N7 - Fractional indices (Higher) | AQA GCSE Mathematics

A fraction in an index combines a root with a power. You will read each part of the index, choose a short order of calculation and handle negative fractional indices without a calculator. The same index laws connect all these forms.

Unit fractions as root instructions (Higher)

What does $81^{\frac34}$ mean, and can you work it out without a decimal approximation? Its denominator gives the first clue: we need a fourth root.

Begin with the simpler index $\frac12$ . The index law says

a^{\frac12}\times a^{\frac12}=a^1=a

So the positive number $a^{\frac12}$ squares to $a$ . That is exactly what $\sqrt a$ does.

You already know

For the same base, multiplication adds indices and a power of a power multiplies indices. A root undoes its matching power, for example $\sqrt[3]{64}=4$ .

Three copies of $a^{\frac13}$ give $a^{\frac13+\frac13+\frac13}=a$ , so the same reasoning makes $\frac13$ a cube-root instruction. The denominator names the root; it is not a number to divide the base by.

Unit fractional indices

For $a>0$ and a positive integer $n$ ,

a^{\frac1n}=\sqrt[n]{a}

In particular, $a^{\frac12}=\sqrt a$ and $a^{\frac13}=\sqrt[3]{a}$ . With an even root, use its non-negative value.

Not on the formulae sheet: learn it.

Use exact power facts before reaching for a calculator. An awkward-looking index often asks for a very familiar number.

Worked example 1: a half index

Question. Work out $64^{\frac12}$ . Non-calculator.

Plan. Read the denominator $2$ as a square-root instruction.

\begin{aligned} 64^{\frac12}&=\sqrt{64} && \text{half means square root} \\ &=8 && 8^2=64 \end{aligned}

Why a root rather than half of $64$ ? The fraction is an index, so its meaning must fit the index laws.

Answer. $8$ .

Check. $8^2=64$ , whereas $32^2\ne64$ .

Change only the denominator from $2$ to $3$ . The base stays $64$ , but we must now undo cubing.

Worked example 2: a third index

Question. Work out $64^{\frac13}$ . Non-calculator.

Plan. Find the cube root of $64$ .

\begin{aligned} 64^{\frac13}&=\sqrt[3]{64} && \text{third means cube root} \\ &=4 && 4^3=64 \end{aligned}

Why does the answer change? A square root asks for two equal factors; a cube root asks for three.

Answer. $4$ .

Check. $4\times4\times4=64$ .

The same idea gives $81^{\frac14}=3$ , because four factors of $3$ make $81$ , and $32^{\frac15}=2$ .

Watch out

Wrong: $64^{\frac13}=64\div3$ . Right: $64^{\frac13}=4$ . The index law requires three copies of the answer to multiply to $64$ .

A new base should change the number fact, while the root instruction stays the same.

Combining a root with a power (Higher)

In an index such as $\frac34$ , the denominator $4$ asks for a fourth root and the numerator $3$ asks for a cube.

Taking the root first usually keeps the numbers small. For $81^{\frac34}$ , take the fourth root $3$ , then cube it. Cubing $81$ first would reach the same result, but via a much larger number.

General positive fractional indices

For $a>0$ and positive integers $m$ , $n$ ,

a^{\frac mn}=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}

The numerator gives the power; the denominator gives the root. Equivalent fractions give the same index.

Not on the formulae sheet: learn it.

This follows from the power-of-a-power law: $(a^{\frac1n})^m=a^{\frac mn}$ . Reduce an index fraction when that makes the root easier, for example $64^{\frac26}=64^{\frac13}=4$ .

Worked example 3: a root before a power

Question. Work out $81^{\frac34}$ . Non-calculator.

Plan. Take the fourth root first, then cube the result.

\begin{aligned} 81^{\frac34}&=\left(\sqrt[4]{81}\right)^3 && \text{root }4\text{, power }3 \\ &=3^3 && 3^4=81 \\ &=27 && \text{cube }3 \end{aligned}

Why root first? It replaces $81$ by $3$ before the cubing, so every calculation stays small.

Answer. $27$ .

Check. $27^4=531\,441=81^3$ , as the original fractional power requires.

Change only the numerator from $3$ to $5$ . The fourth root stays $3$ , but now raise it to the fifth power.

Worked example 4: an index greater than one

Question. Work out $81^{\frac54}$ . Non-calculator.

Plan. Use the same fourth root, then take the fifth power.

\begin{aligned} 81^{\frac54}&=\left(\sqrt[4]{81}\right)^5 && \text{root }4\text{, power }5 \\ &=3^5 && \text{fourth root is }3 \\ &=243 && \text{five factors of }3 \end{aligned}

Why is the answer now larger than $81$ ? The base is greater than $1$ , and the index $\frac54$ is greater than $1$ .

Answer. $243$ .

Check. $\frac54=1+\frac14$ , so $81^{\frac54}=81\times3=243$ .

A neat connection goes both ways: $16^{\frac34}=8$ , and $8^{\frac43}=16$ . The root and power instructions reverse each other because $\frac34\times\frac43=1$ .

Watch out

Wrong: $81^{\frac34}=(\sqrt[3]{81})^4$ . That swaps the jobs of numerator and denominator. Right: the bottom number names the root, so $81^{\frac34}=(\sqrt[4]{81})^3=27$ .

Keep the root-first order with a fifth-root number. You should never need to write the enormous power $32^3$ .

Fractional bases and negative indices (Higher)

A negative fractional index adds one more instruction: take the reciprocal of the positive power.

Keep its three jobs separate. In $-\frac32$ , the $2$ means square root, the $3$ means cube and the minus sign means reciprocal. None of those instructions makes a positive base produce a negative answer.

Negative fractional indices

For $a>0$ and positive integers $m$ , $n$ ,

a^{-\frac mn}=\frac1{a^{\frac mn}} =\left(\frac1a\right)^{\frac mn}

For a positive fraction $\frac pq$ , taking the reciprocal swaps $p$ and $q$ . You may take that reciprocal before or after the root and power.

Not on the formulae sheet: learn it.

Roots of fractions work on numerator and denominator: $\sqrt{\frac{25}{36}}=\frac56$ . Choose the order that makes the arithmetic easiest, while retaining every instruction.

Worked example 5: a positive index on a fraction

Question. Work out $\left(\frac{25}{36}\right)^{\frac32}$ . Non-calculator.

Plan. Take the square root of the fraction, then cube.

\begin{aligned} \left(\frac{25}{36}\right)^{\frac32} &=\left(\frac56\right)^3 && \text{square root top and bottom} \\ &=\frac{5^3}{6^3} && \text{cube top and bottom} \\ &=\frac{125}{216} && \text{evaluate} \end{aligned}

Why keep a fraction? It preserves the exact value, whereas rounding the root to a decimal would lose accuracy.

Answer. $\frac{125}{216}$ .

Check. The base is between $0$ and $1$ . Raising it to a power greater than $1$ should make it smaller; $\frac{125}{216}<\frac{25}{36}$ .

Change only the sign of the index. The positive-power calculation stays useful; the final value must be its reciprocal.

Worked example 6: adding the reciprocal instruction

Question. Work out $\left(\frac{25}{36}\right)^{-\frac32}$ . Non-calculator.

Plan. Turn the base upside down first, then take its square root and cube.

\begin{aligned} \left(\frac{25}{36}\right)^{-\frac32} &=\left(\frac{36}{25}\right)^{\frac32} && \text{reciprocal removes minus} \\ &=\left(\frac65\right)^3 && \text{square root} \\ &=\frac{216}{125} && \text{cube} \end{aligned}

Why take the reciprocal first? It removes the minus sign from the index without changing its square-root and cubing instructions.

Answer. $\frac{216}{125}$ .

Check. $\frac{216}{125}\times\frac{125}{216}=1$ , so the two worked examples are reciprocals.

For a mixed calculation, evaluate each power before combining them. For instance,

27^{\frac23}\times9^{-\frac12} =3^2\times\frac13 =3

All three instructions may appear in one short AQA “Work out” question. Showing an exact intermediate fraction helps preserve credit if a later calculation slips.

Watch out

Wrong: $\left(\frac{25}{36}\right)^{-\frac32}=-\frac{125}{216}$ . Right: the negative index takes a reciprocal, giving $\frac{216}{125}$ . A minus sign in the index is not a minus sign in front of the value.

The opening expression is now settled: $81^{\frac34}=27$ , and its negative-index partner is $81^{-\frac34}=\frac1{27}$ . Neither needs a decimal approximation.

Quick Recap

The denominator of a fractional index tells you the root.
The numerator tells you the power; taking the root first often keeps numbers small.
A negative index takes the reciprocal. It does not make a positive base negative.
Index laws still apply: equivalent fractions give equivalent indices, and a power of a power multiplies indices.

The opening value is $81^{\frac34}=3^3=27$ .

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N7 - Fractional indices (Higher)

Unit fractions as root instructions (Higher)

Combining a root with a power (Higher)

Fractional bases and negative indices (Higher)

Quick Recap

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