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