Fractional and Negative Indices

What you'll learn

How negative indices turn powers into reciprocals.
How fractional indices mean roots, such as square roots and cube roots.
How to evaluate expressions like $34323343^{\frac{2}{3}}$ and $(1625)−32\left(\frac{16}{25}\right)^{-\frac{3}{2}}$ .
How to use index laws when everything can be written with the same base.

The basics: powers, roots and reciprocals

An expression like $a^n$ is called a power. You read it as “ $a$ to the power of $n$ ”.

Definition

Index

In $a^n$ , the number or expression being repeatedly multiplied is the base $a$ , and the small raised number is the index or exponent $n$ .

Before negative indices make sense, you need the idea of a reciprocal.

Definition

Reciprocal

The reciprocal of a non-zero number is one divided by that number. For example, the reciprocal of 8 is $18\frac{1}{8}$ , and the reciprocal of $56\frac{5}{6}$ is $65\frac{6}{5}$ .

Example

Finding reciprocals

To find the reciprocal of 6, write one divided by 6: $16\frac{1}{6}$ .
To find the reciprocal of $49\frac{4}{9}$ , swap the numerator and denominator.
$149=94\frac{1}{\frac{4}{9}}=\frac{9}{4}$
So the reciprocal of 6 is $16\frac{1}{6}$ , and the reciprocal of $49\frac{4}{9}$ is $94\frac{9}{4}$ .

Negative indices

A negative index does not make the answer negative. It tells you to take a reciprocal.

Key Idea

Negative indices mean reciprocal

For any non-zero base $a$ , the rule is $a−n=1ana^{-n}=\frac{1}{a^n}$ . In particular, $a−1=1aa^{-1}=\frac{1}{a}$ .

Common Mistake

Zero cannot have a negative index

Expressions such as $0^{-1}$ are undefined, because they would require division by zero.

Example

Evaluating negative indices

For $5^{-2}$ , the negative index means put the power in the denominator.
$5−2=1525^{-2}=\frac{1}{5^2}$
Now square 5: $5−2=1255^{-2}=\frac{1}{25}$ .
For $(38)−1\left(\frac{3}{8}\right)^{-1}$ , the index -1 means “take the reciprocal”.
$(38)−1=83\left(\frac{3}{8}\right)^{-1}=\frac{8}{3}$
Notice that neither answer is negative.

Common Mistake

Negative index, not negative number

Writing $4^{-1}=-4$ is wrong. The correct idea is reciprocal: $4−1=144^{-1}=\frac{1}{4}$ .

Fractional indices: roots first

A fractional index is an index written as a fraction. The denominator of the fraction tells you which root to take.

Definition

Fractional index

For positive values of $a$ , $a12=aa^{\frac{1}{2}}=\sqrt{a}$ and $a13=a3a^{\frac{1}{3}}=\sqrt[3]{a}$ . More generally, $a1n=ana^{\frac{1}{n}}=\sqrt[n]{a}$ .

For a fraction like $mn\frac{m}{n}$ , the denominator tells you the root, and the numerator tells you the power.

Key Idea

Denominator root, numerator power

For suitable values, $amn=(an)ma^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m$ . It is usually easiest to take the root first, then apply the power.

The diagram below is a useful memory aid for combining fractional and negative indices.

$Flow diagram for evaluating a negative fractional index$

Example

Evaluating fractional indices

For $14412144^{\frac{1}{2}}$ , the denominator 2 means square root.
$14412=144=12144^{\frac{1}{2}}=\sqrt{144}=12$
For $34323343^{\frac{2}{3}}$ , the denominator 3 means cube root first.
$3433=7\sqrt[3]{343}=7$
The numerator 2 means square the result: $7^2=49$ .
So $34323=49343^{\frac{2}{3}}=49$ .

Common Mistake

A half power is not halving

$14412144^{\frac{1}{2}}$ means $144\sqrt{144}$ , not 144 divided by 2.

Negative fractional indices

Now combine the two ideas:

the negative sign means reciprocal;
the denominator means root;
the numerator means power.

Key Idea

The combined rule

For non-zero $a$ , $a−mn=1amn=1(an)ma^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}=\frac{1}{\left(\sqrt[n]{a}\right)^m}$ .

Example

Evaluating a negative fractional index

Start with $32−2532^{-\frac{2}{5}}$ . The negative sign means take the reciprocal of $322532^{\frac{2}{5}}$ .
$32−25=1322532^{-\frac{2}{5}}=\frac{1}{32^{\frac{2}{5}}}$
The denominator 5 means take the fifth root: $325=2\sqrt[5]{32}=2$ .
The numerator 2 means square the result: $2^2=4$ .
Therefore $32−25=1432^{-\frac{2}{5}}=\frac{1}{4}$ .

Fraction bases

When the base is a fraction and the index is negative, it is often quickest to invert the fraction first.

Tip

Invert first when helpful

For expressions like $(ab)−mn\left(\frac{a}{b}\right)^{-\frac{m}{n}}$ , first change the base to $(ba)mn\left(\frac{b}{a}\right)^{\frac{m}{n}}$ , then deal with the root and power.

Example

A fraction with a negative fractional index

Deal with the negative sign by taking the reciprocal of the base.
$(1625)−32=(2516)32\left(\frac{16}{25}\right)^{-\frac{3}{2}}=\left(\frac{25}{16}\right)^{\frac{3}{2}}$
The denominator 2 means take the square root of the fraction.
$2516=54\sqrt{\frac{25}{16}}=\frac{5}{4}$
The numerator 3 means cube the result.
$(54)3=12564\left(\frac{5}{4}\right)^3=\frac{125}{64}$
So $(1625)−32=12564\left(\frac{16}{25}\right)^{-\frac{3}{2}}=\frac{125}{64}$ .

Algebraic expressions with fractional indices

The same index laws apply to algebraic terms. The most useful one here is the power of a power rule:

a^p)^q=a^{pq}

You can also take roots of powers of variables when the exponents divide neatly.

Example

Simplifying an algebraic fractional power

Consider $(27x6)23(27x^6)^{\frac{2}{3}}$ . The denominator 3 means take the cube root of each part.
$27x63=3x2\sqrt[3]{27x^6}=3x^2$
The numerator 2 means square the result.
$3x^2)^2=9x^4$
Therefore $(27x6)23=9x4(27x^6)^{\frac{2}{3}}=9x^4$ .

Rewriting everything with the same base

You will often be asked to find an unknown index. The strategy is to rewrite both sides using the same base, then compare the indices.

The key laws are:

$ap×aq=ap+qa^p \times a^q=a^{p+q}$
$a^p)^q=a^{pq}$
$ap÷aq=ap−qa^p \div a^q=a^{p-q}$

Example

Finding an unknown index

Suppose $\times \sqrt{32}=2^n$ . First rewrite 32 as a power of 2: $32=2^5$ .
Convert the square root into a half power.
$32=(25)12=252\sqrt{32}=(2^5)^{\frac{1}{2}}=2^{\frac{5}{2}}$
Rewrite the full left-hand side using powers of 2.
$\times \sqrt{32}=2^1 \times 2^{\frac{5}{2}}$
Add the indices because the bases are the same.
$21×252=2722^1 \times 2^{\frac{5}{2}}=2^{\frac{7}{2}}$
Since $\times \sqrt{32}=2^n$ , we have $n=72n=\frac{7}{2}$ .

Example

Expressing powers in terms of letters

Let $r=5^p$ and $s=5^q$ . Then $5p+q=5p×5q=rs5^{p+q}=5^p \times 5^q=rs$ .
For $5^{2p}$ , use the power of a power rule: $5^{2p}=(5^p)^2=r^2$ .
For $5^{q-1}$ , split the index: $5q−1=5q÷5=s55^{q-1}=5^q \div 5=\frac{s}{5}$ .

Large roots written as products

Sometimes a root looks awkward, but the product has been chosen to make a perfect power.

Example

Evaluating a fourth root product

Consider $5×125×1084\sqrt[4]{5 \times 125 \times 10^8}$ .
Multiply the first two factors: $\times 125=625=5^4$ .
Rewrite the power of 10 as a fourth power: $10^8=(10^2)^4$ .
Combine the factors into one fourth power.
$54×(102)4=(5×102)45^4 \times (10^2)^4=(5 \times 10^2)^4$
Taking the fourth root gives $\times 10^2=500$ .

Exam technique

In the exam

Check whether the index is negative first; if it is, plan for a reciprocal.
For fractional indices, read the denominator as the root and the numerator as the power.
If an equation has powers, try to rewrite everything using the same base before comparing indices.

Self review

Check yourself

Can you explain why $a^{-1}$ means $1a\frac{1}{a}$ rather than $- a$ ?
What root is shown by the denominator in $a23a^{\frac{2}{3}}$ ?
How would you start simplifying $(916)−32\left(\frac{9}{16}\right)^{-\frac{3}{2}}$ ?

Fractional and Negative Indices

What you'll learn

How negative indices turn powers into reciprocals.
How fractional indices mean roots, such as square roots and cube roots.
How to evaluate expressions like $34323343^{\frac{2}{3}}$ and $(1625)−32\left(\frac{16}{25}\right)^{-\frac{3}{2}}$ .
How to use index laws when everything can be written with the same base.

The basics: powers, roots and reciprocals

An expression like $a^n$ is called a power. You read it as “ $a$ to the power of $n$ ”.

Definition

Index

In $a^n$ , the number or expression being repeatedly multiplied is the base $a$ , and the small raised number is the index or exponent $n$ .

Before negative indices make sense, you need the idea of a reciprocal.

Definition

Reciprocal

The reciprocal of a non-zero number is one divided by that number. For example, the reciprocal of 8 is $18\frac{1}{8}$ , and the reciprocal of $56\frac{5}{6}$ is $65\frac{6}{5}$ .

Example

Finding reciprocals

To find the reciprocal of 6, write one divided by 6: $16\frac{1}{6}$ .
To find the reciprocal of $49\frac{4}{9}$ , swap the numerator and denominator.
$149=94\frac{1}{\frac{4}{9}}=\frac{9}{4}$
So the reciprocal of 6 is $16\frac{1}{6}$ , and the reciprocal of $49\frac{4}{9}$ is $94\frac{9}{4}$ .

Negative indices

A negative index does not make the answer negative. It tells you to take a reciprocal.

Key Idea

Negative indices mean reciprocal

For any non-zero base $a$ , the rule is $a−n=1ana^{-n}=\frac{1}{a^n}$ . In particular, $a−1=1aa^{-1}=\frac{1}{a}$ .

Common Mistake

Zero cannot have a negative index

Expressions such as $0^{-1}$ are undefined, because they would require division by zero.

Example

Evaluating negative indices

For $5^{-2}$ , the negative index means put the power in the denominator.
$5−2=1525^{-2}=\frac{1}{5^2}$
Now square 5: $5−2=1255^{-2}=\frac{1}{25}$ .
For $(38)−1\left(\frac{3}{8}\right)^{-1}$ , the index -1 means “take the reciprocal”.
$(38)−1=83\left(\frac{3}{8}\right)^{-1}=\frac{8}{3}$
Notice that neither answer is negative.

Common Mistake

Negative index, not negative number

Writing $4^{-1}=-4$ is wrong. The correct idea is reciprocal: $4−1=144^{-1}=\frac{1}{4}$ .

Fractional indices: roots first

A fractional index is an index written as a fraction. The denominator of the fraction tells you which root to take.

Definition

Fractional index

For positive values of $a$ , $a12=aa^{\frac{1}{2}}=\sqrt{a}$ and $a13=a3a^{\frac{1}{3}}=\sqrt[3]{a}$ . More generally, $a1n=ana^{\frac{1}{n}}=\sqrt[n]{a}$ .

For a fraction like $mn\frac{m}{n}$ , the denominator tells you the root, and the numerator tells you the power.

Key Idea

Denominator root, numerator power

For suitable values, $amn=(an)ma^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m$ . It is usually easiest to take the root first, then apply the power.

The diagram below is a useful memory aid for combining fractional and negative indices.

$Flow diagram for evaluating a negative fractional index$

Example

Evaluating fractional indices

For $14412144^{\frac{1}{2}}$ , the denominator 2 means square root.
$14412=144=12144^{\frac{1}{2}}=\sqrt{144}=12$
For $34323343^{\frac{2}{3}}$ , the denominator 3 means cube root first.
$3433=7\sqrt[3]{343}=7$
The numerator 2 means square the result: $7^2=49$ .
So $34323=49343^{\frac{2}{3}}=49$ .

Common Mistake

A half power is not halving

$14412144^{\frac{1}{2}}$ means $144\sqrt{144}$ , not 144 divided by 2.

Negative fractional indices

Now combine the two ideas:

the negative sign means reciprocal;
the denominator means root;
the numerator means power.

Key Idea

The combined rule

For non-zero $a$ , $a−mn=1amn=1(an)ma^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}=\frac{1}{\left(\sqrt[n]{a}\right)^m}$ .

Example

Evaluating a negative fractional index

Start with $32−2532^{-\frac{2}{5}}$ . The negative sign means take the reciprocal of $322532^{\frac{2}{5}}$ .
$32−25=1322532^{-\frac{2}{5}}=\frac{1}{32^{\frac{2}{5}}}$
The denominator 5 means take the fifth root: $325=2\sqrt[5]{32}=2$ .
The numerator 2 means square the result: $2^2=4$ .
Therefore $32−25=1432^{-\frac{2}{5}}=\frac{1}{4}$ .

Fraction bases

When the base is a fraction and the index is negative, it is often quickest to invert the fraction first.

Tip

Invert first when helpful

For expressions like $(ab)−mn\left(\frac{a}{b}\right)^{-\frac{m}{n}}$ , first change the base to $(ba)mn\left(\frac{b}{a}\right)^{\frac{m}{n}}$ , then deal with the root and power.

Example

A fraction with a negative fractional index

Deal with the negative sign by taking the reciprocal of the base.
$(1625)−32=(2516)32\left(\frac{16}{25}\right)^{-\frac{3}{2}}=\left(\frac{25}{16}\right)^{\frac{3}{2}}$
The denominator 2 means take the square root of the fraction.
$2516=54\sqrt{\frac{25}{16}}=\frac{5}{4}$
The numerator 3 means cube the result.
$(54)3=12564\left(\frac{5}{4}\right)^3=\frac{125}{64}$
So $(1625)−32=12564\left(\frac{16}{25}\right)^{-\frac{3}{2}}=\frac{125}{64}$ .

Algebraic expressions with fractional indices

The same index laws apply to algebraic terms. The most useful one here is the power of a power rule:

a^p)^q=a^{pq}

You can also take roots of powers of variables when the exponents divide neatly.

Example

Simplifying an algebraic fractional power

Consider $(27x6)23(27x^6)^{\frac{2}{3}}$ . The denominator 3 means take the cube root of each part.
$27x63=3x2\sqrt[3]{27x^6}=3x^2$
The numerator 2 means square the result.
$3x^2)^2=9x^4$
Therefore $(27x6)23=9x4(27x^6)^{\frac{2}{3}}=9x^4$ .

Rewriting everything with the same base

You will often be asked to find an unknown index. The strategy is to rewrite both sides using the same base, then compare the indices.

The key laws are:

$ap×aq=ap+qa^p \times a^q=a^{p+q}$
$a^p)^q=a^{pq}$
$ap÷aq=ap−qa^p \div a^q=a^{p-q}$

Example

Finding an unknown index

Suppose $\times \sqrt{32}=2^n$ . First rewrite 32 as a power of 2: $32=2^5$ .
Convert the square root into a half power.
$32=(25)12=252\sqrt{32}=(2^5)^{\frac{1}{2}}=2^{\frac{5}{2}}$
Rewrite the full left-hand side using powers of 2.
$\times \sqrt{32}=2^1 \times 2^{\frac{5}{2}}$
Add the indices because the bases are the same.
$21×252=2722^1 \times 2^{\frac{5}{2}}=2^{\frac{7}{2}}$
Since $\times \sqrt{32}=2^n$ , we have $n=72n=\frac{7}{2}$ .

Example

Expressing powers in terms of letters

Let $r=5^p$ and $s=5^q$ . Then $5p+q=5p×5q=rs5^{p+q}=5^p \times 5^q=rs$ .
For $5^{2p}$ , use the power of a power rule: $5^{2p}=(5^p)^2=r^2$ .
For $5^{q-1}$ , split the index: $5q−1=5q÷5=s55^{q-1}=5^q \div 5=\frac{s}{5}$ .

Large roots written as products

Sometimes a root looks awkward, but the product has been chosen to make a perfect power.

Example

Evaluating a fourth root product

Consider $5×125×1084\sqrt[4]{5 \times 125 \times 10^8}$ .
Multiply the first two factors: $\times 125=625=5^4$ .
Rewrite the power of 10 as a fourth power: $10^8=(10^2)^4$ .
Combine the factors into one fourth power.
$54×(102)4=(5×102)45^4 \times (10^2)^4=(5 \times 10^2)^4$
Taking the fourth root gives $\times 10^2=500$ .

Exam technique

In the exam

Check whether the index is negative first; if it is, plan for a reciprocal.
For fractional indices, read the denominator as the root and the numerator as the power.
If an equation has powers, try to rewrite everything using the same base before comparing indices.

Self review

Check yourself

Can you explain why $a^{-1}$ means $1a\frac{1}{a}$ rather than $- a$ ?
What root is shown by the denominator in $a23a^{\frac{2}{3}}$ ?
How would you start simplifying $(916)−32\left(\frac{9}{16}\right)^{-\frac{3}{2}}$ ?

Fractional and Negative Indices

What you'll learn

The basics: powers, roots and reciprocals

Negative indices

Fractional indices: roots first

Negative fractional indices

Fraction bases

Algebraic expressions with fractional indices

Rewriting everything with the same base

Large roots written as products

Recap questions

You've reached the end

Fractional and Negative Indices

What you'll learn

The basics: powers, roots and reciprocals

Negative indices

Fractional indices: roots first

Negative fractional indices

Fraction bases

Algebraic expressions with fractional indices

Rewriting everything with the same base

Large roots written as products

Recap questions

You've reached the end

Fractional and Negative Indices

What you'll learn

The basics: powers, roots and reciprocals

Negative indices

Fractional indices: roots first

Negative fractional indices

Fraction bases

Algebraic expressions with fractional indices

Rewriting everything with the same base

Large roots written as products

Recap questions

You've reached the end

Fractional and Negative Indices Revision Guide

Fractional and Negative Indices

What you'll learn

The basics: powers, roots and reciprocals

Negative indices

Fractional indices: roots first

Negative fractional indices

Fraction bases

Algebraic expressions with fractional indices

Rewriting everything with the same base

Large roots written as products

Recap questions

You've reached the end

Fractional and Negative Indices Revision Guide