Rationalisation (mathematics)

In

are eliminated.

If the denominator is a monomial in some radical, say ${\displaystyle a{\sqrt[{n}]{x}}^{k},}$ with k < n, rationalisation consists of multiplying the numerator and the denominator by ${\displaystyle {\sqrt[{n}]{x}}^{n-k},}$ and replacing ${\displaystyle {\sqrt[{n}]{x}}^{n}}$ by x (this is allowed, as, by definition, a nth root of x is a number that has x as its nth power). If k ≥ n, one writes k = qn + r with 0 ≤ r < n (Euclidean division), and ${\displaystyle {\sqrt[{n}]{x}}^{k}=x^{q}{\sqrt[{n}]{x}}^{r};}$ then one proceeds as above by multiplying by ${\displaystyle {\sqrt[{n}]{x}}^{n-r}.}$

If the denominator is linear in some square root, say ${\displaystyle a+b{\sqrt {x}},}$ rationalisation consists of multiplying the numerator and the denominator by ${\displaystyle a-b{\sqrt {x}},}$ and expanding the product in the denominator.

This technique may be extended to any algebraic denominator, by multiplying the numerator and the denominator by all

of the old denominator. However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.

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:

${\displaystyle {\frac {10}{\sqrt {5}}}}$

To rationalise this kind of expression, bring in the factor ${\displaystyle {\sqrt {5}}}$:

${\displaystyle {\frac {10}{\sqrt {5}}}={\frac {10}{\sqrt {5}}}\cdot {\frac {\sqrt {5}}{\sqrt {5}}}={\frac {10{\sqrt {5}}}{\left({\sqrt {5}}\right)^{2}}}}$

The square root disappears from the denominator, because ${\displaystyle \left({\sqrt {5}}\right)^{2}=5}$ by definition of a square root:

${\displaystyle {\frac {10{\sqrt {5}}}{\left({\sqrt {5}}\right)^{2}}}={\frac {10{\sqrt {5}}}{5}},}$

which is the result of the rationalisation.

Example 2:

${\displaystyle {\frac {10}{\sqrt[{3}]{a}}}}$

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

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

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

${\displaystyle {\frac {10{\sqrt[{3}]{a}}^{2}}{{\sqrt[{3}]{a}}^{3}}}={\frac {10{\sqrt[{3}]{a}}^{2}}{a}},}$

which is the result of the rationalisation.

Dealing with more square roots

For a

that is:

${\displaystyle {\sqrt {2}}\pm {\sqrt {3}}\,}$

Rationalisation can be achieved by multiplying by the

:

${\displaystyle {\sqrt {2}}\mp {\sqrt {3}}\,}$

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

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

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

${\displaystyle x\pm {\sqrt {y}}\,}$

by multiplication by

${\displaystyle x\mp {\sqrt {y}}}$

Example:

${\displaystyle {\frac {3}{{\sqrt {3}}\pm {\sqrt {5}}}}}$

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

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

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

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

Example 2:

This process also works with

with ${\displaystyle i={\sqrt {-1}}}$

${\displaystyle {\frac {7}{1\pm {\sqrt {-5}}}}}$

The fraction must be multiplied by a quotient containing ${\displaystyle {1\mp {\sqrt {-5}}}}$.

${\displaystyle {\frac {7}{1+{\sqrt {-5}}}}\cdot {\frac {1-{\sqrt {-5}}}{1-{\sqrt {-5}}}}={\frac {7(1-{\sqrt {-5}})}{1^{2}-{\sqrt {-5}}^{2}}}={\frac {7(1-{\sqrt {-5}})}{1-(-5)}}={\frac {7-7{\sqrt {5}}i}{6}}}$

Generalizations

Rationalisation can be extended to all

should be used, or equivalently a quadratic factor.

References

This material is carried in classic algebra texts. For example:

• ISBN 1402159072
  ); a trinomial example with square roots is on p. 256, while a general theory of rationalising factors for surds is on pp. 189–199.