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

Let $\sqrt{3}$ be a rational number, then

$\sqrt{3} = \dfrac{p}{q},$

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

$\Rightarrow 3 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 3q^2 \qquad \text{….(i)}$

As 3 divides 3q², so 3 divides p² but 3 is prime

$\Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 3m, where m is an integer.

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

$(3m)^2 = 3q^2 \\[0.5em] \Rightarrow 9m^2 = 3q^2 \\[0.5em] \Rightarrow 3m^2 = q^2 \\[0.5em]$

As 3 divides 3m², so 3 divides q² but 3 is prime

$\Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{3}$ is not a rational number. So, we conclude that $\sqrt{3}$ is an irrational number.

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

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

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

As r is rational and r ≠ 0, so $\dfrac{5}{2}r$ is rational
[∵ System of rational numbers is closed under all four fundamental arithmetic operations (except division by zero)]

$\Rightarrow \sqrt{3}$ is rational

But this contradicts that $\sqrt{3}$ is irrational. Hence, our supposition is wrong.

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