Show that if is even, then is even.
I think I've got it:
By definition, we can write for some integer . This implies that . But, this means that has a factor of 2. Therefore, has a factor of 2. So, by definition, is even.
Now, my gut instinct is that I butchered this. Can someone clean it up?
Isn't this circular reasoning. You are assuming m is even, that is what we want to prove. I think a cleaner way is to prove the logically equivalent statement the contra-positive.
Remember that if p then q is equivalent to not q implies not p. The reformulation is
If m is not even then is not even
Now just set is odd. QED
I think the original proof is fine, but could possibly use another sentence or two. You can always just say that if m is odd, then m and 2 are relatively prime. Since 2 divides the LHS but doesn't divide m, it must divide the other factor, which is m. Contradiction.
The contrapositive still seems fastest and most basic though.