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The original question is to prove that the product of any two consecutive integers is even.

I can do this using direct proof: P implies Q.

I was curious to try to prove it by contraposition (not Q implies not P), but in order to do that i need to negate "the product of any two consecutive integers", which I couldn't find out how to do.
Any help would be great. (first time posting so tell if I am doing anything wrong :)

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    I think what you're looking for is: assume that the product is odd and show that the integers cannot be consecutive – 2017-02-08
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    Also there is an easier way to prove this by just considering the case of even and odd numbers and knowing that odd times even is even. – 2017-02-08
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    Awesome!!! You guy rock! I did not expect such fast and effective answers. – 2017-02-08

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The negation of "Every product of two consecutive integers is even" is "There exists two consecutive integers whose product is odd".

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    This is the negation, but the OP asks for the contrapositive, which is different. – 2017-02-08
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    The OP asks for both a negation so they can find a contrapostive. – 2017-02-08
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    What "both" did you negate? It seems to me you negated the entire implication at once. – 2017-02-08
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    @DavidK Typo there shouldn't be a "both" in there. – 2017-02-08
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    OK, but a contrapositive requires the negations of two separate statements. Which two statements did your answer negate, and how did you negate them? – 2017-02-08
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Rephrase the statement as follows:

If $n$ is the product of two consecutive integers, then $n$ is even.

So $P$ is "$n$ is the product of two consecutive integers" and $Q$ is "$n$ is even." Therefore the contrapositive, $\lnot Q \implies \lnot P,$ is

If $n$ is not even then $n$ is not the product of two consecutive integers.

To do this a little more formally we should quantify $n,$ "for all $n$," or perhaps more specifically "for every integer $n$."