Prove that $\mathbb Q(\sqrt 2) \neq \mathbb Q(\sqrt 3)$ I've tried writing out the contents of each and attempting to get a contradiction by equating arbitrary elements but can't get this to work. I can't think of any counterexamples as everything I come up with seems as though it would work by letting both the coefficient of $\sqrt 2$ and $\sqrt 2$ itself be the coefficient of $\sqrt 3$ and vice versa. Help!

If $\mathbb Q(\sqrt 2) = \mathbb Q(\sqrt 3)$ then there is $\alpha \in \mathbb Q(\sqrt 2)$ such that $\alpha^2=3$. Writing $\alpha=a+b\sqrt 2$ we get $a^2+2ab\sqrt 2+2b^2=3$. This would imply that $\sqrt 2$ is rational, unless $ab=0$. But in this case, we'd have that $3$ is a rational square or twice a rational square, which it isn't.
