OxfordAQA International GCSE Mathematics: Index laws, surds, standard form and sets (9260)

This guide is aligned to the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). It covers specification statements N6 to N9 in section 3.1.1 (Number: structure and calculation). Core content is also assessable on the Extension tier. Two statements are Extension only and are labelled that way below: fractional powers (N6) and calculating exactly with surds, including rationalising a denominator (N7).

Related pages: revision notes, practice questions, the number overview, the course hub and the printable checklist. To find your weak spots first, try the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
N6	Use index laws for multiplication and division with integer powers	Core
N6	Use fractional powers	Extension only
N7	Calculate exactly with fractions	Core
N7	Calculate exactly with surds; simplify surds; rationalise a denominator	Extension only
N8	Calculate with and interpret standard form, including calculator displays	Core
N9	Use set language and notation n(A), A′, A ∪ B, A ∩ B, ξ; use Venn diagrams to solve problems	Core

N6: Index laws

In aⁿ, a is the base and n is the index (power). Two laws cover multiplication and division of the same base:

aᵐ × aⁿ = aᵐ⁺ⁿ
aᵐ ÷ aⁿ = aᵐ⁻ⁿ

Integer powers include zero and negative numbers:

a⁰ = 1 (for a ≠ 0)
a⁻ⁿ = 1/aⁿ

A negative power means “reciprocal”, not “negative number”. So 5⁻² = 1/25, which is positive.

When terms have coefficients, deal with the numbers and each letter separately.

Worked example 1. Simplify (4p⁵q²) × (3p⁻⁶q) ÷ (6pq⁴).

Numbers:  4 × 3 ÷ 6 = 2
p:        5 + (−6) − 1 = −2   →  p⁻²
q:        2 + 1 − 4 = −1      →  q⁻¹
Answer:   2p⁻²q⁻¹ = 2/(p²q)

Worked example 2. Simplify 12y⁷ ÷ 8y³. The numbers give 12/8 = 3/2 and the powers give y⁷⁻³ = y⁴. The answer is 3y⁴/2.

Worked example 3. Solve 3²ˣ⁻¹ = 1/27. Write 1/27 = 3⁻³. Equal bases mean equal indices, so 2x − 1 = −3, giving x = −1.

Fractional powers (Extension only)

A power of 1/n means the nth root, and a power of m/n means “root, then power”:

a^(1/n) = ⁿ√a        a^(m/n) = (ⁿ√a)ᵐ

To use this, you also need (aᵐ)ⁿ = aᵐⁿ, which is why a^(m/n) can be built from the root and the power. Take the root first, because it keeps the numbers small.

Worked example 4 (Extension only). Evaluate 625^(3/4).

625^(1/4) = 5   (since 5⁴ = 625)
5³ = 125

Worked example 5 (Extension only). Evaluate 8^(−2/3). The negative sign gives a reciprocal: 8^(−2/3) = 1 ÷ 8^(2/3). Then 8^(1/3) = 2 and 2² = 4, so the answer is 1/4.

Worked example 6 (Extension only). Simplify (16x⁸)^(3/4). Apply the power to each factor: 16^(3/4) = 2³ = 8, and (x⁸)^(3/4) = x⁶. The answer is 8x⁶.

Worked example 7 (Extension only). Solve x^(3/2) = 125. Raise both sides to the power 2/3, because (x^(3/2))^(2/3) = x. So x = 125^(2/3) = 5² = 25.

N7: Exact calculation

“Exact” means you do not round part-way. Keep fractions as fractions (and surds as surds) until the end. A decimal such as 0.333 stands in for 1/3 but is not equal to it, so repeated rounding drifts away from the true answer.

Fractions (Core)

• Add or subtract: use a common denominator, then add or subtract the numerators.
• Multiply: numerators together and denominators together. Cancel common factors first to keep numbers small.
• Divide: turn the second fraction upside down and multiply.
• Mixed numbers: convert to improper fractions before you start.

Worked example 8. Calculate 3²⁄₅ − 1³⁄₄.

17/5 − 7/4 = 68/20 − 35/20 = 33/20 = 1 13/20

Worked example 9. Calculate 2²⁄₃ × 1⁴⁄₅. Convert: 8/3 × 9/5. Cancel the 3 with the 9: 8/1 × 3/5 = 24/5 = 4⁴⁄₅.

Worked example 10. Calculate 5/6 ÷ 10/21. Multiply by the reciprocal: 5/6 × 21/10. Cancel 5 with 10, and 21 with 6 (a factor of 3): (1 × 7)/(2 × 2) = 7/4.

Worked example 11. Calculate (3/5 + 1/2) ÷ (7/10 − 1/4). Work out each bracket first.

3/5 + 1/2 = 6/10 + 5/10 = 11/10
7/10 − 1/4 = 14/20 − 5/20 = 9/20
11/10 ÷ 9/20 = 11/10 × 20/9 = 22/9 = 2 4/9

Surds (Extension only)

Roots like √2 and √7 are irrational: no whole number or fraction equals them. Leaving them as √2 or √7 keeps them exact, and these are called surds. These rules let you simplify:

√(ab) = √a × √b        √(a/b) = √a / √b

To simplify, pull out the largest square factor.

Worked example 12 (Extension only). Simplify √200 and 3√12 + √75.

√200 = √(100 × 2) = 10√2
3√12 = 3 × 2√3 = 6√3
√75 = 5√3
Sum = 6√3 + 5√3 = 11√3

You can only add or subtract surds that have the same root after simplifying, like terms in algebra.

Expanding brackets works as in algebra, with √a × √a = a.

Worked example 13 (Extension only). Expand (2 + √5)(3 − √5).

= 6 − 2√5 + 3√5 − (√5 × √5)
= 6 + √5 − 5
= 1 + √5

Worked example 14 (Extension only). Expand (4 + √3)² = 16 + 8√3 + 3 = 19 + 8√3. Note that (√7 + 2)(√7 − 2) = 7 − 4 = 3 because the middle terms cancel (difference of two squares).

Rationalising a denominator (Extension only)

Rationalising means rewriting a fraction so the denominator is a whole number.

• For a single surd, multiply top and bottom by that surd.
• For a denominator a + √b, use the conjugate a − √b on both top and bottom. The two denominators multiply to a² − b, which is rational.

Worked example 15 (Extension only). Rationalise 6/√3 and 5/(2√5).

6/√3 = 6√3/3 = 2√3
5/(2√5) = 5√5/(2 × 5) = √5/2

Worked example 16 (Extension only). Write 10/(3 + √2) with a rational denominator.

10(3 − √2) / [(3 + √2)(3 − √2)] = (30 − 10√2)/(9 − 2) = (30 − 10√2)/7

Worked example 17 (Extension only). Write (1 + √3)/(2 − √3) in the form a + b√3. Multiply by (2 + √3)/(2 + √3). The denominator is 4 − 3 = 1. The numerator is (1 + √3)(2 + √3) = 2 + √3 + 2√3 + 3 = 5 + 3√3. So the answer is 5 + 3√3.

N8: Standard form

Standard form writes a number as A × 10ⁿ. Here A is at least 1 but less than 10, and n is a whole number (positive or negative). Large numbers have positive n and numbers between 0 and 1 have negative n.

Worked example 18. Write 45 000 000 and 0.000 062 in standard form. The decimal point moves 7 places left for the first and 5 places right for the second.

45 000 000 = 4.5 × 10⁷
0.000 062 = 6.2 × 10⁻⁵

Calculating

• Multiply or divide: do the A values first, then the powers of 10, then tidy up so that 1 ≤ A < 10.
• Add or subtract: first rewrite both numbers so they share one power of 10.

Worked example 19. Calculate (3.2 × 10⁵) × (4 × 10⁻²).

3.2 × 4 = 12.8        10⁵ × 10⁻² = 10³
12.8 × 10³ = 1.28 × 10⁴

Worked example 20. Calculate (7.2 × 10⁸) ÷ (1.8 × 10³). The A parts give 7.2 ÷ 1.8 = 4 and the powers give 10⁸⁻³ = 10⁵, so the answer is 4 × 10⁵.

Worked example 21. Calculate (6.5 × 10⁴) + (8 × 10³). Write 8 × 10³ as 0.8 × 10⁴. Then 6.5 × 10⁴ + 0.8 × 10⁴ = 7.3 × 10⁴.

Worked example 22. Light takes how long to travel 1.5 × 10¹¹ m at 3 × 10⁸ m/s? Time = distance ÷ speed = (1.5 × 10¹¹) ÷ (3 × 10⁸) = 0.5 × 10³ = 5 × 10² seconds.

Calculator displays

Your calculator may show standard form as 3.6E9 or as 3.6 × 10⁹, and 4.5E-3 means 4.5 × 10⁻³. “E” stands for “× 10 to the power”. Do not copy the display as written: in your answer, write the full form with the × and the power of 10. Check that the answer you write has the right size, for example that a negative power matches a small number.

N9: Sets

A set is a collection of objects, called elements, written in curly brackets. These are the symbols:

Symbol	Meaning
ξ	the universal set (everything under consideration)
n(A)	the number of elements in A
A′	the complement of A: elements of ξ not in A
A ∪ B	union: in A, or in B, or in both
A ∩ B	intersection: in both A and B

In a Venn diagram, a rectangle represents ξ and each set is a circle. Overlapping circles show A ∩ B.

Worked example 23. ξ = {integers from 1 to 18}, A = {multiples of 4}, B = {factors of 24}.

A = {4, 8, 12, 16}      B = {1, 2, 3, 4, 6, 8, 12}
A ∩ B = {4, 8, 12}
A ∪ B = {1, 2, 3, 4, 6, 8, 12, 16}, so n(A ∪ B) = 8
A′ = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18}
(A ∪ B)′ = {5, 7, 9, 10, 11, 13, 14, 15, 17, 18}, so n((A ∪ B)′) = 10

For two sets, add the sizes of A and B, then take away the size of the overlap, so that the overlap is counted once. Check: n(A ∪ B) = 4 + 7 − 3 = 8.

Solving problems with Venn diagrams

Start with the most-overlapped region and work outwards. Fill the middle first, then subtract from each set total.

Worked example 24. Of 40 students, 24 play football, 18 play tennis and 7 play neither. How many play both?

Play at least one sport: 40 − 7 = 33
n(F ∩ T) = 24 + 18 − 33 = 9
Football only = 24 − 9 = 15     Tennis only = 18 − 9 = 9
Check: 15 + 9 + 9 + 7 = 40

Worked example 25. A club of 60 members: 25 play chess, 20 act, 22 play music. 8 do chess and acting, 9 do chess and music, 7 do acting and music, and 4 do all three.

Centre (all three) = 4
Chess and acting only = 8 − 4 = 4
Chess and music only = 9 − 4 = 5
Acting and music only = 7 − 4 = 3
Chess only = 25 − 4 − 4 − 5 = 12
Acting only = 20 − 4 − 4 − 3 = 9
Music only = 22 − 5 − 4 − 3 = 10
Total in circles = 12 + 9 + 10 + 4 + 5 + 3 + 4 = 47
Outside the circles = 60 − 47 = 13

Common errors

• Writing 5⁻² as −25 or −10. It is 1/25.
• Adding indices when multiplying different bases: 2³ × 3² is not 6⁵.
• Applying a fractional power to the numbers but forgetting the letters in (16x⁸)^(3/4).
• Adding surds with different roots: √2 + √3 is not √5.
• Forgetting the middle terms when squaring (4 + √3)²; the answer is not 16 + 3.
• Leaving 12.8 × 10³ as it is. A must satisfy 1 ≤ A < 10.
• Adding standard-form numbers by adding the powers of 10.
• Counting A ∩ B twice in n(A ∪ B), or filling the outer regions of a Venn diagram before the centre.

Next, work through the revision notes and then the practice questions. The algebra notation and manipulation guide covers index laws in algebraic expressions, and the diagnostics help you decide what to revise.

Official syllabus

Source: OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.1.1, statements N6 to N9.