CIS 1910 Lecture Notes - Lecture 8: Luiza, If And Only If

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Follow a series of logical steps to conclude that r r f for some proposition. Theorem: for all integers n, if n2 is odd, then n is odd. P (n) : n2 is an odd integer. Assume n2 is and odd integer, and n is an even integer. An integer n is even iff n = 2k for some integer k. This concluded that n2 is even, but this contradicts the assumption that n2 is an odd integer. Theorem: if two integers a, b then a2 4b = 2. Q(a, b) : a2 4b = 2. By contradiction; assume a b (a2 4b = 2) An integer n is even iff n = 2k, k z. We have, a2 4b = 2 a2 = 2 + 4b. From a previously proven theorem, we know that if a2 is an even integer, then a is an even integer.

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