Prove that $\sqrt{7}$ is an irrational number.

Let $\sqrt{7}$ be a rational number, then

$\sqrt{7} = \dfrac{p}{q},$

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

$\Rightarrow 7 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 7q^2 \qquad \text{….(i)}$

As 7 divides 7q², so 7 divides p² but 7 is prime

$\Rightarrow 7 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 7m, where m is an integer.

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

$(7m)^2 = 7q^2 \\[0.5em] \Rightarrow 49m^2 = 7q^2 \\[0.5em] \Rightarrow 7m^2 = q^2 \\[0.5em]$

As 7 divides 7m², so 7 divides q² but 7 is prime

$\Rightarrow 7 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 7. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{7}$ is not a rational number. So, we conclude that $\sqrt{7}$ is an irrational number.