Question

Classify the number as rational or irrational : $\sqrt{23}$

Answer 1

Let $\sqrt{23}$ be rational number

$\Rightarrow \sqrt{23}=\frac{p}{q}[p\in 2,q\in 2,q\ne 0HCFofp\&q=1]$

$\Rightarrow (\sqrt{23}{)}^2={\left(\frac{p}{q}\right)}^2[sbs]$

$\Rightarrow 23=\frac{p^2}{q^2}$

$\Rightarrow 23q^2=p^2$

$\Rightarrow p^2=23q^2$

$\Rightarrow 23$ is a factor of $p^2$

$\Rightarrow 23$ is also a factor of $p$.....I

Let $p^2=23n$

$\Rightarrow (p{)}^2=(23n{)}^2\left[sbs\right]$

$\Rightarrow p^2=529n$

$\Rightarrow 23q^2=529n[\therefore p^2=23q^2]$

$\Rightarrow q^2=\frac{529}{23}$

$\Rightarrow q^2=23n$

$\Rightarrow 23$ is a factor of $q^2$

$\Rightarrow 23$ is also a factor of $q$......II

From this, I & II equations we got $23$ as a common factor of $p$ & $q$.

so, one assumption is wrong

$\Rightarrow \sqrt{23}$ is not a rational number.

$\Rightarrow \sqrt{23}$ is an irrational number