I'm having a really hard time learning proofs, and cannot pick up on why this is actually proved. This isn't a homework problem, it's straight from the solutions manual (it just doesn't tell me why the proof is actually solved).
If $x$ is an even integer, then $x^2$ is even.
Proof. Suppose $x$ is even. Thus $x=2a$ for some $a\in\mathbb Z$.
Therefore $x^2=2b$, where $b$ is the integer $2a^2$.
Thus $x^2$ is even by definition of an even number.
I'm confused on the third line. I understand that $x^2=2b$, but why do we need to know "where $b=2a^2$, and why do we need to know it? How did we even get that value?
Shoot, I think I just got it as I typed it out... What it's really saying is that $x^2=2(2a^2)$, and since $x^2=(2a)^2$ factors down to $2(2a^2)$, $b$ is a valid even integer $2a^2$ when it's value is for '$b$' in $x^2=2b$.