Good solutions were submitted by Hyeyoun from St Paul's Girls' School, London, Michal from Daramalan College, and Ling Xiang Ning from Raffles Institution, Singapore.

This is Hyeyoun's solution.

Suppose that

\sqrt{(} a)

is rational. Therefore, where

x

and

y

are coprime integers and

y \neq 0

, we have

a = \frac{x^{2}}{y^{2}}

We can always write the rational number so that x and y are coprime, that is they have no common factors except 1 so, as $\sqrt{a}$ is rational, it follows that $x$ and $y$ are coprime.

If $a$ is an integer, $y^{2}$ must be a factor of $x^{2}$ and so $y = 1$ .Therefore $a$ is a square number.

It follows (as the contrapositive) that if $a$ is an integer and is not a square number then $\sqrt{(} a)$ is irrational.