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.
 A: 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 .
A: 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.
A: Note that $1$ and $\sqrt{2}$ is a basis for $\mathbb{Q}(\sqrt{2}).$
Hence from
$$3=a^2+b^2+2\sqrt{2}ab$$
We have $a^2+b^2=3$ and $2ab=0$.
From there, you should be able to see a contradiction.
A: I would start the same:
$$\sqrt3 = a + b\sqrt{2},\ \ a,b\in\mathbb{Q}$$
then square both to get
$$3 = a^2 + 2ab\sqrt2 + b^2$$
just like you and then rearrange to get
$$\frac{3 - a^2 - b^2}{2ab} = \sqrt{2}$$
meaning that $\sqrt2$ would have to be rational, which we know is not true.
