Lesson 1: Number

Cambridge O Level Mathematics 4024 / IGCSE Mathematics 0580 — Syllabus Reference: Number

Lesson 1 of 10
10% complete

1. Types of Number

Before solving any mathematical problem, you must understand the different types of numbers and how they relate to each other.

TypeDefinitionExamples
Natural NumbersPositive counting numbers — used for counting objects.1, 2, 3, 4, 5, ...
IntegersAll whole numbers — positive, negative, and zero. No fractions or decimals...., −3, −2, −1, 0, 1, 2, 3, ...
Prime NumbersA number greater than 1 with exactly two factors — 1 and itself.2, 3, 5, 7, 11, 13, 17, 19, 23, ...
Composite NumbersA positive integer with more than two factors — not prime.4, 6, 8, 9, 10, 12, ...
Square NumbersThe result of multiplying an integer by itself.1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Cube NumbersThe result of multiplying an integer by itself three times.1, 8, 27, 64, 125, 216
Rational NumbersAny number that can be written as a fraction p/q where p and q are integers and q ≠ 0.½, 0.75, −3, 0, 2.333...
Irrational NumbersNumbers that cannot be written as a fraction — their decimal expansion is non-terminating and non-repeating.√2, √3, π, √5
⚠ Common Mistakes:
• 1 is NOT a prime number — it has only one factor (itself).
• 2 is the ONLY even prime number.
• 0 is neither positive nor negative — it is an integer but not a natural number.

Factors and Multiples

Factor: A whole number that divides exactly into another number with no remainder.
Factors of 12: 1, 2, 3, 4, 6, 12
Multiple: The result of multiplying a number by a positive integer.
Multiples of 4: 4, 8, 12, 16, 20, ...

Highest Common Factor (HCF) and Lowest Common Multiple (LCM)

HCF — Highest Common Factor

The largest factor that divides exactly into two or more numbers.

Method — Prime Factorisation:
Write each number as a product of prime factors. The HCF is the product of the common prime factors using the lowest power of each.

LCM — Lowest Common Multiple

The smallest number that is a multiple of two or more numbers.

Method — Prime Factorisation:
Write each number as a product of prime factors. The LCM is the product of all prime factors using the highest power of each.

📐 Worked Example 1 — Find the HCF and LCM of 36 and 84

1
Write each number as a product of prime factors:
36 = 2² × 3²
84 = 2² × 3 × 7
2
HCF = product of COMMON factors at LOWEST power:
Common factors: 2² and 3¹
HCF = 2² × 3 = 4 × 3 = 12
3
LCM = product of ALL factors at HIGHEST power:
All factors: 2², 3², 7
LCM = 2² × 3² × 7 = 4 × 9 × 7 = 252

2. Fractions, Decimals and Percentages

Fractions

OperationRuleExample
Add / SubtractFind a common denominator first, then add/subtract numerators.⅓ + ¼ = 4/12 + 3/12 = 7/12
MultiplyMultiply numerators together and denominators together. Simplify.⅔ × ¾ = 6/12 = ½
DivideMultiply by the reciprocal of the second fraction (flip and multiply).⅔ ÷ ¼ = ⅔ × 4/1 = 8/3 = 2⅔
Mixed NumbersConvert to improper fractions first, then operate.2½ + 1¾ = 5/2 + 7/4 = 10/4 + 7/4 = 17/4 = 4¼

Converting Between Fractions, Decimals and Percentages

ConvertMethodExample
Fraction → DecimalDivide numerator by denominator.3/8 = 3 ÷ 8 = 0.375
Decimal → FractionWrite as fraction over appropriate power of 10. Simplify.0.45 = 45/100 = 9/20
Fraction → PercentageMultiply by 100.3/8 × 100 = 37.5%
Percentage → DecimalDivide by 100.37.5% ÷ 100 = 0.375
Percentage → FractionWrite over 100. Simplify.45% = 45/100 = 9/20

Percentage Calculations

Finding a Percentage of an Amount

Multiply the amount by the percentage as a decimal.

Find 35% of $240:
0.35 × 240 = $84

Percentage Increase / Decrease

Percentage change = (Change ÷ Original) × 100

Price rises from $80 to $92:
Change = 12. Percentage = (12/80) × 100 = 15%

Reverse Percentage

Find the original value before a percentage change.

After 20% increase, price = $120. Find original:
Original = 120 ÷ 1.20 = $100

Compound Interest

Amount = P × (1 + r/100)ⁿ
where P = principal, r = rate %, n = years.

$500 at 8% for 3 years:
500 × (1.08)³ = 500 × 1.2597 = $629.86

📐 Worked Example 2 — Reverse Percentage

A jacket is sold for $68 after a 15% discount. Find the original price.

1
After a 15% discount, the price represents 85% of the original.
85% of Original = $68
2
Divide by 0.85 to find 100%:
Original = 68 ÷ 0.85 = $80
3
Check: 15% of $80 = $12. $80 − $12 = $68 ✓

3. Standard Form (Scientific Notation)

Standard Form: A number written in the form A × 10ⁿ, where 1 ≤ A < 10 and n is an integer (positive or negative). Used to express very large or very small numbers concisely.

Large Numbers (positive n)

Move the decimal point LEFT. Count the moves = n.

4,750,000 = 4.75 × 10⁶

302,000 = 3.02 × 10⁵

Small Numbers (negative n)

Move the decimal point RIGHT. Count the moves = n (negative).

0.000047 = 4.7 × 10⁻⁵

0.00302 = 3.02 × 10⁻³

Calculating with Standard Form

📐 Worked Example 3 — Multiplying in Standard Form

Calculate (3.2 × 10⁴) × (2.5 × 10³). Give your answer in standard form.

1
Multiply the A values together:
3.2 × 2.5 = 8.0
2
Add the powers of 10:
10⁴ × 10³ = 10⁷
3
Combine: 8.0 × 10⁷
Check: A = 8.0, which satisfies 1 ≤ A < 10 ✓

📐 Worked Example 4 — Dividing in Standard Form

Calculate (6.6 × 10⁵) ÷ (3 × 10⁻²). Give your answer in standard form.

1
Divide the A values:
6.6 ÷ 3 = 2.2
2
Subtract the powers (dividing = subtract indices):
10⁵ ÷ 10⁻² = 10⁵⁻⁽⁻²⁾ = 10⁷
3
Answer: 2.2 × 10⁷
⚠ After multiplying/dividing: Always check that the A value is between 1 and 10. If the result is e.g. 12.4 × 10³, rewrite as 1.24 × 10⁴. If it is 0.6 × 10⁵, rewrite as 6.0 × 10⁴.

4. Ratio and Proportion

Ratio: A comparison of two or more quantities of the same kind. Written as a : b. Always simplify ratios by dividing by the HCF.

Dividing in a Given Ratio

📐 Worked Example 5 — Sharing in a Ratio

Divide $360 between Ali and Sara in the ratio 5 : 4.

1
Find the total number of parts:
5 + 4 = 9 parts
2
Find the value of one part:
$360 ÷ 9 = $40 per part
3
Multiply each share:
Ali = 5 × $40 = $200
Sara = 4 × $40 = $160
Check: $200 + $160 = $360 ✓

Direct and Inverse Proportion

Direct Proportion

As one quantity increases, the other increases at the same rate.

y ∝ x → y = kx

Example: If 5 metres of fabric costs $15, find the cost of 8 metres.
k = 15/5 = 3. Cost = 3 × 8 = $24

Inverse Proportion

As one quantity increases, the other decreases at the same rate.

y ∝ 1/x → y = k/x → xy = k

Example: 4 workers take 15 days. How long for 6 workers?
k = 4 × 15 = 60. Time = 60/6 = 10 days

5. Rate — Speed, Density, and Pressure

Speed

Speed = Distance ÷ Time
Distance = Speed × Time
Time = Distance ÷ Speed

Density

Density = Mass ÷ Volume
Mass = Density × Volume
Volume = Mass ÷ Density

Pressure

Pressure = Force ÷ Area
Force = Pressure × Area
Area = Force ÷ Pressure
Units Matter! Always check units are consistent before calculating.
Speed: km/h or m/s. If speed is in km/h and time in minutes → convert minutes to hours first.
Converting m/s to km/h: multiply by 3.6. Converting km/h to m/s: divide by 3.6.

📐 Worked Example 6 — Average Speed

A car travels 120 km in 1 hour 30 minutes. Find the average speed in km/h.

1
Convert time to hours:
1 hour 30 minutes = 1.5 hours
2
Apply the formula:
Speed = 120 ÷ 1.5 = 80 km/h

6. Estimation, Rounding, and Approximation

Rounding Methods

MethodRuleExample: 0.036482
Decimal Places (d.p.) Count digits after the decimal point. Look at the next digit — if 5 or more, round up. 2 d.p. → 0.04  |  3 d.p. → 0.036  |  4 d.p. → 0.0365
Significant Figures (s.f.) Start counting from the first non-zero digit. Zeros between non-zero digits count. Leading zeros do NOT count. 1 s.f. → 0.04  |  2 s.f. → 0.036  |  3 s.f. → 0.0365
Significant Figures — Key Rules:
• Leading zeros (before first non-zero digit) are NOT significant: 0.0045 has 2 s.f.
• Zeros between non-zero digits ARE significant: 3.04 has 3 s.f.
• Trailing zeros after a decimal point ARE significant: 2.50 has 3 s.f.
• Trailing zeros in a whole number are ambiguous — use standard form to clarify: 3.00 × 10³ has 3 s.f.

Limits of Accuracy

Limits of Accuracy: When a measurement is rounded to a given degree of accuracy, the true value lies within a range called the limits of accuracy.

If a value x is rounded to the nearest unit u:
Lower Bound = x − ½u    Upper Bound = x + ½u

📐 Worked Example 7 — Limits of Accuracy

A length is measured as 8.4 cm to the nearest 0.1 cm. Write down the upper and lower bounds.

1
The unit of rounding = 0.1 cm. Half of this = 0.05 cm.
2
Lower Bound = 8.4 − 0.05 = 8.35 cm
Upper Bound = 8.4 + 0.05 = 8.45 cm
3
The true length satisfies: 8.35 ≤ length < 8.45
Note: the upper bound uses strict inequality (<) not ≤.
Upper/Lower Bounds in Calculations:
Maximum result = Upper Bound of numerator ÷ Lower Bound of denominator
Minimum result = Lower Bound of numerator ÷ Upper Bound of denominator

7. Surds

Surd: An irrational number left in square root (or cube root) form because it cannot be simplified to a rational number. Examples: √2, √3, √5, 3√7.

Rules for Surds

Surd Laws:
√a × √b = √(ab)     Example: √3 × √5 = √15
√a ÷ √b = √(a/b)     Example: √12 ÷ √3 = √4 = 2
√(a²) = a     Example: √25 = 5
(√a)² = a     Example: (√7)² = 7

Simplifying Surds

📐 Worked Example 8 — Simplifying Surds

Simplify: (a) √48   (b) 3√2 × 4√6

1
Part (a): Find the largest perfect square factor of 48.
48 = 16 × 3
√48 = √(16 × 3) = √16 × √3 = 4√3
2
Part (b): Multiply the integer parts and the surd parts separately.
3√2 × 4√6 = (3 × 4) × (√2 × √6) = 12 × √12
√12 = √(4 × 3) = 2√3
= 12 × 2√3 = 24√3

Rationalising the Denominator

Rationalising: Eliminating surds from the denominator of a fraction. Multiply both numerator and denominator by the surd in the denominator.

📐 Worked Example 9 — Rationalising the Denominator

Rationalise: 6/√3

1
Multiply top and bottom by √3:
6/√3 × √3/√3 = 6√3/(√3 × √3) = 6√3/3
2
Simplify:
6√3/3 = 2√3

8. Number Sequences

A sequence is an ordered list of numbers following a pattern. The nth term formula allows you to find any term without listing all previous terms.

Types of Sequences

TypePatternExamplenth Term
ArithmeticAdd a fixed number (common difference d) each time.3, 7, 11, 15, ... (d = 4)4n − 1
GeometricMultiply by a fixed number (common ratio r) each time.2, 6, 18, 54, ... (r = 3)2 × 3ⁿ⁻¹
Square NumbersSequence of n².1, 4, 9, 16, 25, ...
FibonacciEach term is the sum of the two previous terms.1, 1, 2, 3, 5, 8, 13, ...No simple formula

Finding the nth Term of an Arithmetic Sequence

nth Term of Arithmetic Sequence

nth term = a + (n − 1)d

where a = first term, d = common difference

Simplified form: nth term = dn + (a − d)

📐 Worked Example 10 — nth Term

Find the nth term of the sequence: 5, 9, 13, 17, ...

1
Find the common difference d:
d = 9 − 5 = 4
2
First term a = 5. Apply the formula:
nth term = 4n + (5 − 4) = 4n + 1
3
Check: n=1: 4(1)+1=5 ✓   n=2: 4(2)+1=9 ✓   n=4: 4(4)+1=17 ✓

📝 Exam Practice Questions

Q1 [2 marks] — Find the HCF and LCM of 60 and 90.

Working:
60 = 2² × 3 × 5
90 = 2 × 3² × 5
HCF = 2¹ × 3¹ × 5¹ = 30
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180

Q2 [2 marks] — Write 0.00000823 in standard form.

Answer: Move decimal 6 places right → A = 8.23
8.23 × 10⁻⁶
Exam Tip: For small numbers, the power is negative. Count how many places you move the decimal RIGHT to get A between 1 and 10.

Q3 [3 marks] — A price increases from $240 to $276. Calculate the percentage increase.

Working:
Change = 276 − 240 = $36
Percentage increase = (36 ÷ 240) × 100 = 15%

Q4 [3 marks] — A length is measured as 12.6 cm correct to 1 decimal place. Write down the lower and upper bounds of this length.

Working:
Unit = 0.1 cm → half unit = 0.05 cm
Lower bound = 12.6 − 0.05 = 12.55 cm
Upper bound = 12.6 + 0.05 = 12.65 cm
So: 12.55 ≤ length < 12.65
Exam Tip: The upper bound uses strict < (not ≤) because the value would round UP to the next value at exactly 12.65.

Q5 [3 marks] — Simplify fully: √75 − √27

Working:
√75 = √(25 × 3) = 5√3
√27 = √(9 × 3) = 3√3
5√3 − 3√3 = 2√3
Exam Tip: You can only add or subtract surds if they have the same surd part (like terms). Always simplify each surd fully first.

Q6 [2 marks] — Calculate (4.5 × 10³) + (6 × 10²). Give your answer in standard form.

Working:
Convert to ordinary numbers: 4500 + 600 = 5100
OR: 4.5 × 10³ = 45 × 10² → (45 + 6) × 10² = 51 × 10² = 5.1 × 10³
Exam Tip: When adding/subtracting in standard form, convert to the same power of 10 first, OR convert to ordinary numbers and then back to standard form.

Q7 [3 marks] — The nth term of a sequence is 3n − 5. (a) Find the 8th term. (b) Find the value of n for which the nth term equals 40. (c) Is 50 a term in this sequence? Justify your answer.

(a) n = 8: 3(8) − 5 = 24 − 5 = 19

(b) 3n − 5 = 40 → 3n = 45 → n = 15

(c) 3n − 5 = 50 → 3n = 55 → n = 55/3 = 18.33...
Since n is not a whole number (integer), 50 is NOT a term in this sequence.
Exam Tip: To test if a value is in a sequence, set the nth term equal to the value and solve for n. If n is a positive integer → yes. If not → no.

Q8 [4 marks] — Ali invests $2,500 at a compound interest rate of 6% per year. Calculate the total amount after 4 years. Give your answer correct to the nearest dollar.

Working:
Amount = P × (1 + r/100)ⁿ
Amount = 2500 × (1.06)⁴
(1.06)⁴ = 1.26248...
Amount = 2500 × 1.26248 = 3156.19...
Amount = $3,156 (to nearest dollar)
Exam Tip: Always use the compound interest formula for "per year" problems. Never add 6% × 4 = 24% — that gives simple interest, not compound interest. The difference grows significantly over many years.
← Back to Maths Hub Lesson 2: Algebra I →