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

Assume, for contradiction, that $\sqrt{5}$ is rational.

Then $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are co-prime integers and $q \neq 0$.

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

$$\Rightarrow p^2 = 5q^2 \quad ...(i)$$

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

Since 5 is prime, $p$ must also be divisible by 5.

Let $p = 5k$.

Substituting in (i),

$$(5k)^2 = 5q^2$$

$$\Rightarrow 25k^2 = 5q^2$$

$$\Rightarrow q^2 = 5k^2.$$

So, $q^2$ is divisible by 5, hence $q$ is divisible by 5. But then 5 divides both $p$ and $q$, contradicting the fact that they are co-prime.

Hence, our assumption is false, and $\sqrt{5}$ is irrational.