3.01a, 3.01b - Integer powers and roots | OCR GCSE Mathematics

Powers compress repeated multiplication; roots recover equal factors. In 3.01a you will use positive and negative integer indices, and in 3.01b you will calculate powers and roots and recognise familiar powers of two, three, four and five.

Positive integer powers

Can one number be both a cube and a sixth power? We will discover why $64$ can wear both labels after making the notation precise.

Four equal factors of three can be written $3\times3\times3\times3=3^4=81$ . This is repeated multiplication, whereas $3\times4=12$ is repeated addition.

Base and index

In $a^n$ , $a$ is the base and $n$ is the index or exponent. For a positive integer $n$ , $a^n$ is the product of $n$ equal factors of $a$ . In particular, $a^1=a$ .

A negative base needs brackets: $(-3)^2=(-3)\times(-3)=9$ . An even number of negative factors can be paired into positive products. An odd number leaves one negative factor unpaired, so the power is negative. Zero to any positive integer power is zero; one to any such power is one.

Worked example 1: an even power of a negative base

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

Plan. Write four negative factors and pair them.

\begin{aligned} (-3)^4&=(-3)\times(-3)\times(-3)\times(-3) &&\text{four factors}\\ &=9\times9 &&\text{pair negative factors}\\ &=81 &&\text{multiply the pair products} \end{aligned}

Answer. $81$ .

Check. Successive multiplication gives $-3,9,-27,81$ ; the fourth product agrees and is positive.

Change only the exponent from four to three. One negative factor is now unpaired.

Worked example 2: an odd power of a negative base

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

Plan. Multiply three factors, keeping the sign of each.

\begin{aligned} (-3)^3&=(-3)\times(-3)\times(-3) &&\text{three factors}\\ &=9\times(-3) &&\text{multiply the first pair}\\ &=-27 &&\text{one negative factor remains} \end{aligned}

Answer. $-27$ .

Check. $(-27)\times(-3)=81$ , the fourth power already found.

Brackets specify the base. In $-3^2$ , the base of the power is $3$ , and the minus sign sits outside it: $-3^2=-(3\times3)=-9$ . In $(-3)^2$ , the whole negative number is squared, giving $9$ .

Watch out

$(-3)^4=81$ , not $-12$ . An exponent counts factors. Read the brackets before deciding which number is the base.

Use an odd number of negative factors.

Exact square and cube roots

An exact root can be found by recognising equal factors rather than using a decimal approximation.

Since $12^2=144$ , the principal square root $\sqrt{144}=12$ . Since $(-3)^3=-27$ , the cube root $\sqrt[3]{-27}=-3$ . Evaluate each root before combining their values.

You already know

$\sqrt a$ selects the nonnegative square root for $a\ge0$ . A negative number has no real square root, but it does have a real cube root. Both $\sqrt0$ and $\sqrt[3]0$ are zero.

A useful way to check a root is to apply the matching power: square a proposed square root or cube a proposed cube root. Also check the nonnegative condition for the principal square root.

Worked example 3: combining exact roots

Question. Work out $\sqrt{144}+\sqrt[3]{-27}$ .

Plan. Evaluate the two roots separately, then add their values.

\begin{aligned} \sqrt{144}&=12 &&\text{since }12\times12=144\\ \sqrt[3]{-27}&=-3 &&\text{since }(-3)^3=-27\\ \sqrt{144}+\sqrt[3]{-27}&=12+(-3)=9 &&\text{combine the values} \end{aligned}

Answer. $9$ .

Check. $12$ is nonnegative and squares to $144$ ; $-3$ cubes to $-27$ . Their sum is three below twelve.

Change only the sign of the number inside the cube root. The square-root contribution stays twelve.

Worked example 4: changing the cube root's sign

Question. Work out $\sqrt{144}+\sqrt[3]{27}$ .

Plan. Keep the square root and recalculate the cube root.

\begin{aligned} \sqrt{144}&=12 &&\text{same principal root}\\ \sqrt[3]{27}&=3 &&\text{since }3^3=27\\ \sqrt{144}+\sqrt[3]{27}&=12+3=15 &&\text{add the roots} \end{aligned}

Answer. $15$ .

Check. Both contributions are positive. Changing $-3$ to $3$ increases the previous result by six: $9+6=15$ .

Keep roots attached to their own numbers. For example, $\sqrt9+\sqrt{16}=3+4=7$ , but $\sqrt{9+16}=\sqrt{25}=5$ . Adding before taking the root changes the calculation.

Watch out

$\sqrt{144}=12$ , not $\pm12$ . If asked for all numbers whose square is $144$ , give $12$ and $-12$ ; the root symbol itself selects the nonnegative answer.

Evaluate the new roots separately.

Negative indices and reciprocals

A negative index changes a power into a reciprocal; it does not automatically make its value negative.

Look at successive powers of two. Each step down in the index divides the value by two:

2^3=8,\quad2^2=4,\quad2^1=2,\quad2^0=1,\quad2^{-1}=\frac12,\quad2^{-2}=\frac14

The notation extends the same pattern. A reciprocal multiplies the original nonzero number to give one: the reciprocal of $8$ is $\frac18$ , because $8\times\frac18=1$ .

Negative integer index

For $a\ne0$ and a positive integer $n$ , $a^{-n}=\frac1{a^n}$ . Also, $a^0=1$ for $a\ne0$ . Zero has no reciprocal, so negative powers of zero are undefined; the zero-index rule here also excludes a zero base.

Calculate the positive power in the denominator, including its sign, then take its reciprocal. A reciprocal keeps the sign of its nonzero input.

Worked example 5: a positive reciprocal

Question. Work out $2^{-3}$ . Give an exact answer.

Plan. Find the third power, then take its reciprocal.

\begin{aligned} 2^3&=2\times2\times2=8 &&\text{positive power}\\ 2^{-3}&=\frac1{2^3}=\frac18 &&\text{reciprocal} \end{aligned}

Answer. $\frac18$ .

Check. $\frac18\times8=1$ . Also, three divisions by two take $1$ to $\frac18$ .

Change only the base's sign. The cube is now negative, so its reciprocal is negative too.

Worked example 6: a negative reciprocal

Question. Work out $(-2)^{-3}$ . Give an exact answer.

Plan. Cube the negative base inside the brackets, then take the reciprocal.

\begin{aligned} (-2)^3&=(-2)\times(-2)\times(-2)=-8 &&\text{odd power}\\ (-2)^{-3}&=\frac1{(-2)^3}=-\frac18 &&\text{reciprocal of }-8 \end{aligned}

Answer. $-\frac18$ .

Check. $(-\frac18)\times(-8)=1$ , as required for reciprocals.

An even power of a negative base is positive, so $(-2)^{-2}=\frac14$ . The negative index tells you to take a reciprocal; the base and whether the positive exponent is odd or even determine the sign.

Watch out

$2^{-3}=\frac18$ , not $-8$ . A negative index does not mean “put a minus sign on the positive power”.

Keep the brackets when finding the denominator.

Recognising familiar powers

Recognising a familiar power lets you move between a large number and its smaller repeated factor.

Build each sequence by multiplying by the base again:

Base	Successive positive integer powers
$2$	$2,4,8,16,32,64,128,256$
$3$	$3,9,27,81,243$
$4$	$4,16,64,256,1024$
$5$	$5,25,125,625,3125$

For example, $27=3^3$ and $625=5^4$ . You can reconstruct a forgotten value rather than guess: $5^4=5\times125=625$ . Repeated division by a proposed base checks which index produces a number.

Worked example 7: one number, two bases

Question. Write $64$ as a power of $2$ and as a power of $4$ .

Plan. Count the divisions by each base needed to reach one.

\begin{aligned} 64\div2\div2\div2\div2\div2\div2&=1 &&\text{six divisions}\\ 64&=2^6 &&\text{six equal factors of two}\\ 64\div4\div4\div4&=1 &&\text{three divisions}\\ 64&=4^3 &&\text{three equal factors of four} \end{aligned}

Answer. $64=2^6=4^3$ .

Check. Six twos can be grouped into three pairs: $(2\times2)(2\times2)(2\times2)=4\times4\times4=64$ .

Change only the number to $256$ . Continue each familiar sequence far enough to reach it.

Worked example 8: continuing both patterns

Question. Write $256$ as a power of $2$ and as a power of $4$ .

Plan. Extend the known powers of each base.

\begin{aligned} 2^6&=64,\quad2^7=128,\quad2^8=256 &&\text{double each time}\\ 4^3&=64,\quad4^4=256 &&\text{multiply by four}\\ 256&=2^8=4^4 &&\text{read the indices} \end{aligned}

Answer. $256=2^8=4^4$ .

Check. Four pairs of twos make four fours, so the two repeated products are equal. Also, $16\times16=256$ .

This answers the opening question: $64$ is both a cube and a sixth power because pairing its six factors of two makes three factors of four. Recognising the repeated factors is more useful than memorising isolated answers.

Watch out

$64=2^6$ does not mean $64=6^2$ . The base and index have different jobs; $6^2=36$ .

Use the same two-base connection with a smaller number.

Quick Recap

A positive integer index counts equal factors. Brackets determine whether a negative sign belongs to the base.
Check square and cube roots by applying the matching power; a principal square root is nonnegative.
For a nonzero base, a negative index gives a reciprocal, and a zero index gives one.
Build familiar power sequences by multiplying by the base; check them by repeated division.

One number can have several power descriptions: $64=2^6=4^3$ .

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3.01a, 3.01b - Integer powers and roots

Positive integer powers

Exact square and cube roots

Negative indices and reciprocals

Recognising familiar powers

Quick Recap

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