Question

Prove:

Answer 1

Given the statement, "If $n^2$ is odd, then $n$ is odd," its contrapositive claims that, "If $n$ is not odd, then $n^2$ is not odd."

So assume $n$ is not odd, i.e. $n$ is even. This means there is an integer $k$ for which $n=2k$. Squaring this gives $n^2=(2k)^2=4k^2$.

Well, we can write $4k^2=2(2k^2)$, and $2k^2$ is just another integer, which means $4k^2=(2k)^2=n^2$ must be even.