MHB Contrapositive Proof: Ints $m$ & $n$ - Even/Odd Combinations

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To prove the contrapositive statement, assume one integer, $m$, is even and the other, $n$, is odd. The sum of an even and an odd integer is always odd, which demonstrates that if $m$ and $n$ are not both even or both odd, then $m + n$ cannot be even. This confirms the contrapositive: if $m + n$ is even, then both integers must be either even or odd. Thus, the proof is complete by showing the necessary condition for the original statement.
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For all integers $m$ and $n$, if $m+ n$ is even then $m$ and $n$ are both even or both odd.

For a contrapositive proof, I need to show that for all ints $m$ and $n$ if $m$ and $n$ and not both even and not both odd, then $ m + n $ is not even.

How do I go about doing this?
 
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The negation of the conclusion is that exactly ONE of $m,n$ is odd.

In a proof of this type, you may assume $m$ is even, and $n$ is odd (or else we may "switch them").

Your mission, should you decide to accept it, is to prove that in this case, we have the negation of the premise:

that is, to show that $m+n$ is thus odd.
 

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