1.02aPureFoundation

Laws of Indices

Q: Why is any non-zero number to the power zero equal to 1?

Because the division law leaves no alternative. a^na^n is 1, and the law rewrites it as a^n-n = a^0. Setting a^0 = 1 is the only value that keeps the laws consistent — and it is where the halving pattern in the powers table naturally lands.

Q: Is 0^0 equal to 1?

The law is stated for a 0, and 0^0 is left undefined at A-Level. The derivation divides a^n by itself, which is not allowed when a = 0.

OCR expects you to use the laws of indices for $all rational exponents$ — not just the whole-number ones. That phrase is doing a lot of work: it means $x^{0}$ , $x^{- 3}$ and $x^{2/3}$ all have to obey the same three rules that $x^{2} \times x^{5}$ does. This lesson builds every one of them out of a single table.

25 min Video by TLMaths Indices and Surds

The big picture

Almost nothing in A-Level Pure is written without indices. You will differentiate $x^{n}$ , integrate $x^{n}$ , expand $(1 + x)^{n}$ , solve $a^{x} = b$ with logarithms, and model growth and decay — and each of those assumes you can move fluently between $3 x^{2}$ , $\frac{1}{x ^{4}}$ and $x^{p / q}$ without thinking about it. Students who meet a $\frac{1}{x}$ in a calculus question and stall have almost always never rewritten it as $x^{- 1/2}$ . Index fluency is not a topic you finish; it is the notation the rest of the course is written in.

A-Level Maths B1-01: The Laws of IndicesCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Apply the multiplication, division and power-of-a-power laws to any rational index
Derive $a^{0} = 1$ and $a^{- n} = \frac{1}{a ^{n}}$ by continuing a pattern rather than memorising
Convert between root notation and fractional index notation in both directions
Simplify expressions where a coefficient sits inside a bracketed power

Start with a table, not a rule

Write out the powers of 2, working downwards:

$2^{4} = 16$ , $2^{3} = 8$ , $2^{2} = 4$ , $2^{1} = 2$ .

Every step down the list $halves$ the value — that is, divides by the base. There is nothing special about 2: powers of any base $a$ fall by a factor of $a$ each time the index drops by one. Keep that single observation in view. Almost every "strange" index rule is just this pattern refusing to stop.

Why this works

Reading the table upwards gives multiplication, downwards gives division. That symmetry is why the laws come in pairs: multiplying adds indices because it climbs the table, dividing subtracts because it descends.

The three laws for whole-number indices

For positive whole numbers the laws are easy to see by writing the powers out in full. $a^{3} \times a^{2}$ is three $a$ s multiplied by two more $a$ s, which is five $a$ s: the indices $add$ . Division cancels matching factors top and bottom, so the indices $subtract$ . And $(a^{3})^{2}$ means $a^{3}$ written down twice and multiplied, giving six $a$ s, so the indices $multiply$ .

OCR states these for all rational exponents, so once you have them you are entitled to use them on fractions and negatives too — which is exactly how the rest of this lesson pins those cases down.

a^{m} \times a^{n} = a^{m + n}

Same base, multiplying → add the indices.

\frac{a ^{m}}{a ^{n}} = a^{m - n}

Same base, dividing → subtract the indices.

(a^{m})^{n} = a^{mn}

A power raised to a power → multiply the indices.

1Top first — multiplying, so add:

p^{9} \times p^{2} = p^{11}

2Now divide, so subtract:

p^{11} \div p^{5} = p^{11 - 5}

Answer

p^{6}

Tip — These laws only ever apply to $\times$ and $\div$ between powers of the $same base$ . $a^{4} + a^{2}$ does not become $a^{6}$ , and $2^{3} \times 3^{2}$ cannot be combined at all.

Letting the table run past zero

Go back to the powers of 2 and keep stepping down, halving each time: after $2^{1} = 2$ comes $2^{0} = 1$ , then $2^{- 1} = \frac{1}{2}$ , then $2^{- 2} = \frac{1}{4}$ . Nothing was decided here — the pattern simply continued, and it landed on 1 at the zero index and on reciprocals below it.

The division law says the same thing more formally. $\frac{a ^{n}}{a ^{n}}$ is 1 because anything non-zero over itself is 1, and it is $a^{n - n} = a^{0}$ by the law. Both are true, so $a^{0} = 1$ . Similarly $\frac{a ^{2}}{a ^{5}}$ cancels to $\frac{1}{a ^{3}}$ and equals $a^{- 3}$ by the law, so a negative index must mean a reciprocal.

Read the minus sign as an instruction to flip, never as a sign on the answer. $2^{- 3}$ is $\frac{1}{8}$ , a positive number.

a^{0} = 1 (a \neq = 0)

Where the halving pattern crosses the axis.

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

Below zero the pattern gives reciprocals.

1A negative index flips the fraction:

(\frac{3}{4})^{- 2} = (\frac{4}{3})^{2}

2Now square top and bottom:

\frac{16}{9}

Answer

\frac{16}{9}

Why this works

Notice what was not done: nobody defined $a^{0}$ to be 1 by decree. The value was forced — it is the only number that keeps the division law true. Extending a rule by insisting the existing pattern survives is a habit worth carrying into the rest of the course.

Fractions in the index mean roots

The power-of-a-power law fixes fractional indices too. Whatever $a^{1/2}$ is, squaring it gives $(a^{1/2})^{2} = a^{1} = a$ . The number that squares to $a$ is $a$ , so $a^{1/2} = a$ . The same argument with an $n$ in place of the 2 gives $a^{1/ n} = n a$ .

A general fraction $\frac{m}{n}$ then does both jobs: the denominator takes the root, the numerator raises to the power. Take the root first whenever you are working without a calculator — $3 2 7^{2}$ means cubing 27 to 19683 before rooting, whereas $(327)^{2}$ is just $3^{2}$ .

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

The denominator of the index is the root.

a^{\frac{m}{n}} = (n a)^{m}

Root by the bottom, then power by the top.

1Denominator 3 → cube root:

327 = 3

2Numerator 4 → fourth power:

3^{4} = 81

Answer

81

Tip — "Write in the form $a x^{n}$ " is OCR shorthand for "clear every root sign and fraction bar into one index". It is nearly always the first line of a differentiation or integration question.

Think like an examiner

A bracketed power hits the coefficient as well as the variable — $(3 x)^{2}$ is $9 x^{2}$ , not $3 x^{2}$ . This is the single most common lost mark in the topic.
When asked for the form $a x^{n}$ , state $a$ and $n$ explicitly if the question asks for their values; leaving the answer as a root will not score.
Split a fractional power into its root step and its power step on the page. If the arithmetic slips, the method mark still stands.
Negative and fractional indices are examined together with calculus far more often than on their own — expect to meet them as the setup line of a longer question.

Common misconceptions

Writing

x^{5} + x^{2} = x^{7}

Indices combine under multiplication, not addition.

x^{5} + x^{2}

has no simpler form.

Reading

5^{- 2}

- 25

5^{- 2} = \frac{1}{5 ^{2}} = \frac{1}{25}

. The minus sign flips; it does not change the sign of the value.

Simplifying

(4 x^{6})^{1/2}

4 x^{3}

The half applies to the 4 as well:

4^{1/2} x^{3} = 2 x^{3}

Evaluating

8^{2/3}

by squaring first and then guessing the root.

Cube root first:

38 = 2

, then

2^{2} = 4

. Rooting first keeps the numbers manageable.

The laws of indices

a^{m} \times a^{n} = a^{m + n}

Multiplying adds.

a^{m} \div a^{n} = a^{m - n}

Dividing subtracts.

(a^{m})^{n} = a^{mn}

Power of a power multiplies.

(ab)^{n} = a^{n} b^{n}

The index reaches every factor.

a^{0} = 1

Where the pattern crosses zero.

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

Negative index → reciprocal.

a^{\frac{m}{n}} = (n a)^{m}

Root by the bottom, power by the top.

Stretch yourself

Given that $9^{x + 1} = 2 7^{2 x - 1}$ , find the exact value of $x$ .

Hint — Both 9 and 27 are powers of 3. Rewrite each side as a single power of 3, then compare indices.

Questions students ask

Why is any non-zero number to the power zero equal to 1?

Because the division law leaves no alternative.

\frac{a ^{n}}{a ^{n}}

is 1, and the law rewrites it as

a^{n - n} = a^{0}

. Setting

a^{0} = 1

is the only value that keeps the laws consistent — and it is where the halving pattern in the powers table naturally lands.

0^{0}

equal to 1?

The law is stated for

a \neq = 0

, and

0^{0}

is left undefined at A-Level. The derivation divides

a^{n}

by itself, which is not allowed when

a = 0

For

a^{m / n}

, does it matter whether I root or power first?

Mathematically no, the answers agree. Practically yes: rooting first keeps the numbers small.

3 2^{3/5}

2^{3} = 8

if you take the fifth root first, but requires you to find the fifth root of 32768 if you do not.

What does OCR mean by "all rational exponents"?

Any index that can be written as a fraction of whole numbers — which covers positives, negatives, zero and fractions. In practice it is the exam board saying the three laws are never suspended, whatever the index looks like.

Key takeaways

Stepping down a column of powers divides by the base each time; every index law is that pattern continuing.
Multiplying adds indices, dividing subtracts, a power of a power multiplies.
A bracketed index applies to coefficients as well as variables.
$a^{0} = 1$ and $a^{- n} = \frac{1}{a ^{n}}$ are forced by the division law, not chosen.
A fractional index is a root: denominator roots, numerator powers — root first for easy arithmetic.

How this fits the course

Build on

Indices and roots (GCSE)soon

Fractional and Negative Indices

Leads to

Surds and Rationalisation

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Real past-paper questions on Laws of Indices, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Laws of Indices

Start with a table, not a rule

The three laws for whole-number indices

Letting the table run past zero

Fractions in the index mean roots

Practise this topic

Related topics

Laws of Indices

Start with a table, not a rule

The three laws for whole-number indices

Letting the table run past zero

Fractions in the index mean roots

Practise this topic

Related topics