Question

Given that $\sqrt{2}$ is irrational, prove that $\left(5+3\sqrt{2}\right)$ is an irrational number.

Answer 1

Let us assume, to the contrary that $5+3\sqrt{2}$ is rational
That is, we can find coprime a and b $(b\ne 0)$ such that $5+3\sqrt{2}=\frac{a}{b}$
Therefore,
$$\begin{gathered} 5-\frac{a}{b}=-3\sqrt{2} \\ \Rightarrow \frac{5}{3}-\frac{a}{3b}=-\sqrt{2} \\ \Rightarrow \frac{a-5b}{3b}=\sqrt{2} \end{gathered}$$
Since, a and b are integers, we get $\frac{a-5b}{3b}$ is rational, and so $\sqrt{2}$ is rational.
which contradicts the fact that $\sqrt{2}$ is irrational.
So, our assumption was wrong that $5+3\sqrt{2}$ is rational.
Hence, we conclude that $5+3\sqrt{2}$ is irrational.