Assume, for contradiction, there exists $p, q \in \mathbb{Z}$ satisfying $$ \left( \frac{p}{q} \right)^2 = 2 $$ where $p$ and $q$ are relatively prime and $q \neq 0$. Then, rearranging, we have $p^2 = 2q^2$ and so $p$ is even or $p = 2r$ where $r\in \mathbb{Z}$. But, substituting it above, we also have $q^2 = 2r^2$ and $q$ is even as well. This is a contradiction to being relatively prime.

In the lecture on Field automorphism (at 6:35), Professor Macauley gives another proof based on the following proposition. If $\phi$ is an automorphism of $\mathbb{Q}(\sqrt{2})$ ($\mathbb{Q}$ adjoining $\sqrt{2}$), then, $$ \phi(q) = q, \quad \forall q\in \mathbb{Q} .$$ One of such automorphisms is $\phi(a+b\sqrt{2}) = a-b\sqrt{2}$. But, now, $\phi(\sqrt{2}) = -\sqrt{2} \neq \sqrt{2}$. Thus, $\sqrt{2}$ is not rational. Constructing the extension field $\mathbb{Q}(\sqrt{2})$, it is tautological but nonetheless cute.

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