How can you prove that the square root of two is irrational? I have read a few proofs that $\sqrt{2}$ is irrational.
I have never, however, been able to really grasp what they were talking about.
Is there a simplified proof that $\sqrt{2}$ is irrational?

Here's a short algebraic proof. It nowhere uses rules about primes or even numbers.
You need to first show that $1<\sqrt{2}<2,$ but that is obvious.
We first assume that $\sqrt{2}$ is rational. Then pick the smallest positive $q$ so that $p=q\sqrt{2}$ is an integer. Then $q<p<2q.$
Now compute:
$$\left(\frac{2q-p}{p-q}\right)^2 = \frac{4q^2-4pq+p^2}{p^2-2pq+q^2}=\frac{6q^2-4pq}{3q^2-2pq}=2$$
But $q<p<2q$ means $0<p-q<q$, and $\frac{2q-p}{p-q}=\sqrt{2},$ contradicting the assumption that $q$ was the least possible positive denominator.

More generally
We can prove, more generally, that if $n$ is an integer and $n^2<D<(n+1)^2$ then $\sqrt{D}$ is irrational. In the case $D=2$ we have $n=1$.
If $\sqrt{D}$ is rational, find the least positive $q$ such that there is a $p$ such that $\frac{p}{q}=\sqrt{D}$. So $p^2=Dq^2$ means $n^2q^2<p^2<(n+1)^2q^2$ and hence $nq<p<(n+1)q$.
Therefore $0<p-nq<q$.
But then:
$$\left(\frac{Dq-pn}{p-qn}\right)^2=\frac{D^2q^2-2Dpqn + p^2n^2}{p^2-2pqn+q^2n^2}=\frac{D^2q^2-2Dpqn + Dq^2n^2}{Dq^2-2pqn+q^2n^2}=D\tag{*}$$
contradicting the fact that $q$ was the smallest positive denominator for $\sqrt{D}$.

You can prove if $D\geq 0$ is an integer, then there is exactly one non-negative integer $n$ such that $n^2\leq D<(n+1)^2$. We first prove $n$ exists:

Since $(1+D)^2=D+(1+D+D^2)$, we know that $D<(1+D)^2$ and hence there is a least positive $m$ such that $D<m^2.$ We know $m\neq 0$ because $D\geq 0^2$,  so $m\geq 1$. Let $n=m-1$. Then $n^2\leq D<(n+1)^2$.

Uniqueness follows from:

If $0\leq m<n$ then $1\leq m+1\leq n$ and thus $(m+1)^2\leq n^2$.
So if $n^2\leq D< (n+1)^2$ and $m^2\leq D< (m+1)^2$, then we can't have $m<n$ or we'd have $D<(m+1)^2\leq n^2\leq D$. Similarly, we can't have $m>n$. So we must have $m=n$.

Together, the above say that if $D\geq 0$ then $\sqrt{D}$ is rational if and only if $\sqrt{D}$ is an integer.

(*) The magic trick in the above computation is that If $\frac{np}{nq}=\sqrt{D}$ then $\frac{Dq}{p}=\sqrt{D}.$ And if $\frac{a}{b}=\frac{c}{d}$ then $d\neq b$ then $$\frac {a-c}{b-d}=\frac{a}{b}.$$

The expression is arrived by computing (using that $p=q\sqrt{D}):$
$$0<\left(p+q\sqrt{D}\right)\left(\sqrt{D}-n\right)=(qD-np)+(p-nq)\sqrt{D}$$
From this we see $(qD-np)^2-D(p-nq)^2=0.$
And $p-nq=q(\sqrt{D}-n)<q.$ since $0<\sqrt{D}-n<1.$
