🔢 The Real Number System
Core Relationship
Real Numbers (ℝ) = Rational (ℚ) ∪ Irrational (ℚ′)
ℚ and ℚ′ are completely disjoint sets — a number cannot be both rational and irrational.
ℝ
Real Numbers
Every point on the number line. Contains all rational AND irrational numbers.
√2, π, 3/4, −7, 0, e ∈ ℝ
ℚ
Rational Numbers
Any number = p/q where p,q ∈ ℤ and q ≠ 0
3/4, −7/2, 5, 0, 0.333…
ℚ′
Irrational Numbers
Cannot be written as p/q. Non-terminating non-recurring decimal.
√2, √3, π, e, √5
ℤ
Integers
All whole numbers and their negatives. Subset of ℚ (every integer = n/1).
…−3, −2, −1, 0, 1, 2, 3…
ℤ⁻
Negative Integers
−1, −2, −3, …
𝕎
Whole Numbers
0, 1, 2, 3, …
ℕ
Natural Numbers
Counting numbers. The smallest subset. N ⊂ W ⊂ Z ⊂ Q ⊂ R
1, 2, 3, 4, 5, …
Key Fact
22/7 ≠ π
22/7 ≈ 3.14285… is a rational number. π ≈ 3.14159… is irrational. They are not equal — 22/7 is just a common approximation of π used in calculations.
Number Line
Locating √2 geometrically
Draw a right triangle with both legs = 1 unit. By Pythagoras, hypotenuse = √2. Use a compass to mark this length on the number line at a point right of 1. This is the geometric home of √2.
Density
Infinitely many between any two
Between any two distinct real numbers there are infinitely many rationals AND infinitely many irrationals. Surprisingly, there are far more irrationals than rationals (Cantor, 1874).
💡 Remember the subset chain with: "Nice Wild Zebras Quickly Roam" → N ⊂ W ⊂ Z ⊂ Q ⊂ R
📐 Euclid's Division Algorithm
Euclid's Division Lemma (statement)
a = bq + r where 0 ≤ r < b
For any positive integers a and b, there exist unique integers q and r satisfying the above. (a = dividend, b = divisor, q = quotient, r = remainder)
🔢 Interactive HCF Calculator
Step 1
Write the division equation
Divide the larger number by the smaller. Write in the form a = bq + r. Identify the remainder r.
Step 2
Is remainder = 0?
If r = 0: the divisor b is the HCF — you're done!
If r ≠ 0: continue to Step 3.
If r ≠ 0: continue to Step 3.
Step 3
Replace a ← b, b ← r
The old divisor becomes the new dividend. The remainder becomes the new divisor. Go back to Step 1.
Step 4
HCF = last non-zero remainder
Keep repeating until remainder = 0. The last non-zero remainder (which is also the last non-zero divisor) is the HCF.
📝 Worked Example — HCF(870, 225)
870 = 225 × 3 + 195
225 = 195 × 1 + 30
195 = 30 × 6 + 15
30 = 15 × 2 + 0
225 = 195 × 1 + 30
195 = 30 × 6 + 15
30 = 15 × 2 + 0
∴ HCF(870, 225) = 15
4 steps only! Compare this to listing all factors manually — much more efficient for large numbers.
Application
Word Problem — Equal Pieces
Ropes of 840 cm and 560 cm need to be cut into equal lengths with no waste. Find the maximum length of each piece.
HCF(840, 560)
840 = 560×1+280
560 = 280×2+0
∴ Max length = 280 cm
Application
Three Numbers: HCF(12, 18, 24)
Apply algorithm twice:
HCF(12,18):
18=12×1+6, 12=6×2+0 → 6
HCF(6,24):
24=6×4+0 → 6
∴ HCF(12,18,24) = 6
🌳 Fundamental Theorem of Arithmetic
The Theorem
Every composite number = product of primes in a UNIQUE way
The prime factorisation of any natural number n > 1 is unique (ignoring the order of factors). Primes are the indivisible "building blocks" of all integers.
Factor Trees — Visual Prime Factorisation
EXAMPLE 1 — Factorise 360
360
╱ ╲
2 180
╱ ╲
2 90
╱ ╲
2 45
╱ ╲
3 15
╱ ╲
3 5
╱ ╲
2 180
╱ ╲
2 90
╱ ╲
2 45
╱ ╲
3 15
╱ ╲
3 5
360 = 2³ × 3² × 5
EXAMPLE 2 — Factorise 420
420
╱ ╲
2 210
╱ ╲
2 105
╱ ╲
3 35
╱ ╲
5 7
╱ ╲
2 210
╱ ╲
2 105
╱ ╲
3 35
╱ ╲
5 7
420 = 2² × 3 × 5 × 7
📝 HCF and LCM using FTA — HCF(360, 420)
360 = 2³ × 3² × 5¹
420 = 2² × 3¹ × 5¹ × 7¹
HCF = lowest power of common primes
Common primes: 2, 3, 5
HCF = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
LCM = highest power of all primes
All primes: 2, 3, 5, 7
LCM = 2³ × 3² × 5¹ × 7¹ = 8 × 9 × 5 × 7 = 2520
420 = 2² × 3¹ × 5¹ × 7¹
HCF = lowest power of common primes
Common primes: 2, 3, 5
HCF = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
LCM = highest power of all primes
All primes: 2, 3, 5, 7
LCM = 2³ × 3² × 5¹ × 7¹ = 8 × 9 × 5 × 7 = 2520
Verify: HCF × LCM = 60 × 2520 = 151,200 = 360 × 420 ✓
💡 HCF → pick LOWEST powers of COMMON primes. LCM → pick HIGHEST powers of ALL primes. Easy mnemonic: HCF = "Humble" (low), LCM = "Large" (high).
Why is this "Fundamental"?
- Without uniqueness of prime factorisation, HCF and LCM calculations would be ambiguous
- All of cryptography (RSA encryption) relies on the difficulty of finding prime factorisations of very large numbers
- The theorem was proven by Euclid but was made rigorous by Gauss in 1801
- Every rational number p/q (in lowest terms) has a unique prime factorisation for both p and q
⚖️ HCF and LCM
Golden Formula (for two numbers only)
HCF(a, b) × LCM(a, b) = a × b
Proof: In prime factorisation, for each prime pᵢ: min(power in a, power in b) + max(power in a, power in b) = (power in a) + (power in b). So product of all min powers × product of all max powers = a × b.
| Feature | Euclid's Division | Prime Factorisation (FTA) |
|---|---|---|
| Finds | HCF only | HCF and LCM both |
| Best for | Large numbers | Smaller numbers; when LCM needed |
| Steps | Division steps until remainder = 0 | Factor tree + compare powers |
| Three numbers | Apply twice sequentially | Same formula works directly |
| Works if one number is prime? | Yes | Yes |
| Can make arithmetic errors? | Less likely (only division) | More likely (factorisation of large numbers) |
Co-prime
HCF = 1 → Co-prime
If HCF(a,b) = 1, then a and b are co-prime (relatively prime). In this case LCM = a × b.
Example: HCF(8,15) = 1
→ LCM = 8 × 15 = 120
Note: 8 and 15 are not prime — but they are co-prime!
Caution!
Formula for 3 numbers?
HCF(a,b,c) × LCM(a,b,c) ≠ a × b × c in general!
For 3 numbers, always use prime factorisation. The product formula ONLY holds for exactly two numbers.
📝 Word Problem — Bells Ringing Together
Three bells ring every 6, 8, and 12 minutes respectively.
If they all ring together at 8:00 am,
when will they next ring together?
LCM(6, 8, 12):
6 = 2 × 3
8 = 2³
12 = 2² × 3
LCM = 2³ × 3 = 24 minutes
If they all ring together at 8:00 am,
when will they next ring together?
LCM(6, 8, 12):
6 = 2 × 3
8 = 2³
12 = 2² × 3
LCM = 2³ × 3 = 24 minutes
They will next ring together at 8:24 am
📝 Word Problem — Remainder Questions
Find the largest number that divides 626, 3127, and 15628,
leaving remainders 1, 2, and 3 respectively.
Required: largest divisor of (626−1), (3127−2), (15628−3)
= HCF(625, 3125, 15625)
625 = 5⁴, 3125 = 5⁵, 15625 = 5⁶
HCF = 5⁴ = 625
leaving remainders 1, 2, and 3 respectively.
Required: largest divisor of (626−1), (3127−2), (15628−3)
= HCF(625, 3125, 15625)
625 = 5⁴, 3125 = 5⁵, 15625 = 5⁶
HCF = 5⁴ = 625
Answer: 625
Key insight: subtract the given remainders before finding HCF.
💡 "HCF when dividing equally; LCM when combining cycles." — The best mental rule for choosing between them in word problems.
🔣 Decimal Expansions of Rational Numbers
The Rule for Terminating Decimals
p/q terminates ⟺ q = 2ⁿ × 5ᵐ (in lowest terms)
The denominator (after fully reducing the fraction) must have only 2s and/or 5s as prime factors. Any other prime factor in q → non-terminating decimal.
✓ Terminating
Decimal ends after a finite number of digits.
3/8 = 0.375
7/20 = 0.35
1/4 = 0.25
Denominator = 2ⁿ × 5ᵐ only.
8 = 2³ ✓ | 20 = 2² × 5 ✓
↻ Non-term. Recurring
Decimal goes on forever but a block repeats.
1/3 = 0.333… = 0.3̄
1/7 = 0.142857142857…
5/6 = 0.8333… = 0.83̄
Denominator has other prime factors.
3 has factor 3 | 6 = 2 × 3 ✓/✗
∞ Non-term. Non-recurring
Decimal goes on forever with NO repeating block.
√2 = 1.41421356…
π = 3.14159265…
e = 2.71828182…
Always irrational — cannot be p/q. Not possible for any rational number.
| Fraction | Denominator factors | Type | Decimal |
|---|---|---|---|
| 3/8 | 2³ only ✓ | Terminating | 0.375 |
| 7/20 | 2² × 5 only ✓ | Terminating | 0.35 |
| 1/6 | 2 × 3 — has 3 ✗ | Non-term. Recurring | 0.1666… = 0.16̄ |
| 2/11 | 11 — prime ✗ | Non-term. Recurring | 0.181818… = 0.18̄ |
| 7/12 | 2² × 3 — has 3 ✗ | Non-term. Recurring | 0.5833… = 0.583̄ |
| √2 | Not p/q form | Non-term. Non-recurring | 1.41421356… |
Proof: √2 is Irrational (Proof by Contradiction)
- Assume √2 = p/q where p, q are integers, q ≠ 0, and p/q is in its lowest terms (HCF(p,q) = 1)
- Square both sides: 2 = p²/q² → p² = 2q²
- p² is even → p must be even (if p were odd, p² would be odd) → let p = 2m
- Substitute: (2m)² = 2q² → 4m² = 2q² → q² = 2m² → q² is even → q is even
- Contradiction! Both p and q are even, so HCF(p,q) ≥ 2. But we assumed HCF(p,q) = 1.
∴ Our assumption was false → √2 is irrational ■
💡 Exam shortcut: To check if p/q terminates — simplify the fraction fully, then check if the denominator = 2ⁿ × 5ᵐ. If yes: terminating. If no: non-terminating recurring.