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Prove that $(\sqrt{2} + \sqrt{3})^2$ is an irrational number, given that $\sqrt{6}$ is an irrational number.
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$(\sqrt{2}+\sqrt{3})^2 = 2+3+2\sqrt{6} = 5 + 2\sqrt{6}$
Let us assume, to the contrary, that $5 + 2\sqrt{6}$ is rational
$\therefore 5 + 2\sqrt{6} = \frac{a}{b}$; $a, b$ are integers, $b \neq 0$
$\sqrt{6} = \frac{a-5b}{2b}$
RHS is a rational number, whereas LHS is an irrational number.
$\therefore$ Our assumption is wrong.
$\Rightarrow 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2$ is an irrational number
Let us assume, to the contrary, that $5 + 2\sqrt{6}$ is rational
$\therefore 5 + 2\sqrt{6} = \frac{a}{b}$; $a, b$ are integers, $b \neq 0$
$\sqrt{6} = \frac{a-5b}{2b}$
RHS is a rational number, whereas LHS is an irrational number.
$\therefore$ Our assumption is wrong.
$\Rightarrow 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2$ is an irrational number