Maths revision: Indices and surds including ‘rationalising the denominator’

Ahh, indices and surds, they look so messy and many find them hard to deal with… but they’re perhaps some of the easiest marks in your A Level paper.

Indices

Let’s begin with the basics: An index (indices is the plural) is the power to which something is raised, for example in the expression 23 the index is 3 and 2 is referred to as the base.

Rules of indices

RuleExample
$a^0\text{ = 1}$$x^0\text{ = 1}$
$a^n\text{ x }a^m\text{ = }a^{n + m}$$x^2\text{ x }x^4\text{ = }x^6$
$\frac{a^n}{a^m}\text{ = }a^{n - m}$$\frac{x^7}{x^4}\text{ = }x^3$
$(a^n)^m\text{ = }a^{nm}$$(x^2)^3\text{ = }x^6$
$a^{-n}\text{ = }\frac{1}{a^n}$$x^{-2}\text{ = }\frac{1}{x^2}$
$a^{\frac{1}{n}}\text{ = }\sqrt[n]{a}$$2^{\frac{1}{2}}\text{ = }\sqrt[2]{2}\text{ = }\sqrt{2}$
$a^{\frac{n}{m}}\text{ = }(a^{\frac{1}{m}})^n$$16^{\frac{3}{2}}\text{ = }(16^{\frac{1}{2}})^3\text{ = }4^3\text{ = }64$

Rational and irrational numbers

A rational number is one which can be written as either can integer or as the ratio of two integers (i.e. a fraction). For example, 2, $sqrt{16}$, $\frac{2}{9}$.

An irrational number is one which cannot be expressed as a fraction, for example $\pi$ or $sqrt{2}$ these numbers do not end or have a recurring pattern.

Surds

Surds are expressions for irrational numbers, for example $sqrt{2}$

Simplifying surds

To simplify a surd, such as $sqrt{27}$ firstly write it as the product of other integers, in this case $sqrt{9\text{ x }3}$. This can be expressed as $sqrt{9}sqrt{3}$. You know the square root of 9 is 3 and hence can simplify the expression further: $3\sqrt{3}$

Examples

Simplify the following surds:
1. $sqrt{24}$
2. $sqrt{50}$
3. $sqrt{96}$
4. $sqrt{29}$

Here are the following worked solutions:
1. $sqrt{24}$ = $sqrt{4\text{ x }6}$ = $sqrt{4}sqrt{6}$ = $2\sqrt{6}$

2. $sqrt{50}$ = $sqrt{25\text{ x }2}$ = $sqrt{25}sqrt{2}$ = $5\sqrt{2}$

3. $sqrt{96}$ = $sqrt{24\text{ x }4}$ = $sqrt{24}sqrt{4}$ = $2\sqrt{6}\text{ x }2$ = $4\sqrt{6}$
4. $sqrt{29}$ = $sqrt{29}$

Rationalising the denominator

Knowing (x + y)(x – y) = x2 – y2 can be used to solve more complex surds

For example:

Simplify the following surd: $\fs4\frac{3}{2 - \sqrt{2}}$

By multiplying both sides by $2 + \sqrt{2}$ you can rationalise the denominator (i.e. turn it into a rational number):

$\fs5 \frac{3}{2 - \sqrt{2}} = \frac{3(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})}$

Using the rule mentioned above:

$\fs5 \frac{3(2 + \sqrt{2})}{2^2 - (\sqrt{2})^2}\text{ = } \frac{3(2 + \sqrt{2})}{4 - 2}\text{ = } \frac{3(2 + \sqrt{2})}{2}$

$\fs5 \frac{6 + 3\sqrt{2}}{2}$