We want to show P (n) is true for all integers

Transcription

1 Generalized Induction Proof: Let P (n) be the proposition n = 2 n+1 1. We want to show P (n) is true for all integers n 0.

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3 Generalized Induction Example: Use generalized induction to prove that the sum of the first n + 1 terms in a geometric progression with initial term a and common ratio r 1 is given by a + ar + ar ar n = arn+1 a r 1

4 Generalized Induction Proof: Let P (n) be the proposition a + ar + ar ar n = arn+1 a r 1 We want to show P (n) is true for all integers n 0.

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6 Generalized Induction Example: Use generalized induction to prove the inequality 2 n < n! for all integers n 4.

7 Generalized Induction Proof: Let P (n) be the proposition 2 n < n!. We want to prove that P (n) is true for all integers n 4.

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9 Generalized Induction Example: Use mathematical induction to prove that 7 n n+1 is divisible by 57 for all n 0.

10 Generalized Induction Proof: Let P (n) be the proposition m (m Z 7 n n+1 = 57m). We want to show P (n) is true for all n 0.

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12 Generalized Induction Generalized De Morgan s Law Let A 1, A 2,..., A n be subsets of a universal set U, and let P (n) be the proposition c n A j j=1 = n A c j j=1 Prove P (n) is true for all n 2. Proof: Base Step: The statement P (2) asserts (A 1 A 2 ) c = (A 1 ) c (A 2 ) c, which is true by De Morgan s Law. Inductive Step: Assume P (k) is true for some fixed integer k 2. That is, assume c k A j j=1 = k A c j j=1

13 for any collection of k sets A 1, A 2,..., A k. Then, k+1 c A j j=1 = = = k A j j=1 c k A j j=1 k A c j j=1 A k+1 A c k+1 A c k+1 c = k+1 A c j j=1 This completes the inductive step. Therefore, P (n) is true for all n 2.

14 Complete Induction and the Well- Ordering Principle Complete Induction as a Rule of Inference In mathematical proofs, complete induction (PCI) is a rule of inference of the form P (a) P (a + 1) P (b) k b ([P (a) P (a + 1) P (k)] P (k + 1)) n a (P (n)) where a and b are positive integers with a b, and P is any propositional function with domain n a.

15 Complete Induction To prove that a propositional function P (n) is true for all positive integers n a, we complete two steps: 1. (Base Case) Verify the truth of P (a) P (a + 1) P (b) 2. (Inductive Step) Prove the conditional statement [P (a) P (a + 1) P (k)] P (k + 1) is true for all positive integers k b. To complete the inductive step, we assume P (m) is true for all m = a, a+1, a+2,..., k, then show P (k + 1) must also be true for all k b.

16 Complete Induction Special Case: a = b = 1 To prove that a propositional function P (n) is true for all positive integers n, we complete two steps: 1. (Base Case) Verify that P (1) is true. 2. (Inductive Step) Prove the conditional statement [P (1) P (2) P (3) P (k)] P (k + 1) is true for all positive integers k.

17 Complete Induction Special Case: a = b = 1 Example: Prove that every positive integer n has a binary expansion of the form n = a 0 + a a a j 2 j, where j is a nonnegative integer, a j = 1, and a i {0, 1} for all i.

18 Complete Induction Proof: Let P (n) be the proposition: n = a 0 + a a a j 2 j, where j is a nonnegative integer, a j = 1, and a i {0, 1} for all i. We want to show P (n) is true for all n 1. Base Step: P (1) is true, since 1 = a where j = 0 and a 0 = 1. Inductive Step: Assume P (m) is true for all positive integers m k. Then, consider the integer k + 1. We have two cases:

19 Case 1: (k+1 is even) If k + 1 is even, then k + 1 = 2m where m k. Therefore, P (m) is true and we have k + 1 = 2m = 2(a 0 + a a a j 2 j ) = 0 + a a a a j 2 j+1 = â 0 + â â â â j+1 2 j+1 where â j+1 = a j = 1, â 0 = 0, and â i = a i 1 {0, 1} for all 1 i j. Therefore, P (k + 1) is true in this case. Case 2: (k+1 is odd) If k + 1 is odd, then k = 2m where m < k.

20 Therefore, P (m) is true and we have k + 1 = 1 + 2m = 1 + 2(a 0 + a a a j 2 j ) = 1 + a a a a j 2 j+1 = â 0 + â â â â j+1 2 j+1 where â j+1 = a j = 1, â 0 = 1, and â i = a i 1 {0, 1} for all 1 i j. Therefore, P (k + 1) is true in either case. This completes the inductive step. Therefore, it follows by complete induction that P (n) is true for all n 1.

21 Complete Induction Example: Prove that every positive integer n 2 can be expressed as a product of prime numbers.

22 Complete Induction Proof: Let P (n) be the proposition: n = p 1 p 2 p j, where j 1 and p i is prime for all j. We want to show P (n) is true for all n 2. We will use complete induction corresponding to the case a = b = 2. Base Step: P (2) is true, since is a prime number. 2 = p 1 Inductive Step: Assume P (m) is true for all integers m = a, a + 1,..., k. That is, assume all integers m = 2, 3,..., k have a prime factorization.

23 Then, for k b = 2, we have two cases. Case 1: (k+1 is prime) If k +1 = p 1 is prime, then P (k +1) is true. Case 2: (k+1 is composite) If k + 1 is composite, then there exist integers c and d with 1 < c d < k + 1, such that k + 1 = c d Then, 2 c k and 2 d k, and it follows by the induction hypothesis that c = p 1 p 2 p j d = q 1 q 2 q l where p i and q i are prime for all i. Thus, k + 1 = (p 1 p 2 p j ) (q 1 q 2 q l )

24 and we conclude that P (k + 1) is true. This completes the inductive step, and it follows by complete induction that P (n) is true for all n 2.

25 Well-Ordering Principle Axiom: Every nonempty subset of Z + has a least element. That is, if S Z + and S, then S has a smallest element.

26 Well-Ordering Principle Example: Use well-ordering property to prove the division algorithm: If a is an integer and d is a positive integer, then there exist (unique) integers q and r with 0 r < d such that a = dq + r.

27 Well-Ordering Principle Proof: Consider the set S = {n Z + n = a dq +1 where q Z} This set is nonempty because dq can be made as large as desired (by taking q < 0 with q sufficiently large). By the wellordering principle, S has a least element n 0 = (a dq 0 ) + 1 where q 0 Z. We claim n 0 d. If not, then n 1 = (a d(q 0 + 1)) + 1 = n 0 d > 0, which implies n 1 S and n 1 < n 0. This contradicts the assumption that n 0 is the least element of S. Therefore, 1 n 0 d. This proves a = dq 0 + r where r = n 0 1 is an integer such that 0 r < d.

28 The proof that q 0 and r are unique is left as an exercise.

29 Well-Ordering Principle Theorem: The Well-Ordering Principle (WOP) and the Principle of Mathematical Induction (PMI) are equivalent.

30 WOP PMI Proof: Assume WOP holds. We want to prove PMI holds. Let S Z + such that (i) 1 S, (ii) k S k + 1 S for all k Z +. We want to prove S = Z +. Assume for the sake of contradiction that S Z +. Then, T = Z + S is a non-empty subset of Z +. Therefore, by the well-ordering principle, T has a least element m. That is, there exists m T such that m n for all n T. Since 1 S, we know m 2. Therefore m 1 is a positive integer. Clearly m 1 / T (otherwise m 1 would be the least element

31 of T ). This means m 1 S, and by assumption (ii) (m 1) S (m 1) + 1 S. Therefore, m S which contradicts the fact that m T. We conclude that S = Z +. Therefore PMI holds.