TITLE: Question involving proof by contrapositive of: If $n$ is even, then $nk$ is even, where $k \in \mathbb Z$.
QUESTION [0 upvotes]: The contrapositive of the statement is: If $nk$ is odd, then $n$ is odd; where $k \in \mathbb Z$.
I am using the cases where $k$ is odd and $k$ is even.
If $k$ is odd:
$$nk=2l+1$$
$$n(2m+1)=2l+1$$
$$2mn+n=2l+1$$
$$2mn-2l-1=-n$$
$$2(mn-l)-1=-n$$
$$2(l-mn)+1=n$$
So, $n$ is odd.
Now, if $k$ is even, then 
$$nk=2l+1$$
$$n(2m)=2l+1$$
$$2(mn)=2l+1$$
So, we have even is equal to odd. So, $k$ cannot be even.
Does that change the domain in the contrapositive, hence making the two statements not equal?

REPLY [1 votes]: When we make the claim 'if n is even, then nk is even', we are not restricting the domain to where $n$ is even: the claim as a whole is about the whole domain. As another example: If I say 'All even integers greater than 2 are the sum of two prime numbers', I am making that claim within the domain of all numbers ... I am just claiming that some of the numbers from that domain have some interesting property. 
So in your case, once we specify that $n$ and $k$ are integers, we have fixed our domain, and that is not going to change. Even if the 'if' part of a conditional claim like seems to restrict the domain .. it does not. Likewise, the contrapositive 'if nk is odd, then n is even' still assumes that the domain for $n$ and $k$ is all integers.
And otherwise, this is indeed a perfectly good proof by contraposition ... Maybe clean it up a little bit by first showing that $k$ cannot be even, and then showing that $n$ has to be odd (all under the assumption that $nk$ is odd, of course). But again, none of that changes the domain that the claims were about.
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