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Let us assume that $2-\sqrt{3}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$2-\sqrt{3}=\frac{a}{b}$

$\sqrt{3}=2-\frac{a}{b}$

This implies, $\sqrt{3}$ is a rational number, which is a contradiction.

Hence, $2-\sqrt{3}$ is irrational number.