All notesMathTalk
CAMBRIDGE O LEVEL · 2025–2027

Indices I.

A few consistent rules connect positive, zero, negative and fractional powers.

Syllabus 1.7Visual notesWorked examples
BY THE END

Interpret every type of index · evaluate fractional powers · apply the laws of indices

01

Extend the pattern beyond positive powers

Start with a pattern you understand: 2³ = 8 and 2² = 4. Each time the index decreases by 1, the value divides by 2. Continue: 2¹ = 2, so the next value must be 2⁰ = 1.

Continuing the same pattern gives 2⁻¹ = 12 and 2⁻² = 14. A negative index therefore tells us to use a reciprocal; it does not make the number negative. This pattern explains the rules before we apply them to unfamiliar bases.

An index can be positive, zero, negative or fractional. Each extension keeps the rules of powers consistent.

238
224
212
201
2−112
2−214
2−318

Each step to the right lowers the index by 1 and divides the value by 2.

Zero index

a⁰ = 1   (a ≠ 0)

For example, 7⁰ = 1 and (−3)⁰ = 1.

Negative index

a−n = 1an   (a ≠ 0)

7⁻² = 17² = 149. A negative index means reciprocal, not a negative value.

Pause and explainWhat is the difference between 3⁻¹ and −3?

3⁻¹ is the reciprocal 13, which is positive. −3 is a negative number.

02

Fractional indices are roots

If an index of 2 means squaring, an index of 12 should undo squaring. This connects fractional indices to roots. For positive numbers, raising to 12 is another way to write a square root.

In an index such as 23, the denominator tells us to take a cube root and the numerator tells us to square. Work with the root first when that creates a smaller exact number. A negative sign in the index adds a reciprocal after those steps.

a1n = ⁿ√a   ·   amn = (ⁿ√a)m

For a positive base, the denominator tells you the root; the numerator tells you the power. Taking the root first often keeps numbers small.

WORKED EXAMPLE

Evaluate 8112, 1614 and 8−23.

  1. 8112 = √81 = 9.
  2. 1614 = ∜16 = 2.
  3. 8−23 = 1(∛8)² = 12² = 14.
WORKED EXAMPLE

Evaluate (278)23.

(278)23 = (32)² = 94

Take the cube root of the numerator and denominator, then square.

Pause and explainHow would you calculate 2723 without large numbers?

Take the cube root of 27 to get 3, then square 3 to get 9.

03

Use the rules with the right structure

Write 2³ × 2² as repeated factors: three twos followed by two more twos. There are five twos altogether, which explains why the indices add. The base stays 2 because the repeated factor has not changed.

Division cancels equal factors and therefore subtracts indices. Raising a power to another power repeats a group of factors, so it multiplies indices. These are rules about multiplication and division; there is no matching rule that combines powers across addition.

Rules of indices, where the expressions are defined
StructureRuleExample
Same base, multiplyam × an = am+n2⁻³ × 2⁴ = 2¹ = 2
Same base, divideam ÷ an = am−n2³ ÷ 2⁴ = 2⁻¹ = 12
Power of a power(am)n = amn(2³)² = 2⁶ = 64
Power of a product(ab)n = anbn(2 × 3)² = 2² × 3² = 36
Power of a quotient(ab)n = anbn(23)² = 49
Pause and explainCan 2³ + 2² be replaced by 2⁵?

No. The values are 8 + 4 = 12, not 32. Adding indices applies to a product with the same base.

04

Your turn. Name the rule.

Read an index in stages: identify its sign, its numerator and its denominator. Decide whether the calculation involves a power, a root, a reciprocal, or a combination of them.

For a product or quotient, check that the bases match before combining indices. Expand one short example into repeated factors if you are unsure which rule applies. Keep exact fractions rather than rounding their decimals.

1

Evaluate 5⁰, 4⁻², 12513 and 1634.

Reveal worked answer
5⁰ = 1; 4⁻² = 116; 12513 = 5; 1634 = (∜16)³ = 2³ = 8.
2

Simplify 3⁻² × 3⁵ ÷ 3².

Reveal worked answer
Same base: add indices for multiplication, subtract for division. 3−2 + 5 − 2 = 3¹ = 3.
3

Find the value of (2⁻²)³ and 32−25.

Reveal worked answer
(2⁻²)³ = 2⁻⁶ = 164. 32−25 = 1(⁵√32)² = 12² = 14.
4

Explain the error in 4² × 4³ = 16⁵.

Reveal worked answer
The base stays 4; add only the indices. 4² × 4³ = 4⁵ = 1024.