Prove that $\sqrt{6}$ is an irrational number.

Suppose that $\sqrt{6}$ is a rational number, then

$\sqrt{6} = \dfrac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$\Rightarrow 6 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 6q^2 \qquad \text{….(i)}$

As 2 divides 6q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 2k, where k is some integer.

Substituting this value of p in (i), we get

$(2k)^2 = 6q^2 \\[0.5em] \Rightarrow 4k^2 = 6q^2 \\[0.5em] \Rightarrow 2k^2 = 3q^2 \\[0.5em]$

As 2 divides 2k², so 2 divides 3q²

$\Rightarrow$ 2 divides 3 or 2 divides q²

But 2 does not divide 3, therefore, 2 divides q²

$\Rightarrow$ 2 divides q      (Theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. Therefore, $\sqrt{6}$ is not a rational number. So, we conclude that $\sqrt{6}$ is an irrational number.