Prove that $(\sqrt{2} + \sqrt{3})^2$ is an irrational number, given that $\sqrt{6}$ is an irrational number.

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Step 1: $(\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6}$
Step 2: Assume to the contrary that $5 + 2\sqrt{6}$ is rational $= \frac{a}{b}$, where $a$, $b$ are integers, $b 
eq 0$
Step 3: $\sqrt{6} = \frac{a - 5b}{2b}$ which is rational. But this contradicts that $\sqrt{6}$ is irrational.
Step 4: $	herefore 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2$ is irrational.

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