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Prove by contradiction that for any integer n if n2 is odd then n is odd.

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Consider the next statement: for any integer if n^{2} is odd then n is non-odd (if n is non-odd integer then n is even)

So, if n^{2} is odd then n is even

Let n^{2} = 2k+1, k \in Z

n = 2m, m \isin N ∪ {0}

n^{2} = 2k+1 \to n=\sqrt{2k+1}

\sqrt{2k+1} = 2m \to 2k+1=4m^{2}

Since 2k is even integer, then 2k+1 is odd. 4m^{2} is also even. We came to the conclusion that an odd number is equal to the even integer number, which is false. So, assumption that ' for any integer if n^{2} is odd then n is non-odd ' is false, which means that for any integer if n^{2} is odd then n is odd. The statement has been proved.

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Question ID: mtid-5-stid-8-sqid-1399-qpid-1137