Indices, roots and surds

Indices provide a compact language for repeated multiplication. Roots reverse powers, and surds keep irrational roots exact. These ideas connect directly:

$$a^{1/n}=\sqrt[n]{a}.$$

Fluency with them is assumed in algebra, trigonometry, exponentials, logarithms and calculus. The aim is not merely to remember rules, but to know when each rule is valid.

Prerequisites

You should be able to:

• find integer factors and prime factorisations;
• calculate with fractions and negative numbers;
• distinguish a term from a factor;
• expand a pair of brackets.

Review exact arithmetic with fractions and pi or algebraic manipulation and factorisation if either skill is insecure.

What an index means

For a positive integer $n$,

$a^n=\underbrace{a\times a\times\cdots\times a}_{n\text{ factors}}.$

Here $a$ is the base and $n$ is the index, also called the exponent or power. Thus

$3^4=3\times3\times3\times3=81.$

Brackets determine the base:

\qquad -3^4=-(3^4)=-81.$$ In the second expression, the index applies only to $3$. The minus sign is applied afterwards. ## The laws of indices For a nonzero base $a$, the central laws are

\begin{aligned} a^m a^n&=a^{m+n},\ \frac{a^m}{a^n}&=a^{m-n},\ (a^m)^n&=a^{mn}. \end{aligned}

These rules apply to powers of the **same base**. They follow by counting factors. For example, $$a^3a^2=(aaa)(aa)=a^5.$$ Powers also distribute across multiplication and division:

(ab)^n=a^nb^n, \qquad \left(\frac ab\right)^n=\frac{a^n}{b^n},

provided the expressions are defined. ### Worked example 1: combine several laws Simplify $$\frac{(x^3)^4x^2}{x^5},\qquad x\ne0.$$ Apply the power rule first, then combine powers of $x$:

\begin{aligned} \frac{(x^3)^4x^2}{x^5} &=\frac{x^{12}x^2}{x^5}\ &=x^{12+2-5}\ &=x^9. \end{aligned}

The restriction $x\ne0$ comes from the original denominator. ### Worked example 2: coefficients and variables Simplify $$\frac{12a^5b^2}{3a^2b^6},\qquad a\ne0, b\ne0.$$ Treat numerical and variable factors separately:

\begin{aligned} \frac{12a^5b^2}{3a^2b^6} &=4a^{5-2}b^{2-6}\ &=4a^3b^{-4}\ &=\frac{4a^3}{b^4}. \end{aligned}

The last form uses positive indices and is usually preferred. ### Misconception: powers do not distribute across addition In general, $$(a+b)^n\ne a^n+b^n.$$ For example, $(2+3)^2=25$, but $2^2+3^2=13$. Expanding gives the missing middle term: $$(a+b)^2=a^2+2ab+b^2.$$ ### Self-check 1 Simplify, using positive indices in the final answers. 1. $p^7p^{-3}$ 2. $(y^4)^3\div y^5$, where $y\ne0$ 3. $\dfrac{15m^2n^7}{5m^6n^3}$, where $m,n\ne0$ <details> <summary>Answers</summary> 1. $p^{7-3}=p^4$. 2. $y^{12-5}=y^7$. 3. $3m^{-4}n^4=\dfrac{3n^4}{m^4}$. </details> ## Zero and negative indices If $a\ne0$, then $$\frac{a^m}{a^m}=1.$$ The quotient law gives the same expression as $a^{m-m}=a^0$, so $$a^0=1,\qquad a\ne0.$$ Similarly, $$a^{-n}=\frac{1}{a^n}.$$ A negative index means **reciprocal**, not a negative value: $$2^{-3}=\frac{1}{2^3}=\frac18.$$ ### Worked example 3: evaluate negative powers Evaluate $$\left(\frac23\right)^{-3}.$$ The negative index takes the reciprocal of the whole bracket:

\left(\frac23\right)^{-3} =\left(\frac32\right)^3 =\frac{27}{8}.

Compare this with $$-\left(\frac23\right)^3=-\frac8{27},$$ which has a minus sign outside the power and is a different expression. ### Why zero needs care $0^n=0$ for every positive integer $n$, but $0^{-1}$ would mean $1/0$, which is undefined. The expression $0^0$ is not assigned a value by these index laws. Never use $a^0=1$ without checking that $a\ne0$. ## Fractional indices and roots The value $a^{1/n}$ is defined so that raising it to the power $n$ gives $a$. Therefore $$a^{1/n}=\sqrt[n]{a}.$$ More generally,

a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m.

For real numbers, an even root requires $a\ge0$. An odd root can have a negative input: $$\sqrt[3]{-8}=-2, \qquad \sqrt{-8}\text{ is not real}.

Worked example 4: choose an efficient order

Evaluate $64^{2/3}$ exactly.

Take the cube root first because it keeps the numbers small:

$$64^{2/3} =\left(64^{1/3}\right)^2 =4^2 =16.$$

Worked example 5: a negative fractional index

Evaluate $81^{-3/4}$ exactly.

The negative sign means reciprocal. The denominator $4$ means fourth root, and the numerator $3$ means cube:

$$\begin{aligned} 81^{-3/4} &=\frac{1}{81^{3/4}}\\ &=\frac{1}{(\sqrt[4]{81})^3}\\ &=\frac{1}{3^3}\\ &=\frac1{27}. \end{aligned}$$

A vital sign distinction

The symbol $\sqrt a$ means the nonnegative square root of $a$. Hence

$\sqrt{25}=5,$

not $\pm5$. However, the equation $x^2=25$ has two solutions because both $5^2$ and $(-5)^2$ equal $25$:

$x=\pm5.$

For the same reason,

$\sqrt{x^2}=|x|,$

not always $x$. If $x=-4$, then $\sqrt{x^2}=\sqrt{16}=4=|x|$.

Self-check 2

Evaluate exactly.

1. $16^{3/4}$
2. $32^{-2/5}$
3. $\sqrt{(-7)^2}$
4. Solve $t^2=49$.

Answers

1. $(\sqrt[4]{16})^3=2^3=8$.
2. $1/(\sqrt[5]{32})^2=1/2^2=1/4$.
3. $7$. A square root is nonnegative.
4. $t=\pm7$. An equation may have both roots.

What is a surd?

A surd is an irrational root written exactly, such as $\sqrt2$, $3\sqrt5$ or $\sqrt[3]{7}$. By contrast, $\sqrt9=3$ is not a surd because it simplifies to a rational number.

Keeping surds exact avoids premature rounding:

$\sqrt2=1.414213\ldots$

The dots show that the decimal has been truncated. Unless a question requests a decimal, retain the surd form.

Simplifying surds

For $a,b\ge0$,

$\sqrt{ab}=\sqrt a\sqrt b.$

To simplify a square root, extract its largest square factor.

Worked example 6: extract square factors

Simplify $\sqrt{180}$.

The largest square factor of $180$ is $36$:

$$\begin{aligned} \sqrt{180} &=\sqrt{36\times5}\\ &=\sqrt{36}\sqrt5\\ &=6\sqrt5. \end{aligned}$$

If the largest square factor is not obvious, use prime factors:

$180=2^2\times3^2\times5,$

so one $2$ and one $3$ leave the square root.

Misconception: roots do not split across addition

Although $\sqrt{ab}=\sqrt a\sqrt b$ for nonnegative $a$ and $b$,

$\sqrt{a+b}\ne\sqrt a+\sqrt b$

in general. For example, $\sqrt{9+16}=5$, whereas $\sqrt9+\sqrt{16}=7$.

Calculating with surds

Like surds can be collected in the same way as like algebraic terms:

$4\sqrt3-\sqrt3=3\sqrt3.$

Unlike surds cannot be collected until they have been simplified.

Worked example 7: simplify before collecting

Simplify $\sqrt{48}+2\sqrt{27}-\sqrt3$.

$$\begin{aligned} \sqrt{48}+2\sqrt{27}-\sqrt3 &=4\sqrt3+2(3\sqrt3)-\sqrt3\\ &=4\sqrt3+6\sqrt3-\sqrt3\\ &=9\sqrt3. \end{aligned}$$

Worked example 8: multiply surds

Simplify $(2\sqrt3)(5\sqrt6)$.

$$\begin{aligned} (2\sqrt3)(5\sqrt6) &=10\sqrt{18}\\ &=10(3\sqrt2)\\ &=30\sqrt2. \end{aligned}$$

Brackets containing surds expand by the usual distributive law:

$$\begin{aligned} (3+\sqrt2)(4-\sqrt2) &=12-3\sqrt2+4\sqrt2-2\\ &=10+\sqrt2. \end{aligned}$$

Conjugates

The expressions $a+b\sqrt c$ and $a-b\sqrt c$ are conjugates. Their product is rational because it is a difference of two squares:

$$(a+b\sqrt c)(a-b\sqrt c)=a^2-b^2c.$$

For example,

$(5+\sqrt3)(5-\sqrt3)=25-3=22.$

Rationalising denominators

An exact answer is conventionally written without a surd in its denominator. To rationalise a denominator, multiply the numerator and denominator by a suitable expression equal to the surd factor or its conjugate.

Worked example 9: one surd in the denominator

Simplify $\dfrac{7}{2\sqrt5}$.

Multiply by $\sqrt5/\sqrt5$, which equals $1$:

$$\begin{aligned} \frac{7}{2\sqrt5} &=\frac{7}{2\sqrt5}\times\frac{\sqrt5}{\sqrt5}\\ &=\frac{7\sqrt5}{2\times5}\\ &=\frac{7\sqrt5}{10}. \end{aligned}$$

Worked example 10: a two term denominator

Rationalise and simplify

$\frac{3}{2+\sqrt5}.$

Multiply by the conjugate $2-\sqrt5$:

$$\begin{aligned} \frac{3}{2+\sqrt5} &=\frac{3}{2+\sqrt5}\times\frac{2-\sqrt5}{2-\sqrt5}\\ &=\frac{3(2-\sqrt5)}{2^2-(\sqrt5)^2}\\ &=\frac{6-3\sqrt5}{4-5}\\ &=3\sqrt5-6. \end{aligned}$$

The negative denominator is not an error. A size check confirms the result: $2+\sqrt5$ is positive and small, so the original value is positive; $3\sqrt5-6\approx0.708$ is also positive.

Worked example 11: exact cancellation hidden by surds

Simplify

$\frac{1}{\sqrt3-1}-\frac{1}{\sqrt3+1}.$

Use a common denominator. The conjugate factors multiply to a rational number:

$$\begin{aligned} \frac{1}{\sqrt3-1}-\frac{1}{\sqrt3+1} &=\frac{(\sqrt3+1)-(\sqrt3-1)}{(\sqrt3-1)(\sqrt3+1)}\\ &=\frac{2}{3-1}\\ &=1. \end{aligned}$$

Mixed self-check

1. Simplify $\dfrac{(2x^3)^2}{4x^{-1}}$, where $x\ne0$.
2. Evaluate $\left(\dfrac{1}{27}\right)^{-2/3}$.
3. Simplify $3\sqrt{20}-\sqrt{45}+2\sqrt5$.
4. Expand and simplify $(\sqrt7+2)^2$.
5. Rationalise $\dfrac{4}{3-\sqrt2}$.
6. A student writes $\sqrt{x^2}=x$. Give a value of $x$ showing that this is not always true.

Answers

1. $(2x^3)^2=4x^6$, so $4x^6/(4x^{-1})=x^{6-(-1)}=x^7$.
2. $(1/27)^{-2/3}=27^{2/3}=(\sqrt[3]{27})^2=9$.
3. $3(2\sqrt5)-3\sqrt5+2\sqrt5=5\sqrt5$.
4. $7+4\sqrt7+4=11+4\sqrt7$.
5. $\dfrac{4(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)}=\dfrac{12+4\sqrt2}{7}$.
6. For example, $x=-3$: $\sqrt{x^2}=\sqrt9=3$, but $x=-3$. The correct identity is $\sqrt{x^2}=|x|$.

Exam strategy and common errors

• Check the base before applying an index law. You may combine $x^3x^5$, but not $x^3y^5$.
• A negative index creates a reciprocal: $a^{-n}=1/a^n$.
• The denominator of a fractional index gives the root; the numerator gives the power.
• Simplify each surd before deciding whether terms are like.
• Never split a root across addition or subtraction.
• Keep exact values through the working and round only at the end if requested.
• When rationalising a two term denominator, use its conjugate and expand both numerator and denominator.
• Substitute a simple value to test a doubtful identity. One counterexample disproves a false general rule.

Next steps

These skills feed directly into algebraic fractions, quadratic equations and standard form and accuracy. At A level, continue to indices before studying exponential functions and logarithms.