I think I might be beginning to wrap my head around some simpler proofs, but I'm a little stumped on this one from my textbook:
Use a direct proof to show that if two integers have opposite parity,
then their product is even.
If I have integers $(m,n)$ with even parity, I would then have (from what I've gathered) an integer $a = 2m$ and an integer $b = 2n$. I'm not sure where I go from here in looking for a product?
In my mind I would do $a \cdot b = (2m)(2n) = ???$
I know I'm dealing with integers, and I use $m$ and $n$ to respectively denote different integer values I'm dealing with.
Can anyone walk me through how to finish this out?
Best Answer
If two integers have opposite parity, then one is even and the other is odd. So, the product is even::
Let $a$, $b$ with opposite parity, say $a$ even, then $a=2n$ and $b=2m+1$. Therefore $ab=2n(2m+1)$ which is even