sqrt(2) + sqrt(3) is irrational

Prove that \sqrt{2} + \sqrt{3} is irrational.


Note that \sqrt{2} and \sqrt{3} are algebraic over \mathbb{Q} with minimal polynomials x^2-2 and x^2-3, respectively, and with conjugates \pm \sqrt{2} and \pm \sqrt{3}. Since \sqrt{2} - \sqrt{3}, -\sqrt{2}+\sqrt{3}, and -\sqrt{2}-\sqrt{3} are distinct from \sqrt{2}+\sqrt{3}, we have \mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3}).

Suppose \sqrt{2}+\sqrt{3} \in \mathbb{Q}. Then \mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}, and so \mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}. In particular, \sqrt{2} \in \mathbb{Q}– a contradiction.

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