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Cambridge IGCSE Mathematics 0580 · Number

Indices

What a power means, why $a^{0} = 1$ and $a^{- n} = \frac{1}{a^{n}}$ , the three rules of indices, and (Extended) fractional indices such as $8^{- \frac{2}{3}}$ . Every example here is worked without a calculator, because that is how Papers 1 and 2 will ask it.

C1.7E1.7Core and ExtendedFree lesson

What the syllabus says

C1.7 (Core)

1 Understand and use indices (positive, zero and negative integers).
2 Understand and use the rules of indices.

Notes and examples

e.g. find the value of $7^{- 2}$ .
e.g. find the value of $2^{- 3} \times 2^{4}$ , $(2^{3})^{2}$ , $2^{3} \div 2^{4}$ .

E1.7 (Extended)

1 Understand and use indices (positive, zero, negative, and fractional).
2 Understand and use the rules of indices.

Notes and examples

Examples include:
$6^{\frac{1}{2}} = \sqrt{6}$
$16^{\frac{1}{4}} = \sqrt[4]{16}$
find the value of $7^{- 2}$ , $81^{\frac{1}{2}}$ , $8^{- \frac{2}{3}}$ .
e.g. find the value of $2^{- 3} \times 2^{4}$ , $(2^{3})^{2}$ , $2^{3} \div 2^{4}$ .

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

What an index means

An index (also called a power or exponent) tells you how many times a number, the base, is multiplied by itself.

$5^{3} = 5 \times 5 \times 5 = 125$

Here 5 is the base and 3 is the index. You should know the squares from $1^{2}$ to $15^{2}$ and the cubes $1^{3}$ , $2^{3}$ , $3^{3}$ , $4^{3}$ , $5^{3}$ and $10^{3}$ by heart, because the non-calculator papers use them constantly.

Two special cases are worth saying out loud:

$a^{1} = a$ : one copy of the base is just the base.
$(- 2)^{2} = 4$ but $- 2^{2} = - 4$ . The index applies only to what it touches: in $- 2^{2}$ the square is worked out first and the minus sign is applied afterwards.

Zero and negative indices

Look at the powers of 2 going down one step at a time. Each step divides by 2:

$2^{3} = 8, 2^{2} = 4, 2^{1} = 2, 2^{0} = 1, 2^{- 1} = \frac{1}{2}, 2^{- 2} = \frac{1}{4}$

Keeping the pattern going is the reason for both definitions:

A zero index gives 1: $a^{0} = 1$ for any non-zero number $a$ . So $7^{0} = 1$ and $1000^{0} = 1$ .
A negative index means the reciprocal: $a^{- n} = \frac{1}{a^{n}}$ . So $7^{- 2} = \frac{1}{7^{2}} = \frac{1}{49}$ .

A negative index never makes the answer negative. $7^{- 2}$ is a small positive number, not $- 49$ and not $- 14$ .

For a fraction, the reciprocal turns it upside down: $(\frac{2}{3})^{- 2} = (\frac{3}{2})^{2} = \frac{9}{4}$ .

The rules of indices

When the base is the same, three rules let you combine powers without writing them all out:

Multiplying: add the indices. $a^{m} \times a^{n} = a^{m + n}$
Dividing: subtract the indices. $a^{m} \div a^{n} = a^{m - n}$
A power of a power: multiply the indices. $(a^{m})^{n} = a^{m n}$

Each rule is just counting. $2^{3} \times 2^{2}$ is three 2s multiplied by two more 2s, which is five 2s, so $2^{5}$ . And $(2^{3})^{2} = 2^{3} \times 2^{3} = 2^{6}$ .

The rules only work when the bases match. $2^{3} \times 3^{2}$ cannot be combined into one power: work out each part, $8 \times 9 = 72$ .

Worked example 1

Find the value of $2^{- 3} \times 2^{4}$ .

The bases are the same, so add the indices: $- 3 + 4 = 1$ .
$2^{- 3} \times 2^{4} = 2^{1} = 2$ .
Check another way: $2^{- 3} = \frac{1}{8}$ and $2^{4} = 16$ , and $\frac{1}{8} \times 16 = 2$ .

Answer: $2$

Worked example 2

Find the value of (a) $(2^{3})^{2}$ and (b) $2^{3} \div 2^{4}$ .

(a) A power of a power: multiply the indices, $3 \times 2 = 6$ . So $(2^{3})^{2} = 2^{6} = 64$ .
(b) Dividing: subtract the indices, $3 - 4 = - 1$ . So $2^{3} \div 2^{4} = 2^{- 1} = \frac{1}{2}$ .
A common slip in (a) is to add instead and get $2^{5} = 32$ . Adding is only for multiplying powers side by side.

Answer: (a) $64$ (b) $\frac{1}{2}$

Check yourself

Find the value of $5^{- 2}$ . Give your answer as a fraction.

Answer it in the portal, where your answer is checked as you type →

Check yourself

Write $3^{4} \times 3^{- 6}$ as a fraction.

Answer it in the portal, where your answer is checked as you type →

Fractional indices Extended

Extended candidates also need fractional indices. Start from the rule for a power of a power. If $(a^{\frac{1}{2}})^{2} = a^{\frac{1}{2} \times 2} = a^{1} = a$ , then $a^{\frac{1}{2}}$ is the number that squares to give $a$ . That is the square root.

$a^{\frac{1}{n}} = \sqrt[n]{a}$ , so $6^{\frac{1}{2}} = \sqrt{6}$ and $16^{\frac{1}{4}} = \sqrt[4]{16} = 2$ .
$a^{\frac{m}{n}} = (\sqrt[n]{a})^{m}$ : take the root first, then the power.
A negative fractional index does all three jobs: root, power, then reciprocal.

Take the root first. $27^{\frac{2}{3}}$ is $(\sqrt[3]{27})^{2} = 3^{2} = 9$ . Squaring first, $27^{2} = 729$ , and then finding $\sqrt[3]{729}$ gives the same 9, but with far harder numbers. On Paper 2 the root-first route is the one that works.

Worked example 3 Extended

Find the value of (a) $81^{\frac{1}{2}}$ and (b) $8^{- \frac{2}{3}}$ .

(a) A power of $\frac{1}{2}$ is the square root: $81^{\frac{1}{2}} = \sqrt{81} = 9$ .
(b) Deal with the three parts of the index in turn. The denominator 3 means cube root: $\sqrt[3]{8} = 2$ .
The numerator 2 means square: $2^{2} = 4$ , so $8^{\frac{2}{3}} = 4$ .
The minus sign means reciprocal: $8^{- \frac{2}{3}} = \frac{1}{4}$ .

Answer: (a) $9$ (b) $\frac{1}{4}$

Worked example 4 Extended

Find the value of $(\frac{27}{8})^{- \frac{1}{3}}$ .

The negative index means take the reciprocal: $(\frac{27}{8})^{- \frac{1}{3}} = (\frac{8}{27})^{\frac{1}{3}}$ .
A power of $\frac{1}{3}$ is the cube root of the top and of the bottom: $\frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}$ .

Answer: $\frac{2}{3}$

Check yourself Extended

Find the value of $125^{\frac{2}{3}}$ .

Answer it in the portal, where your answer is checked as you type →

Check yourself Extended

Find the value of $4^{- \frac{3}{2}}$ . Give your answer as a fraction.

Answer it in the portal, where your answer is checked as you type →

Key points

$a^{0} = 1$ and $a^{- n} = \frac{1}{a^{n}}$ . A negative index makes a reciprocal, never a negative number.
With the same base: multiply means add the indices, divide means subtract them, and a power of a power means multiply them.
Different bases cannot be combined into one power; work each one out.
Extended: $a^{\frac{m}{n}} = (\sqrt[n]{a})^{m}$ . Take the root first, then the power, then the reciprocal if the index is negative.

Where marks are lost

Writing $7^{- 2} = - 49$ or $- 14$ . The minus sign in an index means reciprocal.
Adding indices for a power of a power: $(2^{3})^{2}$ is $2^{6}$ , not $2^{5}$ .
Multiplying the bases as well as adding the indices: $3^{2} \times 3^{4} = 3^{6}$ , not $9^{6}$ .
Combining different bases: $2^{3} \times 3^{2}$ is not $6^{5}$ .
Leaving an answer as $2^{- 1}$ when the question says find the value: give $\frac{1}{2}$ .
Extended: applying the power before the root on a non-calculator paper, which turns an easy question into a very hard one.

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An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Indices

What an index means

Zero and negative indices

The rules of indices

Worked example 1

Worked example 2

Check yourself

Check yourself

Fractional indices Extended

Worked example 3 Extended

Worked example 4 Extended

Check yourself Extended

Check yourself Extended

Key points

Where marks are lost

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