Prove that $\sqrt{5}$ is irrational.

Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.

If $\sqrt{5}$ is rational, that means it can be written in the form of $\dfrac{a}{b}$, where a and b are integers that have no common factor other than 1 and b ≠ 0. i.e., a and b are co-prime numbers.

$\Rightarrow \dfrac{\sqrt{5}}{1} = \dfrac{a}{b} \\[1em] \Rightarrow \sqrt{5}b = a$

Squaring both sides,

⇒ 5b² = a² ………..(1)

⇒ $b^2 = \dfrac{a^2}{5}$

This means 5 divides a².

From this, 5 also divides a.

Then a = 5c, for some integer 'c'.

On squaring, we get

⇒ a² = 25c²

Substituting above value of a² in equation (1)

⇒ 5b² = 25c²

⇒ b² = $\dfrac{25c^2}{5}$

⇒ b² = 5c²

⇒ c² = $\dfrac{b^2}{5}$

This means b² is divisible by 5 and so b is also divisible by 5. Therefore, a and b have 5 as common factor but this contradicts the fact that a and b are co-prime. This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is a rational number.

Hence, proved that $\sqrt{5}$ is irrational.