Prove n is odd if and only if n^3 is odd. So we want to prove if n is odd then n^3 is not odd. Which is the contradiction.

If n is odd then there exist a number in which n=2m+1. Where do I go from here to finish the proof?

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- May 4th 2010, 07:47 PMbaker11108Contradiction proof
Prove n is odd if and only if n^3 is odd. So we want to prove if n is odd then n^3 is not odd. Which is the contradiction.

If n is odd then there exist a number in which n=2m+1. Where do I go from here to finish the proof? - May 4th 2010, 08:20 PMjakncoke
ok, e n is odd and $\displaystyle n^3$ is even. By def of odd and even, there

$\displaystyle \exists \: m \in \mathbb{Z}$ such that $\displaystyle n = 2m + 1 $ similarly, $\displaystyle \exists \: k \in \mathbb{Z}$ such that $\displaystyle n^3 = 2k$ so $\displaystyle (2m+1)^3 = 2k $ if you remember what division meant, if a divides b then there exists an integer g such that b = ag. since k is an integer, this means 2 divides (2m+1)^3 which means 2 divides (2m+1) which is a contradiction.