Show that $\sqrt{2} \notin \mathbb{Q}(a)$ for real numbers a.

Question

As the title says, I am trying to prove that $\sqrt{2} \notin \mathbb{Q}(a)$, for all $ a \neq \sqrt[k]{2}$. My intuition says that this is true but I haven't made much progress.

Answer 1

This isn't true. For instance, $$ \sqrt 2\in \Bbb Q\left(\sqrt{5 + \sqrt2}\right) $$ yet $\left(\sqrt{5 + \sqrt2}\right)^k$ never becomes $2$ for any integer $k$.

Even simpler, we could have $\Bbb Q(\sqrt 2 + 1)$, or just $\Bbb Q(2\sqrt 2)$.