Proving two integers of opposite parity have an even product?
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?
$\endgroup$ 12 Answers
$\begingroup$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
$\endgroup$ $\begingroup$If $m$ and $n$ have opposite parity, then one must be odd and one must be even. Without loss of generality, let $m$ be even and $n$ be odd. Then use the definition of odd/even integers, calculate $mn$, and proceed from there.
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