If the square of a number is even, then the number is even. Is that true for 2? I'll quickly go over my understanding of it:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that if $n$ is not even, then $n^2$ is not even.
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(2p^2+2p) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Now here's my question, set $n^2=2$. $n^2$ is even but $n=\sqrt2$ isn't. Or is it? I'm confused
 A: Both $n^2$ and $n$ must be integers for this theorem to hold.
A: Everything you wrote is correct up until the last line.  If $n^2$ is even, then
$$
n^2 = 2m \quad\text{ for some } m.
$$
When you wrote $n^2 = 2$, you were (falsely) assuming that $m=1$.
By the way the even/odd language is only used when you're talking about integers (whole numbers), so there is no square root of $2$.
A: The original statement about "the square of a number" is true in a context where "number" means "integer." If that context is not clear, you had better say explicitly "the square of an integer". The square of any even integer can be written $2m$ where $m$ is an integer, but not every number that can be written
as $2m$ where $m$ is an integer is a square of an integer of any kind.
Examples of squares of even integers written in the form $2m$:
$$\begin{align}
2^2 &=  4 = 2m &   \text{where }\  & m=2. \\
4^2 &= 16 = 2m &   \text{where }\  & m=8. \\
6^2 &= 36 = 2m &   \text{where }\  & m=18. \\
8^2 &= 64 = 2m &   \text{where }\  & m=32. \\
\end{align}$$
Examples of $2m$ where $m$ is an integer,
but $2m$ is not the square of any integer:
$$\begin{align}
 2 = 2m &  & \text{where }\  & m=1. & & \text{Not the square of any integer: }\
1^2 < 2 < 2^2.\\
 6 = 2m &  & \text{where }\  & m=3. & & \text{Not the square of any integer: }\
2^2 < 6 < 3^2.\\
 8 = 2m &  & \text{where }\  & m=4. & & \text{Not the square of any integer: }\
2^2 < 8 < 3^2.\\
10 = 2m &  & \text{where }\  & m=5. & & \text{Not the square of any integer: }\
3^2 < 10 < 4^2.\\
\end{align}$$
A: 
if n is not even (odd)

Not being even is the same as being odd only if n is an integer. Concretely, in your case, sqrt(2) is certainly not even. But that doesn't mean it's odd. :-)
A: [This answer was sparked by a question $4$ years later about the above question].
Almost surely the intended context is the ring of integers, so $\,n\in \Bbb Z,\,$ which excludes $\,n = \sqrt 2$.
But it is instructive to examine what occurs in the more general number ring that you consider.
In fact it remains true if we adjoin $\,\sqrt 2\,$ to $\,\Bbb Z\,$ to obtain $\,\Bbb Z[\sqrt 2] = \{ j + k \sqrt 2\ :\ j,k\in\Bbb Z\}$
As I explain here this ring has a sense of parity: $\,\alpha = j+k\sqrt 2\,$ is even $\iff \sqrt 2\mid \alpha \iff 2\mid j$ 
which immediately yields $\ \alpha^2\,$ even $\iff \alpha$ even.  As explained in the linked post, integer parity results immediately generalize to any ring which has $\,\Bbb Z/2 = $ integers $\!\bmod 2\,$ as an image.
As above, many results from elementary number theory generalize to algebraic numbers. These topics are covered in any course on algebraic number theory.
