integers n, if n2 is odd, then n is odd. Proof: Assume that n2 is odd but n is not odd, so n must be even. By definition, there are some integers p and q such that n2 = 2p + 1 and n = %3D n? = (2q)² = 2q. Therefore 2p + 1 4q?. Rewriting above equation, we have 1 = 4q? - 2p = 2(2q? - p), which is even since 2q² - p is an integer. %3D Which proof method is used in above proof. Proof by counterexample Direct proof O Proof by contraposition Dreof b y Contradiction

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