I am stuck at the following seemingly simple problem: is Is $\sqrt{2 + \sqrt{2}} \in \mathbb{Q}(\sqrt{2})$?
(Context: I want to show that $\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2 + \sqrt{2}})$ is an extension of degree 2).
So far, I assumed it is not, and argued with contradiction. Suppose $\sqrt{2 + \sqrt{2}} = a + b\sqrt{2}$ for some $a,b \in \mathbb{Q}$. Then squaring both sides we have that $$ 2 + \sqrt{2} = a^2 + 2ab\sqrt{2} + 2b^2, $$ and hence $$ 2 - a^2 - 2b^2 = (2ab-1)\sqrt{2}. $$ Now if $2ab-1 \neq 0$, we get $\sqrt{2} \in \mathbb{Q}$, a contradiction. So suppose we are in the case $2ab = 1$, I don't know how to proceed to derive a contradiction in this case. Is there a general strategy to do this? I notice that then $$ a^2 = 2-2b^2, $$ so $a = \sqrt{2(1-b^2)} = \sqrt{2} \sqrt{1-b^2} \in \mathbb{Q}$. Not sure how to proceed from here.