Question

prove that 3 root 6 is an irrational number.

Answer 1

Lets assume that $\sf{3\sqrt{6} }$ is an irrational number.

$\sf\Large{\implies 3\sqrt{6} = \frac{p}{q}}$

where p and q are integers.

$\sf\Large{\implies \sqrt{6} = \frac{p}{3q}}$

Since, p and 3q are integers,

$\sf{\therefore \sqrt{6}}$ is a rational number, which is a contradiction.

Therefore, our assumption is wrong.

Hence, $\bf{3\sqrt{6} }$ is an irrational number.

_____________

@zaqwertyuioplm :)