IGCSE Mathematics: Number & Algebra
Number and Algebra form the backbone of IGCSE Mathematics, covering roughly half of the marks across both Cambridge (0580) and Edexcel (4MA1) specifications. This guide covers every major topic with clear explanations, fully worked examples, and exam-style practice problems.
Exam Board Note
This guide covers content from Cambridge IGCSE Mathematics (0580) and Edexcel International GCSE Mathematics (4MA1). Both boards assess the same core topics, though Cambridge distinguishes Extended (A*–E) and Core (C–G) tiers. Extended-only content is marked accordingly.
1. Number Types and Sets
Understanding which type of number you are dealing with is fundamental to IGCSE Number work. The main sets of numbers are nested within one another.
Key Number Sets
Natural numbers $\mathbb{N}$: $\{1, 2, 3, 4, \ldots\}$ (positive integers; some definitions include 0).
Integers $\mathbb{Z}$: $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$
Rational numbers $\mathbb{Q}$: numbers expressible as $\frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$. Includes all terminating and recurring decimals.
Irrational numbers: cannot be expressed as a fraction; decimal expansion is non-terminating and non-recurring. Examples: $\sqrt{2},\, \pi,\, e$.
Real numbers $\mathbb{R}$: all rational and irrational numbers.
Factors, Multiples, Primes
A prime number has exactly two factors: 1 and itself. The primes begin: 2, 3, 5, 7, 11, 13, 17, 19, 23, … Note that 1 is not prime.
Every integer greater than 1 has a unique prime factorisation (Fundamental Theorem of Arithmetic). We write it using index notation.
Worked Example 1.1 — Prime Factorisation, HCF and LCM
Find the HCF and LCM of 360 and 504 by prime factorisation.
Step 1 Factorise each number:
$$360 = 2^3 \times 3^2 \times 5$$ $$504 = 2^3 \times 3^2 \times 7$$
Step 2 HCF: take the lowest power of each shared prime factor:
$$\text{HCF}(360, 504) = 2^3 \times 3^2 = 8 \times 9 = 72$$
Step 3 LCM: take the highest power of every prime factor that appears:
$$\text{LCM}(360, 504) = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$$
Surds (Extended)
Surds are irrational square roots that cannot be simplified to an integer. Key rules:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}, \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}, \qquad (\sqrt{a})^2 = a$$
Rationalising the denominator: multiply numerator and denominator by the conjugate to remove surds from the denominator.
Worked Example 1.2 — Simplifying Surds
Simplify: (a) $\sqrt{72}$, (b) $\dfrac{5}{\sqrt{3}-1}$.
(a) Find the largest perfect square factor of 72:
$$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$$
(b) Multiply by the conjugate $(\sqrt{3}+1)$:
$$\frac{5}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{5(\sqrt{3}+1)}{(\sqrt{3})^2 - 1^2} = \frac{5(\sqrt{3}+1)}{3 - 1} = \frac{5(\sqrt{3}+1)}{2}$$
2. Fractions, Decimals and Percentages
Fluency with converting between fractions, decimals, and percentages is tested heavily in both tiers.
Conversion Rules
Fraction to decimal: divide numerator by denominator.
Decimal to percentage: multiply by 100.
Percentage to fraction: write over 100 and simplify.
Recurring decimal to fraction: use the algebraic method below.
Worked Example 2.1 — Recurring Decimal to Fraction
Express $0.\overline{36} = 0.363636\ldots$ as a fraction in its simplest form.
Step 1 Let $x = 0.363636\ldots$
Step 2 The recurring block has 2 digits, so multiply by $100$:
$$100x = 36.363636\ldots$$
Step 3 Subtract: $100x - x = 36$, so $99x = 36$.
Step 4 $x = \dfrac{36}{99} = \dfrac{4}{11}$.
Percentage Problems
The most common percentage calculations at IGCSE involve percentage increase/decrease, reverse percentage, and compound interest.
Worked Example 2.2 — Compound Interest
A car is bought for £12,000. Its value depreciates by 15% per year. Find its value after 3 years, to the nearest pound.
Step 1 Each year, the car retains $100\% - 15\% = 85\%$ of its value. The multiplier is $0.85$.
Step 2 After 3 years:
$$V = 12000 \times 0.85^3 = 12000 \times 0.614125 = 7369.50$$
Result The car is worth £7,370 to the nearest pound.
Worked Example 2.3 — Reverse Percentage
After a 20% increase, a jacket costs £96. Find the original price.
Step 1 £96 represents $100\% + 20\% = 120\%$ of the original price.
Step 2 Original price $= \dfrac{96}{1.20} = £80$.
3. Standard Form (Scientific Notation)
Standard form expresses any number as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer. This is tested on both non-calculator and calculator papers.
Worked Example 3.1 — Operations in Standard Form
Calculate $(3.2 \times 10^5) \times (4.0 \times 10^{-3})$, giving your answer in standard form.
Step 1 Multiply the numbers and powers of 10 separately:
$$3.2 \times 4.0 = 12.8, \qquad 10^5 \times 10^{-3} = 10^{5+(-3)} = 10^2$$
Step 2 Combine: $12.8 \times 10^2$. Since $12.8 \geq 10$, adjust:
$$12.8 \times 10^2 = 1.28 \times 10^1 \times 10^2 = 1.28 \times 10^3$$
Worked Example 3.2 — Division in Standard Form
Calculate $\dfrac{9.6 \times 10^{-2}}{1.2 \times 10^4}$, giving your answer in standard form.
Step 1 $\dfrac{9.6}{1.2} = 8.0$, $\quad 10^{-2} \div 10^4 = 10^{-2-4} = 10^{-6}$
Result $8.0 \times 10^{-6}$
4. Ratio and Proportion
Ratio
Ratios compare quantities of the same kind. Simplify by dividing by the HCF. To share in a ratio $a : b$, divide the total into $a + b$ equal parts.
Worked Example 4.1 — Sharing in a Ratio
Share £350 among Alice, Bob, and Carol in the ratio $3 : 2 : 2$.
Step 1 Total parts: $3 + 2 + 2 = 7$. Each part $= \dfrac{350}{7} = £50$.
Step 2 Alice gets $3 \times 50 = £150$, Bob gets $2 \times 50 = £100$, Carol gets $2 \times 50 = £100$.
Direct and Inverse Proportion
Two quantities $y$ and $x$ are in direct proportion if $y = kx$ for some constant $k$. They are in inverse proportion if $y = \dfrac{k}{x}$.
Worked Example 4.2 — Inverse Proportion
$y$ is inversely proportional to $x^2$. When $x = 3$, $y = 4$. Find $y$ when $x = 6$.
Step 1 Write $y = \dfrac{k}{x^2}$. Substitute: $4 = \dfrac{k}{9}$, so $k = 36$.
Step 2 When $x = 6$: $y = \dfrac{36}{36} = 1$.
5. Algebraic Manipulation
Expanding and Factorising
Key identities to memorise:
$$(a+b)^2 = a^2 + 2ab + b^2$$ $$(a-b)^2 = a^2 - 2ab + b^2$$ $$(a+b)(a-b) = a^2 - b^2$$
Worked Example 5.1 — Factorising Quadratics
Factorise completely: (a) $6x^2 - x - 2$, (b) $4x^2 - 25$.
(a) We need two numbers that multiply to $6 \times (-2) = -12$ and add to $-1$: those are $-4$ and $3$. Split the middle term:
$$6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x + 1)(3x - 2)$$
(b) Difference of two squares: $4x^2 - 25 = (2x)^2 - 5^2 = (2x+5)(2x-5)$.
Algebraic Fractions (Extended)
Worked Example 5.2 — Simplifying Algebraic Fractions
Simplify $\dfrac{x^2 - 4}{x^2 - x - 6}$.
Step 1 Factorise numerator: $x^2 - 4 = (x-2)(x+2)$.
Step 2 Factorise denominator: $x^2 - x - 6 = (x-3)(x+2)$.
Step 3 Cancel common factor $(x+2)$:
$$\frac{(x-2)(x+2)}{(x-3)(x+2)} = \frac{x-2}{x-3}$$
Changing the Subject of a Formula
Worked Example 5.3 — Rearranging Formulae
Make $r$ the subject of $V = \dfrac{4}{3}\pi r^3$.
Step 1 Multiply both sides by $\dfrac{3}{4\pi}$: $\quad r^3 = \dfrac{3V}{4\pi}$
Step 2 Cube root both sides: $\quad r = \sqrt[3]{\dfrac{3V}{4\pi}}$
6. Equations and Inequalities
Solving Quadratic Equations
Methods: factorising, completing the square, and the quadratic formula. For $ax^2 + bx + c = 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The discriminant $\Delta = b^2 - 4ac$ tells you the number of real solutions: $\Delta > 0$ gives two distinct roots, $\Delta = 0$ gives one repeated root, $\Delta < 0$ gives no real roots.
Worked Example 6.1 — Completing the Square
Solve $x^2 - 6x + 7 = 0$ by completing the square.
Step 1 Write $x^2 - 6x = (x-3)^2 - 9$. So the equation becomes:
$$(x-3)^2 - 9 + 7 = 0 \implies (x-3)^2 = 2$$
Step 2 Take square roots: $x - 3 = \pm\sqrt{2}$, giving $x = 3 \pm \sqrt{2}$.
Simultaneous Equations
Worked Example 6.2 — One Linear, One Quadratic
Solve simultaneously: $y = 2x - 1$ and $x^2 + y^2 = 10$.
Step 1 Substitute $y = 2x - 1$ into the circle equation:
$$x^2 + (2x-1)^2 = 10 \implies x^2 + 4x^2 - 4x + 1 = 10 \implies 5x^2 - 4x - 9 = 0$$
Step 2 Factorise: $(5x - 9)(x + 1) = 0$, so $x = \dfrac{9}{5}$ or $x = -1$.
Step 3 Find corresponding $y$ values:
- $x = \frac{9}{5}$: $y = 2 \cdot \frac{9}{5} - 1 = \frac{13}{5}$
- $x = -1$: $y = 2(-1) - 1 = -3$
Solutions $\left(\dfrac{9}{5}, \dfrac{13}{5}\right)$ and $(-1, -3)$.
Inequalities
Solve inequalities like equations, but reverse the inequality sign when multiplying or dividing by a negative number. For quadratic inequalities, always sketch the parabola.
Worked Example 6.3 — Quadratic Inequality
Solve $x^2 - 5x + 6 < 0$.
Step 1 Factorise: $(x-2)(x-3) < 0$. Roots are $x = 2$ and $x = 3$.
Step 2 The parabola opens upward ($a > 0$), so it is below zero between the roots.
Solution $2 < x < 3$.
7. Sequences
Arithmetic Sequences
Arithmetic Sequence Formulae
For a sequence with first term $a$ and common difference $d$:
$$n\text{th term: } u_n = a + (n-1)d$$
$$\text{Sum of first } n \text{ terms: } S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(a + l)$$
where $l$ is the last term.
Worked Example 7.1 — Finding the nth Term
An arithmetic sequence has 3rd term 11 and 8th term 31. Find the $n$th term and the sum of the first 20 terms.
Step 1 $u_3 = a + 2d = 11$ and $u_8 = a + 7d = 31$. Subtracting: $5d = 20$, so $d = 4$. Then $a = 11 - 8 = 3$.
Step 2 $n$th term: $u_n = 3 + (n-1) \times 4 = 4n - 1$.
Step 3 $S_{20} = \dfrac{20}{2}(2 \times 3 + 19 \times 4) = 10(6 + 76) = 10 \times 82 = 820$.
Geometric Sequences
Geometric Sequence Formulae
For a sequence with first term $a$ and common ratio $r$:
$$n\text{th term: } u_n = ar^{n-1}$$
$$\text{Sum of first } n \text{ terms: } S_n = \frac{a(r^n - 1)}{r - 1} \quad (r \neq 1)$$
Quadratic Sequences (Extended)
If the second differences are constant, the sequence is quadratic: $u_n = an^2 + bn + c$. Find $a$ from half the second difference, then use simultaneous equations for $b$ and $c$.
8. Functions
A function maps each input in the domain to exactly one output. Notation: $f : x \mapsto 2x + 3$ or $f(x) = 2x + 3$.
Worked Example 8.1 — Composite and Inverse Functions
Given $f(x) = 3x - 1$ and $g(x) = x^2 + 2$, find: (a) $fg(x)$, (b) $f^{-1}(x)$.
(a) $fg(x) = f(g(x)) = f(x^2 + 2) = 3(x^2 + 2) - 1 = 3x^2 + 5$
(b) To find $f^{-1}$, write $y = 3x - 1$ and solve for $x$:
$$x = \frac{y + 1}{3} \implies f^{-1}(x) = \frac{x + 1}{3}$$
9. Practice Problems
Problem 1
Find the HCF and LCM of 84 and 126 using prime factorisation.
Show Solution
$84 = 2^2 \times 3 \times 7$, $\quad 126 = 2 \times 3^2 \times 7$
$\text{HCF} = 2^1 \times 3^1 \times 7^1 = 42$
$\text{LCM} = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252$
Problem 2
Express $0.\overline{142857}$ as a fraction. (Hint: this is $\frac{1}{7}$.)
Show Solution
Let $x = 0.\overline{142857}$. The block has 6 digits, so multiply by $10^6 = 1{,}000{,}000$:
$1{,}000{,}000\,x = 142857.\overline{142857}$
Subtract: $999{,}999\,x = 142857$, so $x = \dfrac{142857}{999999} = \dfrac{1}{7}$.
Problem 3
Rationalise the denominator of $\dfrac{3}{\sqrt{5} + \sqrt{2}}$ and simplify.
Show Solution
Multiply by the conjugate $\dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$:
$$\frac{3(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{3(\sqrt{5} - \sqrt{2})}{5 - 2} = \frac{3(\sqrt{5} - \sqrt{2})}{3} = \sqrt{5} - \sqrt{2}$$
Problem 4
Write $(7.5 \times 10^3) + (2.5 \times 10^2)$ in standard form.
Show Solution
$7500 + 250 = 7750 = 7.75 \times 10^3$
Problem 5
$y$ is directly proportional to $\sqrt{x}$. When $x = 16$, $y = 12$. Find $y$ when $x = 25$.
Show Solution
$y = k\sqrt{x}$. Substitute: $12 = 4k$, so $k = 3$.
When $x = 25$: $y = 3\sqrt{25} = 3 \times 5 = 15$.
Problem 6
Factorise $6x^2 + 7x - 3$.
Show Solution
Find two numbers with product $6 \times (-3) = -18$ and sum $7$: those are $9$ and $-2$.
$6x^2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3)$
Problem 7
Solve $2x^2 - 7x + 3 = 0$ using the quadratic formula.
Show Solution
$a = 2, b = -7, c = 3$. Discriminant: $\Delta = 49 - 24 = 25$.
$$x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$$
$x = \dfrac{12}{4} = 3$ or $x = \dfrac{2}{4} = \dfrac{1}{2}$.
Problem 8
An arithmetic sequence has first term $5$ and common difference $3$. Which term is equal to $98$?
Show Solution
$u_n = 5 + (n-1) \times 3 = 3n + 2$. Set $3n + 2 = 98 \Rightarrow 3n = 96 \Rightarrow n = 32$.
The 32nd term equals 98.
Problem 9
Make $h$ the subject of $T = 2\pi\sqrt{\dfrac{h}{g}}$.
Show Solution
$\dfrac{T}{2\pi} = \sqrt{\dfrac{h}{g}}$
Square both sides: $\dfrac{T^2}{4\pi^2} = \dfrac{h}{g}$
$h = \dfrac{gT^2}{4\pi^2}$
Problem 10
Given $f(x) = \dfrac{x+2}{x-1}$, $x \neq 1$, find $f^{-1}(x)$ and state the value of $x$ that must be excluded from its domain.
Show Solution
Let $y = \dfrac{x+2}{x-1}$. Rearrange: $y(x-1) = x + 2 \Rightarrow xy - y = x + 2 \Rightarrow x(y-1) = y + 2$.
$$f^{-1}(x) = \frac{x+2}{x-1}$$
Interesting: this function is self-inverse. Exclude $x = 1$ from the domain of $f^{-1}$.