The number $\sqrt{2}$ is an irrational.

Theorem: The number $\sqrt{2}$ is an irrational. 

Proof:  Suppose on the contrary that $\sqrt{2}$ is rational number.   Therefore there exist integers $a$ and $b$ such that $\sqrt{2} = \frac{a}{b}$ with $gcd(a, b) = 1$. By Theorem,  there must exist integers $r$ and $s$ such that $ ar + bs = 1$. Hence $ \sqrt{2} = \sqrt{2}~ (ar + bs) = \sqrt{2} ar + \sqrt{2} bs =  2br +as $.

This show that $\sqrt{2}$ is an integer, which is absurd. Therefore  The number $\sqrt{2}$ is an irrational. 

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