Mathematics · AQA 8300 §N1

Types of Number

Spec: §N1 ⭐⭐ 40 mins AQA · Edexcel · OCR Grade 9

Classify numbers as natural, integer, rational, irrational or real
Find prime factorisations using factor trees and repeated division
Find HCF and LCM using prime factorisation and Venn diagrams
Apply HCF and LCM to solve real-world and algebraic problems
Prove whether a given number is prime using systematic testing

🔑 Core Concepts

Number Sets

Every number belongs to one or more number sets that form a nested hierarchy. Understanding these classifications is essential throughout GCSE Mathematics — they underpin algebra, surds, and proof.

📖

DEFINITION — Natural Numbers (ℕ)

The positive counting numbers: $\mathbb{N} = \{1, 2, 3, 4, 5, \ldots\}$. At GCSE, natural numbers start at 1 (some university-level definitions include 0). They are the numbers we use to count discrete objects.

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DEFINITION — Integers (ℤ)

All whole numbers, including negatives and zero: $\mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$. Every natural number is an integer, but integers also include negative values and zero. The symbol ℤ comes from the German Zahlen (numbers).

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DEFINITION — Rational Numbers (ℚ)

Any number expressible in the form $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. This includes:

All integers (e.g. $-3 = \frac{-3}{1}$)
Terminating decimals (e.g. $0.25 = \frac{1}{4}$)
Recurring decimals (e.g. $0.\overline{3} = \frac{1}{3}$, $\ 0.\overline{142857} = \frac{1}{7}$)

The key test: can the decimal be written as a fraction? If yes, it is rational.

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DEFINITION — Irrational Numbers

Numbers that cannot be written as $\frac{p}{q}$. Their decimal form is non-terminating and non-repeating. They fill the "gaps" in the rational number line. Examples include: $\sqrt{2} = 1.41421356\ldots$, $\quad \sqrt{3}$, $\quad \pi = 3.14159265\ldots$, $\quad e = 2.71828\ldots$

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DEFINITION — Real Numbers (ℝ)

The complete set of all numbers on the number line — every rational and every irrational number. The hierarchy is: $$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$$ Every natural number is an integer; every integer is rational; every rational is real. The real numbers do not include imaginary numbers (e.g. $\sqrt{-1}$), which are beyond GCSE.

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EXAM TIP — Identifying Irrationals Quickly

$\sqrt{n}$ is irrational unless $n$ is a perfect square. Perfect squares to memorise: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$. So $\sqrt{16} = 4$ (rational), but $\sqrt{17}$ is irrational. $\pi$ and $e$ are always irrational — they cannot be expressed as exact fractions.

✗

COMMON MISTAKE — Confusing Integer and Natural Number

Students often write that integers are "all positive numbers." Integers include negative numbers and zero. The number $-7$ is an integer but not a natural number. The number $0$ is an integer but not a natural number.

Prime Numbers

Prime numbers are the fundamental building blocks of all integers. The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be written as a unique product of primes — this is why primes matter so deeply in mathematics.

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DEFINITION — Prime Number

A prime number is a natural number greater than 1 that has exactly two distinct factors: 1 and itself. The first ten primes are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.

⚠️

IMPORTANT — 1 is NOT Prime

The number 1 has only one factor (itself). A prime must have exactly two factors. If 1 were prime, the Fundamental Theorem of Arithmetic would break down — for example, $6$ could be written as $2 \times 3$ or as $1 \times 2 \times 3$ or $1^{100} \times 2 \times 3$, destroying the uniqueness of prime factorisation.

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EXAM TIP — The Only Even Prime

2 is the only even prime number. Every other even number is divisible by 2, giving it at least three factors: 1, 2, and itself. In any question asking you to list primes, remember to include 2.

Primality Testing

To prove that a number $n$ is prime, we must show it has no factors other than 1 and $n$. The key insight is that we only need to test primes up to $\sqrt{n}$.

Why $\sqrt{n}$? If $n = a \times b$ where $a \leq b$, then $a \leq \sqrt{n}$. So every factor pair has one member that is $\leq \sqrt{n}$. If no prime up to $\sqrt{n}$ divides $n$, there can be no factorisation, and $n$ is prime.

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EXAM TIP — Primality Test Procedure

To test whether $n$ is prime:

Calculate $\sqrt{n}$ and round up to the next whole number.
List all prime numbers up to that value.
Check whether any of those primes divides $n$ exactly.
If none divide $n$, conclude clearly: "$n$ is prime."

Always state $\sqrt{n}$, list the primes tested, and write the conclusion — all three steps earn marks.

Prime Factorisation

Every composite number can be expressed as a unique product of primes. This is called its prime factorisation. We always express the result in index notation.

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DEFINITION — Prime Factorisation (Index Notation)

Writing a number as a product of its prime factors, grouping repeated primes as powers. For example: $$360 = 2^3 \times 3^2 \times 5$$ This means $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$.

Method 1: Factor Tree

Split the number into any two factors (neither being 1). Keep splitting composite factors until every branch ends in a prime. Circle each prime as you reach it. This method works from any starting split — the final answer is always the same (Fundamental Theorem of Arithmetic).

✏️

WORKED EXAMPLE — Factor Tree for 180

Choose any split, e.g. $180 = 4 \times 45$:

         180
        /    \
       4      45
      / \    /  \
     2   2  9    5
           / \
          3   3

Primes collected: $2, 2, 3, 3, 5$

Group and write in index notation: $\mathbf{180 = 2^2 \times 3^2 \times 5}$

Method 2: Repeated Division (Ladder Method)

Always divide by the smallest prime that divides evenly. Continue with each result until you reach 1. This method is more systematic and less prone to errors.

✏️

WORKED EXAMPLE — Repeated Division for 180

$180 \div 2 = 90$

$90 \div 2 = 45$

$45 \div 3 = 15$

$15 \div 3 = 5$

$5 \div 5 = 1$

Divisors used (reading down): $2, 2, 3, 3, 5$

Therefore: $\mathbf{180 = 2^2 \times 3^2 \times 5}$

✗

COMMON MISTAKE — Forgetting Index Notation

Writing $180 = 2 \times 2 \times 3 \times 3 \times 5$ loses the final mark when the question says "express in index notation." Always group repeated primes: $180 = 2^2 \times 3^2 \times 5$.

Highest Common Factor (HCF)

The HCF is the largest number that divides exactly into all the given numbers. Once we have prime factorisations, we select the primes that appear in both factorisations, taking the lower power.

HCF Formula

$$\text{HCF}(a,\,b) = \text{product of SHARED prime factors at LOWEST power}$$

SHARED = the prime appears in both factorisations LOWEST power = take the minimum exponent seen in either HCF is always ≤ the smaller of the two numbers

✏️

WORKED EXAMPLE — HCF of 24 and 36

$24 = 2^3 \times 3^1 \qquad 36 = 2^2 \times 3^2$

Shared prime 2: lowest power is $\min(3,2) = 2$, so contribute $2^2$
Shared prime 3: lowest power is $\min(1,2) = 1$, so contribute $3^1$

$$\text{HCF} = 2^2 \times 3 = 4 \times 3 = \mathbf{12}$$

Lowest Common Multiple (LCM)

The LCM is the smallest positive number that is a multiple of all the given numbers — the first number where their "times tables" coincide. We include every prime that appears in either factorisation, taking the higher power.

LCM Formula

$$\text{LCM}(a,\,b) = \text{product of ALL prime factors at HIGHEST power}$$

ALL = include every prime from either factorisation HIGHEST power = take the maximum exponent seen in either LCM is always ≥ the larger of the two numbers

✏️

WORKED EXAMPLE — LCM of 24 and 36

$24 = 2^3 \times 3^1 \qquad 36 = 2^2 \times 3^2$

Prime 2: highest power is $\max(3,2) = 3$, so contribute $2^3$
Prime 3: highest power is $\max(1,2) = 2$, so contribute $3^2$

$$\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = \mathbf{72}$$

Venn Diagram Method — Both HCF and LCM in One Diagram

The Venn diagram method is the most efficient exam technique: one diagram gives both answers simultaneously and is highly mark-scheme-friendly. The two overlapping circles represent the two numbers; prime factors are placed individually (not as powers).

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DEFINITION — Prime Factor Venn Diagram

Left region only: prime factors appearing in the first number but not the second
Intersection: prime factors shared by both numbers (list each occurrence individually)
Right region only: prime factors appearing in the second number but not the first

$$\text{HCF} = \text{product of the intersection}$$ $$\text{LCM} = \text{product of everything in both circles}$$

🎯

EXAM TIP — Venn Diagram Step-by-Step

Write out the full prime factorisation of both numbers, listing each prime individually (e.g. $60 = 2 \times 2 \times 3 \times 5$, not $2^2 \times 3 \times 5$).
Match up shared primes and place them in the intersection (one at a time).
Place any leftover primes from number 1 in the left region; leftover primes from number 2 in the right region.
HCF = multiply only the intersection; LCM = multiply all primes across both circles.

Key Relationship: HCF × LCM

$$\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$$

Use this to check your answers every time Example: HCF(24,36) × LCM(24,36) = 12 × 72 = 864 = 24 × 36 ✓

🧠

MEMORY TRICK — HCF vs LCM

HCF is Humble — it makes things smaller. Use the lowest power of only the shared primes.
LCM is Large — it makes things bigger. Use the highest power of all primes.

Venn shortcut: HCF = intersection, LCM = everything.

🗺️ Visual Notes

Types of Number

Number Sets

ℕ — positive counting: 1, 2, 3 ...
ℤ — integers: ..., −1, 0, 1, ...
ℚ — rational: any p/q (q ≠ 0)
ℝ — real: rational + irrational

Prime Numbers

Exactly 2 factors: 1 and itself
1 is NOT prime (only one factor)
2 is the only even prime
First 10: 2,3,5,7,11,13,17,19,23,29

Prime Factorisation

Factor tree — split until all prime
Repeated division — divide by smallest prime
Always write in index notation
Result is unique (FTA)

HCF

Shared primes at LOWEST power
HCF ≤ smaller number
Intersection of Venn diagram
Used to simplify fractions

LCM

All primes at HIGHEST power
LCM ≥ larger number
Everything in both Venn circles
Common denominator for fractions

Primality Test

Test all primes up to √n only
Factors come in pairs — one ≤ √n
List primes tested in working
State conclusion clearly for marks

Number Sets: Comparison Table

Set	Symbol	Definition	Examples	NOT in This Set
Natural	ℕ	Positive counting numbers	1, 2, 3, 100	0, −3, ½, π
Integer	ℤ	All whole numbers	−5, 0, 7, 100	½, π, √2
Rational	ℚ	Writable as p/q, q ≠ 0	½, 0.3̄, −4, √4	√2, π, e
Irrational	—	Cannot write as p/q	√2, π, √5, e	All fractions & integers
Real	ℝ	All rational + irrational	Any point on number line	√−1 (imaginary)

Venn Diagram: HCF and LCM of 60 and 84

$60 = 2 \times 2 \times 3 \times 5$ and $84 = 2 \times 2 \times 3 \times 7$. Place each prime factor individually:

Process: Finding Prime Factorisation

Write the number

→

Divide by smallest prime that works

→

Divide result again

→

Repeat until result = 1

→

Group primes in index notation

Process: HCF and LCM via Venn Diagram

Prime factorise both numbers (list each prime individually)

→

Pair up matching primes → place in intersection

→

Remaining primes → outer regions

→

HCF = product of intersection

→

LCM = product of all factors in both circles

✏️ Worked Examples

Grade 4–5

Express 72 as a product of its prime factors. Write your answer in index notation.

Divide by the smallest prime repeatedly

Start with 72 and keep dividing by the smallest prime that divides evenly:

$72 \div 2 = 36$

$36 \div 2 = 18$

$18 \div 2 = 9$

$9 \div 3 = 3$

$3 \div 3 = 1$

Collect all the prime divisors

Reading the divisors used: $2, 2, 2, 3, 3$

Group repeated primes using index notation

$2$ appears three times → $2^3$. $\quad 3$ appears twice → $3^2$. $$72 = 2^3 \times 3^2$$

$$72 = 2^3 \times 3^2$$

Grade 6–7

Using a Venn diagram, find the HCF and LCM of 60 and 84. Show all your working.

Prime factorise both numbers (list each prime individually)

$60: \quad 60 \div 2 = 30,\ 30 \div 2 = 15,\ 15 \div 3 = 5,\ 5 \div 5 = 1$
So $60 = 2 \times 2 \times 3 \times 5$

$84: \quad 84 \div 2 = 42,\ 42 \div 2 = 21,\ 21 \div 3 = 7,\ 7 \div 7 = 1$
So $84 = 2 \times 2 \times 3 \times 7$

Sort primes into the Venn diagram regions

Both lists share: $2, 2, 3$ → place in the intersection
Leftover from 60: $5$ → place in the left region
Leftover from 84: $7$ → place in the right region

Calculate HCF from the intersection

$$\text{HCF} = 2 \times 2 \times 3 = 12$$

Calculate LCM from all primes in both circles

$$\text{LCM} = 5 \times 2 \times 2 \times 3 \times 7 = 5 \times 12 \times 7 = 420$$

Verify using HCF × LCM = a × b

$12 \times 420 = 5040$ and $60 \times 84 = 5040$ ✓

HCF = 12 LCM = 420

Grade 9

(a) Prove that 127 is a prime number. [3]

(b) Find the HCF and LCM of $12a^2b^3$ and $18ab^4$. [4]

(c) Hence simplify $\dfrac{1}{12a^2b^3} + \dfrac{1}{18ab^4}$. [3]

Part (a) — Find the upper bound for testing

$\sqrt{127} \approx 11.27$, so we need only test primes up to and including 11: $\{2,\ 3,\ 5,\ 7,\ 11\}$.

Part (a) — Test each prime systematically

Divisible by 2? No — 127 is odd.
Divisible by 3? No — digit sum $1 + 2 + 7 = 10$, and 10 is not divisible by 3.
Divisible by 5? No — 127 does not end in 0 or 5.
Divisible by 7? No — $127 \div 7 = 18.14\ldots$ (not exact).
Divisible by 11? No — $127 \div 11 = 11.54\ldots$ (not exact).

Part (a) — State the conclusion

Since no prime up to $\sqrt{127}$ divides 127 exactly, 127 has no factors other than 1 and itself. Therefore 127 is prime.

Part (b) — Prime factorise both expressions

Treat numerical coefficients and variables separately: $$12a^2b^3 = 2^2 \times 3 \times a^2 \times b^3$$ $$18ab^4 = 2 \times 3^2 \times a^1 \times b^4$$ Variables $a$ and $b$ behave exactly like prime numbers.

Part (b) — Find HCF (shared factors at lowest power)

Factor	In $12a^2b^3$	In $18ab^4$	Min power
2	$2^2$	$2^1$	$2^1$
3	$3^1$	$3^2$	$3^1$
a	$a^2$	$a^1$	$a^1$
b	$b^3$	$b^4$	$b^3$

$$\text{HCF} = 2^1 \times 3^1 \times a^1 \times b^3 = 6ab^3$$

Part (b) — Find LCM (all factors at highest power)

Factor	In $12a^2b^3$	In $18ab^4$	Max power
2	$2^2$	$2^1$	$2^2$
3	$3^1$	$3^2$	$3^2$
a	$a^2$	$a^1$	$a^2$
b	$b^3$	$b^4$	$b^4$

$$\text{LCM} = 2^2 \times 3^2 \times a^2 \times b^4 = 4 \times 9 \times a^2 \times b^4 = 36a^2b^4$$

Part (c) — Use LCM as lowest common denominator

Find the multiplier for each fraction: $$\frac{36a^2b^4}{12a^2b^3} = 3b \qquad \frac{36a^2b^4}{18ab^4} = 2a$$ Now rewrite with the common denominator: $$\frac{1}{12a^2b^3} + \frac{1}{18ab^4} = \frac{1 \times 3b}{36a^2b^4} + \frac{1 \times 2a}{36a^2b^4} = \frac{3b + 2a}{36a^2b^4}$$ Check: can this be simplified further? $\gcd(3b+2a,\, 36a^2b^4)$ — since $3b+2a$ is a linear sum and does not share a factor with $36a^2b^4$ in general, the answer is already fully simplified.

(a) 127 is prime — no prime up to $\sqrt{127} \approx 11.3$ divides it.
(b) $\text{HCF} = 6ab^3$; $\quad \text{LCM} = 36a^2b^4$
(c) $\dfrac{1}{12a^2b^3} + \dfrac{1}{18ab^4} = \dfrac{3b + 2a}{36a^2b^4}$

❓ Exam Questions

Q1 3 marks

State whether each of the following is rational or irrational. Give a reason for each.

(a) $\sqrt{25}$ (b) $\sqrt{11}$ (c) $0.\overline{36}$

(a) Rational [1]. $\sqrt{25} = 5$, which is an integer and can be written as $\frac{5}{1}$. Since 25 is a perfect square, its square root is rational.

(b) Irrational [1]. 11 is not a perfect square, so $\sqrt{11}$ cannot be expressed as $\frac{p}{q}$. Its decimal $3.31662\ldots$ is non-terminating and non-repeating.

(c) Rational [1]. All recurring decimals are rational. $0.\overline{36} = \frac{36}{99} = \frac{4}{11}$.

Q2 2 marks

Express 360 as a product of its prime factors. Write your answer in index notation.

$360 \div 2 = 180,\ 180 \div 2 = 90,\ 90 \div 2 = 45,\ 45 \div 3 = 15,\ 15 \div 3 = 5,\ 5 \div 5 = 1$

[1] for at least three correct prime factors identified; [1] for full correct answer:

$$360 = 2^3 \times 3^2 \times 5$$

Q3 4 marks

Find the HCF and LCM of 48 and 72. Show all your working clearly.

$48 = 2^4 \times 3$ [1]

$72 = 2^3 \times 3^2$ [1]

HCF = shared primes at lowest power: $2^{\min(4,3)} \times 3^{\min(1,2)} = 2^3 \times 3 = 8 \times 3 = \mathbf{24}$ [1]

LCM = all primes at highest power: $2^{\max(4,3)} \times 3^{\max(1,2)} = 2^4 \times 3^2 = 16 \times 9 = \mathbf{144}$ [1]

Check: $24 \times 144 = 3456 = 48 \times 72$ ✓

Q4 4 marks

Using a Venn diagram, find the HCF and LCM of 120 and 180.

$120 = 2 \times 2 \times 2 \times 3 \times 5$ [1]

$180 = 2 \times 2 \times 3 \times 3 \times 5$ [1]

Venn diagram sorting:

Left only (in 120, not 180): one extra factor of 2
Intersection: 2, 2, 3, 5 (matched from both lists)
Right only (in 180, not 120): one extra factor of 3

HCF = product of intersection = $2 \times 2 \times 3 \times 5 = \mathbf{60}$ [1]

LCM = product of all = $2 \times 2 \times 2 \times 3 \times 3 \times 5 = \mathbf{360}$ [1]

Q5 3 marks

Prove that 97 is a prime number. You must show all steps of your reasoning.

$\sqrt{97} \approx 9.85$, so test all primes up to 9: $\{2, 3, 5, 7\}$ [1]

97 is odd → not divisible by 2
Digit sum $9 + 7 = 16$ → not divisible by 3
Does not end in 0 or 5 → not divisible by 5
$97 \div 7 = 13.857\ldots$ → not divisible by 7

[1] for all four tests correct and shown [1] Since no prime up to $\sqrt{97}$ divides 97, 97 is prime.

Q6 6 marks

Let $p = 2^3 \times 3^2 \times 5$ and $q = 2^2 \times 3^3 \times 7$.

(a) Find the HCF of $p$ and $q$. [2]

(b) Find the LCM of $p$ and $q$. [2]

(a) Shared primes: 2 (min power = 2) and 3 (min power = 2). No 5 or 7 in the other.

HCF $= 2^2 \times 3^2 = 4 \times 9 = \mathbf{36}$ [2]

(b) All primes at highest power: $2^3, 3^3, 5^1, 7^1$.

LCM $= 2^3 \times 3^3 \times 5 \times 7 = 8 \times 27 \times 5 \times 7 = \mathbf{7560}$ [2]

(c) Use LCM = 7560 as the common denominator:

Multiplier for $\frac{1}{p}$: $\dfrac{7560}{360} = 21$ Multiplier for $\frac{1}{q}$: $\dfrac{7560}{756} = 10$

$$\frac{1}{p} + \frac{1}{q} = \frac{21}{7560} + \frac{10}{7560} = \frac{31}{7560} = \frac{31}{2^3 \times 3^3 \times 5 \times 7}$$

Check: 31 is prime; $7560 = 2^3 \times 3^3 \times 5 \times 7$ does not contain 31, so the fraction cannot be simplified. [2]

⭐ Grade 9 Model Answers

The following is a fully annotated model answer for Q6 — the highest-demand question. Each annotation explains why the step earns marks and what examiners are looking for.

⚠️

Q6 (a) — HCF: Grade 9 Annotation

Given: $p = 2^3 \times 3^2 \times 5$ and $q = 2^2 \times 3^3 \times 7$

A Grade 9 student identifies which primes are shared and why we use the minimum power:

Prime 2 is in both ($2^3$ in $p$, $2^2$ in $q$). The HCF cannot contain more than $2^2$ because $q$ only has $2^2$ — using $2^3$ would not divide $q$. So we take $2^{\min(3,2)} = 2^2$.
Prime 3 is in both ($3^2$ in $p$, $3^3$ in $q$). We take $3^{\min(2,3)} = 3^2$.
Prime 5 is only in $p$, not $q$. So 5 cannot be in the HCF.
Prime 7 is only in $q$, not $p$. So 7 cannot be in the HCF.

$$\text{HCF} = 2^2 \times 3^2 = 4 \times 9 = \mathbf{36}$$

Mark-earning insight: stating "5 and 7 are not shared" explicitly shows you understand why they are excluded — this is the Grade 9 distinction.

⚠️

Q6 (b) — LCM: Grade 9 Annotation

For LCM, we include every prime that appears in either number, at its highest power. A common Grade 6 error is to include only the shared primes — that gives the HCF, not the LCM.

Prime 2: $\max(3,2) = 3$ → take $2^3$
Prime 3: $\max(2,3) = 3$ → take $3^3$
Prime 5: only in $p$, so include $5^1$
Prime 7: only in $q$, so include $7^1$

$$\text{LCM} = 2^3 \times 3^3 \times 5 \times 7 = 8 \times 27 \times 35 = \mathbf{7560}$$

Mark-earning insight: including 5 and 7 (the "unshared" primes) is the key move that distinguishes LCM from HCF. Failing to include them is the single most common error in this question type.

⚠️

Q6 (c) — Adding Fractions: Grade 9 Annotation

Using the LCM (not $p \times q$) as the denominator is the efficient, Grade 9 approach. Using $p \times q = 360 \times 756 = 272{,}160$ would work but produce a fraction needing simplification — wasteful under exam conditions.

Calculate each "top-up multiplier" by dividing the LCM by each denominator:

For $\frac{1}{p} = \frac{1}{360}$: $\dfrac{7560}{360} = 21$ (so multiply top and bottom by 21)

For $\frac{1}{q} = \frac{1}{756}$: $\dfrac{7560}{756} = 10$ (so multiply top and bottom by 10)

$$\frac{1}{p} + \frac{1}{q} = \frac{21}{7560} + \frac{10}{7560} = \frac{31}{7560}$$

Finally, check simplification: 31 is prime (test primes up to $\sqrt{31} \approx 5.6$: not divisible by 2, 3, or 5). Is 31 a factor of 7560? $7560 = 2^3 \times 3^3 \times 5 \times 7$ — 31 does not appear, so the fraction is already in its simplest form.

Mark-earning insight: the explicit simplification check is a Grade 9 habit — it shows mathematical rigour and earns the second accuracy mark.

🎯

EXAM TIP — Show Method Even When Uncertain of Answer

On a 6-mark question, method marks (M marks) are independent of the final answer. If you correctly state "HCF = shared primes at lowest power" and set up the structure, you earn method marks even with an arithmetic error. Never leave a multi-mark question blank.

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EXAM TIP — Algebraic HCF/LCM

When variables appear, treat each variable as a prime factor. The rules are identical: for HCF use the minimum power of each variable; for LCM use the maximum power of each variable. In $12a^2b^3$ and $18ab^4$: the variable $a$ plays the same role as a prime — it has power 2 in the first expression and power 1 in the second, so HCF takes $a^{\min(2,1)} = a^1$ and LCM takes $a^{\max(2,1)} = a^2$.

📋 Revision Sheet

Key Definitions

Term	Meaning
Natural (ℕ)	Positive integers: 1, 2, 3, ...
Integer (ℤ)	All whole numbers: ..., −1, 0, 1, ...
Rational (ℚ)	Expressible as p/q (q ≠ 0)
Irrational	Non-terminating, non-repeating; cannot write as p/q
Real (ℝ)	All rational + irrational numbers
Prime	Exactly 2 factors: 1 and itself
HCF	Largest number dividing both exactly
LCM	Smallest positive multiple of both

Essential Formulae

$$\text{HCF} = \text{SHARED primes at LOWEST power}$$

$$\text{LCM} = \text{ALL primes at HIGHEST power}$$

$$\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$$

Primality: test all primes $\leq \sqrt{n}$

Index notation: $a^m \times a^n = a^{m+n}$

Venn: HCF = intersection; LCM = everything

Memory Hooks

HCF is Humble (smaller) — LOWEST + SHARED only
LCM is Large (bigger) — HIGHEST + ALL
Venn intersection = HCF; entire diagram = LCM
HCF × LCM = a × b — always use to check!
Only test primes to √n — factors come in pairs
1 has ONE factor → NOT prime
2 is the only EVEN prime
$\sqrt{\text{perfect square}}$ is rational; $\sqrt{\text{non-perfect}}$ is irrational

Exam Tips

Always use index notation: $2^3 \times 3^2$, not $2 \times 2 \times 2 \times 3 \times 3$
State $\sqrt{n}$, list primes tested, then conclude — all three earn marks in primality questions
Draw the Venn diagram for method marks — even a labelled sketch counts
Use HCF × LCM = a × b to verify before moving on
For algebraic expressions, treat each variable as a prime factor
For fractions, use LCM as common denominator (not a × b — it gives the same answer but needs simplifying)
Show all divisibility checks, not just the final result

🔄 Flashcards

Click each card to reveal the answer. Tap again to flip back. Work through all 15 before your exam.

✗ Common Mistakes

✗

MISTAKE 1 — Swapping HCF and LCM

What students do wrong: Using the "highest power" rule for the HCF and/or the "shared only" rule for the LCM — effectively computing the wrong value for both.

Why marks are lost: Both final answers are wrong, losing all accuracy marks even when the prime factorisations are perfectly correct.

How to avoid it: Remember the mnemonic: HCF is Humble (smaller — use LOWEST and SHARED); LCM is Large (bigger — use HIGHEST and ALL). After computing, check: HCF must be ≤ the smaller number, and LCM must be ≥ the larger number.

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MISTAKE 2 — Treating 1 as a Prime

What students do wrong: Listing 1 as a prime factor, or writing "the prime factors of 12 are 1, 2, 3."

Why marks are lost: The mark scheme for prime factorisation requires writing 12 = $2^2 \times 3$, not $1 \times 2^2 \times 3$. Including 1 in a list of primes loses marks on classification questions.

How to avoid it: Learn the exact definition: a prime must have exactly two distinct factors. The number 1 has only one. The smallest prime is 2.

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MISTAKE 3 — Stopping Primality Tests Too Early

What students do wrong: Only checking 2, 3, and 5 to prove a number is prime, without testing all primes up to $\sqrt{n}$.

Why marks are lost: A mark is awarded specifically for "testing all primes up to $\sqrt{n}$." A partial test is not a complete proof — for example, $91 = 7 \times 13$ would slip through if you only test 2, 3, and 5.

How to avoid it: Always compute $\sqrt{n}$ first, round up, and list every prime up to that bound. For $n = 127$: $\sqrt{127} \approx 11.3$, so the list is $\{2, 3, 5, 7, 11\}$ — all five must be tested.

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MISTAKE 4 — Wrong Venn Diagram Placement

What students do wrong: Placing the entire prime factorisation of each number in its circle (including the shared factors on both sides), rather than putting shared factors exclusively in the intersection.

Why marks are lost: An incorrect Venn diagram leads to wrong HCF and LCM. The intersection must contain only factors that appear in both factorisations.

How to avoid it: List primes individually (not as powers). Match them up one by one: for every prime in the first number, check if it also appears in the second. Matched pairs go in the intersection; unmatched primes go in the outer regions.

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MISTAKE 5 — Assuming All Square Roots Are Irrational

What students do wrong: Classifying $\sqrt{9}$, $\sqrt{16}$, or $\sqrt{100}$ as irrational because "square roots are irrational."

Why marks are lost: $\sqrt{9} = 3$, $\sqrt{16} = 4$ — these are integers and thus rational. Misclassifying them loses classification marks.

How to avoid it: Memorise the first 12 perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$. If the number under the root sign is in this list, the result is rational.

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MISTAKE 6 — Not Simplifying the Final Fraction After Adding

What students do wrong: Giving $\frac{21}{7560} + \frac{10}{7560} = \frac{31}{7560}$ and not checking whether this can be simplified — or worse, incorrectly simplifying $\frac{31}{7560}$ when 31 is prime and does not divide 7560.

Why marks are lost: The question may ask for the "simplified" or "simplest" form. Failing to check loses the final accuracy mark.

How to avoid it: Always test whether the numerator is prime (or check whether the numerator's prime factors appear in the denominator). $31$ is prime and $7560 = 2^3 \times 3^3 \times 5 \times 7$ — 31 is not a factor, so no further simplification is possible.

✅ Final Checklist

Click each item to tick it off. Your progress is saved in your browser.

I can classify any number as natural, integer, rational, irrational or real
I know that 1 is NOT prime (one factor) and 2 is the only even prime
I can express any integer as a product of prime factors using a factor tree
I can express any integer as a product of prime factors using repeated division
I always write prime factorisations in index notation ($2^3 \times 3^2$, not $2 \times 2 \times 2 \times 3 \times 3$)
I can find the HCF by taking shared prime factors at the LOWEST power
I can find the LCM by taking all prime factors at the HIGHEST power
I can draw a Venn diagram to find HCF and LCM simultaneously
I use HCF × LCM = a × b to verify my answers
I can prove a number is prime by testing all primes up to its square root
I understand WHY we only need to test primes up to √n (factors come in pairs)
I can find the HCF and LCM of algebraic expressions (e.g. $12a^2b^3$ and $18ab^4$)
I can use the LCM as a common denominator to add algebraic fractions
I know that $\sqrt{4} = 2$ is rational, but $\sqrt{2}$, $\pi$, and $e$ are irrational
I am confident tackling 6-mark HCF/LCM and primality questions under exam conditions

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