Prove that $\sqrt{2}$ is an irrational number. Hence, show that $3 - \sqrt{2}$ is an irrational number.

Let $\sqrt{2}$ be a rational number, then

$\sqrt{2} = \dfrac{p}{q},$

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

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

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

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

Let p = 2m, where m is an integer.

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

$(2m)^2 = 2q^2 \\[0.5em] \Rightarrow 4m^2 = 2q^2 \\[0.5em] \Rightarrow 2m^2 = q^2 \\[0.5em]$

As 2 divides 2m², so 2 divides q² but 2 is prime

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

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

Hence, $\sqrt{2}$ is not a rational number. So, we conclude that $\sqrt{2}$ is an irrational number.

Suppose that $3 - \sqrt{2}$ is a rational number, say r.

Then, $3 - \sqrt{2}$ = r (note that r ≠ 0)

$\Rightarrow - \sqrt{2} = r - 3 \\[0.5em] \Rightarrow \sqrt{2} = 3 - r \\[0.5em]$

As r is rational and r ≠ 0, so 3 - r is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts that $\sqrt{2}$ is irrational. Hence, our supposition is wrong.

∴ $3 - \sqrt{2}$ is an irrational number.