Prove that

Let us assume that $2 \sqrt{3}-1$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$2 \sqrt{3}-1=\frac{a}{b}$

$2 \sqrt{3}=\frac{a}{b}+1$

$\sqrt{3}=\frac{\frac{a}{b}+1}{2}$

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

This contradicts the fact that $\sqrt{3}$ is an irrational

Hence $2 \sqrt{3}-1$ is irrational