Rationalising the denominator

​​In a nutshell

Rationalising surds eliminates the irrational number in the denominator. You can think of this as converting to equivalent fractions with a rational denominator. Surd fractions with more complex denominators will need the conjugate surd to rationalise them.

Definitions

Rationalising the denominator

Rationalising the denominator means removing the surd from the bottom of a fraction, by converting to an equivalent fraction. This is done by multiplying the numerator and the denominator by the same term.

To rationalise a surd, you multiply the numerator and the denominator by the surd in the denominator.

​​Example 1

Rationalise the denominator $$\dfrac{1}{\sqrt7}$$​.

​

Multiply both the numerator and the denominator by the surd in the denominator.

​$$\begin {aligned} \frac{1}{\sqrt7} &= \frac{1}{\sqrt7}\times {\frac{\sqrt7}{\sqrt7}}\\ &= \underline{\frac{\sqrt7}{7}} \end {aligned}$$

Rationalising more complex denominators

Sometimes you can have more complex denominators like:

​$$\dfrac{1}{2+\sqrt2}$$​​

Multiplying top and bottom by $$\sqrt2$$​ will still leave a surd in the denominator. You need the conjugate expression of the denominator to eliminate the surd. In this case, the conjugate is:

​$$2-\sqrt2$$​​

procedure

Example 2

Rationalise the denominator $$\dfrac{2}{3+\sqrt3}$$​.

Find the conjugate.

 $$3-\sqrt3$$.

Multiply both the numerator and the denominator by the conjugate. 

​$$\begin{aligned}\begin{split} \dfrac{2}{3+\sqrt{3}} &= \dfrac{2}{(3+\sqrt{3})} \times \dfrac{(3-\sqrt{3})}{(3-\sqrt{3})} \\\newline&=\dfrac{6 - 2\sqrt{3}}{9 - 3\sqrt{3} + 3\sqrt{3} - 3} \\\newline&=\dfrac{6-2\sqrt{3}}{6} \\\newline&=\dfrac{2(3-\sqrt{3})}{6} \\\newline&=\underline{\dfrac{3-\sqrt{3}}{3}} \\\end{split}\end{aligned}$$​​