Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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The Fundamental Theorem of Arithmetic
Every composite number can be broken down into prime factors — and there is only ONE way to do it (ignoring the order)!
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🌟 Think of it this way
Every pizza order (composite number) is made of specific toppings (prime factors). No matter who assembles the pizza, the same toppings always appear — you can’t secretly swap mushrooms for olives. That’s the FTA: the prime “ingredients” of any number are always unique!
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Every composite number…
…can be written as a product of prime numbers. No exceptions!
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…in a unique way
The set of prime factors is always the same — only the order might differ.
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…using a factor tree
Break the number into branches until every branch ends in a prime number.
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Factor Trees — How to Factorise
A factor tree splits a number into two factors at each step, until every branch is a prime number (circle them!). Let’s see three examples.
Factorise 60
2
30
2
15
3
5
60 = 2² × 3 × 5
Primes: 2, 2, 3, 5
Factorise 36
2
18
2
9
3
3
36 = 2² × 3²
Primes: 2, 2, 3, 3
Factorise 48
2
24
2
12
2
6
2
3
48 = 2⁴ × 3
Primes: 2, 2, 2, 2, 3
Purple circles = prime numbers (can’t be split further). Keep splitting the blue circles until ALL branches are prime. Once every branch is prime, collect them all — that’s your prime factorisation!
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HCF & LCM Using Prime Factorisation
Once you have prime factorisations, finding HCF and LCM becomes a simple comparison of powers!
Rule: Take the smallest power of each common prime factor.
SMALLEST powers of
SHARED prime factors
Rule: Take the largest power of all prime factors that appear.
LARGEST powers of
ALL prime factors
HCF (a, b) × LCM (a, b) = a × b
This works for ANY two positive integers! Use it to check your answers.
✏️ Worked Example 1 — HCF and LCM of 12 and 18
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Prime factorise both numbersBreak each number into its prime factors:
12 = 2² × 3¹
18 = 2¹ × 3² -
Find HCF — smallest power of COMMON primesCommon primes: 2 and 3. Take the smallest power of each:
HCF = 2¹ × 3¹ = 2 × 3 = 6
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Find LCM — largest power of ALL primesAll primes: 2 and 3. Take the largest power of each:
LCM = 2² × 3² = 4 × 9 = 36
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Verify using the golden relationshipHCF × LCM should equal 12 × 18:
6 × 36 = 216 ✓ | 12 × 18 = 216 ✓ Match!
✏️ Worked Example 2 — HCF and LCM of 8, 12 and 24 (Three Numbers!)
8 = 2³
12 = 2² × 3
24 = 2³ × 3
Common to all three: 2
Smallest power of 2: 2²
HCF = 4
Largest 2: 2³ = 8
Largest 3: 3¹ = 3
LCM = 8 × 3 = 24
✏️ Worked Example 3 — Does 4ⁿ ever end with digit 0?
A number ends in 0 if and only if it is divisible by 10 = 2 × 5. So we need to check whether 4ⁿ has 5 as a prime factor.
Only prime factor = 2
The prime 5 never appears!
∴ 4ⁿ can NEVER end in digit 0
The FTA guarantees no “hidden” prime factors exist — if 5 doesn’t appear in the factorisation, 4ⁿ is not divisible by 5.
∞
Irrational Numbers
Numbers that cannot be written as a fraction p/q (where p, q are integers, q ≠ 0) — they go on forever without repeating!
Rational Numbers
Can be written as p/q where q ≠ 0
1/3 = 0.333… (repeats)
7/4 = 1.75 (terminates)
22/7 ≈ 3.142857142857…
Irrational Numbers
Cannot be written as p/q — decimals never terminate or repeat!
√3 = 1.73205080…
√5 = 2.23606797…
π = 3.14159265…
✏️ Proving √2 is Irrational — Step by Step
We use a technique called Proof by Contradiction: assume the opposite of what we want to prove, show it leads to an impossible situation, so our assumption must be wrong!
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Assume √2 IS rational (our “wrong” assumption)
Then √2 = a/b, where a and b are integers with no common factor (coprime), and b ≠ 0.
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Square both sides
2 = a²/b² → a² = 2b²
So 2 divides a². By the Key Tool, 2 also divides a. Write a = 2c.
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Substitute a = 2c back in
(2c)² = 2b² → 4c² = 2b² → b² = 2c²
So 2 divides b². By the Key Tool, 2 also divides b.
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💥 Contradiction!
Both a and b are divisible by 2. But we said a and b have NO common factors. This is impossible!
∴ Our assumption was WRONG. √2 is IRRATIONAL. ✓
✏️ Worked Example 4 — Prove √3 is Irrational
Same technique — assume √3 = a/b (coprime, b ≠ 0).
So 3 | a² → (Key Tool) 3 | a
Write a = 3c → 3b² = 9c² → b² = 3c²
So 3 | b² → 3 | b
💥 Both a and b divisible by 3 — contradicts coprime!
∴ √3 is IRRATIONAL ✓
✏️ Worked Example 5 — Show that 5 − √3 is Irrational
Strategy: Assume it IS rational, then rearrange to show √3 must be rational — which we already know is false!
Rearrange: √3 = 5 − a/b = (5b − a)/b
(5b − a)/b is rational (integers divided by integer)
So √3 would be rational — but we proved √3 is irrational!
💥 Contradiction!
∴ 5 − √3 is IRRATIONAL ✓
✏️ Worked Example 6 — Show that 3√2 is Irrational
Rearrange: √2 = a/(3b)
a/(3b) is rational (integer over integer)
So √2 would be rational — but √2 is irrational!
💥 Contradiction!
∴ 3√2 is IRRATIONAL ✓
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Memory Tricks for Exam!
Quick reminders to recall when you’re in the exam hall.
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Factor Tree
Split until all branches are primes. Primes cannot be split further — they are the “atoms” of numbers.
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HCF = SMALL
Highest but smallest powers! Take only COMMON primes with their smallest exponents.
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LCM = BIG
Lowest but largest powers! Take ALL primes with their largest exponents.
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H × L = a × b
For TWO numbers: HCF × LCM = product of the numbers. Use this to verify!
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Proof by Contradiction
Assume the OPPOSITE → find a contradiction → conclude original statement is TRUE.
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Key Theorem
If prime p divides a², then p divides a. This is the engine behind all irrational proofs!
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Extra Practice with Solutions
Try these yourself first, then check the solutions!
🔷 Practice 1 — Find HCF and LCM of 26 and 91
👁️ Click to see solution
91 = 7 × 13HCF = 13 (smallest power of common prime 13)
LCM = 2 × 7 × 13 = 182
✅ Verify: HCF × LCM = 13 × 182 = 2366
26 × 91 = 2366 ✓
🔷 Practice 2 — Find HCF and LCM of 336 and 54
👁️ Click to see solution
54 = 2 × 3³Common primes: 2 and 3
HCF = 2¹ × 3¹ = 6
LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024
✅ Verify: 6 × 3024 = 18144
336 × 54 = 18144 ✓
🔷 Practice 3 — Given HCF(306, 657) = 9, find LCM(306, 657)
👁️ Click to see solution
9 × LCM = 306 × 657
9 × LCM = 201,042
LCM = 201,042 ÷ 9
LCM = 22,338
🔷 Practice 4 — Prove that √5 is irrational
👁️ Click to see solution
Squaring: 5 = a²/b² → a² = 5b²
So 5 | a² → 5 | a (Key Theorem)
Write a = 5c
→ 5b² = 25c² → b² = 5c²
So 5 | b² → 5 | b
Both a and b divisible by 5 — contradicts coprime!
∴ √5 is IRRATIONAL ✓
🔷 Practice 5 — Sonia & Ravi on a Circular Track
Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round. They start together. After how many minutes will they meet again at the starting point?
👁️ Click to see solution
This happens at the LCM of their times.18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36
They meet again after 36 minutes ✓
🎯 Quick Test Yourself!
Can you get all 5 right? 🏆
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Chapter Summary
Everything you need to know at a glance!
| Concept | Key Rule | Example |
|---|---|---|
| FTA | Every composite = unique product of primes | 60 = 2² × 3 × 5 |
| HCF | Smallest power of COMMON primes | HCF(12,18) = 6 |
| LCM | Largest power of ALL primes | LCM(12,18) = 36 |
| Golden Rule | HCF × LCM = a × b (two numbers only) | 6 × 36 = 216 = 12×18 ✓ |
| Irrational | Cannot write as p/q; non-terminating, non-repeating | √2, √3, √5, π |
| Key Theorem | If prime p divides a², then p divides a | 2|a² → 2|a |
| Proof method | Proof by contradiction — assume opposite, find impossibility | Used to prove √2, √3 irrational |
The Golden Rule HCF × LCM = a × b works only for TWO numbers. For three or more numbers, you cannot use this shortcut — you must find HCF and LCM separately using prime factorisation.
🔢 Real Numbers · Grade 10 Mathematics · Chapter 1
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