Chapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1

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1 Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1

2 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide

3 Recall that denotes the set of positive integers, also called the natural numbers. = {1,, 3, 4,... } In this text we do not attempt to develop the properties of the natural numbers in a rigorous way from set theory. Rather, we assume familiarity with the usual arithmetic operations of addition and multiplication and the relation of less than. There is one additional property of that we assume as an axiom. It is based on the intuitive idea that each nonempty subset of must have a least member. Axiom (Th 6..5) If S is a nonempty subset of m k for all k S. The Well-Ordering Property of, then there exists an element m S such that In proving theorems about the natural numbers, we need a technique that does not depend on verifying the validity one number at a time. This is provided by the Principle of Mathematical Induction. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 3

4 Section 6.3 MATHEMATICAL INDUCTION 013 Pearson Education, Inc.1 Slide 4

5 Theorem The Principle of Mathematical Induction Let P (n) be a statement that is either true or false for each n for all n, provided that (a) P (1) is true, and (b) for each k, if P (k) is true, then P (k + 1) is true.. Then P (n) is true Proof: Our argument is a proof by contradiction using tautology (g). That is, we suppose that (a) and (b) hold but that P (n) is false for some n. Let S = {n : P (n) is false}. Then S is not empty and the well-ordering property guarantees the existence of an element m S that is a least element of S. Since P (1) is true by hypothesis (a), 1 S, so that m > 1. It follows that m 1 is also a natural number, and since m is the least element in S, we must have m 1 S. But since m 1 S, it must be that P (m 1) is true. We now apply hypothesis (b) with k = m 1 to conclude that P (k + 1) = P (m) is true. This implies that m S, which contradicts our original choice of m. We conclude that P (n) must be true for all n. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 5

6 Theorem 3.1. The Principle of Mathematical Induction Let P (n) be a statement that is either true or false for each n. Then P (n) is true for all n, provided that (a) P (1) is true, and (b) for each k, if P (k) is true, then P (k + 1) is true. In applying the Principle of Mathematical Induction, there are two steps: Verifying condition (a) is called the basis for induction. Verifying condition (b) is called the induction step. The assumption that P (k) is true in verifying part (b) is known as the induction hypothesis. Let s look at a couple of examples of using mathematical induction. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 6

7 Example 1 Prove that n= 1 nn ( + 1) for every natural number n. Proof: Let P(n) be the statement n= 1 nn ( + 1). Then P (1) asserts that 1 = 11(1 + 1), P () asserts that 1 + = 1 ( + 1), and so on. In particular, we see that P(1) is true, and this establishes the basis for induction. To verify the induction step, we suppose that P(k) is true, where k. That is, we assume k= 1 kk ( + 1). Since we wish to conclude that P (k + 1) is true, we add k + 1 to both sides to obtain k+ ( k+ 1) = kk ( + 1) + ( k+ 1) Now do some algebraic manipulation on the right-hand side of the equation: k+ ( k+ 1) = [ kk ( + 1) + ( k+ 1)] = [( k + 1)( k + )] = ( k + 1)[( k + 1) + 1] This last equation is the statement P (k + 1). So P (k + 1) is true whenever P (k) is true, and by the principle of mathematical induction we conclude that P (n) is true for all n. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 7

8 Since the format of a proof using mathematical induction always consists of the same two steps (establishing the basis for induction and verifying the induction step), it is common practice to reduce some of the formalism by omitting explicit reference to the statement P (n). Example Prove by induction that 7 n 4 n is a multiple of 3, for all n. Proof: Clearly, this is true when n = 1, since = 3. Now let k that 7 k 4 k is a multiple of 3. That is, 7 k 4 k = 3m for some m. Comment: We now want to show that 7 k k + 1 is a multiple of 3. we use algebra to show how 7 k k + 1 depends on 7 k 4 k. It follows that 7 k k + 1 = 7 k k k 4 4 k = 7(7 k 4 k ) k = 7(3m ) k = 3(7m + 4 k ) and suppose To do this, Since m and k are natural numbers, so is 7m + 4 k. Thus 7 k +1 4 k +1 is also a multiple of 3, and by induction we conclude that 7 n 4 n is a multiple of 3 for all n. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 8

9 There is a generalization of the principle of mathematical induction that enables us to conclude that a given statement is true for all natural numbers sufficiently large. Theorem 6.3.6, Principle of Mathematical Induction Variant 1 Let m and let P(n) be a statement that is either true or false for each n m. Then P(n) is true for all n m, provided that (a) P (m) is true, and (b) for each k m, if P (k) is true, then P (k + 1) is true. Proof: The proof will use the original principle of induction. For each r, let Q (r) be the statement P(r + m 1) is true. Then from (a) we know that Q (1) holds. Now let j That is, P ( j + m 1) is true. Since j, j + m 1 = m + ( j 1) m, and suppose that Q ( j) holds. so by (b), P ( j + m) must be true. Thus Q ( j + 1) holds and the induction step is verified. We conclude that Q (r) holds for all r. Now if n m, let r = n m + 1, so that r But P (r + m 1) is the same as P (n), so P (n) is true for all n m.. Since Q (r) holds, P (r + m 1) is true. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 9

10 Exercise For which values of n N does the inequality n 9n + 19 > 0 hold? Prove your answer by induction = 11 > = 5 > = 1 > = 1 < = 1 < 0 Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 10

11 = 1 > = 5 > = 11 > 0 P(n): For n 6, the inequality n 9n + 19 > 0 hold. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 11

12 Theorem The Principle of Mathematical Induction-Variant Let P (n) be a statement that is either true or false for each n. Then P (n) is true for all n, provided that (a) P (1) is true, and (b) for each k, if P (1),, P (k) are true, then P (k + 1) is true. Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1

13 Theorem Let n N. Suppose that n. Then n is either a prime number or a product of finitely many prime numbers. Proof (using the Principle of Mathematical Induction-Variant ) Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 13

14 Exercise (3) n 3 = n n+1 4 (4) n 1 3 = n (n 1) Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 14

15 hw Exercise 6.3.1, (1), (), (5). Exercise 6.3., Exercise Due Tue. Nov. 19 Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 15

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