Let $5-\sqrt{3}$ is a rational number.
We have to find out two integers a and b such as
$5-\sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}}$
$-\sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}}-5$
$\sqrt{3}=5-\frac{\mathrm{a}}{\mathrm{b}}$
$\sqrt{3}=\frac{5\mathrm{b}-\mathrm{a}}{\mathrm{b}}$
a,b and 5 all are integers
$\frac{5\mathrm{b}-\mathrm{a}}{\mathrm{b}}$ is a rational number.
$\sqrt{3}$ will be also rational number.
But this contradicts the fact that $\sqrt{3}$ is an irrational number.
So our hypothesis is wrong.
Hence, $5-\sqrt{3}$ is an irrational number.
Hence proved.