Question

Use division method to show that $\sqrt{3}$ and $\sqrt{5}$ are irrational numbers.

Answer 1

Suppose for the sake of contradiction that $\sqrt{3}$ is rational

We know that rational numbers are those numbers which can be expressed in $\frac{p}{q}$ form ,where $p$ and $q$ are integers and $q\ne 0$

$⟹\sqrt{3}=\frac{p}{q}$

Squaring on both sides

$3=\frac{p^2}{q^2}$

$⟹p^2=3q^2$

$\because p^2$ is a multiple of $3⟹p$ must be a multiple of $3$

let $p=3n⟹p^2=9n^2⟹q^2=3n^2$

This means $q$ is also a multiple of $3$,which contradicts the fact that $p$ and $q$ had no common factor

Hence $\sqrt{3}$ is an irrational number

Suppose for the sake of contradiction that $\sqrt{5}$ is rational

We know that rational numbers are those numbers which can be expressed in $\frac{p}{q}$ form ,where $p$ and $q$ are integers and $q\ne 0$

$⟹\sqrt{5}=\frac{p}{q}$

Squaring on both sides

$5=\frac{p^2}{q^2}$

$⟹p^2=5q^2$

$\because p^2$ is a multiple of $5⟹p$ must be a multiple of $5$

let $p=5n⟹p^2=25n^2⟹q^2=5n^2$

This means $q$ is also a multiple of $5$,which contradicts the fact that $p$ and $q$ had no common factor

Hence $\sqrt{5}$ is an irrational number