4.1 Rationalising the Denominator
When a fraction has a surd in its denominator we usually rewrite it so that the denominator is rational. This is called rationalising the denominator, and it is done by multiplying the numerator and the denominator by the same well-chosen quantity — which changes the form of the fraction without changing its value, since we are multiplying by \(1\).
For a single surd, multiply by that surd: \[\frac {3}{\sqrt {7}}=\frac {3}{\sqrt {7}}\times \frac {\sqrt {7}}{\sqrt {7}} =\frac {3\sqrt {7}}{7}.\]
For a denominator of the form \(a+b\sqrt {c}\), multiply by its conjugate \(a-b\sqrt {c}\). The product is a difference of two squares, \[\left (a+b\sqrt {c}\right )\left (a-b\sqrt {c}\right )=a^{2}-b^{2}c,\] in which the surd has vanished — which is the whole point.
Note 4.4. This is the same manoeuvre used to divide complex numbers in the previous chapter, and for the same reason: multiply by the conjugate and the awkward part squares away. There \(\left (x+iy\right )\left (x-iy\right )=x^{2}+y^{2}\); here \(\left (a+b\sqrt {c}\right )\left (a-b\sqrt {c}\right )=a^{2}-b^{2}c\). The sign differs because \(i^{2}=-1\) while \(\left (\sqrt {c}\right )^{2}=+c\), but the idea is identical.
Why bother? Partly convention, but there is a practical reason from before calculators: \(\frac {3}{\sqrt {7}}\) requires dividing by \(2.6457\ldots \), whereas \(\frac {3\sqrt {7}}{7}\) requires only multiplying and then dividing by \(7\). It also makes two expressions easy to compare, since a rationalised form is essentially unique.
Rationalise the denominator for \(\frac {3}{5+\sqrt {7}}\)
\begin {align*} \frac {3}{5+\sqrt {7}} & =\frac {3}{5+\sqrt {7}}\times \frac {5-\sqrt {7}}{5-\sqrt {7}}\\\\ & = \frac {15-3\sqrt {7}}{25-7}\\\\ & =\frac {15}{18}-\frac {3\sqrt {7}}{18}\\\\ & = \frac {5}{6}-\frac {\sqrt {7}}{6}\\\\ \end {align*}
Express \(\frac {1-5\sqrt {2}}{2\sqrt {2}+5}\) in the form \(a+b\sqrt {2}\) where \(a\) and \(b\) are rational numbers.
\begin {align*} \frac {1-5\sqrt {2}}{2\sqrt {2}+5} & =\frac {(1-5\sqrt {2})(2\sqrt {2}-5)}{(2\sqrt {2}+5)(2\sqrt {2}-5)}\\\\ &=\frac {2\sqrt {2}-5-10(2)+25(\sqrt {2})}{4(2)-25}\\\\ &=\frac {-5-20+2\sqrt {2}+25\sqrt {2}}{8-25}\\\\ &=\frac {-25+27\sqrt {2}}{-17}\\\\ & =\frac {-25}{-17}+\frac {27\sqrt {2}}{-17}\\\\ & =\frac {25}{17}-\frac {27\sqrt {2}}{17}\\\\ \end {align*}
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