Real Numbers
Euclid's Division Lemma and HCF
What is Euclid’s Division Lemma? Euclid’s Division Lemma states that for any two positive integers a (dividend) and b (divisor), there exist unique integers q (quotient) and r (remainder) such that:
a = b × q + r, where 0 ≤ r < b.
This means: when you divide a by b, you get a quotient and a remainder that is always smaller than the divisor.
Real-life example: If you have 23 candies (a = 23) and want to share equally among 5 friends (b = 5), each friend gets 4 candies (q = 4), and 3 candies are left over (r = 3). Check: 23 = 5 × 4 + 3 ✓
Key properties:
- q is the floor value of a ÷ b (greatest integer less than or equal to a/b)
- r is always a non‑negative number less than b
- For every pair (a, b), q and r are unique — no other pair works
- If r = 0, then b divides a exactly (a is a multiple of b)
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│ EUCLID'S DIVISION LEMMA - VISUAL REPRESENTATION │
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CASE 1: a = 23, b = 5 (r = 3 ≠ 0)
QUOTIENT (q = 4)
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▼
┌─────────────────────┐
│ a │ = │ b │ × │ q │ + │ r │
└─────────────────────┘
│ │ │ │
▼ ▼ ▼ ▼
23 = 5 × 4 + 3
│ │
└──── 5 × 4 = 20 ────┘
│
└── 20 + 3 = 23 ✓
REMAINDER (r = 3) is less than DIVISOR (b = 5)
CASE 2: a = 20, b = 5 (r = 0)
20 = 5 × 4 + 0
│
└── b divides a exactly
REAL-LIFE APPLICATION:
[23 Pizza Slices] → [5 Tables]
│ │
▼ ▼
┌─────────┐ ┌─────────┐
│ Each │ │ Each │
│ Table │ = │ Table │
│ gets 4 │ │ has 4 │
│ slices │ │ slices │
└─────────┘ └─────────┘
│ │
└────── 3 slices leftover ──┘
(r = 3)Example 1: Apply Euclid’s Division Lemma to find q and r when a = 47 and b = 6.
- Step 1: Divide 47 by 6 → 6 × 7 = 42, remainder = 5
- Step 2: Write in form a = b × q + r → 47 = 6 × 7 + 5
- Step 3: Check remainder condition: 0 ≤ 5 < 6 ✓
Answer: q = 7, r = 5
Example 2: A teacher has 125 notebooks. She wants to pack them into boxes that hold 12 notebooks each. Find the number of full boxes and the number of notebooks left unpacked.
- Step 1: Here a = 125 (total notebooks), b = 12 (capacity per box)
- Step 2: 125 ÷ 12 = 10 remainder 5 (since 12 × 10 = 120, 125 − 120 = 5)
- Step 3: 125 = 12 × 10 + 5
Answer: 10 full boxes, 5 notebooks leftover
Example 3: For what value of a does Euclid’s Division Lemma give q = 8, b = 7, and r = 3? Also verify the uniqueness.
- Step 1: Use formula a = b × q + r → a = 7 × 8 + 3
- Step 2: Calculate: 7 × 8 = 56, 56 + 3 = 59
- Step 3: Verify: 59 ÷ 7 = 8 remainder 3 (since 7 × 8 = 56, 59 − 56 = 3)
- Step 4: Check if any other (q, r) works? 7 × 7 = 49 gives r = 10 (but 10 ≥ 7, invalid); 7 × 9 = 63 gives r = −4 (invalid)
Answer: a = 59, and q, r are unique
Quick recap
- Euclid’s Division Lemma: a = b × q + r, where 0 ≤ r < b for any positive integers a and b
- q is the quotient (floor of a ÷ b), r is the remainder
- The remainder is always less than the divisor b
- When r = 0, b divides a exactly — this forms the basis of the division algorithm
- q and r are unique for given a and b
- This lemma is the foundation for the Euclidean algorithm (finding HCF)
✓ Quick check
After how many decimal places will the expansion of 13/3125 terminate?
3125 = 5⁵. Multiplying numerator and denominator by 2⁵ gives 416/100000, which terminates after 5 places.
The value of 23/(2³ × 5²) as a decimal is:
2³ × 5² = 8 × 25 = 200, and 23 ÷ 200 = 0.115.
Fundamental Theorem of Arithmetic
What is the Fundamental Theorem of Arithmetic? Every composite number (greater than 1) can be expressed as a product of prime numbers in exactly one way, except for the order of the primes. This unique factorization is called the prime factorization of the number.
Example: 84 = 2 × 2 × 3 × 7 = \(2^{2} \times 3 \times 7\) (written with exponents)
Real-life analogy: Think of prime numbers as "atoms" or "building blocks". Just as every word is made from a unique set of letters (order doesn't matter), every number is made from a unique set of prime factors.
Key properties:
- Prime numbers (2, 3, 5, 7, 11, …) cannot be broken further — they are the "atoms"
- Composite numbers are made by multiplying primes together
- The order in which you write the primes does NOT matter (2 × 3 × 7 = 3 × 2 × 7)
- This theorem guarantees that prime factorization is unique — no number has two different prime factorizations
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│ FUNDAMENTAL THEOREM OF ARITHMETIC - TREE DIAGRAMS │
└─────────────────────────────────────────────────────────────┘
PRIME FACTORIZATION TREE (METHOD 1):
┌─────┐
│ 84 │
└──┬──┘
│
┌────┴────┐
▼ ▼
┌───┐ ┌───┐
│ 2 │ │ 42│
└───┘ └─┬─┘
│
┌────┴────┐
▼ ▼
┌───┐ ┌───┐
│ 2 │ │ 21│
└───┘ └─┬─┘
│
┌────┴────┐
▼ ▼
┌───┐ ┌───┐
│ 3 │ │ 7 │ ← ALL PRIMES
└───┘ └───┘
84 = 2 × 2 × 3 × 7 = 2² × 3 × 7
LIST OF PRIME FACTORS (METHOD 2):
84 ÷ 2 = 42 (2)
42 ÷ 2 = 21 (2)
21 ÷ 3 = 7 (3)
7 ÷ 7 = 1 (7)
Factors: 2, 2, 3, 7
UNIQUENESS DEMONSTRATION:
┌──────────────────────────────────┐
│ 84 = 2 × 2 × 3 × 7 │
│ │ │ │ │ │
│ ANY ORDER is same set: │
│ 7 × 3 × 2 × 2 = 84 │
└──────────────────────────────────┘
No other combination of primes gives 84!Example 1: Find the prime factorization of 240.
- Step 1: Divide by smallest prime (2): 240 ÷ 2 = 120
- Step 2: 120 ÷ 2 = 60
- Step 3: 60 ÷ 2 = 30
- Step 4: 30 ÷ 2 = 15
- Step 5: 15 ÷ 3 = 5
- Step 6: 5 ÷ 5 = 1 (stop when you reach 1)
- Step 7: Write as product: 2 × 2 × 2 × 2 × 3 × 5 = \(2^{4} \times 3 \times 5\)
Answer: 240 = \(2^{4} \times 3 \times 5\)
Example 2: Show that 432 has the same prime factors as 72 × 6, verifying uniqueness.
- Step 1: Prime factor 72: 72 = 8 × 9 = \(2^{3} \times 3^{2}
- Step\) 2: Prime factor 6: 6 = 2 × 3
- Step 3: Multiply: 72 × 6 = (\(2^{3} \times 3^{2}\)) × (2 × 3) = \(2^{4} \times 3^{3}
- Step\) 4: Prime factor 432 directly: 432 ÷ 2 = 216, ÷ 2 = 108, ÷ 2 = 54, ÷ 2 = 27, ÷ 3 = 9, ÷ 3 = 3, ÷ 3 = 1 → \(2^{4} \times 3^{3}
- Step\) 5: Both match → uniqueness confirmed
Answer: 432 = \(2^{4} \times 3^{3}\) (same either way)
Example 3: A number is \(2^{3} \times 3^{2} \times 5\). If it is multiplied by 2 × 3 × 5, what is the prime factorization of the result?
- Step 1: Original number N = \(2^{3} \times 3^{2} \times 5^{1}
- Step\) 2: Multiply by 2 × 3 × 5 = \(2^{1} \times 3^{1} \times 5^{1}
- Step\) 3: Add exponents for same prime bases:
- 2: 3 + 1 = 4 → \(2^{4}\)
- 3: 2 + 1 = 3 → \(3^{3}\)
- 5: 1 + 1 = 2 → \(5^{2}
- Step\) 4: New number = \(2^{4} \times 3^{3} \times 5^{2}\)
Answer: \(2^{4} \times 3^{3} \times 5^{2}\)
Quick recap
- Every composite number > 1 has a unique prime factorization (order doesn’t matter)
- Prime numbers are the "building blocks" of all natural numbers > 1
- To find prime factors, repeatedly divide by the smallest prime until you reach 1
- Exponents show how many times a prime is multiplied
- This theorem guarantees no number has two different prime factorizations
- It helps find HCF, LCM, and simplify fractions easily
✓ Quick check
Can 6ⁿ end with the digit 0 for any natural number n?
6ⁿ = 2ⁿ × 3ⁿ has no factor of 5, and a number ends in 0 only if it is divisible by both 2 and 5.
If a prime number p divides a², then:
By the fundamental theorem of arithmetic, if a prime divides a² it must appear in the factorisation of a, so p divides a.
Irrational Numbers and Decimals
What are irrational numbers? Irrational numbers are real numbers that cannot be expressed as a fraction p/q, where p and q are integers and q ≠ 0. Their decimal expansions are non‑terminating and non‑repeating.
Examples: \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), π, e
What makes a number irrational?
- Its decimal form goes on forever without repeating a pattern
- It cannot be written exactly as a ratio of two integers
- Rational numbers (like 1/2 = 0.5, 2/3 = 0.666… repeating) have terminating or repeating decimals
Proof that \(\sqrt{2}\) is irrational (classic method): Assume \(\sqrt{2}\) = p/q in lowest terms. Squaring: 2 = \(p^{2}\)/\(q^{2}\) → \(p^{2}\) = \(2q^{2}\). So \(p^{2}\) is even → p is even → p = 2k. Then (2k)\(^{2}\) = \(2q^{2}\) → \(4k^{2}\) = \(2q^{2}\) → \(2k^{2}\) = \(q^{2}\) → q is also even. Contradiction (p and q would share factor 2, not in lowest terms). Hence \(\sqrt{2}\) is irrational.
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│ IRRATIONAL NUMBERS - VISUAL MAP │
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REAL NUMBER SYSTEM:
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│ REAL NUMBERS │
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│
┌───────────────┴───────────────┐
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┌───────────────┐ ┌───────────────┐
│ RATIONAL │ │ IRRATIONAL │
│ NUMBERS │ │ NUMBERS │
└───────┬───────┘ └───────┬───────┘
│ │
┌───────┴────────┐ ┌────────┴────────┐
▼ ▼ ▼ ▼
┌───────┐ ┌──────────┐ ┌──────────┐ ┌─────────┐
│ p/q │ │ Integers │ │ √2, √3 │ │ π, e │
│form │ │ ...-1,0,1│ │ √5, √7 │ │ Golden │
└───────┘ └──────────┘ └──────────┘ │ Ratio φ │
└─────────┘
DECIMAL EXPANSIONS:
RATIONAL (1/3) → 0.333333… (REPEATING)
RATIONAL (1/2) → 0.500000… (TERMINATING)
IRRATIONAL (√2) → 1.414213562… (NON-REPEATING)
IRRATIONAL (π) → 3.141592653… (NON-REPEATING)
PROVING √2 IS IRRATIONAL - PROOF FLOW:
START: Assume √2 = p/q (in simplest form)
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Square both sides
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p² = 2q² (p² is even)
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p is even → p = 2k
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(2k)² = 2q² → 4k² = 2q²
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2k² = q² (q² is even)
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q is even → p and q both even
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CONTRADICTION (p/q not in lowest terms)
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∴ √2 is IRRATIONALExample 1: Identify whether 0.101001000100001… (each time one more zero) is rational or irrational.
- Step 1: Observe the decimal pattern: 0.10 100 1000 10000 …
- Step 2: The pattern never repeats the same block of digits cyclically
- Step 3: Terminating or repeating decimals are rational
- Step 4: This decimal is non‑terminating AND non‑repeating
Answer: It is irrational
Example 2: Prove that 5 + \(\sqrt{3}\) is irrational, given that \(\sqrt{3}\) is irrational.
- Step 1: Assume 5 + \(\sqrt{3}\) is rational → 5 + \(\sqrt{3}\) = p/q (q ≠ 0, in simplest form)
- Step 2: Then \(\sqrt{3}\) = p/q − 5 = (p − 5q)/q
- Step 3: p and q are integers → (p − 5q) is an integer → (p − 5q)/q is rational
- Step 4: This would mean \(\sqrt{3}\) is rational — but we know \(\sqrt{3}\) is irrational
- Step 5: Contradiction → our assumption is false
Answer: 5 + \(\sqrt{3}\) is irrational
Example 3: Determine whether 2\(\sqrt{3}\) ÷ 3\(\sqrt{3}\) is rational or irrational.
- Step 1: Simplify: (2\(\sqrt{3}\))/(3\(\sqrt{3}\)) = 2/3 (since \(\sqrt{3}\) cancels)
- Step 2: 2/3 is in the form p/q (p = 2, q = 3, q ≠ 0)
- Step 3: This is a terminating decimal (0.666…) = repeating decimal (rational)
Answer: 2/3 is rational (the irrational \(\sqrt{3}\) canceled out)
Quick recap
- Irrational numbers cannot be written as p/q (q ≠ 0)
- Their decimal expansions are non‑terminating and non‑repeating
- \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), π, e are common examples
- Sum/difference of a rational and an irrational is irrational
- Product/quotient of a non‑zero rational and an irrational is irrational
- To prove a number is irrational, use the contradiction method (assume it is rational and derive a contradiction)
✓ Quick check
A shopkeeper arranges 420 red and 130 blue beads into rows so each row has the same number of beads of one colour, as many as possible. Each row has:
HCF(420, 130) = 10 beads per row.
Two bells ring at intervals of 9 minutes and 12 minutes. If they ring together now, they will next ring together after:
They coincide after LCM(9, 12) = 36 minutes.
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