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IGCSE Mathematics – Chapter 1: Number

Cambridge IGCSE 0580 & Edexcel 4MA1 · Updated March 2026

This chapter covers the entire Number strand of the IGCSE Mathematics specification. Number is the foundation of all other topics: a secure understanding of primes, indices, standard form, surds, ratio and percentage enables you to tackle geometry, algebra and statistics with confidence. Work through each section in order, paying close attention to the worked examples before attempting the practice problems.

Specification Note

Content labelled Extended is required for the Extended tier (Cambridge) or Higher tier (Edexcel) only. Core/Foundation tier students should still read these sections for enrichment.

1. Types of Numbers

1.1 Number Sets

Mathematics organises numbers into overlapping sets, each with specific properties.

Key Number Sets

Example 1.1 — Classifying numbers

Classify each of the following: $-7,\; \dfrac{3}{4},\; \sqrt{9},\; \sqrt{10},\; 0.\overline{6},\; \pi$.

Step 1 $-7$: A whole number that is negative. It is an integer (also rational and real).

Step 2 $\dfrac{3}{4} = 0.75$: Written as a ratio of two integers. It is rational (and real).

Step 3 $\sqrt{9} = 3$: Equal to a natural number. It is a natural number, integer, rational and real.

Step 4 $\sqrt{10} \approx 3.16227\ldots$: Cannot be written as $\dfrac{p}{q}$. It is irrational (real).

Step 5 $0.\overline{6} = \dfrac{2}{3}$: A recurring decimal. It is rational.

Step 6 $\pi \approx 3.14159\ldots$: Non-terminating, non-recurring. It is irrational.

1.2 Primes, Factors and Multiples

Definitions

1.3 HCF and LCM

The Highest Common Factor (HCF) of two or more numbers is the largest factor shared by all of them. The Lowest Common Multiple (LCM) is the smallest multiple shared by all of them.

Method: Prime Factorisation

Example 1.2 — HCF and LCM by prime factorisation

Find the HCF and LCM of $84$ and $120$.

Step 1 Factorise each number:
$84 = 2^2 \times 3 \times 7$
$120 = 2^3 \times 3 \times 5$

Step 2 HCF — take each common prime with the lower power:
$\text{HCF} = 2^2 \times 3 = 12$

Step 3 LCM — take each prime with the higher power:
$\text{LCM} = 2^3 \times 3 \times 5 \times 7 = 840$

Exam Tip

A useful relationship: $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$. Use this to check your answers. For the example above: $12 \times 840 = 10\,080 = 84 \times 120$. ✓

Practice 1A

(a) Write $360$ as a product of prime factors.
(b) Find the HCF and LCM of $360$ and $504$.

Show Solution

(a) $360 = 2^3 \times 3^2 \times 5$

(b) $504 = 2^3 \times 3^2 \times 7$
HCF $= 2^3 \times 3^2 = 72$
LCM $= 2^3 \times 3^2 \times 5 \times 7 = 2520$

Practice 1B

Three lighthouse beams flash every $12$ seconds, $18$ seconds, and $30$ seconds respectively. They all flash together at time $t = 0$. After how many seconds will they next all flash together?

Show Solution

We need the LCM of $12$, $18$ and $30$.
$12 = 2^2 \times 3$, $\;18 = 2 \times 3^2$, $\;30 = 2 \times 3 \times 5$
LCM $= 2^2 \times 3^2 \times 5 = 180$ seconds.

2. Fractions, Decimals and Percentages

2.1 Converting Between Forms

Conversion Rules

2.2 Recurring Decimals to Fractions

A recurring decimal is rational. To convert it to a fraction, use the algebraic method.

Example 2.1 — Converting $0.\overline{36}$ to a fraction

Step 1 Let $x = 0.363636\ldots$

Step 2 Multiply by $100$ (since two digits recur): $100x = 36.363636\ldots$

Step 3 Subtract: $100x - x = 36.363636\ldots - 0.363636\ldots$
$99x = 36$

Step 4 $x = \dfrac{36}{99} = \dfrac{4}{11}$

Example 2.2 — Converting $0.1\overline{8}$ to a fraction

Step 1 Let $x = 0.1888\ldots$

Step 2 $10x = 1.888\ldots$ and $100x = 18.888\ldots$

Step 3 Subtract: $100x - 10x = 18.888\ldots - 1.888\ldots$
$90x = 17$

Step 4 $x = \dfrac{17}{90}$

2.3 Ordering Fractions and Decimals

To order fractions, convert them to a common denominator or convert all to decimals.

Example 2.3 — Ordering $\dfrac{5}{8},\; 0.6,\; \dfrac{7}{12},\; 62\%$

Convert to decimals: $\dfrac{5}{8} = 0.625$, $\;0.6$, $\;\dfrac{7}{12} \approx 0.583$, $\;62\% = 0.62$.

Ascending order: $\dfrac{7}{12} < 0.6 < 62\% < \dfrac{5}{8}$

Practice 2A

(a) Convert $0.\overline{7}$ to a fraction in its simplest form.
(b) Convert $0.2\overline{16}$ to a fraction.

Show Solution

(a) Let $x = 0.777\ldots$, so $10x = 7.777\ldots$
$9x = 7 \Rightarrow x = \dfrac{7}{9}$

(b) Let $x = 0.21616\ldots$
$10x = 2.1616\ldots$ and $1000x = 216.1616\ldots$
$990x = 214 \Rightarrow x = \dfrac{214}{990} = \dfrac{107}{495}$

3. Indices and Standard Form

3.1 Index Laws

The Index Laws

For any base $a \neq 0$ and indices $m, n$:

Example 3.1 — Evaluating with index laws

Evaluate: (a) $3^{-2}$   (b) $16^{3/4}$   (c) $\left(\dfrac{27}{8}\right)^{-2/3}$

(a) $3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$

(b) $16^{3/4} = \left(16^{1/4}\right)^3 = 2^3 = 8$

(c) $\left(\dfrac{27}{8}\right)^{-2/3} = \left(\dfrac{8}{27}\right)^{2/3} = \left(\dfrac{8^{1/3}}{27^{1/3}}\right)^2 = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$

Example 3.2 — Simplifying algebraic expressions

Simplify: (a) $\dfrac{x^5 \times x^{-2}}{x^4}$   (b) $(2a^3b^{-1})^4$

(a) $\dfrac{x^5 \times x^{-2}}{x^4} = \dfrac{x^{5+(-2)}}{x^4} = \dfrac{x^3}{x^4} = x^{3-4} = x^{-1} = \dfrac{1}{x}$

(b) $(2a^3b^{-1})^4 = 2^4 \cdot a^{12} \cdot b^{-4} = \dfrac{16a^{12}}{b^4}$

3.2 Standard Form

Standard form (scientific notation) expresses a number as $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

Example 3.3 — Writing in standard form

(a) $4\,700\,000$   (b) $0.000\,38$   (c) $3.2 \times 10^4 + 4.5 \times 10^3$

(a) Move decimal point $6$ places left: $4.7 \times 10^6$

(b) Move decimal point $4$ places right: $3.8 \times 10^{-4}$

(c) Convert to same power: $32\,000 + 4\,500 = 36\,500 = 3.65 \times 10^4$

Example 3.4 — Multiplying and dividing in standard form

Calculate $(3.6 \times 10^5) \times (2.5 \times 10^{-3})$, giving your answer in standard form.

Step 1 Multiply the number parts: $3.6 \times 2.5 = 9.0$

Step 2 Multiply the powers: $10^5 \times 10^{-3} = 10^2$

Result $9.0 \times 10^2 = 9.0 \times 10^2$

Practice 3A

(a) Evaluate $\left(\dfrac{8}{125}\right)^{-2/3}$.
(b) Write $0.000\,0456$ in standard form.
(c) Calculate $\dfrac{6 \times 10^7}{1.5 \times 10^{-2}}$ in standard form.

Show Solution

(a) $\left(\dfrac{125}{8}\right)^{2/3} = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$

(b) $4.56 \times 10^{-5}$

(c) $\dfrac{6}{1.5} \times 10^{7-(-2)} = 4 \times 10^9$

4. Surds Extended

A surd is an irrational root that cannot be simplified to a rational number. Working with surds allows exact answers without rounding.

4.1 Simplifying Surds

Rules for Surds

Example 4.1 — Simplifying surds

Simplify: (a) $\sqrt{72}$   (b) $3\sqrt{8} + 2\sqrt{18} - \sqrt{50}$

(a) $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

(b) $3\sqrt{8} = 3 \times 2\sqrt{2} = 6\sqrt{2}$
$2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$
$\sqrt{50} = 5\sqrt{2}$
$\therefore \; 6\sqrt{2} + 6\sqrt{2} - 5\sqrt{2} = 7\sqrt{2}$

4.2 Expanding Surd Expressions

Example 4.2 — Expanding and collecting

Expand and simplify $(2 + \sqrt{3})(5 - 2\sqrt{3})$.

$= 2(5) + 2(-2\sqrt{3}) + \sqrt{3}(5) + \sqrt{3}(-2\sqrt{3})$
$= 10 - 4\sqrt{3} + 5\sqrt{3} - 2 \times 3$
$= 10 - 4\sqrt{3} + 5\sqrt{3} - 6$
$= 4 + \sqrt{3}$

4.3 Rationalising the Denominator

A fraction with a surd in the denominator should be simplified by rationalising: multiplying numerator and denominator by the conjugate or by the surd itself.

Example 4.3 — Rationalising

(a) $\dfrac{6}{\sqrt{3}}$   (b) $\dfrac{4}{2 + \sqrt{5}}$

(a) $\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$

(b) Multiply by conjugate $(2 - \sqrt{5})$:
$\dfrac{4(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})} = \dfrac{8 - 4\sqrt{5}}{4 - 5} = \dfrac{8 - 4\sqrt{5}}{-1} = -8 + 4\sqrt{5}$

Practice 4A

(a) Simplify $\sqrt{180}$.
(b) Expand and simplify $(3\sqrt{2} - 1)^2$.
(c) Rationalise $\dfrac{10}{3 - \sqrt{2}}$, giving your answer in the form $a + b\sqrt{2}$.

Show Solution

(a) $\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$

(b) $(3\sqrt{2} - 1)^2 = 18 - 6\sqrt{2} + 1 = 19 - 6\sqrt{2}$

(c) $\dfrac{10(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \dfrac{30 + 10\sqrt{2}}{9 - 2} = \dfrac{30 + 10\sqrt{2}}{7}$
$= \dfrac{30}{7} + \dfrac{10}{7}\sqrt{2}$. So $a = \dfrac{30}{7}$, $b = \dfrac{10}{7}$.

5. Ratio and Proportion

5.1 Simplifying Ratios

A ratio compares two or more quantities of the same kind. Simplify by dividing all parts by their HCF, or by converting to a common unit first.

Example 5.1 — Simplifying a ratio with units

Simplify the ratio $45\text{ cm} : 1.2\text{ m}$.

Step 1 Convert to the same unit: $1.2\text{ m} = 120\text{ cm}$

Step 2 $45 : 120 = \dfrac{45}{\text{HCF}} : \dfrac{120}{\text{HCF}}$. HCF$(45, 120) = 15$.

Result $3 : 8$

5.2 Dividing in a Given Ratio

Example 5.2 — Sharing £720 in the ratio 3 : 5

Step 1 Total parts: $3 + 5 = 8$

Step 2 Value of one part: $720 \div 8 = £90$

Step 3 Shares: $3 \times 90 = £270$ and $5 \times 90 = £450$

Check: $270 + 450 = 720$ ✓

5.3 Direct Proportion

Two quantities are in direct proportion if their ratio is constant: $y = kx$ for some constant $k > 0$. The graph is a straight line through the origin.

Example 5.3 — Direct proportion

$y$ is directly proportional to $x$. When $x = 5$, $y = 35$. Find $y$ when $x = 12$.

$y = kx \Rightarrow 35 = 5k \Rightarrow k = 7$
When $x = 12$: $y = 7 \times 12 = 84$

5.4 Inverse Proportion

Two quantities are in inverse proportion if their product is constant: $y = \dfrac{k}{x}$. As one increases, the other decreases proportionally.

Example 5.4 — Inverse proportion

$y$ is inversely proportional to $x$. When $x = 4$, $y = 9$. Find $y$ when $x = 12$.

$y = \dfrac{k}{x} \Rightarrow 9 = \dfrac{k}{4} \Rightarrow k = 36$
When $x = 12$: $y = \dfrac{36}{12} = 3$

Practice 5A

(a) A recipe uses flour, butter and sugar in the ratio $8 : 3 : 2$. If $240\text{ g}$ of flour is used, how much butter and sugar are needed?
(b) $y$ is inversely proportional to $x^2$. When $x = 3$, $y = 4$. Find $y$ when $x = 6$.

Show Solution

(a) One part $= 240 \div 8 = 30\text{ g}$.
Butter $= 3 \times 30 = 90\text{ g}$; Sugar $= 2 \times 30 = 60\text{ g}$.

(b) $y = \dfrac{k}{x^2}$. $4 = \dfrac{k}{9} \Rightarrow k = 36$.
When $x = 6$: $y = \dfrac{36}{36} = 1$.

6. Percentages in Context

6.1 Percentage Increase and Decrease

Multiplier Method

An increase of $r\%$ corresponds to multiplying by $\left(1 + \dfrac{r}{100}\right)$.
A decrease of $r\%$ corresponds to multiplying by $\left(1 - \dfrac{r}{100}\right)$.

Example 6.1 — Percentage change

(a) Increase £350 by 12%.   (b) A jacket costing £180 is reduced by 35%. Find the sale price.

(a) $350 \times 1.12 = £392$

(b) $180 \times 0.65 = £117$

6.2 Reverse Percentages

Given the value after a percentage change, find the original value by dividing by the multiplier.

Example 6.2 — Reverse percentage

After a 15% increase, a salary is £32\,200. What was the original salary?

Original $= 32\,200 \div 1.15 = £28\,000$

Common Mistake

Do not find 15% of £32\,200 and subtract. The 15% was applied to the original value, not the final value.

6.3 Simple and Compound Interest

Formulae

Simple interest: $I = \dfrac{PRT}{100}$, where $P$ = principal, $R$ = rate per year, $T$ = time in years. Total = $P + I$.

Compound interest: $A = P\left(1 + \dfrac{r}{100}\right)^n$, where $n$ = number of compounding periods.

Example 6.3 — Comparing simple and compound interest

£2\,000 is invested at 5% per annum for 6 years. Calculate the total amount under (a) simple interest (b) compound interest.

(a) $I = \dfrac{2000 \times 5 \times 6}{100} = £600$. Total $= £2\,600$.

(b) $A = 2000 \times 1.05^6 = 2000 \times 1.340095\ldots \approx £2\,680.19$

Fig 1: Comparison of simple interest (blue) and compound interest (red) on £2 000 at 5% p.a. over 20 years. Drag the time slider to explore.

6.4 Depreciation

Depreciation uses the compound interest formula with a reduction: $V = P\left(1 - \dfrac{r}{100}\right)^n$.

Example 6.4 — Depreciation

A car bought for £14\,500 depreciates at 18% per year. Find its value after 3 years.

$V = 14\,500 \times 0.82^3 = 14\,500 \times 0.551368 \approx £7\,995$

Practice 6A

(a) A television is sold for £306 after a 15% reduction. Find the original price.
(b) £5\,000 is invested at 3.2% compound interest per year. How much is it worth after 4 years? Give your answer correct to the nearest penny.
(c) A motorbike worth £8\,200 depreciates at 12% per year. After how many complete years will its value first fall below £5\,000?

Show Solution

(a) $\text{Original} = 306 \div 0.85 = £360$

(b) $A = 5000 \times 1.032^4 = 5000 \times 1.13469\ldots \approx £5\,673.45$

(c) We need $8200 \times 0.88^n < 5000$, so $0.88^n < \dfrac{5000}{8200} \approx 0.6098$.
$n = 4$: $0.88^4 \approx 0.5997 < 0.6098$ ✓
$n = 3$: $0.88^3 \approx 0.6815 > 0.6098$ ✗
After 4 complete years.

Fig 2: Ratio visualisation — the line segment from 0 to 13 divided in the ratio 5 : 8. The blue point marks the division at $x = 5$.

7. Mixed Practice Problems

Question 1

Express $\sqrt{12} + \sqrt{75} - \sqrt{48}$ in the form $k\sqrt{3}$.

Show Solution

$\sqrt{12} = 2\sqrt{3}$, $\;\sqrt{75} = 5\sqrt{3}$, $\;\sqrt{48} = 4\sqrt{3}$
$2\sqrt{3} + 5\sqrt{3} - 4\sqrt{3} = 3\sqrt{3}$. So $k = 3$.

Question 2

Find the HCF and LCM of $2^3 \times 3^2 \times 5$ and $2^2 \times 3 \times 7$.

Show Solution

HCF $= 2^2 \times 3 = 12$
LCM $= 2^3 \times 3^2 \times 5 \times 7 = 2520$

Question 3

Write $0.000\,0304$ in standard form, then multiply it by $(6 \times 10^4)$. Give your answer in standard form.

Show Solution

$0.000\,0304 = 3.04 \times 10^{-5}$
$(3.04 \times 10^{-5}) \times (6 \times 10^4) = 18.24 \times 10^{-1} = 1.824 \times 10^0 = 1.824$

Question 4

$y$ is proportional to $\sqrt{x}$. When $x = 16$, $y = 10$. Find $x$ when $y = 25$.

Show Solution

$y = k\sqrt{x}$. $10 = k\sqrt{16} = 4k \Rightarrow k = 2.5$
$25 = 2.5\sqrt{x} \Rightarrow \sqrt{x} = 10 \Rightarrow x = 100$

Question 5

Evaluate $\left(\dfrac{4}{9}\right)^{-3/2}$ without a calculator.

Show Solution

$\left(\dfrac{4}{9}\right)^{-3/2} = \left(\dfrac{9}{4}\right)^{3/2} = \left(\sqrt{\dfrac{9}{4}}\right)^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}$

Question 6

Three friends share a prize of £1\,350 in the ratio $2 : 3 : 4$. How much does each receive?

Show Solution

Total parts $= 9$. One part $= £150$.
Shares: £300, £450, £600.

Question 7

Convert $0.\overline{153}$ to a fraction in its simplest form. Extended

Show Solution

Let $x = 0.153153\ldots$
$1000x = 153.153\ldots$
$999x = 153 \Rightarrow x = \dfrac{153}{999} = \dfrac{17}{111}$

Question 8

After increasing by 20%, then decreasing by 20%, is a quantity back to its original value? Show working to justify your answer.

Show Solution

Let original $= P$. After increase: $1.2P$. After decrease: $1.2P \times 0.8 = 0.96P$.
This is $96\%$ of the original — a net decrease of $4\%$. It is not back to the original value.

Question 9

Rationalise the denominator of $\dfrac{3\sqrt{2} + 1}{\sqrt{2} - 3}$. Extended

Show Solution

Multiply by $\dfrac{\sqrt{2}+3}{\sqrt{2}+3}$:
Numerator: $(3\sqrt{2}+1)(\sqrt{2}+3) = 6 + 9\sqrt{2} + \sqrt{2} + 3 = 9 + 10\sqrt{2}$
Denominator: $(\sqrt{2})^2 - 9 = 2 - 9 = -7$
Result: $\dfrac{9 + 10\sqrt{2}}{-7} = -\dfrac{9}{7} - \dfrac{10\sqrt{2}}{7}$

Question 10

A population of bacteria increases by 8% every hour. Starting with 2\,500 bacteria, after how many complete hours will the population first exceed 5\,000?

Show Solution

We need $2500 \times 1.08^n > 5000$, so $1.08^n > 2$.
Trial: $n = 9$: $1.08^9 \approx 1.999$ (just under) ✗
$n = 10$: $1.08^{10} \approx 2.159 > 2$ ✓
After 10 complete hours.