TITLE: A proof using the contrapositive
QUESTION [10 upvotes]: I am trying to prove the following conjecture:

Prove that if $m$ and $n$ are integers and $mn$ is even, then $m$
  is even or $n$ is even.

Proof by contraposition:
Assume $m$ and $n$ are odd. Then $m = 2k + 1$ and $n = 2l + 1$. So
$$mn = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1$$
QED
Is there anything else I need to do in order to prove this conjecture? Thank you!

REPLY [3 votes]: Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.