Chap 1 The Number System

# Suppose next that pm is true for every natural number

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Unformatted text preview: at the conclusion of (PIS) does not hold. Then the subset S = {n ∈ N : p(n) is false} of N is non-empty. By (WO), S has a least element, n0 say. If n0 = 1, then clearly (PIS1) does not hold. If n0 > 1, then p(m) is true for all m ≤ n0 − 1 but p(n0 ) is false, contradicting (PIS2). ((PIS) ⇒ (WO)) Suppose that a non-empty subset S of N does not have a least element. Consider the statement p(n), given by n ∈ S . Then p(1) is true, otherwise 1 would be the least element of S . Suppose next that p(m) is true for every natural number m ≤ n, so that none of the numbers 1, 2, 3, . . . , n belongs to S . Then p(n + 1) must also be true, for otherwise n + 1 would be the least element of S . It now follows from (PIS) that S does not contain any element of N, contradicting the assumption that S is a non-empty subset of N. Next, we complete the proof by showing that the Principle of induction (weak form) (PIW) is equivalent to the Principle of induction (strong form) (PIS). ((PIS) ⇒ (PIW)) Suppose that (PIW1) and (PIW2) both hold. Then clearly (PIS1) holds, since it is the same as (PIW1). On the other hand, if p(m) is true for all...
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