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      <text>4.0.6 (Fractional indices)</text>
    </name>
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      <text><![CDATA[<h3>Fractional indices</h3>

<p>How are surds connected to the topic of indices? Well, although so far we have only met indices which are whole numbers, it is possible that to have an index which is in fact a fraction. We shall see that fractional indices correspond to roots. Consider the expression \((16^{1\over 2})^2\). We shall see that we can re-write this as:</p>
\[
(16^{1\over 2})^2 = 16^{{1\over 2}\times 2} = 16^1 =16.
\]</p>
<p>So \(16^{1\over 2}\) is a number which when squared equals 16, that is \(4\) or \(-4\). In other words \(16^{1\over 2}\) is a square root of 16. There are always two square roots of a non-zero positive number, and we write \( 16^{1\over 2} = \pm 4\).
In general, 
\[
a^{\frac{1}{2}} = \sqrt{a}, \qquad \mbox{ for } a > 0.
\]
</p>

<p>In fact, we can go further than this. Whilst indices involving fractions over two correspond to square roots, indices with fractions over different denominators correspond to the different roots. For example \( (8^{1\over 3})^3 = 8^{{1\over 3}\times 3} = 8^1 = 8\), so that \(8^{1\over 3}\) is a number which when cubed equals \(8\). Thus \(8^{1\over 3}\) is the cube root of \(8\), that is \(\sqrt[3]{8}\), namely \(2\). Each number has only one cube root, and so \( 8^{1\over 3} = 2\). In general, 
\[
a^{\frac{1}{3}} \mbox{ is the cube root of } a. 
\]
</p>

<p>We can generalise this further still by noting that the \(n\)th root of \(a\) is denoted by \(a^{\frac{1}{n}}\). If \(a&gt;0\) the positive \(n\)th root is denoted by \(\sqrt[n]{a}\)</p>]]></text>
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