C1.7; E1.7 - Integer index laws | Cambridge IGCSE International Mathematics

The powers $2^3$ , $2^2$ and $2^1$ halve each time the index falls by 1. What values should $2^0$ and $2^{-1}$ have if the pattern continues? This lesson develops the integer indices in C1.7 and E1.7, explaining how repeated multiplication leads to rules for positive, zero and negative powers.

Index notation

An index counts repeated equal factors. For example, $2^3=2\times2\times2=8$ : the base is 2 and the index is 3. This is different from $2\times3=6$ , where there are only two factors.

Positive integer index

For a positive integer $n$ , $a^n$ is a product of $n$ copies of the base $a$ . The number $n$ is the index or exponent. In particular, $a^1=a$ .

The base is the complete expression immediately before the index. Brackets make a signed base clear: $(-3)^4$ uses four factors of $-3$ , whereas $-3^4$ means the negative of $3^4$ .

Worked example 1: expanding a power

Question. Work out $3^4$ .

Plan. Write four factors of 3 and multiply in stages.

\begin{aligned} 3^4&=3\times3\times3\times3\\ &=9\times9\\ &=81 \end{aligned}

Answer. 81.

Check. Starting from $3^1=3$ , three further multiplications by 3 give $9,\ 27,\ 81$ . There are four factors in total, not three.

The next example changes the base to a negative number. Count the negative factors as well as their magnitudes.

Worked example 2: a negative base

Question. Work out $(-3)^4$ .

Plan. Expand the bracketed base and pair the negative factors.

\begin{aligned} (-3)^4&=(-3)\times(-3)\times(-3)\times(-3)\\ &=9\times9 &&\text{each pair has a positive product}\\ &=81 \end{aligned}

Answer. 81.

Check. Four negative factors make two positive pairs. The magnitude is $3^4=81$ , so the positive sign is consistent.

An odd number of negative factors leaves one negative factor unpaired: $(-3)^3=-27$ . Thus an even positive power of a negative base is positive, and an odd positive power is negative. The brackets matter: $-3^4=-81$ , because the power is evaluated before the leading minus.

Watch out

$4^3$ means $4\times4\times4=64$ , not $4\times3=12$ . The index counts factors, not the amount to multiply the base by once.

Expand the signed base before evaluating.

Multiplication and division laws

Equal bases let you count factors instead of multiplying them out. In $2^3\times2^4$ , the first group contains three factors of 2 and the second contains four. Together there are seven, so the result is $2^7$ .

Division cancels equal factors. In $2^5\div2^3$ , three of the five factors cancel, leaving two. A power raised to a power repeats a whole group: $(2^3)^4$ has four groups of three factors, so it contains twelve factors of 2.

Index laws

For a non-zero base $a$ and integer indices $m$ and $n$ :

\begin{aligned} a^m\times a^n&=a^{m+n}\\ a^m\div a^n&=a^{m-n}\\ (a^m)^n&=a^{mn} \end{aligned}

The first two laws require the same base. The next part explains the zero and negative indices that can arise.

Not in the supplied formula list: learn these rules.

The non-zero condition ensures division is defined. For example, cancelling powers of zero would involve division by zero and is invalid.

Worked example 3: multiplying and dividing equal bases

Question. Work out $2^3\times2^4\div2^5$ .

Plan. Keep base 2, adding indices for multiplication and subtracting for division.

\begin{aligned} 2^3\times2^4\div2^5&=2^{3+4-5}\\ &=2^2\\ &=4 \end{aligned}

Answer. 4.

Check. Evaluating the powers separately gives $8\times16\div32=128\div32=4$ .

The next expression replaces the product of two powers by a power of a power. Count repeated groups by multiplying their indices before applying the division law.

Worked example 4: a power of a power

Question. Work out $(2^3)^4\div2^5$ .

Plan. Multiply the two nested indices, then subtract the index of the divisor.

\begin{aligned} (2^3)^4\div2^5&=2^{3\times4}\div2^5\\ &=2^{12-5}\\ &=2^7=128 \end{aligned}

Answer. 128.

Check. $(2^3)^4=8^4=4096$ , and $4096\div32=128$ . Unlike the previous example, the bracket repeats a complete group four times.

Adding powers does not join their factors: $2^3+2^4=8+16=24$ , not $2^7$ . Different bases cannot use the equal-base rule either: $2^3\times3^2=8\times9=72$ , not a single power obtained by adding 3 and 2.

Watch out

Distinguish $a^m\times a^n$ , which adds the indices, from $(a^m)^n$ , which multiplies them. Write the relevant intermediate power so the operation is clear.

Simplify the nested power before the product and quotient.

Zero and negative powers

Continue the opening pattern by dividing by 2 at each step:

Power	$2^3$	$2^2$	$2^1$	$2^0$	$2^{-1}$	$2^{-2}$
Value	8	4	2	1	$\frac12$	$\frac14$

Zero and negative indices preserve the division law. For a non-zero base, $a^3\div a^3=1$ , while subtracting indices gives $a^{3-3}=a^0$ . Therefore $a^0=1$ . Dividing once more by $a$ gives $a^{-1}=\frac1a$ .

Zero and negative integer powers

For $a\ne0$ and a positive integer $n$ :

a^0=1,\qquad a^{-n}=\frac1{a^n}.

A negative index means a reciprocal, not a negative value. The expressions $0^0$ and negative powers of zero are not defined by these rules.

Not in the supplied formula list: learn these rules.

You can also see the reciprocal by cancellation: $a^2\div a^5=\frac1{a^3}$ , so the same result written with the index law is $a^{-3}$ .

Worked example 5: evaluating zero and negative powers

Question. Write down $6^0$ and work out $6^{-2}$ as a fraction in its simplest form.

Plan. Apply the zero-power rule and replace the negative power by a reciprocal.

\begin{aligned} 6^0&=1\\ 6^{-2}&=\frac1{6^2}=\frac1{36} \end{aligned}

Answer. 1 and $\frac1{36}$ .

Check. $6^2\div6^2=1$ . Also, $\frac1{36}\times6^2=1$ , so the negative power is the reciprocal of the positive power.

The next calculation combines negative and positive indices. Use the same laws; the signs belong in the exponent arithmetic.

Worked example 6: a calculation with a negative index

Question. Work out $2^{-3}\times2^4\div2^2$ , giving an exact fraction.

Plan. Combine the equal-base powers, then interpret the final negative index.

\begin{aligned} 2^{-3}\times2^4\div2^2&=2^{-3+4-2}\\ &=2^{-1}\\ &=\frac12 \end{aligned}

Answer. $\frac12$ .

Check. Direct evaluation gives $\frac18\times16\div4=2\div4=\frac12$ .

The base still controls the sign: $(-2)^{-3}=\frac1{(-2)^3}=-\frac18$ , whereas $2^{-3}=\frac18$ . If a non-zero fraction is the base, its reciprocal swaps numerator and denominator: $\left(\frac23\right)^{-2}=\left(\frac32\right)^2=\frac94$ . This follows from taking the reciprocal of $\frac49$ .

Watch out

$5^{-2}=\frac1{25}$ , not $-25$ and not $-10$ . The minus in an index requests a reciprocal; it does not multiply the base or the answer by $-1$ .

Combine the indices before deciding the final numerical value.

Quick Recap

A positive integer index counts repeated factors; brackets identify the complete base.
For equal non-zero bases, add indices when multiplying and subtract them when dividing.
A power of a power multiplies the indices.
A non-zero base to power zero is 1; a negative index gives a reciprocal.

The opening halving pattern continues with $2^0=1$ and $2^{-1}=\frac12$ .

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C1.7; E1.7 - Integer index laws

Index notation

Multiplication and division laws

Zero and negative powers

Quick Recap

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