Question

Prove that $\sqrt{6}$ is an irrational number.

Answer 1

As it is Given number $\sqrt{6}$

We need to prove that $\sqrt{6}$ is irrational

Let us assume that $\sqrt{6}$ is a rational number.

So it can be expressed in the form $\frac{p}{q}$ where $p,q$ are co-prime integers and $q\ne 0$
$\Rightarrow \sqrt{6}=\frac{p}{q}$

On squaring both the sides we get,
$$\begin{gathered} \Rightarrow 6=p²/q² \\ \Rightarrow 6q²=p²—————–\left(i\right) \\ \frac{p²}{6}=q² \\ So6divides{p}^2,pisamultipleof6 \\ \Rightarrow p=6m \\ \Rightarrow p²=36m²————-\left(ii\right) \\ Fromequations\left(i\right)and\left(ii\right),weget, \\ 6q²=36m² \\ \Rightarrow q²=6m² \\ q²isamultipleof6So, \\ qisamultipleof6 \end{gathered}$$
Hence,$p,q$ have a common factor $6$. This contradicts our assumption that $p,q$ are co-primes. Therefore,$\frac{p}{q}$ is not a rational number

$\sqrt{6}$ is an irrational number.