# A calculus of the absurd

##### 4.2.2 Rationalising the denominator

Sometimes surds can be the same, even if it’s not immediately apparent. For example

$$\frac {1}{\sqrt {2}} = \frac {\sqrt {2}}{2}$$

Why? We can just multiply by $$1$$,

\begin{align} \frac {1}{\sqrt {2}} & = \frac {1}{\sqrt {2}} \cdot \frac {\sqrt {2}}{\sqrt {2}} \\ & = \frac {\sqrt {2}}{\sqrt {2}\sqrt {2}} \\ & = \frac {\sqrt {2}}{(\sqrt {2})^2} \\ & = \frac {\sqrt {2}}{2} \end{align}

Note that through this process we have removed the irrational part from the denominator or, alternatively, we have rationalised the denominator.

We can also do this in the case where we have slightly more complex expressions in the form

$$\frac {a \pm \sqrt {b}}{c\pm \sqrt {d}}$$

For example, we might want to make the denominator of one of these expressions rational

\begin{align} & \frac {3}{7-\sqrt {11}} \\ & \frac {4 - \sqrt {2}}{\sqrt {3} + 44} \\ & \frac {1 + \sqrt {5}}{1 + \sqrt {2}} \end{align}

The solution here is to use the difference of two squares1313 That is, $$a^2 - b^2 = (a+b)(a-b)$$.

\begin{align} \frac {1 + \sqrt {5}}{1 + \sqrt {2}} & = \frac {1 + \sqrt {5}}{1 + \sqrt {2}} \frac {1 - \sqrt {2}}{1 - \sqrt {2}} \\ & = \frac {(1+\sqrt {5})(1-\sqrt {2})}{(1+\sqrt {2})(1-\sqrt {2})} \\ & = \frac {1 - \sqrt {2} + \sqrt {5} - \sqrt {10}}{1^2 - (\sqrt {2})^2} \\ & = \frac {1 - \sqrt {2} + \sqrt {5} - \sqrt {10}}{-1} \end{align}