Prove that the following number is irrational: $\frac{1}{\sqrt{3}}$

(N/A) Assume,to the contrary,that $\frac{1}{\sqrt{3}}$ is a rational number.
Then,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers,$q \neq 0$,and $p, q$ are coprime.
$\frac{1}{\sqrt{3}} = \frac{p}{q}$
Rearranging the equation,we get $\sqrt{3} = \frac{q}{p}$.
Since $p$ and $q$ are integers,$\frac{q}{p}$ must be a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
Therefore,our initial assumption is false.
Hence,$\frac{1}{\sqrt{3}}$ is an irrational number.