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## 'The Root Cause' printed from http://nrich.maths.org/

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 $y \ne 0$, 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.