Mathematics · IGCSE 0580 · §E1.7–E1.12

Powers, Form & Proportion

Number at every scale: how to compress it with indices, write it honestly in standard form, admit what a measurement cannot tell you, and compare quantities fairly.

Mathematics · 0580 Extended Topic 2 of 12
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Indices I: the three laws

10−7 10−6 10−5 10−4 10−3 10−2 10−1 100 101 102 103 104 105 106 length in metres a virus 1 × 10−7 m a coin 2.3 × 10−2 m a family car 4.5 × 100 m a stadium 2.5 × 102 m Mount Everest 8.8 × 103 m
FIG 1.2.0 Thirteen decades of length on one line: standard form is what lets a virus and a mountain be written, and compared, in the same notation.

An index is shorthand for repeated multiplication: aⁿ means n copies of a multiplied together. Three laws let you combine powers of the same base without ever writing them out.

Definition
Index and base
The index (or power, or exponent) is the small raised number showing how many times the base is multiplied by itself. In 5³ the base is 5 and the index is 3.

The three laws

Multiplying powers adds the indices: aᵐ × aⁿ = aᵐ⁺ⁿ — counting the copies is the same as adding the indices. Dividing cancels copies, so the indices subtract: aᵐ ÷ aⁿ = aᵐ⁻ⁿ. A power of a power multiplies the indices: (aᵐ)ⁿ = aᵐⁿ, because the outer index says how many times the inner power appears.

All three laws require the same base. 2³ × 3² cannot be combined into a single power — it is simply 8 × 9 = 72.

Worked example: simplify 12p⁷ × 3p² ÷ 4p⁵. Deal with the coefficients separately: 12 × 3 ÷ 4 = 9. Then the indices: p⁷ × p² = p⁹, and p⁹ ÷ p⁵ = p⁴. Answer: 9p⁴. Keeping the two streams apart is what stops the working going wrong.

Examiner note
Read the verb. 'Simplify' wants an expression in letters back; 'find the value of' wants a number. Answering in the wrong form loses the accuracy mark even when the working is right.
Why this matters
The same three laws let a scientist handle the diameter of an atom and the distance to a galaxy without changing method.

Indices II: zero, negative and fractional

Counting copies only ever explained whole positive indices. Extending the same three laws to zero, negative and fractional indices turns a counting trick into a system.

The zero index

By the division law, a³ ÷ a³ = a⁰. But anything non-zero divided by itself is 1. So a⁰ = 1 for every base except zero.

Negative indices

The same law gives a² ÷ a⁵ = a⁻³, and cancelling directly gives 1/a³. A negative index is therefore an instruction to take the reciprocal: a⁻ⁿ = 1/aⁿ (a ≠ 0). It changes the position of the power, never the sign of the answer.

Fractional indices

By the power law, (a^(1/2))² = a¹ = a. The quantity that squares to give a is √a — so a^(1/2) = √a. The same argument gives every root: a^(1/n) = ⁿ√a, and a^(m/n) = (ⁿ√a)ᵐ — the denominator is the root, the numerator the power. Take the root first and the numbers stay small: 27^(2/3) becomes 3², not the cube root of 729.

Worked example: find 27^(−2/3). The negative index says take the reciprocal: 27^(−2/3) = 1 / 27^(2/3). The denominator of the fraction is the root: ∛27 = 3. The numerator is the power: 3² = 9. So 27^(−2/3) = 1/9.

Examiner note
A negative index means "reciprocal of" — it never means "negative number". 2⁻³ is 1/8, not −8. This is the single most common error on this objective.
Examiner note
The zero index is usually hidden inside a longer expression, not asked directly: in 4bc⁰, only the c becomes 1, giving 4b.
Why this matters
Fractional indices are the bridge between powers and roots. The same idea returns in exponential graphs and in growth and decay later in the course.

Standard form

Standard form separates a number into two independent questions: what are its digits, and how big is it? The digits live in A, the size lives in the power of ten.

Definition
Standard form
A number written as A × 10ⁿ, where A is at least 1 and less than 10, and n is an integer. A positive n means a large number, a negative n a small one.

Converting into and out of standard form

The index counts the places the decimal point moves. Moving it left gives a positive index; moving it right gives a negative one. 4 700 000 = 4.7 × 10⁶ — the point moved 6 places left. 0.000 82 = 8.2 × 10⁻⁴ — the point moved 4 places right. 3.06 × 10⁻³ = 0.003 06 — read the index backwards.

Calculating with standard form

Handle the two parts separately: multiply or divide the A values as ordinary numbers, and combine the powers of ten with the index laws. Then — always last — check that A still lies between 1 and 10, and correct it if not.

Worked example: (4.5 × 10⁵) × (6 × 10⁻⁸). Multiply the A values: 4.5 × 6 = 27. Add the indices: 10⁵ × 10⁻⁸ = 10⁻³, giving 27 × 10⁻³. A = 27 is outside 1 ≤ A < 10, so rewrite 27 as 2.7 × 10¹: the answer is 2.7 × 10⁻². That final check is where the accuracy mark sits.

Examiner note
Marks are lost for leaving A outside the range: 25 × 10⁴ is arithmetically true but is not standard form. Always check the leading digit last.
Why this matters
Standard form returns wherever measurements get very large or very small — in mensuration, in trigonometry, and in every science paper you will sit.

Estimation and limits of accuracy

A rounded number is not a value but an interval. Estimation exploits that deliberately; limits of accuracy measure it exactly.

Definition
Upper and lower bounds
The upper bound is half a unit of accuracy above the rounded value; the true value is always strictly below it. The lower bound is half a unit below, and the true value may equal it: LB ≤ true value < UB.

Estimating a calculation

Round every number to 1 significant figure first, then calculate with the rounded values. Estimate 61.2 × 9.83 ÷ 0.412: this becomes 60 × 10 ÷ 0.4 = 1500, against a calculator value of 1460 to 3 s.f. An estimate is a check on an answer, never a substitute for one.

Bounds of a measurement

A value rounded to a given accuracy sits half a unit either side of that accuracy. The lower bound is included; the upper bound is not, because a value sitting exactly on it would have rounded upwards. A mass given as 4.5 g to 1 d.p. means 4.45 ≤ mass < 4.55.

4.5 g (recorded) 4.45 g 4.55 g lower bound (included) upper bound (excluded) 4.45 ⩽ mass < 4.55
FIG 1.2.1 A mass given as 4.5 g to 1 d.p. names an interval half a unit wide on each side — closed below, open above.

Bounds of a calculated result

To make a result as large as possible, ask what each input must do. Multiplying or adding: use the upper bounds. Dividing: largest numerator, smallest denominator.

Worked example: a cyclist rides 96 km (nearest km) in 1.4 hours (1 d.p.). Bound each: 95.5 ≤ distance < 96.5 and 1.35 ≤ time < 1.45. The largest speed comes from the largest distance over the smallest time: 96.5 ÷ 1.35 = 71.48… = 71.5 km/h (3 s.f.). A quotient pairs opposite bounds, never matching ones.

Examiner note
"Correct to 1 significant figure" is the standard instruction before an estimate. Read which figure it names — rounding to 1 decimal place instead is a different answer entirely.
Why this matters
A mass recorded as 4.5 g could truly be anything from 4.45 g up to (but not including) 4.55 g. Every measurement carries this uncertainty; bounds are how mathematics states it honestly.

Ratio and proportion

A ratio compares quantities of the same kind; a rate, in the next section, compares quantities of different kinds. Everything else about the two is the same.

Definition
Ratio and simplest form
A ratio compares two or more quantities of the same kind, written a : b, and has no units. Simplest form has whole-number parts sharing no common factor other than 1.

Simplest form

Divide every part by the highest common factor. 18 : 24 : 30 all share a factor of 6, so the ratio simplifies to 3 : 4 : 5. If the parts are not whole numbers, multiply through first: 0.5 : 1.5 becomes 1 : 3.

Dividing a quantity in a given ratio

The parts of the ratio tell you how many equal shares the quantity is cut into. Add the parts, find the value of one share, then multiply back up. Worked example: 84 kg split 5 : 7 is 12 shares; one share is 84 ÷ 12 = 7 kg; the heavier part takes 7 × 7 = 49 kg, and 49 + 35 = 84 kg checks.

Proportional reasoning

Best-value questions ask you to compare unlike packages by reducing each to the same unit. A 750 g bag at $2.40 costs $0.32 per 100 g; a 1.2 kg bag at $3.60 costs $0.30 per 100 g. The larger bag is better value — but only because both were reduced to a common basis first.

Examiner note
"Give your answer in its simplest form" is checked as a separate mark. Divide by the highest common factor — the same check you would make when simplifying a fraction.
Why this matters
Ratio is the engine of scale drawings, map scales and similar shapes — all of which return later in the course as geometry rather than number.

Rates

A rate is a ratio between quantities measured in different units. Speed is the one the syllabus expects you to recall; every other rate formula you need will be supplied in the question. Every rate answer is a rounded answer, so the accuracy rules apply.

Definition
Rate
A comparison of two quantities of different kinds, expressed as one per unit of the other — km per hour, g per cm³, $ per litre.

Speed, distance and time

speed = distance ÷ time, with units matching throughout. Rearranged: distance = speed × time, and time = distance ÷ speed.

D S T
FIG 1.2.2 Cover the quantity you want: D = S × T, S = D ÷ T, T = D ÷ S. A memory aid — not a substitute for the formula above.

Average speed

Average speed is total distance divided by total time — never the average of two speeds. Half a journey at 40 km/h and half at 60 km/h does not average 50 km/h: more time is spent at the slower speed.

Chained conversions

Rate questions rarely ask for a single conversion. Convert one step at a time, and keep the unrounded value in the calculator until the final line. Worked example: a train covers 210 km in 1 h 45 min. Time = 1.75 h, so average speed = 210 ÷ 1.75 = 120 km/h. Converting: 120 km/h = 120 × 1000 m per 3600 s = 33.3 m/s (3 s.f.). Rounding earlier would put the answer outside the accepted range.

Examiner note
Any formula you need other than speed, distance and time is given in the question — density and pressure included. The speed relationship is the one you must know cold.
Why this matters
A rate puts unlike quantities on one fair scale, which is how cost and time, or mass and volume, can be compared at all.

Exam advice

Common mistakes

Treating a negative index as a negative number
Writing a⁻ⁿ = −aⁿ instead of 1/aⁿ. The sign of the index changes the position of the power, never the sign of the answer.
Forgetting that a zero index gives 1
Simplifying 4bc⁰ as 4bc instead of 4b. The zero index is nearly always buried inside a longer expression.
Leaving A outside the range in standard form
Writing 25 × 10⁴ instead of 2.5 × 10⁵. The arithmetic is right and the accuracy mark is still lost.
Rounding to 1 decimal place when asked for 1 significant figure
The two agree between 1 and 10 and diverge everywhere else — which is exactly where estimation questions are set.
Rounding an intermediate value in a multi-step rate question
The final answer falls outside the accepted range and the accuracy mark goes, even though the method marks are earned.

Model answer

In Marovia, one gallon of fuel costs 5.40 dinars. In Selka, one litre costs 1.72 kronor. 1 krona = 1.24 dinars, and 1 gallon = 3.785 litres. In which country does one litre of fuel cost more, and by how much (in dinars)?
[3 marks]
M1
Convert the Marovian price per gallon into a price per litre
5.40 ÷ 3.785 = 1.4266… dinars per litre.
M1
Convert the Selkan price per litre into dinars
1.72 × 1.24 = 2.1328 dinars per litre.
A1
Compare and state the country and the difference
Selka is dearer, by 2.1328 − 1.4266 = 0.71 dinars (2 d.p.), working unrounded until the last line.

Recall checklist

  • Apply the index laws to a product, quotient or power.
  • Evaluate a negative or fractional index.
  • Convert into and out of standard form, and calculate with it.
  • Round a value to given decimal places or significant figures.
  • Estimate a calculation by rounding each value to 1 s.f.
  • Find the bounds of a measurement and of a calculated result.
  • Divide a quantity in a given ratio.
  • Solve a rate problem involving average speed and a unit conversion.

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