Good solutions were submitted by Hyeyoun Chung, age 15, St Paul's
Girls' School, London, Michal Szymanski age 16, Daramalan College,
Ling Xiang Ning, Allan, age 13, Raffles Institution, Singapore. 

<p>This is Hyeyoun's solution.</p>

<p>




 
We are given that 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><msqrt><mrow><mo stretchy="false">(</mo></mrow></msqrt><mi>a</mi><mo stretchy="false">)</mo></mrow></math>  is a natural number and  is rational.
Therefore , where 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>x</mi></mrow></math> and 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>y</mi></mrow></math> are integers and 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>y</mi><mo>&#x2260;</mo><mn>0</mn></mrow></math>. Then
.
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
    <mstyle displaystyle="true"><mrow><mi>a</mi><mo>=</mo>
<mfrac><mrow>
<msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow>
</msup>
</mrow>
<mrow>
<msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow>
</msup>
</mrow>
</mfrac>
</mrow>
    </mstyle></math>
</td></tr></table>
<br />

<p/>
We can always write the rational number so that x and y are coprime, that is they have no common factors except 1. As a is a natural number, 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow>
</msup>
</mrow></math> must be a factor of 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow>
</msup>
</mrow></math>. Therefore, y must be a factor of x and therefore y=1
<p/>
Therefore 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>a</mi></mrow></math> is a square number.
<p/>
It follows that if a is not a square number then 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><msqrt><mrow><mo stretchy="false">(</mo></mrow></msqrt><mi>a</mi><mo stretchy="false">)</mo></mrow></math>  is irrational.

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