TITLE: Proving that all integers are even or odd
QUESTION [21 upvotes]: I know that $\mathbb{Z}$ is a group under addition with a multiplication defined. I have just the definition of even and odd integers: $n$ is even if $n = 2k$ for some integer $k$ and $n$ is odd if $n = 2k+1$ for some integer $k$.
Using just this I am wondering how to prove that all integers are either even or odd. That is, how can I prove that given an integer $n$, $n$ must be even or odd?
My problem with this is that it seems so simple. I know that one can divide an integer by $2$ and the remainder will be $0$ or $1$. Using this, it is clear that the even and odd integers make up everything. But how can I prove it without using this fact about remainders and such? I guess one could also use facts about prime numbers, but I am looking for a proof that just uses the definition of odd and even.

REPLY [8 votes]: I prefer appealing to the well-ordering principle.  It's equivalent to induction but in my opinion a tiny bit cleaner.
First, reduce the proposition to the statement that all positive integers are odd or even.  If there exist positive integers that are neither odd nor even, there must be a smallest positive integer $n$ which is neither odd nor even.  Since $1$ is definitely odd, $n \geq 2$, hence $n-1$ is still positive.
Since $n$ is the least positive integer that is neither odd nor even, $n-1$ must be either odd or even.

If $n-1$ is even, $n-1 = 2k$ for some $k$.  But this implies $n=2k+1$, i.e., $n$ is odd.
If $n-1$ is odd, $n-1 = 2k+1$ for some $k$.  But this implies $n=2k+2 = 2(k+1)$, i.e., $n$ is even.

Either way we have a contradiction, so the set of positive integers which are neither even nor odd is empty.