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}.

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