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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.

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    $\begingroup$ You cannot say that $m$ and $n$ must each be odd. Take $\frac{4}{7}$ or $\frac{7}{4}$. Neither can be written as $\frac{m}{n}$ with $m,n$ each odd integers. $\endgroup$ – Carl Schildkraut Sep 21 '16 at 1:26
  • $\begingroup$ Hi there. I think it would be more correct to say I'm letting m,n and be odd, and so it is more of an assumption. I'm simply mimicking Rudin's proof to prove that $ \sqrt{2} $ is irrational (as seen on 1.1 in the book). $\endgroup$ – Nikitau Sep 21 '16 at 1:46
  • $\begingroup$ The problem with that is that's not a correct assumption. If $2+\sqrt{2}$ is rational, it is not necessarily true that it can be written as $\frac{m}{n}$ with odd $m$,$n$. $\endgroup$ – Carl Schildkraut Sep 21 '16 at 1:57
  • $\begingroup$ @CarlSchildkraut Ah, I see your point. Out of curiosity (and I apologize if this is a stupid question), how come the assumption worked for $ \sqrt{2} $? Also, would it work if I assumed m.n were coprime? I remember seeing that as an alternative proof. $\endgroup$ – Nikitau Sep 21 '16 at 2:03
  • $\begingroup$ I'm not sure the assumption worked for $\sqrt{2}$ (in fact, I'm sure that exact one didn't). However, it would work if you assumed that $\gcd(m,n)=1$. $\endgroup$ – Carl Schildkraut Sep 21 '16 at 2:07

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