# Rationalisation (mathematics)

For other uses, see Rationalization.

In elementary algebra, root rationalisation is a process by which surds in the denominator of an irrational fraction are eliminated.

These surds may be monomials or binomials involving square roots, in simple examples. There are wide extensions to the technique.

## Rationalisation of a monomial square root and cube root

For the fundamental technique, the numerator and denominator must be multiplied by the same factor.

Example 1:

$\frac{10}{\sqrt{a}}$

To rationalise this kind of monomial, bring in the factor $\sqrt{a}$:

$\frac{10}{\sqrt{a}} = \frac{10}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{{10\sqrt{a}}}{\sqrt{a}^2}$

The square root disappears from the denominator, because it is squared:

$\frac{{10\sqrt{a}}}{\sqrt{a}^2} = \frac{10\sqrt{a}}{a}$

This gives the result, after simplification:

$\frac{{10\sqrt{a}}}{{a}}$

Example 2:

$\frac{10}{\sqrt[3]{b}}$

To rationalise this radical, bring in the factor $\sqrt[3]{b}^2$:

$\frac{10}{\sqrt[3]{b}} = \frac{10}{\sqrt[3]{b}} \cdot \frac{\sqrt[3]{b}^2}{\sqrt[3]{b}^2} = \frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3}$

The cube root disappears from the denominator, because it is cubed:

$\frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3} = \frac{10\sqrt[3]{b}^2}{b}$

This gives the result, after simplification:

$\frac{{10\sqrt[3]{b}^2}}{{b}}$

## Dealing with more square roots

For a denominator that is:

$\sqrt{2}+\sqrt{3}\,$

Rationalisation can be achieved by multiplying by the Conjugate:

$\sqrt{2}-\sqrt{3}\,$

and applying the difference of two squares identity, which here will yield −1. To get this result, the entire fraction should be multiplied by

$\frac{ \sqrt{2}-\sqrt{3} }{\sqrt{2}-\sqrt{3}} = 1.$

This technique works much more generally. It can easily be adapted to remove one square root at a time, i.e. to rationalise

$x +\sqrt{y}\,$

by multiplication by

$x -\sqrt{y}$

Example:

$\frac{3}{\sqrt{3}+\sqrt{5}}$

The fraction must be multiplied by a quotient containing ${\sqrt{3}-\sqrt{5}}$.

$\frac{3}{\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} = \frac{3(\sqrt{3}-\sqrt{5})}{\sqrt{3}^2 - \sqrt{5}^2}$

Now, we can proceed to remove the square roots in the denominator:

$\frac{{3(\sqrt{3}-\sqrt{5}) }}{\sqrt{3}^2 - \sqrt{5}^2} = \frac{ 3 (\sqrt{3} - \sqrt{5} ) }{ 3 - 5 } = \frac{ 3( \sqrt{3}-\sqrt{5} ) }{-2}$

## Generalisations

Rationalisation can be extended to all algebraic numbers and algebraic functions (as an application of norm forms). For example, to rationalise a cube root, two linear factors involving cube roots of unity should be used, or equivalently a quadratic factor.