Show that $\mathbb{Q}( \sqrt2) \neq \mathbb{Q}( \sqrt3)$ The way that I'm thinking is by showing that the field extension $\mathbb{Q}( \sqrt2) /( \sqrt3) \neq \mathbb{Q}( \sqrt3)$, but is there a simpler way I'm ignoring?
 A: Hint:
$$\sqrt2=a+b\sqrt3\;,\;\;a,b\in\Bbb Q\implies 2=a^2+3b^2+2ab\sqrt3\implies\sqrt3=\frac{2-a^2-3b^2}{2ab}$$
assuming $\;ab\neq0\;$...but what happens if $\;ab=0\;$ ?
A: Overkill answer: 
An abstract idea using Galois theory is to consider $C:=(K^*)^2\cap\mathbb{Q}^*$, where $K=\mathbb{Q}(\sqrt{2})$.
Clearly $S:=(\mathbb{Q}^*)^2\cup 2(\mathbb{Q}^*)^2\subseteq C$, but every element $[c]$
of the quotient group $C/(\mathbb{Q}^*)^2$ induces a homomorphism
$Gal(K/\mathbb{Q})\to\{\pm 1\}$ with $\sigma\mapsto\frac{\sigma(\sqrt{c})}{\sqrt{c}}$ (here the choice of the representative, as well as the choice of the square root, do not matter). 
If $c$ and $c'$ induce the same homomorphism, then one gets $\frac{\sigma(\sqrt{c})}{\sqrt{c}}=\frac{\sigma(\sqrt{c'})}{\sqrt{c'}}$, thus $\sigma\left(\sqrt{\frac{c}{c'}}\right)=\sqrt{\frac{c}{c'}}$, so $\sqrt{\frac{c}{c'}}\in\mathbb{Q}$ (the fixed field) and finally
$c\in c'(\mathbb{Q}^*)^2$. So in fact $[c]=[c']$ in the quotient group. 
So the map $C/(\mathbb{Q}^*)^2\to Hom(Gal(K/\mathbb{Q}),\{\pm 1\})$ is injective, but the target set has only $2$ elements, so $|C/(\mathbb{Q}^*)^2|\le 2$. 
Since $S/(\mathbb{Q}^*)^2$ is a subgroup of $C/(\mathbb{Q}^*)^2$ with two elements, we conclude $S=C$. Now if $K=\mathbb{Q}(\sqrt{3})$ the same argument shows that
$(\mathbb{Q}^*)^2\cup 2(\mathbb{Q}^*)^2=C=(\mathbb{Q}^*)^2\cup 3(\mathbb{Q}^*)^2$, which is absurd. 
I don't really know why I typed this answer, as given the spirit of the question Timbuc's answer is surely the most appropriate one. But I think this idea (from Kummer theory) is very beautiful and general.
