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surds and indices

There are number of quotable rules that allow for the manipulations of indices and surds, all of which can be justified with a little work.

surds and indices - basic rules

Rule 0 $$x^0 = 1, x\neq 0$$

This is a definition, that we require to do anything with indices


Rule 1 $$x^1 = x$$

This is a definition, that we require to do anything with indices


Rule 2 $$x^a \times x^b = x^{a+b}$$ The justification is pretty simple,

$$x^a = \overbrace{x\times x\times \dots \times x}^{\mbox{a times}}$$

$$x^b = \overbrace{x\times x\times \dots \times x}^{\mbox{b times}}$$

$$\begin{align}x^a\times x^b &= \overbrace{x\times x\times \dots \times x}^{\mbox{a times}}\times \overbrace{x\times x\times \dots \times x}^{\mbox{b times}} \\ &= \overbrace{x\times x\times \dots \times x}^{\mbox{a+b times}} \\ &= x^{a+b} \end{align} $$


Rule 3

$$x^{-a} = \frac{1}{x^a}$$

From Rule 2, and then Rule 0 we get $$x^a \times x^{-a} = x^{a + -a}= x^0 = 1$$

Given that $x^a \times x^{-a} = 1$, then it follows that $x^{-a} = \frac{1}{x^a}$


Rule 4 $$\frac{x^a}{x^b} = x^{a-b}$$

By the rules of fractions, and then by Rule 3,and finally Rule 2, we can say $$\frac{x^a}{x^b} = x^a\frac{1}{x^b} = x^a\times x^{-b} = x^{a-b}$$


Rule 5 $$\sqrt[c]{x} = x^{\frac{1}{c}}$$

$$\begin{align}x^1 &= \overbrace{\sqrt[c]{x}\times \sqrt[c]{x}\times \dots \times \sqrt[c]{x}}^{\mbox{c times}} \\ &= \left(x^\alpha\right)^c \\ &= x^{\alpha c}\end{align}$$

This implies $\alpha = \frac{1}{c}$ and hence the result.


Rule 6 $$\left(ab\right)^c = a^cb^c$$

$$\begin{align}\left(ab\right)^c &= ab\times \dots \times ab \\ &= a\times \dots \times a\times b \times \dots \times b \\ &= a^cb^c\end{align}$$


Rule 7 $$\left(a^b\right)^c = a^{b\times c}$$

Surds and Indices - Rationalisation

Rationalisation is a process of rewriting fractions featuring surds. There are two distinct types.

Type 1

Rationalisation of expressions of the form $$\frac{a}{\sqrt{b}}$$. 

In this case, the process is simply that of mulitplication by $\frac{\sqrt{b}}{\sqrt{b}}$ to give

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

Type 2

Rationalisation of expressions of the form $$\frac{a}{b + \sqrt{c}}$$

In this case, the process is that of multiplying the expression by $\frac{b - \sqrt{c}}{b - \sqrt{c}}$

$$\frac{a}{b + \sqrt{c}} = \frac{a}{b + \sqrt{c}}\times \frac{b - \sqrt{c}}{b - \sqrt{c}} = \frac{ab - a\sqrt{c}}{b^2 - c}$$

Surds and indices 1

Express the following in the form $ax^b + cx^d$ $$ \sqrt{16x^4} + \left(\sqrt{9x}\right)^3 $$
solution - press button to display

By rule 6... $$ \sqrt{16x^4} = \sqrt{16}\times\sqrt{x^4} = 4\sqrt{x^4} $$

By rules 5 and 7... $$ \sqrt{x^4} = \left(x^4\right)^\frac{1}{2} = x^{4\times \frac{1}{2}} = x^2 $$

By rules 6 and 7... $$\left(\sqrt{9x}\right)^3 = \left(3\sqrt{x}\right)^3 = 27 \times \sqrt{x}^3 $$

By rules 5 and 7... $$ \sqrt{x}^3 = \left(x^\frac{1}{2}\right)^3 = x^\frac{3}{2}$$

Hence $$  \sqrt{16x^4} + \left(\sqrt{9x}\right)^3 = 4x^2 + 27x^\frac{3}{2} $$

Surds and indices 2

Express the following in the form $a + b\sqrt{2}$ where $a$ and $b$ are rational numbers.

$$ \frac{4}{1+\sqrt{2}} $$
solution - press button to display

$$\begin{align} \frac{4}{1+\sqrt{2}} &= \frac{4}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}\\ &=\frac{4-4\sqrt{2}}{(1+\sqrt{2})(1- \sqrt{2})} \\ &= \frac{4-4\sqrt{2}}{1 - (\sqrt{2})^2} \\ &= -4 + 4\sqrt{2}\end{align}$$