I'm looking for feedback on my proof of the following statement:
"If $m$ and $n$ are integers and $mn$ is even, then $m$ is even or $n$ is even."
I tried using a direct proof method:
(1) Since $mn$ is even, $mn = 2k$ for some integer $k$. The integer $k$ must then be equal to $mn/2$, and 2 divides $mn$.
(2) In order for $k$ to be an integer, $m$ or $n$ must then have a factor of two, and the statement is proved.
I think it might be easier to prove by contrapositive. Proving your statement is the same as proving that if $m$ and $n$ are both odd, then $mn$ is odd.
We have $mn$ is even iff $2\mid mn$. But $2$ is prime, so Euclid's Lemma applies. Hence $2\mid mn$ iff either $2\mid m$ or $2\mid n$. Now $mn$ is even iff either $m$ is even or $n$ is even.
The really serious problem in your argument is here:
(2) In order for $k$ to be an integer, $m$ or $n$ must then have a factor of two, and the statement is proved.
That's a problem because it is just a restatement of what you have been asked to prove.
One good way to start this problem is to suppose that both factors are odd. Then use what you know (and have proved) about the partity of a product of odd numbers.
Your attempt at a direct proof will work once you have proved other important theorems - in particular, the theorem that says that if a prime divides a product it divides one of the factors.
