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1r BTL CCSS · Real numbers Notes 21 Sep 2026

6. Rationalising the denominator

How to get the roots out of the denominator: the three types, with examples of each.

What rationalising means

Definition

Rationalising is a convention: irrational numbers are written without roots in the denominator.

You multiply and divide by the same expression: the fraction keeps its value, because it is like multiplying by $1$, but the root in the denominator disappears.

We will rationalise 3 types of numbers.

Type 1: a square root

How to do it

Multiply and divide by the same root as in the denominator:

$$\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$$

Example

$$\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$$

Type 2: an $n$-th root

How to do it

Multiply and divide by the root that completes the exponent:

$$\frac{c}{\sqrt[n]{a^{m}}} \cdot \frac{\sqrt[n]{a^{p}}}{\sqrt[n]{a^{p}}} \qquad \text{with } m + p = n \text{ or a multiple of } n$$

The denominator becomes $\sqrt[n]{a^{m+p}}$, and the root comes out exactly.

Examples

Since $3 + 4 = 7$, multiply by $\sqrt[7]{2^{4}}$:

$$\frac{5}{\sqrt[7]{2^{3}}} = \frac{5}{\sqrt[7]{2^{3}}} \cdot \frac{\sqrt[7]{2^{4}}}{\sqrt[7]{2^{4}}} = \frac{5\sqrt[7]{2^{4}}}{2}$$

Here $4 + 2 = 6$, a multiple of $3$, and $\sqrt[3]{5^{6}} = 5^{2}$. In the numerator, to join the two roots, both are turned into index $6$:

$$\frac{\sqrt{2}}{\sqrt[3]{5^{4}}} \cdot \frac{\sqrt[3]{5^{2}}}{\sqrt[3]{5^{2}}} = \frac{\sqrt{2} \cdot \sqrt[3]{5^{2}}}{\sqrt[3]{5^{6}}} = \frac{\sqrt[6]{2^{3}} \cdot \sqrt[6]{5^{4}}}{5^{2}} = \frac{\sqrt[6]{2^{3} \cdot 5^{4}}}{25}$$

A negative exponent works the same way: $-1 + 5 = 4$.

$$\frac{3}{\sqrt[4]{2^{-1}}} \cdot \frac{\sqrt[4]{2^{5}}}{\sqrt[4]{2^{5}}} = \frac{3\sqrt[4]{2^{5}}}{2}$$

This last one can still be simplified: $\sqrt[4]{2^{5}} = 2\sqrt[4]{2}$, which leaves $3\sqrt[4]{2}$.

Type 3: a binomial with roots

How to do it

Multiply and divide by the conjugate of the denominator, which is the same binomial with the middle sign changed:

$$\sqrt{a} + \sqrt{b} \;\longleftrightarrow\; \sqrt{a} - \sqrt{b}$$
$$\frac{c}{\sqrt{a} + \sqrt{b}} \cdot \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{c\,(\sqrt{a} - \sqrt{b})}{a - b}$$

Why it works

A binomial times its conjugate is a sum times a difference, which gives the difference of squares:

$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^{2} - (\sqrt{b})^{2} = a - b$$

Done by hand, the two middle terms cancel out:

$$\begin{array}{rrr} & \sqrt{a} & +\,\sqrt{b} \\ \times & \sqrt{a} & -\,\sqrt{b} \\ \hline & -\sqrt{ab} & -\,b \\ a & +\sqrt{ab} & \\ \hline a & & -\,b \end{array}$$

Example

$$\frac{\sqrt{2}}{\sqrt{2} - \sqrt{3}} \cdot \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{2 + \sqrt{6}}{2 - 3} = \frac{2 + \sqrt{6}}{-1} = -2 - \sqrt{6}$$