Question

Prove that $\sqrt{2}$ is on irrational number and also prove that $3+5\sqrt{2}$ is irrational number.

Answer 1

Consider the problem

$\sqrt{2}$ is non terminating, repeating terms so $\sqrt{2}$ is irrational.

For $3+5\sqrt{2}$

Let us consider that the $3+5\sqrt{2}$ is rational

As $3+5\sqrt{2}$ is rational. it must be in the form of $\frac{p}{q}$

Then $3+5\sqrt{2}=\frac{p}{q}$

$5\sqrt{2}=\frac{p}{q}-3$

$\sqrt{2}=\frac{p-3q}{5}$,

And $\sqrt{2}$ irrational

Therefore, $3+5\sqrt{2}$ is irrational.