Prove that

Let $x=5-2 \sqrt{3}$ be a rational number.

$x=5-2 \sqrt{3}$

$\Rightarrow x^{2}=(5-2 \sqrt{3})^{2}$

$\Rightarrow x^{2}=(5)^{2}+(2 \sqrt{3})^{2}-2(5)(2 \sqrt{3})$

$\Rightarrow x^{2}=25+12-20 \sqrt{3}$

$\Rightarrow x^{2}-37=-20 \sqrt{3}$

$\Rightarrow \frac{37-x^{2}}{20}=\sqrt{3}$

Since $x$ is a rational number, $x^{2}$ is also a rational number.

$\Rightarrow 37-x^{2}$ is a rational number

$\Rightarrow \frac{37-x^{2}}{20}$ is a rational number

$\Rightarrow \sqrt{3}$ is a rational number

But $\sqrt{3}$ is an irrational number, which is a contradiction.

Hence, our assumption is wrong.

Thus, $(5-2 \sqrt{3})$ is an irrational number.