Assume $2 + \sqrt{2}$ is rational. Then $2 + \sqrt{2}$ can be expressed as $p=\frac{m}{n}$, where $ p \in \mathbb{Q} \ \& \ m,n \in \mathbb{Z} $. Furthermore, let m, n be odd.
$2 + \sqrt{2} = p $
$\sqrt{2} = p - 2$
$\sqrt{2} = \frac{m}{n} -2 $
$\sqrt{2} = \frac{m-2n}{n} $
Since m, n are odd, then $m - 2n$ will also be odd. Furthermore, since $m,n \in \mathbb Z$, then $m-2n \in \mathbb Z$. Therefore, we can rewrite $m-2n=j$ for some $j \in \mathbb Z$.
$\sqrt{2} = \frac{j}{n} $
$2 = (\frac{j}{n})^2 $
Hence, this shows that $j^2$ must be even. If $j^2$ is even, then this implies that j is even, and thus $j=2k$ for some $k \in \mathbb{Z}$. Therefore, $j^2$ is divisible by 4, and it follows that the right side of the equation is divisible by 4. Therefore $n^2$ is even, which implies that $n$ is even. This contradicts our choice of m,n and therefore no rational exists for $2 + \sqrt{2}$.
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My concerns about the proof is the part where I say $m - 2n$ will be odd if m,n are odd. I'm not sure if I need to formally prove that part.