Proof by "contraposition"? This is a homework problem, so please do not give anything more than hints. I must prove the following:

$\forall m,n\in\mathbb{Z},$ if $mn$ is even then $m\vee n$ is even.

Yes, one of the most basic things you could possibly want to prove. But is the shortest method then the "contraposition"? I do not like this method because it seems I must make tacit logic assumptions to arrive at a "weak" conclusion. For instance, I see here I have three options: Either $m,n$ are both even, both odd, or one is odd and one is even. Is it enough to show that if both $m,n$ are odd, thus resulting in an odd integer, then at least one must be even for the opposite to be true? It is odd to me because I haven't then shown explicity (or even implied directly) that the latter is true, I have merely shown that if both are odd, then I get an odd number. I'm assuming logical equivalence that hasn't been shown..
Any help is appreciated. Thank you for your time,
 A: The contrapositive of what you are proving is: If $m$ and $n$ are both odd, then $mn$ is odd.  [This uses DeMorgan's Law, and the fact that the negation of a number being even is that it is odd.]
So you can prove your result by assuming $m$ and $n$ are odd and showing that $mn$ is odd.
A: The negation of the consequent, "$m$ or $n$ is even", is as follows: "not ($m$ or $n$ is even)" which is equivalent (by DeMorgan's) to "($m$ is not even) and ($n$ is not even)," i.e.,   

Both $m$ and $n$ are odd. 

The negation of the antecedent is "Not($mn$ is even)", i.e., $mn$ is odd.
Then you need to prove that the negation of the consequent implies the negation of the antecedent:  

"If (both $m$ is odd and $n$ is odd), then ($mn$ is odd)."

A: With $E(x)$ meaning "$x$ is even" - and, naturally, $\neg E(x)$ meaning "$x$ is not even" or equivalently "$x$ is odd" - your proposition becomes:
$$\forall m\forall n\big(E(mn)\to E(m)\lor E(n)\big)$$
Thus, by contraposition
$$\forall m\forall n\Big(\neg\big(E(m)\lor E(n)\big)\to\neg E(mn)\Big)$$
$$\Longleftrightarrow \forall m\forall n\Big(\big(\neg E(m)\land\neg E(n)\big)\to\neg E(mn)\Big)$$
Therefore, it is equivalent to show that "if both $m$ and $n$ are odd, then $mn$ must be odd too". As you stated, since the product of two odd integers is another odd integer, then this last proposition is true, and by contraposition, so is the original one.
