Prove that $\sqrt{5}$ is an irrational number. Hence, show that $-3 + 2\sqrt{5}$ is an irrational number.

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \dfrac{p}{q},$

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

$\Rightarrow 5 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2$

As 5 divides 5q², so 5 divides p² but 5 is prime

$\Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 5m, where m is an integer.

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

$(5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em]$

As 5 divides 5m², so 5 divides q² but 5 is prime

$\Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.

Suppose that $-3 + 2\sqrt{5}$ is a rational number, say r.

Then, $-3 + 2\sqrt{5}$ = r (note that r ≠ 0)

$\Rightarrow 2\sqrt{5} = r + 3 \\[0.5em] \Rightarrow \sqrt{5} = \dfrac{r + 3}{2} \\[0.5em]$

As r is rational and r ≠ 0, so $\dfrac{r + 3}{2}$ is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts that $\sqrt{5}$ is irrational. Hence, our supposition is wrong.

∴ $-3 + 2\sqrt{5}$ is an irrational number.