Proving $n^2$ is even whenever $n$ is even via contradiction? I'm trying to understand the basis of contradiction and I feel like I have understood the ground rules of it.
For example: Show that the square of an even number is an even number using a contradiction proof.
What I have is: Let n represent the number. 
n is odd if n = 2k + 1, where k is any number
n is even if n = 2k, where k is any number
We must prove that if n^2 is even, then n is even.
How do I proceed on from here?
 A: We prove the contrapositive.  In this case, we want to prove

$n^2$ even implies $n$ even

which is equivalent to the contrapositive

$n$ not even implies $n^2$ not even

or in other words

$n$ odd implies $n^2$ odd.

If $n$ is odd, then $n=2k+1$ then
\begin{align*}
n^2 &= (2k+1)^2 & \text{substituting in } n=2k+1 \\
 &= 4k^2+4k+1 & \text{expanding} \\
 &= 2(2k^2+2k)+1
\end{align*}
which is odd, since it has the form $2M+1$.
We can essentially turn this into a proof by contradiction by beginning with "If $n$ is odd and $n^2$ is even...", then writing "...giving a contradiction" at the end.  Although this should be regarded as unnecessary.

The other direction, i.e.,

$n$ even implies $n^2$ even

can also be shown in a similar way:  If $n=2k$, then $n^2=(2k)^2=4k^2=2(2k^2)$ which is even.
A: If you insist by contradiction...then consider some $n$ that is even, then:
$$n = 2k$$
Where $k$ is some natural number not $0$. Assume that $n^2$ is not even, but then contradicting the fact that $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$ is even. 
Alternatively, if $n^2$ is even but $n$ is odd, then $n = 2k+1$ so $$(2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$
...contradicting the fact that $n^2$ is assumed to be even. 
