# 4.5 Surds

A "surd" is an irrational square root (i.e. a number which cannot be expressed
in the form $\frac{p}{q}$ where $p$ and $q$ are positive integers^{5}^{5}
5
The
set of positive integers contains $1,2,3,4$, and so on).

For example, $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ are all irrational numbers. We can prove this (and there is in fact a proof of this in 6.4.1).

## 4.5.1 Rationalising the denominator

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

Why? We can just multiply by $1$,

$\displaystyle\frac{1}{\sqrt{2}}$ | $\displaystyle=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}$ | (4.53) | ||

$\displaystyle=\frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}$ | (4.54) | |||

$\displaystyle=\frac{\sqrt{2}}{(\sqrt{2})^{2}}$ | (4.55) | |||

$\displaystyle=\frac{\sqrt{2}}{2}$ | (4.56) |

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

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

$\displaystyle\frac{3}{7-\sqrt{11}}$ | (4.58) | ||

$\displaystyle\frac{4-\sqrt{2}}{\sqrt{3}+44}$ | (4.59) | ||

$\displaystyle\frac{1+\sqrt{5}}{1+\sqrt{2}}$ | (4.60) |

The solution here is to use the difference of two squares (see Section 4.3.3 if unsure on what this is).

$\displaystyle\frac{1+\sqrt{5}}{1+\sqrt{2}}$ | $\displaystyle=\frac{1+\sqrt{5}}{1+\sqrt{2}}\frac{1-\sqrt{2}}{1-\sqrt{2}}$ | (4.61) | ||

$\displaystyle=\frac{(1+\sqrt{5})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$ | (4.62) | |||

$\displaystyle=\frac{1-\sqrt{2}+\sqrt{5}-\sqrt{10}}{1^{2}-(\sqrt{2})^{2}}$ | (4.63) | |||

$\displaystyle=\frac{1-\sqrt{2}+\sqrt{5}-\sqrt{10}}{-1}$ | (4.64) |