Prove that

Let us assume that $(4-\sqrt{3})$ is a rational number.

Thus, $(4-\sqrt{3})$ can be represented in the form of $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0, p$ and $q$ are co-prime numbers.

$4-\sqrt{3}=\frac{p}{q}$

$\Rightarrow-\sqrt{3}=\frac{p}{q}-4$

$\Rightarrow-\sqrt{3}=\frac{p-4 q}{q}$

$\Rightarrow \sqrt{3}=\frac{4 q-p}{q}$

Since, $\frac{4 q-p}{q}$ is rational $\Rightarrow \sqrt{3}$ is rational

But, it is given that $\sqrt{3}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $4-\sqrt{3}$ is an irrational number.