## Compute the degree of a given extension of the rationals

Find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$.

We have $\sqrt{3+2\sqrt{2}} = \sqrt{3 + \sqrt{8}}$, and $3^2 - 8 = 1^2$ is square over $\mathbb{Q}$. By this previous exercise, $\sqrt{3+2\sqrt{2}} = 1 + \sqrt{2}$. So $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}}) = \mathbb{Q}(1+\sqrt{2}) = \mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$.