Question:
Prove that $(2+3 \sqrt{5})$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
Solution:
Let us assume that $(2+3 \sqrt{5})$ is a rational number.
Thus, $(2+3 \sqrt{5})$ 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.
$2+3 \sqrt{5}=\frac{p}{q}$
$\Rightarrow 3 \sqrt{5}=\frac{p}{q}-2$
$\Rightarrow 3 \sqrt{5}=\frac{p-2 q}{q}$
$\Rightarrow \sqrt{5}=\frac{p-2 q}{3 q}$
Since, $\frac{p-2 q}{3 q}$ is rational $\Rightarrow \sqrt{5}$ is rational
But, it is given that $\sqrt{5}$ is an irrational number.
Therefore, our assumption is wrong.
Hence, $2+3 \sqrt{5}$ is an irrational number.