Question

Prove that $\sqrt{5}$ is irrational by the method of Contradiction.

Answer 1

Let $\sqrt{5}$ be a rational number.

then it must be in form of $\frac{p}{q}$ where, $q\ne 0$ ( $p$ and $q$ are co-prime)

$\sqrt{5}=\frac{p}{q}$

$\sqrt{5}\times q=p$

Suaring on both sides,

$5q^2=p^2$ --------------(1)

$p^2$ is divisible by $5$.

So, $p$ is divisible by $5$.

$p=5c$

Suaring on both sides,

$p^2=25c^2$ --------------(2)

Put $p^2$ in eqn.(1)

$5q^2=25(c{)}^2$

$q^2=5c^2$

So, $q$ is divisible by $5$.

.

Thus $p$ and $q$ have a common factor of $5$.

So, there is a contradiction as per our assumption.

We have assumed $p$ and $q$ are co-prime but here they a common factor of $5$.

The above statement contradicts our assumption.

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