Suppose $\sqrt{5}$ is rational, so it can be written as $\dfrac{p}{q}$.

Where $p, q \in \mathbb{Z}$ and $q \ne 0$.

Assume that $\dfrac{p}{q}$ is in its lowest terms.

We can write $\sqrt{5} = \dfrac{p}{q}$

$\Rightarrow 5 = \dfrac{p^2}{q^2}$

$\Rightarrow p^2 = 5q^2$

The L.H.S. must be divisible by 5 since R.H.S is a multiple of 5.

$\Rightarrow p$ is divisible by 5.

If $p$ is divisible by 5, we can say $p = 5p'$ where $p'$ is also an integer.

$\Rightarrow p^2 = 25p'^2$

Substitute $p^2$ with $25p'^2$ in $p^2 = 5q^2$

$\Rightarrow 25p'^2 = 5q^2$

$\Rightarrow 5p'^2 = q^2$

Since L.H.S is divisible by 5, R.H.S must also be divisible by 5.

$\Rightarrow q$ is divisible by 5.

Since both $p$ and $q$ are divisible by 5, they share a common factor of 5.

This contradicts the assumption that $\dfrac{p}{q}$ is in lowest terms.

$\therefore$ Our initial assumption must be false. We cannot write $\sqrt{5}$ in the form of $\dfrac{p}{q}$. Thus, $\sqrt{5}$ is irrational.