Negative and fractional indices — Mathematics

Before this

Index laws
Squares and square roots · coming soon
Standard form

This leads to

What is a logarithm
Exponential functions · coming soon
The power rule in calculus · coming soon

Multiply 2 by itself minus three times

Here are two calculations that break the definition you built in the last lecture.

Type $2^{- 3}$ into a calculator. It says 0.125. Now type $9^{1/2}$ . It says 3.

The definition says that $a^{n}$ means $n$ factors of $a$ multiplied together. So $2^{- 3}$ would mean "minus three factors of 2". You cannot write down minus three factors of anything. And $9^{1/2}$ would mean "half a factor of 9", which is just as meaningless. The definition counts factors, and you cannot count to $- 3$ or to $\frac{1}{2}$ .

So the calculator is using meanings that the definition never gave. Where do they come from? The last lecture ended with exactly this question. It found that $a^{0}$ has to be 1, because that is the only value that keeps the index laws true. This lecture asks the same question twice more. What must a negative index mean? What must a fractional index mean? Each time, the laws leave only one possible answer.

Before you read on, make a guess.

Predict — what is 2⁻³?

Choose the value you think 2⁻³ has. Then read the explanation. The rest of the lecture shows where the answer comes from.

The four answers students most often give for $2^{- 3}$ , and the right one.

$- 8$ . Not right. The minus sign is in the index, not in front of the number. It does not make the answer negative.
$- 6$ . Not right. This is $2 \times (- 3)$ . An index never means "multiply the base by the index".
$- \frac{1}{8}$ . Not right. The $\frac{1}{8}$ part is correct, but the answer is positive.
$\frac{1}{8}$ . Right. A negative index means "one over" the positive power: $2^{- 3} = \frac{1}{2 ^{3}} = \frac{1}{8} = 0.125$ .

The case where the factors ran out

The last lecture promised to come back to $a^{2} \div a^{5}$ . It is the case where the bottom has more factors than the top. Work it out in two ways, exactly as for $a^{0}$ . Take $a$ to be any number except 0.

By cancelling: two pairs of $a$ 's cancel. Three factors are left underneath and nothing is left on top except 1. So $a^{2} \div a^{5} = \frac{1}{a \times a \times a} = \frac{1}{a ^{3}}$ .
By the law $a^{m} \div a^{n} = a^{m - n}$ : $a^{2} \div a^{5} = a^{2 - 5} = a^{- 3}$ .

Both lines describe the same calculation, so both answers must be equal. This means that

a^{- 3} = \frac{1}{a ^{3}}

If $a^{- 3}$ meant anything else, the second law would give the wrong answer for $a^{2} \div a^{5}$ . So once we decide that the laws should keep working, there is no choice left.

Test it with the first law. It says $a^{- 3} \times a^{3} = a^{- 3 + 3} = a^{0} = 1$ . Two numbers that multiply to give 1 are reciprocals of each other. So $a^{- 3}$ must be the reciprocal of $a^{3}$ , which is $\frac{1}{a ^{3}}$ . The first law gives the same answer.

Nothing here depended on the number 3. In general:

Negative indices

For any number $a$ except 0, and any positive whole number $n$ :

$a^{- n} = \frac{1}{a ^{n}}$

A negative index means the reciprocal of the positive power. It is forced by the law $a^{m} \div a^{n} = a^{m - n}$ when $n$ is bigger than $m$ .

So $2^{- 3} = \frac{1}{8}$ , $5^{- 1} = \frac{1}{5}$ , $1 0^{- 2} = \frac{1}{100}$ and $x^{- 4} = \frac{1}{x ^{4}}$ . An index of $- 1$ on its own simply means "one over": $a^{- 1} = \frac{1}{a}$ .

There is a second way to see this, and it is worth holding on to. The last lecture made a table by dividing by the base each time the index went down by 1. Keep going past $2^{0}$ , and the same rule carries you below zero.

The values of

2^{3}

down to

2^{- 3}

. Each time the index goes down by 1, there is one factor of 2 fewer, so the value is divided by 2. Past

2^{0} = 1

, the halving simply carries on:

\frac{1}{2}

\frac{1}{4}

\frac{1}{8}

. Every bar is positive.

The pattern and the law agree: $2^{- 1} = \frac{1}{2}$ , $2^{- 2} = \frac{1}{4}$ and $2^{- 3} = \frac{1}{8}$ . Notice that every value in the picture is positive. A negative index makes a number small, never negative.

The standard form lecture used this pattern with tens. It wrote $1 0^{- 3}$ for one thousandth, and promised that a later lecture would explain negative powers fully. Here is that explanation. $1 0^{- 3} = \frac{1}{1 0 ^{3}} = \frac{1}{1000}$ , because the law $a^{m} \div a^{n} = a^{m - n}$ allows no other value.

The theorem also works backwards, and this is used all the time in algebra. Because $a^{- n} = \frac{1}{a ^{n}}$ , a power underneath a fraction line can be written on top with a negative index: $\frac{1}{x ^{2}} = x^{- 2}$ . And dividing by a negative power is the same as multiplying by the positive one: $\frac{1}{a ^{- n}} = a^{n}$ , because the reciprocal of a reciprocal is the number you started with.

Stop and think

What is $(\frac{1}{2})^{- 3}$ ? And what is $(\frac{2}{3})^{- 2}$ ?

Show the answer

$(\frac{1}{2})^{- 3}$ is one over $(\frac{1}{2})^{3} = \frac{1}{8}$ , and one over $\frac{1}{8}$ is 8. So a negative index on a fraction flips the fraction: $(\frac{1}{2})^{- 3} = 2^{3} = 8$ . In the same way, $(\frac{2}{3})^{- 2} = (\frac{3}{2})^{2} = \frac{9}{4}$ . The reason is the same each time: the reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$ . This is why a negative index can make the answer bigger. It all depends on whether the base is bigger or smaller than 1.

What must a half power mean?

Now the second question from the hook. The last lecture ended by asking what $a^{1/2}$ must be "if $a^{1/2} \times a^{1/2}$ is to equal $a^{1}$ ". Use the first law, which adds the indices:

a^{1/2} \times a^{1/2} = a^{\frac{1}{2} + \frac{1}{2}} = a^{1} = a

So $a^{1/2}$ is a number that gives $a$ when you multiply it by itself. That is exactly what a square root is. For $9^{1/2}$ , the question becomes: which number times itself gives 9? The answer is 3.

But there is a second answer: $(- 3) \times (- 3) = 9$ as well. The laws cannot choose between 3 and $- 3$ . So mathematicians make one agreement. For a positive base, $a^{1/2}$ means the positive square root, $a$ . This is the same agreement the √ sign already follows. So $9^{1/2} = 9 = 3$ , exactly as the calculator said.

The third law, which multiplies indices, gives the same answer from another direction: $(a^{1/2})^{2} = a^{\frac{1}{2} \times 2} = a^{1} = a$ . The square of $a^{1/2}$ is $a$ . Two different laws, one meaning. That is a good sign that the meaning is right.

Thirds work the same way. Three equal factors of $a^{1/3}$ must add to an index of 1:

a^{1/3} \times a^{1/3} \times a^{1/3} = a^{1} = a

So $a^{1/3}$ is the number that gives $a$ when it is cubed. That is the cube root, $3 a$ . For example, $8^{1/3} = 38 = 2$ , because $2^{3} = 8$ . In the same way, $a^{1/ n}$ is the number whose $n$ th power is $a$ . This is called the $n$ th root of $a$ , written $n a$ .

Fractional indices

For a positive number $a$ and a positive whole number $n$ :

$a^{1/ n} = n a$ , the positive number whose $n$ th power is $a$ .

It is forced by the laws $a^{m} \times a^{n} = a^{m + n}$ and $(a^{m})^{n} = a^{mn}$ .

So $1 6^{1/4} = 2$ , because $2^{4} = 16$ . And $3 2^{1/5} = 2$ , because $2^{5} = 32$ . Notice what the denominator of the index does. It tells you which root to take.

Why does the theorem say "positive"? Square roots of negative numbers are not real numbers, so $(- 9)^{1/2}$ has no value at this level. Odd roots of negative numbers do exist: $3 - 8 = - 2$ . But fractional indices with negative bases cause a real problem. The index $\frac{1}{3}$ is the same number as $\frac{2}{6}$ . Yet $(- 8)^{1/3} = - 2$ , while working $(- 8)^{2/6}$ as "square, then take the sixth root" gives $6 4^{1/6} = 2$ . One index, two different answers. So at this level, fractional indices are used only with positive bases. Then every route gives the same answer.

Two parts of one fraction

What about $8^{2/3}$ , where the top of the fraction is not 1? Write the index as $\frac{1}{3} \times 2$ , and use the third law, which multiplies indices:

8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4

Multiplication can be done in either order, so the index is also $2 \times \frac{1}{3}$ . That gives a second route:

8^{2/3} = (8^{2})^{1/3} = 6 4^{1/3} = 4

Both routes give 4. They must, because $\frac{1}{3} \times 2$ and $2 \times \frac{1}{3}$ are the same number. So a fractional index has two jobs. The denominator is a root. The numerator is a power. You may do them in either order.

Indices of the form m/n

For a positive number $a$ and whole numbers $m$ and $n$ , with $n$ positive:

$a^{m / n} = (n a)^{m} = n a^{m}$

The denominator gives the root. The numerator gives the power.

In practice, take the root first. It keeps the numbers small. For $8^{2/3}$ , "root first" needs only the cube root of 8 and then $2^{2}$ . "Power first" needs $8^{2} = 64$ and then the cube root of 64. Both work, but the second one is harder to do in your head.

Stop and think

Are $8^{2/3}$ and $8^{3/2}$ the same number?

Show the answer

No. $8^{2/3}$ is the cube root of 8, squared: $2^{2} = 4$ . $8^{3/2}$ is the square root of 8, cubed: $(8)^{3} = 88 = 162 \approx 22.6$ . The denominator always says which root to take, so swapping the numerator and the denominator changes the question. A quick check catches the mistake: $\frac{2}{3}$ is less than 1, so $8^{2/3}$ must be less than $8^{1} = 8$ . $\frac{3}{2}$ is more than 1, so $8^{3/2}$ must be more than 8.

Do the new meanings fit together?

Each new meaning was forced by the laws. But do they also agree with each other? Here is a strong test. Draw the graph of $y = 2^{x}$ . Plot the whole-number values first, which the old definition already gave. Then add the negative values and the fractional values, which this lecture gave. If the new meanings fit, every point should lie on one smooth curve, with no jumps. Try it on the graph below.

Graph builder — filling in y = 2ˣ

Move x. Each value you visit is plotted as a point, and its exact value is shown. Start with the whole numbers. Then try negative values, then halves and quarters. When you have plotted several, show the curve.

x = 1

Plot all:

The graph of $y = 2^{x}$ from $x = - 3$ to $x = 3$ . The dark points come from the old definition. The teal points come from negative and fractional indices. They all lie on one smooth curve.

$x$	2ˣ, exactly	Value
$- 3$	1 ÷ 2³	0.125
$- \frac{3}{2}$	1 ÷ (√2)³	≈ 0.354
$- 1$	1 ÷ 2	0.5
$\frac{1}{2}$	√2	≈ 1.414
$\frac{3}{4}$	(⁴√2)³	≈ 1.682
$\frac{3}{2}$	(√2)³	≈ 2.828
$3$	2³	8

What should you have seen? The whole-number points rise slowly on the left and steeply on the right. The negative values slide down towards the $x$ -axis but never reach it, because $\frac{1}{8}$ , $\frac{1}{16}$ and the rest are always positive. The halves and quarters fall exactly between their neighbours, on the same smooth curve. Nobody designed the meanings to make that happen. The laws forced each meaning separately, and the meanings still fit together perfectly.

That is it. That is the whole idea. Every meaning in this lecture comes from one demand: the index laws must keep working. Nothing new needs to be memorised.

Using the new meanings

Worked Example 1 — Three jobs in one index

Evaluate $2 7^{- 2/3}$ .

The index $- \frac{2}{3}$ has three parts. The denominator 3 is a cube root. The numerator 2 is a square. The minus sign means "one over". Do them one at a time, root first.

Cube root: $327 = 3$ , because $3^{3} = 27$ . Square: $3^{2} = 9$ . So $2 7^{2/3} = 9$ . Then the minus sign: $2 7^{- 2/3} = \frac{1}{9}$ .

Check: cube the answer and compare with $2 7^{- 2}$ , using the third law. $(\frac{1}{9})^{3} = \frac{1}{729}$ , and $2 7^{- 2} = \frac{1}{2 7 ^{2}} = \frac{1}{729}$ . ✓

Worked Example 2 — A negative index on a fraction

Evaluate $(\frac{16}{81})^{- 3/4}$ .

Deal with the minus sign first, because it flips the fraction: $(\frac{16}{81})^{- 3/4} = (\frac{81}{16})^{3/4}$ .

The index applies to the top and the bottom separately. Fourth roots: $481 = 3$ and $416 = 2$ . Then cube: $(\frac{3}{2})^{3} = \frac{27}{8}$ .

Answer: $\frac{27}{8} = 3.375$ . Check: the base $\frac{16}{81}$ is less than 1 and the index is negative, so the answer should be bigger than 1. It is. ✓

Worked Example 3 — A fractional power of a bracket

Simplify $(16 x^{4})^{3/4}$ , where $x$ is positive.

The bracket is a product, so the index applies to each factor, as the last lecture showed: $1 6^{3/4} \times (x^{4})^{3/4}$ .

For the number, root first: $416 = 2$ , and $2^{3} = 8$ . For the letter, a power of a power multiplies the indices: $(x^{4})^{3/4} = x^{4 \times \frac{3}{4}} = x^{3}$ .

Answer: $8 x^{3}$ . Check with $x = 2$ : the bracket is $16 \times 16 = 256$ , and $25 6^{3/4} = (4256)^{3} = 4^{3} = 64$ . The answer gives $8 \times 2^{3} = 64$ . ✓

Worked Example 4 — Writing an expression as one power

Write $\frac{x x}{x ^{3}}$ as a single power of $x$ , where $x$ is positive.

First turn the root into an index: $x = x^{1/2}$ . On top, $x \times x^{1/2} = x^{1 + \frac{1}{2}} = x^{3/2}$ . Then divide: $x^{3/2} \div x^{3} = x^{\frac{3}{2} - 3} = x^{- 3/2}$ .

Answer: $x^{- 3/2}$ . Check with $x = 4$ : the expression is $\frac{4 \times 2}{64} = \frac{1}{8}$ , and $4^{- 3/2} = \frac{1}{( 4 ) ^{3}} = \frac{1}{8}$ . ✓

Why would anyone want this form? In calculus, the rule for differentiating works on single powers $x^{n}$ . It works for $x^{- 3/2}$ just as well as for $x^{2}$ , but it cannot be used on $\frac{x x}{x ^{3}}$ as it stands.

Worked Example 5 — Finding an unknown index

Solve $4^{x} = \frac{1}{32}$ .

The unknown is in the index. Write both sides as powers of one base, 2. On the left, $4^{x} = (2^{2})^{x} = 2^{2 x}$ . On the right, $\frac{1}{32} = \frac{1}{2 ^{5}} = 2^{- 5}$ .

So $2^{2 x} = 2^{- 5}$ . Two powers of 2 are equal only when their indices are equal. So $2 x = - 5$ , and $x = - \frac{5}{2}$ .

Check: $4^{- 5/2} = \frac{1}{( 4 ) ^{5}} = \frac{1}{2 ^{5}} = \frac{1}{32}$ . ✓

This worked because 4 and 32 are both powers of 2. An equation such as $2^{x} = 10$ cannot be solved this way. Logarithms, in the next lecture, are the tool for that.

Three classic mistakes

Trap 1 — Reading the minus sign as a negative answer

$2^{- 3}$ is not $- 8$ , and it is not $- 6$ . It is $\frac{1}{8}$ . A negative index makes a reciprocal. For any positive base, every power is positive, whatever the index. The bars in the figure above never go below the axis.

Trap 2 — Moving more than the index belongs to

$3 x^{- 2}$ is $\frac{3}{x ^{2}}$ , not $\frac{1}{3 x ^{2}}$ . The index belongs only to the $x$ directly under it, so only the $x^{2}$ moves under the line. With a bracket the whole bracket moves: $(3 x)^{- 2} = \frac{1}{9 x ^{2}}$ . Check with $x = 1$ : $3 \times 1^{- 2} = 3$ , but $(3 \times 1)^{- 2} = \frac{1}{9}$ .

Trap 3 — Reading a fractional index as division

$1 6^{1/2}$ is not $16 \div 2 = 8$ . It is $16 = 4$ . And in $a^{m / n}$ , the denominator is the root, not the power: $8^{2/3} = 4$ , but $8^{3/2} \approx 22.6$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Evaluate: (a) $5^{- 2}$ (b) $(\frac{2}{3})^{- 3}$ (c) $6 4^{1/3}$ (d) $8 1^{- 1/4}$

Hint 1
A minus sign in the index means "one over". The denominator of a fractional index tells you which root to take.

Hint 2
(b) Flip the fraction first. (d) Find the fourth root of 81 first: which number to the power 4 gives 81?

Full answer
(a) $5^{- 2} = \frac{1}{25}$ . (b) $(\frac{2}{3})^{- 3} = (\frac{3}{2})^{3} = \frac{27}{8}$ . (c) $6 4^{1/3} = 4$ , because $4^{3} = 64$ . (d) $481 = 3$ , so $8 1^{1/4} = 3$ and $8 1^{- 1/4} = \frac{1}{3}$ . Check (d): $(\frac{1}{3})^{4} = \frac{1}{81} = 8 1^{- 1}$ . ✓
Evaluate: (a) $12 5^{- 2/3}$ (b) $0.2 5^{- 1/2}$ (c) $(\frac{27}{8})^{2/3}$

Hint 1
Take the root first, then the power, then deal with any minus sign.

Hint 2
(b) Write 0.25 as a fraction, $\frac{1}{4}$ . The minus sign flips it to 4.

Full answer
(a) $3125 = 5$ , $5^{2} = 25$ , so $12 5^{- 2/3} = \frac{1}{25}$ . (b) $0.2 5^{- 1/2} = (\frac{1}{4})^{- 1/2} = 4^{1/2} = 2$ . (c) $327 = 3$ and $38 = 2$ , so the answer is $(\frac{3}{2})^{2} = \frac{9}{4}$ . Check (b): $2^{- 2} = \frac{1}{4} = 0.25$ , which is the base. ✓
Take $x$ to be positive. (a) Write $\frac{x}{x ^{2}}$ as a single power of $x$ . (b) Simplify $(9 x^{6})^{- 1/2}$ . (c) Solve $9^{x} = \frac{1}{27}$ .

Hint 1
(a) and (b): change every root into an index, then use the laws. (c): write both sides as powers of 3.

Hint 2
(b) The index $- \frac{1}{2}$ applies to the 9 and to the $x^{6}$ . (c) $9^{x} = 3^{2 x}$ and $\frac{1}{27} = 3^{- 3}$ .

Full answer
(a) $x^{1/2} \div x^{2} = x^{\frac{1}{2} - 2} = x^{- 3/2}$ . (b) $9^{- 1/2} = \frac{1}{3}$ and $(x^{6})^{- 1/2} = x^{- 3}$ , so the answer is $\frac{1}{3} x^{- 3} = \frac{1}{3 x ^{3}}$ . (c) $3^{2 x} = 3^{- 3}$ , so $2 x = - 3$ and $x = - \frac{3}{2}$ . Check (b) with $x = 1$ : $9^{- 1/2} = \frac{1}{3}$ . ✓ Check (c): $9^{- 3/2} = \frac{1}{( 9 ) ^{3}} = \frac{1}{27}$ . ✓
In 1619 Johannes Kepler found a rule for the planets. If a planet's average distance from the Sun is $r$ (in astronomical units, where Earth's distance is 1), then the time it takes to go round the Sun is $T = r^{3/2}$ years. (a) Mars has $r \approx 1.524$ (NASA). Find $T$ for Mars, in years and in days. (b) Rearrange the rule to give $r$ in terms of $T$ . (c) An imagined asteroid takes 8 years to go round the Sun. How far is it from the Sun, on average?

Hint 1
(a) $r^{3/2}$ is the square root of $r$ , cubed. Use a calculator. One year is about 365.26 days.

Hint 2
(b) Raise both sides to the power $\frac{2}{3}$ . Then use the third law on the right: $\frac{3}{2} \times \frac{2}{3} = 1$ .

Full answer
(a) $T = 1.52 4^{3/2} = (1.524)^{3} \approx 1.881$ years, which is about $1.881 \times 365.26 \approx 687$ days. NASA gives 686.98 days, so the rule fits well. (b) $T^{2/3} = (r^{3/2})^{2/3} = r^{1}$ , so $r = T^{2/3}$ . (c) $r = 8^{2/3} = (38)^{2} = 2^{2} = 4$ astronomical units. Check: $4^{3/2} = (4)^{3} = 8$ years. ✓

The bigger idea

Look back at what happened in these two lectures. The definition of $a^{n}$ worked only for counting numbers. Each time it ran out, at zero, at negative numbers and at fractions, we asked one question: which meaning keeps the laws true? Each time there was only one answer, apart from one agreement, that roots are positive. So the meanings were not invented. They were found.

Mathematicians worked this way long before the rules were written in textbooks. In a letter of 13 June 1676, Isaac Newton described his results using powers with negative and fractional indices, such as $(1 - x^{2})^{3/2}$ . That is the notation you have just built.

There is one gap left. The graph of $y = 2^{x}$ now has points at every fraction, and they lie on a smooth curve. But what about $2^{2}$ , where the index is not a fraction at all? At Pre-University, this gap is filled in the same spirit, so that the curve has no holes. And there is a question going the other way. Given 10, which index $x$ makes $2^{x} = 10$ ? The graph says that $x$ is a little more than 3. The next lecture gives that index a name: a logarithm.

Key concepts in this lesson

A negative index means the reciprocal of the positive power, $a^{- n} = \frac{1}{a ^{n}}$ , because this is the only meaning that keeps $a^{m} \div a^{n} = a^{m - n}$ true when $n$ is bigger than $m$ .
A negative index makes a number small, never negative, and a negative index on a fraction flips the fraction.
For a positive base, $a^{1/ n}$ is the positive $n$ th root of $a$ , because $n$ equal factors of $a^{1/ n}$ must multiply to give $a^{1}$ .
In $a^{m / n}$ , the denominator $n$ is a root and the numerator $m$ is a power, and taking the root first keeps the numbers small.
The new meanings fit together, so the points of $y = 2^{x}$ at whole, negative and fractional values all lie on one smooth curve.