expand (2x + 3y)5. Solution. We use a = 2x, b = 3y, and n = 5 in Theorem 4.9. 54 Ch4-p214

EXAMPLE 4.29 (Cont.) 55 Ch4-p213

EXAMPLE 4.29 (Cont.) (d) Expand (x − y)n. Solution. We use a = x and b = −y in Theorem 4.9. 56 Ch4-p214

Find the coefficient of x40 in (1 + 2x)50. EXAMPLE 4.30 Find the coefficient of x40 in (1 + 2x)50. 57 Ch4-p215

EXAMPLE 4.30 Solution. By the Binomial Theorem, Consequently, the coefficient of x40 (that is, when i = 40) is 58 Ch4-p215

Verify each of the following identities. (a) Proof By Theorem 4.9, EXAMPLE 4.31 Verify each of the following identities. (a) Proof By Theorem 4.9, 59 Ch4-p215

EXAMPLE 4.31 (Cont.) (b) Proof By Theorem 4.9, 60 Ch4-p215

EXAMPLE 4.31 (Cont.) (c) Proof By Theorem 4.9, 61 Ch4-p215