<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] At your step
$$3=a^2+2\sqrt 2 ab +2b^2$$, you can rearrange it to be
$$3-a^2-2b^2=2\sqrt 2 ab$$, the left hand side is a rational number while the right hand side is irrational, which you can't do unless a and b are zero, which does not satisfy the first condition.
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