1.02a - Rational indices and index laws | OCR A-Level Mathematics A

Turn powers, roots and reciprocals into a single language. You will use rational indices accurately, explain their restrictions and simplify expressions that combine several laws, so a forbidding calculation can become a short exact argument.

The three combining laws

Can you simplify $\dfrac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y}$ for $x,y>0$ , without a decimal or a calculator? Start with something smaller: $x^3x^2$ contains five factors of $x$ , so it is $x^5$ . Cancelling two factors from $x^5/x^2$ leaves $x^3$ , provided $x\ne0$ . We will build up to the opening expression in the last part.

You already know

A power such as $x^3$ means repeated multiplication. Expanding a bracket means multiplying every term by the outside factor, and exact fractions need a common denominator before you add them.

The same counting explains a power of a power: $(x^3)^2$ has two groups of three factors. Extending these laws to rational indices gives a consistent way of handling roots as well.

Combining powers

For $x>0$ and rational $a,b$ ,

x^a x^b=x^{a+b},\qquad \frac{x^a}{x^b}=x^{a-b},\qquad (x^a)^b=x^{ab}.

The first two laws require a common base. Positive bases make all three laws safe for rational indices.

Not in the formula sheet: learn it.

Work on one operation at a time: multiply the indices inside a power, then add or subtract them across a product or quotient. Leave fractions exact.

Worked example 1: collecting fractional powers

Question. For $x>0$ , simplify $\dfrac{x^{3/2}x^{1/3}}{x^{5/6}}$ .

Plan. Combine the indices using a common denominator.

\begin{aligned} \frac{x^{3/2}x^{1/3}}{x^{5/6}} &=x^{3/2+1/3-5/6} && \text{product and quotient laws}\\ &=x^{9/6+2/6-5/6} && \text{common denominator}\\ &=x && \text{index is }1 \end{aligned}

Why this step? All three powers have the same base, so their indices can be combined without evaluating the powers.

Answer. $x$ .

Check. At $x=64$ , the expression is $512\times4/32=64$ .

The next example changes one feature: the first power is inside a squared bracket. Deal with that outer power before collecting indices.

Worked example 2: an outer power

Question. For $x>0$ , simplify $\dfrac{(x^{3/4})^2x^{1/3}}{x^{5/6}}$ .

Plan. Multiply the bracket's indices, then use the preceding collection method.

\begin{aligned} \frac{(x^{3/4})^2x^{1/3}}{x^{5/6}} &=x^{(3/4)\times2+1/3-5/6} && \text{power before product}\\ &=x^{9/6+2/6-5/6} && \text{common denominator}\\ &=x && \text{collect} \end{aligned}

Why this step? Squaring a power multiplies its index by two; it does not add two to the index.

Answer. $x$ .

Check. At $x=64$ , $(64^{3/4})^2=512$ , so the remaining check is again $512\times4/32=64$ .

A sum does not join factors into one power. For instance, $2^3+2^2=12$ , whereas $2^{3+2}=32$ .

Watch out

Wrong: $x^a+x^b=x^{a+b}$ . The addition rule for indices belongs to multiplication: $x^a x^b=x^{a+b}$ . Different bases also cannot be collected this way: $2^3\times3^2\ne6^5$ .

Try the same bracket-first route with a changed outer power.

Zero powers and reciprocals

Zero and negative indices keep the quotient law working; they do not make a power vanish or turn negative.

For $x\ne0$ , cancelling $x^a/x^a$ gives $1$ , while subtracting indices gives $x^0$ . Similarly, $x^a x^{-a}=x^0=1$ , so the negative power is the reciprocal.

Zero and negative indices

For a nonzero base, whenever the real powers involved are defined,

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

Neither rule defines $0^0$ , and a negative power of zero would require division by zero.

Not in the formula sheet: learn it.

Here is a useful surprise: the reciprocal of a small positive fraction is larger than one. A negative index can therefore make a positive number larger.

Worked example 3: a negative integer power

Question. Evaluate $4^{-2}$ exactly.

Plan. Take the reciprocal, then square the denominator.

\begin{aligned} 4^{-2}&=\frac1{4^2} && \text{negative means reciprocal}\\ &=\frac1{16} && \text{square }4 \end{aligned}

Why this step? The minus sign belongs to the index. It does not make the value negative.

Answer. $\frac1{16}$ .

Check. $4^2\times4^{-2}=16\times\frac1{16}=1$ , as the zero-index law requires.

Change only the base to a fraction. The reciprocal must now invert the whole fraction.

Worked example 4: a fractional base

Question. Evaluate $\left(\frac23\right)^{-2}$ exactly.

Plan. Invert the fraction, then square it.

\begin{aligned} \left(\frac23\right)^{-2} &=\left(\frac32\right)^2 && \text{take the reciprocal}\\ &=\frac94 && \text{square top and bottom} \end{aligned}

Why this step? The reciprocal of $2/3$ is $3/2$ , so both numerator and denominator change places.

Answer. $\frac94$ .

Check. $\frac49\times\frac94=1$ . The answer exceeds one, as expected for the reciprocal of $4/9$ .

Brackets also tell you whether a negative sign belongs to the base. Thus $-4^2=-16$ , but $(-4)^2=16$ .

Watch out

Wrong: $4^{-2}=-16$ . Right: $4^{-2}=1/16$ . A negative index means reciprocal; a negative sign before the base is a different operation.

Keep the domain even when simplification hides it: $x^2/x^2=1$ still requires $x\ne0$ .

Roots, rational powers and domains

A fractional index tells you which root to take; its numerator then tells you the power.

The positive number whose cube is $8$ is $2$ . To keep $(8^{1/3})^3=8$ , we must have $8^{1/3}=2$ . This is the bridge from powers to roots.

Principal square root

For $u\ge0$ , $\sqrt u$ is the non-negative number whose square is $u$ . For example, $\sqrt9=3$ . It is not $\pm3$ : those are the two solutions of $z^2=9$ .

Reduce a fractional index before interpreting its denominator as the root. For positive bases, the following rule fits the combining laws without ambiguity.

Rational indices

For $x>0$ , integer $m$ and positive integer $n$ ,

x^{1/n}=\sqrt[n]{x},\qquad x^{m/n}=(\sqrt[n]{x})^m.

A negative $m$ also requires a reciprocal. Positive powers extend to $x=0$ ; zero and negative powers exclude it.

Not in the formula sheet: learn it.

Take the root first when it turns a large number into a small integer.

Worked example 5: a root followed by a power

Question. Evaluate $32^{2/5}$ exactly.

Plan. Take the fifth root before squaring.

\begin{aligned} 32^{2/5}&=(\sqrt[5]{32})^2 && \text{denominator gives the root}\\ &=2^2 && 2^5=32\\ &=4 && \text{square} \end{aligned}

Why this step? Taking the fifth root first avoids forming $32^2$ , while giving the same exact value.

Answer. $4$ .

Check. $4^5=1024=32^2$ , and both sides are positive.

The second example changes only the sign of the index. The root and square stay; a reciprocal is added.

Worked example 6: a negative fractional index

Question. Evaluate $32^{-2/5}$ exactly.

Plan. Find the positive power, then invert it.

\begin{aligned} 32^{-2/5}&=\frac1{32^{2/5}} && \text{negative index}\\ &=\frac1{(\sqrt[5]{32})^2} && \text{rational index}\\ &=\frac14 && \text{root then square} \end{aligned}

Why this step? The minus sign asks for the reciprocal of the whole positive power, not a negative root.

Answer. $\frac14$ .

Check. $32^{2/5}\times32^{-2/5}=4\times\frac14=1$ .

Odd roots allow negative inputs: $(-8)^{1/3}=-2$ and $(-8)^{2/3}=4$ , using the reduced index and cube-root interpretation. Even roots of negative numbers are not real. Do not extend every power-of-a-power law to negative bases: $((-1)^2)^{1/2}=1$ , whereas $(-1)^{2\times1/2}=-1$ .

Watch out

For real $x$ , $\sqrt{x^2}=|x|$ . Wrong at $x=-3$ : $\sqrt{x^2}=x$ . Right: $\sqrt9=3=|-3|$ . It equals $x$ only when $x\ge0$ . This identity is not in the formula sheet: learn it.

The positive-base condition is a useful safety rail, not an optional decoration. Check it before combining rational powers.

Combining laws and comparing a common base

Separate the coefficient from each variable, then choose whether you are simplifying an expression or solving an equation.

A power acts on every factor: $(2x^3)^2=4x^6$ , while $2(x^3)^2=2x^6$ . For positive factors the product rule $(uv)^r=u^rv^r$ follows from the same root-and-power interpretation; it is not in the formula sheet: learn it. This lets us return to the opening puzzle.

Worked example 7: the opening expression

Question. For $x,y>0$ , simplify $\dfrac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y}$ .

Plan. Apply the outer power to every factor, then simplify each quotient separately.

\begin{aligned} \frac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y} &=\frac{4x^{-2}y^4}{2x^{-1}y} && \text{multiply each index by }2/3\\ &=2x^{-1}y^3 && \text{subtract denominator indices}\\ &=\frac{2y^3}{x} && \text{positive indices} \end{aligned}

Why this step? Treating the coefficient, $x$ and $y$ separately prevents one variable's index being subtracted from another's.

Answer. $\dfrac{2y^3}{x}$ , for the stated positive inputs.

Check. At $x=2,y=1$ , the starting expression is $1^{2/3}/1=1$ ; the answer is $2/2=1$ .

For the second example, change only the divisor. The numerator already has the same simplified form.

Worked example 8: a different divisor

Question. For $x,y>0$ , simplify $\dfrac{(8x^{-3}y^6)^{2/3}}{4x^{-1}y^2}$ .

Plan. Reuse the numerator calculation, then subtract the new denominator indices.

\begin{aligned} \frac{(8x^{-3}y^6)^{2/3}}{4x^{-1}y^2} &=\frac{4x^{-2}y^4}{4x^{-1}y^2} && \text{evaluate numerator power}\\ &=x^{-1}y^2 && \text{divide like factors}\\ &=\frac{y^2}{x} && \text{remove negative index} \end{aligned}

Why this step? The numerator's algebra is unchanged; only the coefficient and indices subtracted in division change.

Answer. $\dfrac{y^2}{x}$ .

Check. At $x=2,y=1$ , the starting denominator is $2$ , giving $1/2$ , exactly as $y^2/x$ does.

The same laws can expose a common base in an equation. Equal powers of a positive base other than one have equal indices. For example,

\begin{aligned} 8^{2x-1}=4^{x+2} &\ \Longleftrightarrow\ 2^{6x-3}=2^{2x+4} && \text{base }2\\ &\ \Longleftrightarrow\ 6x-3=2x+4 && \text{equal powers}\\ &\ \Longleftrightarrow\ x=\frac74 && \text{solve the linear equation} \end{aligned}

Both indices are then $15/2$ . This common-base principle is not in the formula sheet: learn it. The condition matters: $1^u=1$ for every real $u$ , so base one cannot identify an index. You will use this link again with exponentials and logarithms.

Exam tip

When OCR asks for exact simplification, keep fractional indices exact and give a simplified final expression. For “In this question you must show detailed reasoning”, show the intermediate powers; a calculator value cannot replace that route.

Now change the divisor again and simplify the whole expression, keeping the same positive domain.

Quick Recap

Multiply like bases by adding indices; divide by subtracting; raise a power by multiplying indices.
Zero powers of nonzero bases equal one. Negative indices mean reciprocals.
A rational index combines a root and a power; a principal even root is non-negative.
State domains and keep exact fractions until the end. Positive bases protect the general rational-index laws.

The opening expression simplifies to $2y^3/x$ : the large bracket becomes three small calculations.

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1.02a - Rational indices and index laws

The three combining laws

Zero powers and reciprocals

Roots, rational powers and domains

Combining laws and comparing a common base

Quick Recap

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