<s>[INST] Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.
 [/INST] You can actually continue from your third line by using that $\sqrt 2$ is irrational so if $$a^2+b^2+2\sqrt 2ab=3,$$ then actually $$a^2+b^2=3 \text{ 
 and } 2\sqrt 2ab=0$$ so that either $a$ is zero or $b$ is $0$. Since we can't solve $a^2=3$ or $b^2=3$ in the rationals, you can conclude by using that $\sqrt 3$ is irrational .
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