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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?

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  • $\begingroup$ The answers and comments to math.stackexchange.com/questions/1630733/… didn't show you how to go about this? Your setup is wrong: if the two integers have opposite parity then one is of the form $2n$ and the other is of the form $2m+1$. Now multiply them, and factor... $\endgroup$ – BrianO Jan 28 '16 at 22:08
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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

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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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