Show that a given extension of QQ does not contain sqrt(2)

Show that \mathbb{Q}(\sqrt{5}) does not contain \sqrt{2}.


Recall that every element of \mathbb{Q}(\sqrt{5}) is uniquely of the form a+b\sqrt{5} for some a,b \in \mathbb{Q}. Suppose now that (a+b\sqrt{5})^2 = 2. Comparing coefficients, we see that a^2 + 5b^2 = 2 and 2ab = 0, so either a = 0 or b = 0. If b = 0, then a^2 = 2, and if a = 0, then b^2 = 5/2. Either case leads to a contradiction since these equations have no solutions in \mathbb{Q}.

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