Real Numbers (Class 10) — SOF IMO Study Notes

Overview

Real numbers form the foundation of Class 10 mathematics and are crucial for SOF IMO success. This topic introduces rigorous number theory concepts that underpin algebra, geometry and higher mathematics. You must master three core pillars: Euclid's division lemma (a tool for finding HCF and proving divisibility), the fundamental theorem of arithmetic (unique prime factorization), and properties of irrational numbers (non-terminating, non-repeating decimals).

IMO questions test these concepts through proof-based problems, HCF-LCM calculations, rationality tests and multi-step reasoning puzzles. Unlike school exams that focus on direct formula application, IMO emphasizes understanding why algorithms work and applying them to unfamiliar scenarios. Strong command of divisibility arguments and prime factorization logic will help you tackle both Mathematical Reasoning and Achievers Section problems.

Expect 2–4 questions directly from this topic in the IMO paper, plus indirect applications in number puzzles and logical reasoning sections. Master the theory first, then practice variations to build speed and accuracy.

Key Concepts

• **Euclid's Division Lemma**: For any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b. This is the foundation of the division algorithm and HCF computation.

• **Euclid's Division Algorithm**: To find HCF of two numbers, repeatedly apply the lemma: divide the larger by the smaller, replace the larger with the smaller and the smaller with the remainder, until remainder becomes zero. The last non-zero remainder is the HCF.

• **Fundamental Theorem of Arithmetic**: Every composite number can be expressed as a product of primes in exactly one way (ignoring order). This uniqueness property is central to solving HCF, LCM and divisibility problems.

• **Prime Factorization Method**: Express numbers as products of prime powers (e.g. 72 = 2³ × 3²). HCF is the product of lowest powers of common primes; LCM is the product of highest powers of all primes present.

• **Rational Numbers**: Numbers expressible as p/q where p, q are integers and q ≠ 0. Their decimal expansions either terminate or repeat periodically.

• **Irrational Numbers**: Numbers that cannot be expressed as p/q. Their decimal expansions are non-terminating and non-repeating. Examples: √2, √3, π, e.

• **Proving Irrationality**: Use contradiction — assume the number is rational (p/q in lowest terms), derive that both p and q share a common factor, contradicting the assumption that p/q is in lowest terms.

• **Operations on Irrationals**: Sum or product of rational and irrational is irrational (with exceptions like √2 × √2 = 2). Sum of two irrationals may be rational (√2 + (-√2) = 0) or irrational (√2 + √3).

Formulas / Key Facts

• **Euclid's Division Lemma**: a = bq + r, where 0 ≤ r < b
• **HCF × LCM = Product of two numbers** (for any two positive integers)
• **Prime factorization uniqueness**: If n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ, this representation is unique
• **Terminating decimal condition**: A rational p/q (in lowest terms) has terminating decimal if and only if q = 2ᵐ × 5ⁿ for some non-negative integers m, n
• **√2 is irrational**: Classic proof by contradiction showing √2 cannot equal p/q
• **√p is irrational for any prime p**: Extension of the √2 proof technique
• **Between any two rationals lie infinitely many rationals and irrationals**
• **Any irrational multiplied by non-zero rational gives irrational**

Worked Examples

**Example 1: Euclid's Algorithm for HCF**

Find HCF of 867 and 255 using Euclid's division algorithm.

*Solution*:

• Step 1: 867 = 255 × 3 + 102
• Step 2: 255 = 102 × 2 + 51
• Step 3: 102 = 51 × 2 + 0
• Last non-zero remainder is 51
• **HCF(867, 255) = 51**

**Example 2: Prime Factorization and LCM**

Find LCM of 96 and 120 using prime factorization.

*Solution*:

• 96 = 2⁵ × 3¹
• 120 = 2³ × 3¹ × 5¹
• LCM = product of highest powers = 2⁵ × 3¹ × 5¹ = 32 × 3 × 5 = 480
• **LCM(96, 120) = 480**

Verification: HCF = 2³ × 3¹ = 24, and 96 × 120 = 11520 = 24 × 480 ✓

**Example 3: Proving √3 is Irrational**

Prove that √3 is irrational.

*Solution*:

• Assume √3 is rational, so √3 = p/q where p, q are co-prime integers
• Squaring: 3 = p²/q², hence p² = 3q²
• This means 3 divides p², so 3 divides p (since 3 is prime)
• Let p = 3m, then (3m)² = 3q², so 9m² = 3q², hence q² = 3m²
• This means 3 divides q² and therefore 3 divides q
• But now both p and q are divisible by 3, contradicting that they are co-prime
• **Therefore √3 is irrational**

**Example 4: Decimal Expansion**

Determine whether 147/120 has terminating or non-terminating decimal expansion.

*Solution*:

• Simplify: 147/120 = 49/40 (dividing by HCF 3)
• Check denominator: 40 = 2³ × 5¹
• Since 40 = 2ᵐ × 5ⁿ form, the decimal **terminates**
• Actual value: 49/40 = 1.225

Common Mistakes

**Mistake 1**: Confusing HCF algorithm steps → **Fix**: Always divide larger by smaller, then replace larger with smaller and smaller with remainder. Stop when remainder = 0.

**Mistake 2**: Thinking LCM × HCF formula works for three or more numbers → **Fix**: The product formula (HCF × LCM = product of numbers) applies only to two numbers. For three numbers, use prime factorization.

**Mistake 3**: Concluding √(p/q) is rational if p and q are perfect squares → **Fix**: √(4/9) = 2/3 is rational, but √(2/3) is irrational. Check the result, not just the inputs.

**Mistake 4**: In irrationality proofs, forgetting to mention "co-prime" or "lowest terms" → **Fix**: The contradiction depends on assuming p/q has no common factors. Always state "let p/q be in lowest terms" at the start.

**Mistake 5**: Believing sum of any two irrationals is irrational → **Fix**: √2 + (-√2) = 0, which is rational. Only certain operations guarantee irrationality (rational × irrational with non-zero rational = irrational).

Quick Reference

• **Euclid's lemma**: a = bq + r, 0 ≤ r < b — basis for HCF algorithm
• **HCF via Euclid**: Repeatedly divide, replace, until remainder = 0; last non-zero remainder is HCF
• **Fundamental theorem**: Every composite number has unique prime factorization
• **Terminating decimal test**: Denominator (in lowest terms) must be 2ᵐ5ⁿ only
• **√2, √3, √5 (non-perfect squares)**: All irrational — prove by contradiction
• **HCF × LCM = a × b** for two numbers a, b