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

Let us prove the required result by contradiction.

**Step 1:** Assume $\sqrt{5}$ is a rational number. Hence

$\frac{p}{q}$ where p, q are integers and q $\neq$ 0

Also p, q have no common factors that is $\frac{p}{q}$ is in its simplest form. The only common factor between p and q is 1.

**Step 2:** Square on both sides

($\sqrt{5}$)^{2} = ($\frac{p}{q}$)^{2}

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

5q^{2} = p^{2}

This means p^{2} is a multiple of 5 which in turn implies that p is also a multiple of 5. Let us assume p = 5a

**Step 3:** Substitute p = 5a in 5q^{2} = p^{2}

5q^{2} = p^{2}

5q^{2} = (5a)^{2}

5q^{2} = 25a^{2}

q^{2} = 5a^{2}

This means that q^{2} is a multiple of 5 which in turn implies q is also a multiple of 5.

If both p and q are multiples of 5 that means there is at least one common factor between p and q.

This contradicts our assumption that there is no common factor between p and q except 1.

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