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!

Hint: get inspiration from the proof that $\sqrt 2$ is irrational to show that $\sqrt 2\notin\mathbb Q(\sqrt 3)$..
