Prove that $2+5\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Let us assume that $2+5\sqrt{3}$ is a rational number.

Thus, $2+5\sqrt{3}$ can be represented in the form of $\frac{p}{q}$, where p and q both are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 2+5\sqrt{3}=\frac{p}{q} \\ \Rightarrow 5\sqrt{3}=\frac{p}{q}-2 \\ \Rightarrow 5\sqrt{3}=\frac{p-2q}{q} \\ \Rightarrow \sqrt{3}=\frac{p-2q}{5q} \end{gathered}$$

since, $\frac{p-2q}{5q}$ is rational ⇒ $\sqrt{3}$ is rational.

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

Therefore, our assumption is wrong.

Hence, $2+5\sqrt{3}$ is an irrational number.