Can $\sqrt{2}$ be written as a rational number $\dfrac{p}{q}$?

No, $\sqrt{2}$ cannot be written as a rational number $\dfrac{p}{q}$.

Proof by contradiction :

Assume $\sqrt{2}$ is rational. Then it can be expressed as $\dfrac{p}{q}$, where p and q are integers with q ≠ 0 and gcd(p, q) = 1 (lowest form).

Squaring both sides :

$\Rightarrow \sqrt{2} = \dfrac{p}{q} \\[1em] \Rightarrow 2 = \dfrac{p^2}{q^2} \\[1em] \Rightarrow 2q^2 = p^2.$

This means p² is even, and so p must also be even.

Let p = 2k for some integer k. Then :

$\Rightarrow 2q^2 = (2k)^2 \\[1em] \Rightarrow 2q^2 = 4k^2 \\[1em] \Rightarrow q^2 = 2k^2.$

This means q² is even, so q is also even.

But, both p and q being even contradicts our assumption that gcd(p, q) = 1.

Hence, our assumption is wrong.

Hence, $\sqrt{2}$ cannot be written as $\dfrac{p}{q}$. It is an irrational number.