Download Logical Equivalence of Induction Principles and more Study notes Mathematics in PDF only on Docsity! Variations on the Induction Principle David Radford 10/04/05 The purpose of these notes is to show that there are equivalent formula- tions of the Principle of Induction, Theorem 2 below. Consider the following statements: Theorem 1 (Existence of a Least Element) Every non-empty subset of the positive integers has a least element. Theorem 2 (Induction Principle) For all positive integers n let P (n) be a statement. Suppose that a) P (1) is true and b) for all n ≥ 1 if P (n) is true then P (n + 1) is true. Then P (n) is true for all n ≥ 1. Theorem 3 (Strong Induction Principle) For all positive integers n let P (n) be a statement. Suppose that a) P (1) is true and b) for all n ≥ 1 if P (1), . . . , P (n) are true then P (n + 1) is true. Then P (n) is true for all n ≥ 1. We will show that the preceding theorems are logically equivalent. That is not to say that any of them is true, but any one of them is true then they all are true. Theorem 4 Theorems 1–3 are logically equivalent. Proof: We show that Theorem 1 implies Theorem 2, that Theorem 2 im- plies Theorem 3, and finally that Theorem 3 implies Theorem 1. Thus any one of Theorems 1–3, denoted by A–C respectively, implies any other since “implies” is transitive. This is easily seen from the diagram below, where “−→” denotes “implies”. 1 ¢ ¢ ¢ ¢ ¢ ¢̧ ¾ A A A A A AU C A B Suppose that Theorem 1 is true. We show that Theorem 2 is true. Assume the hypothesis of Theorem 2 and let S be the subset of those positive integers n such that P (n) is false. We need only show that S = ∅. For then P (n) is true for all positive integers n; that is, Theorem 2 is true. Suppose to the contrary that S 6= ∅. Then S has a least element n by Theorem 1. Since P (1) is true by part a) of Theorem 2, 1 6∈ S. Therefore n > 1. Since 0 < n−1 < n it follows that n−1 is a positive integer and n−1 6∈ S. By definition of S the statement P (n − 1) is true. By part b) of Theorem 2 we conclude that P (n) is true. Thus n 6∈ S, a contradiction. Therefore S = ∅ and our proof that Theorem 1 implies Theorem 2 is complete. Suppose that Theorem 2 is true. We show that Theorem 3 is true. Assume the hypothesis of Theorem 3. Let P ′(1) be the statement P (1) and for n > 1 let P ′(n) be the statement P (1) and P (2) · · · and P (n). Observe that P ′(n) is true if and only if P (1), P (2), . . . , P (n) are true. We need only show that the statements P ′(1), P ′(2), . . . satisfy the hy- pothesis of Theorem 2. For then P ′(n) is true for all n ≥ 1. In particular P (n) is true for all n ≥ 1 and thus Theorem 3 is true. Since P (1) is true by part a) of Theorem 3 it follows that P ′(1) is true. Suppose that n ≥ 1 and P ′(n) is true. Then P (1), . . . , P (n) are true. By part b) of Theorem 2 the statement P (n + 1) is true. Therefore P ′(n + 1) is true. We have shown that the hypothesis of Theorem 2 is satisfied for P ′(1), P ′(2), . . . which completes our proof that Theorem 2 implies Theorem 3. Suppose that Theorem 3 is true. We show that Theorem 1 is true. This we do by establishing the contrapositive of the conclusion of Theorem 1; namely that if S is a subset of positive integers with no minimal element then S = ∅. 2
