Indices

A term like $2^{10}$ hides a shortcut: instead of writing $2\times2\times2\times2\times2\times2\times2\times2\times2\times2$ , the raised number $10$ — the index or exponent — tells you how many times the base $2$ multiplies itself. Indices exist so that repeated multiplication, repeated division, roots, and even "no multiplication at all" can all be written and manipulated in one consistent notation. This is the third strand of the ICSE Class 9 Algebra chapter, and it is not just bookkeeping. Once you know the laws of indices — how powers combine under multiplication, division, and further powers — you can simplify expressions that look intimidating into a single clean term, often without a calculator. The same laws also extend naturally to zero, negative, and fractional indices, so $a^0$ , $a^{-3}$ , and $a^{2/3}$ are not new rules to memorise separately but consequences of the same two or three ideas applied consistently. This topic is also the direct foundation of logarithms, the next chapter, since a logarithm is simply an index written in a different form.

The core insight

The laws of indices let you combine, split, and re-express powers of the same base by adding, subtracting, or multiplying their exponents, and the same laws force $a^0=1$ , $a^{-m}=1/a^m$ , and $a^{1/n}=\sqrt[n]{a}$ .

The core idea

For a base $a\neq0$ and a positive whole-number index $m$ , $a^m$ means $a$ multiplied by itself $m$ times. Three laws describe how such powers combine when the base stays the same:

a^m\times a^n=a^{m+n}, \qquad a^m\div a^n=a^{m-n}, \qquad (a^m)^n=a^{mn}

where: $a$ is the base, $m$ and $n$ are indices, multiplying same-base powers adds their indices, dividing subtracts them, and raising a power to a further power multiplies them. Two companion laws handle powers of a product or a quotient, distributing the index over each factor:

(ab)^m=a^mb^m, \qquad \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}

where: $a$ and $b$ are the two factors (or numerator and denominator) sharing the same index $m$ . These five laws were built for positive whole-number indices, but they turn out to force their own extension. Insist that the division law $a^m\div a^n=a^{m-n}$ keeps working even when $m=n$ or $m<n$ , and you are led to zero and negative indices:

a^0=1 \quad (a\neq0), \qquad a^{-m}=\frac{1}{a^m}

where: the zero-index rule requires $a\neq0$ (since $0^0$ is undefined), and $a^{-m}$ is never negative in value — the minus sign in the index means "reciprocal," not "subtract." A fractional index is built the same way, by insisting the power law $(a^m)^n=a^{mn}$ still holds when $n$ is a fraction. This gives the root connection that makes indices and roots two notations for the same idea:

a^{1/n}=\sqrt[n]{a}, \qquad a^{p/q}=\sqrt[q]{a^p}=\left(\sqrt[q]{a}\right)^p

where: $n$ and $q$ are the root's order (square root when $2$ , cube root when $3$ , and so on), and $p$ is a whole-number power applied either before or after the root — both orders give the same answer. Being able to switch between root notation and index notation in either direction is a practical exam skill: $\sqrt[3]{a^2}$ and $a^{2/3}$ are the same quantity, and so are $\dfrac{1}{\sqrt{a}}$ and $a^{-1/2}$ . Recognising which form is easier to simplify — usually the index form — often reveals a shortcut that the root form hides.

Figure

A descending tower of bars showing powers of 2 from 2 cubed down to 2 to the power minus 2, each bar half the height of the one above it, illustrating that dropping the index by 1 divides the value by the base

Derivation AWhy the zero and negative index rules must hold

Show, using nothing but the division law $a^m\div a^n=a^{m-n}$ applied to two cases where it might seem to break down, that $a^0$ is forced to equal $1$ and that $a^{-m}$ is forced to equal the reciprocal $1/a^m$ , so that these are not arbitrary definitions but the only values consistent with the law already established for positive whole-number indices.

Step 1. Apply the division law to $a^m\div a^m$ , where the same power is divided by itself. By the law, $a^m\div a^m=a^{m-m}=a^0$ .

Step 2. Evaluate the same expression directly, without the law. Any nonzero quantity divided by itself equals $1$ , so $a^m\div a^m=1$ .

Step 3. Compare the two results for $a^m\div a^m$ . Step 1 gives $a^0$ and Step 2 gives $1$ , and since both describe the same expression, $a^0=1$ (with $a\neq0$ , since division by $0$ is not defined).

Step 4. Now apply the division law to a case where the second index is larger, say $a^2\div a^5$ . By the law, $a^2\div a^5=a^{2-5}=a^{-3}$ .

Step 5. Evaluate the same expression directly by cancelling factors. Writing out the multiplication, $a^2\div a^5=\dfrac{a\times a}{a\times a\times a\times a\times a}$ , and two of the five $a$ 's in the denominator cancel with the two in the numerator, leaving $\dfrac{1}{a^3}$ .

Step 6. Compare the two results. Step 4 gives $a^{-3}$ and Step 5 gives $\dfrac{1}{a^3}$ , so $a^{-3}=\dfrac{1}{a^3}$ — and the same cancellation argument works for any negative index, giving the general rule $a^{-m}=\dfrac{1}{a^m}$ .

Worked example

Simplify $\left(\dfrac{16}{81}\right)^{-3/4}$ .

Step 1. Apply the negative-index rule to flip the fraction and make the index positive: $\left(\dfrac{16}{81}\right)^{-3/4}=\left(\dfrac{81}{16}\right)^{3/4}$ .

Step 2. Apply the power-of-a-quotient law to split the index across numerator and denominator: $\left(\dfrac{81}{16}\right)^{3/4}=\dfrac{81^{3/4}}{16^{3/4}}$ .

Step 3. Rewrite each base as a power of a small prime, since $81=3^4$ and $16=2^4$ , then apply the power law $(a^m)^n=a^{mn}$ : $\dfrac{(3^4)^{3/4}}{(2^4)^{3/4}}=\dfrac{3^{4\times3/4}}{2^{4\times3/4}}=\dfrac{3^3}{2^3}$ .

Step 4. Evaluate the whole-number powers: $\dfrac{3^3}{2^3}=\dfrac{27}{8}$ . So $\left(\dfrac{16}{81}\right)^{-3/4}=\mathbf{\dfrac{27}{8}}$ . The key move was spotting that $16$ and $81$ are both perfect fourth powers before touching the fractional index — whenever a base looks like it could be a perfect power, rewriting it that way first usually turns an intimidating index into a short whole-number calculation.

Competitive stretch

Simplify $\dfrac{a^5\times a^{-2}}{a^{3}\times a^{-4}}$ using the same law-by-law discipline. Numerator: $a^5\times a^{-2}=a^{5+(-2)}=a^3$ . Denominator: $a^3\times a^{-4}=a^{3+(-4)}=a^{-1}$ . Dividing: $\dfrac{a^3}{a^{-1}}=a^{3-(-1)}=\mathbf{a^4}$ . Applying one law per line, rather than combining several mentally, is exactly what prevents sign errors with negative indices.

Common traps

Treating $a^{-m}$ as a negative number instead of a reciprocal — $a^{-m}=1/a^m$ stays positive whenever $a$ is positive; the minus sign belongs to the index, not the value.
Writing $a^0=0$ instead of $a^0=1$ , or forgetting that the rule needs $a\neq0$ .
Adding indices across different bases, e.g. wrongly simplifying $a^2\times b^3$ using a law — the laws of indices only combine powers that share the same base.
Confusing the power law with the multiplication law, e.g. writing $(a^m)^n=a^{m+n}$ instead of the correct $a^{mn}$ .
Applying an index law to a sum, e.g. treating $a^m+a^n$ as if it simplified to $a^{m+n}$ — the laws apply only to products, quotients, and powers, never to addition.
Losing track of brackets with a negative base and a fractional index, e.g. confusing $(-8)^{1/3}$ (a real cube root, $-2$ ) with $-8^{1/3}$ (the negative of $8^{1/3}$ ) — the bracket decides whether the minus sign is inside or outside the root.
Stopping halfway through a multi-law simplification and treating a partly-simplified expression as final — a fully simplified answer of the form $a^k$ should have every law applied until a single power (or a single number) remains.

Ask the tutor: If the jump from $a^m\div a^m=1$ to $a^0=1$ still feels like a trick, ask your AI tutor to walk through it with a small number like $5^3\div5^3$ side by side with the index law, until the two match up naturally.

Why does a to the power zero equal 1 and not 0, since zero of anything usually means nothing

Because a to the power zero is not "zero copies of $a$ multiplied together" in the everyday sense — indices are defined by the division law, not by counting nothing. Dividing $a^m$ by itself always gives $1$ directly, and the division law says the same expression equals $a^{m-m}=a^0$ . Since one expression cannot have two different values, $a^0$ must equal $1$ for any nonzero $a$ . It is a forced consequence of the rules already in place, not a fresh definition someone chose.

If a negative index does not make the answer negative, what does the minus sign actually do

It tells you to take the reciprocal of the positive-index version. So $a^{-m}$ means "flip $a^m$ upside down," giving $1/a^m$ . For example, $2^{-2}=1/2^2=1/4$ , which is a small positive fraction, not $-4$ . A quick way to check any unfamiliar index rule is to try it on small numbers like this — if $2^{-2}$ came out negative, you would immediately know something was applied wrongly.

How do I decide whether to work with a root or with a fractional index when simplifying

Either form is valid since $a^{p/q}=\sqrt[q]{a^p}=(\sqrt[q]{a})^p$ , but the index form is usually easier to manipulate because you can apply the multiplication, division, and power laws directly to the exponents. Roots are often clearer only at the very last step, when you want a decimal or a recognisable surd. A good habit is to convert everything in a problem to index form first, simplify fully using the laws, and switch back to root notation only if the question asks for it.

Indices

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads