Question

Prove root $5$ is an irrational number.

Answer 1

Let $\sqrt{5}$is a rational number, it would be written in the form ​$\frac{p}{q}$ where, $q\ne 0$( $p$and $q$ are co-prime).

$$\begin{gathered} \sqrt{5}=\frac{p}{q} \\ \Rightarrow \sqrt{5}\times q=p \end{gathered}$$

Squaring on both sides

$5q^2=p^2............................\left(i\right)$

$\Rightarrow p^2$ is divisible by $5$ So, $p$is divisible by $5$.

$p=5c$

Again squaring on both sides,

$p^2=25c^2......................................\left(ii\right)$

Put value of $p^2$from $\left(ii\right)$ into the equation $\left(i\right)$

$$\begin{gathered} 5q^2=25(c{)}^2 \\ \Rightarrow q^2=\frac{25}{5}(c{)}^2 \\ \Rightarrow q^2=5(c{)}^2 \end{gathered}$$

Therefore, $q$ is divisible by $5$, and $p$and $q$ have $5$ as a common factor.

This contradicts the fact that $p$and $q$are co-prime.

Hence, $\sqrt{5}$​is an irrational number.