Number Systems — SOF IMO Study Notes

Overview

Number Systems forms the foundation of algebra and higher mathematics. For SOF IMO, this topic tests your understanding of how real numbers are classified, how decimals behave, and how exponents follow specific laws. Questions typically involve identifying number types, proving irrationality, converting fractions to decimals, simplifying expressions with exponents, and rationalizing denominators.

Mastery here requires both conceptual clarity (Why is √2 irrational? What makes a decimal terminating?) and computational skill (applying exponent laws quickly, rationalizing complex expressions). The topic appears in both direct calculation problems and multi-step reasoning questions. Strong fundamentals in Number Systems directly support your work in Polynomials, Quadratic Equations, and Applied Geometry sections.

Expect 4–6 questions combining classification, decimal analysis, exponent manipulation, and rationalization. The Achievers Section often twists these concepts with unusual bases or nested operations.

Key Concepts

• **Natural Numbers (N)**: Counting numbers {1, 2, 3, ...}. Whole Numbers (W) = N ∪ {0}. Integers (Z) include negative numbers too.
• **Rational Numbers (Q)**: Any number expressible as p/q where p, q are integers and q ≠ 0. Includes terminating and repeating decimals.
• **Irrational Numbers**: Real numbers that cannot be written as p/q. Their decimal expansions are non-terminating and non-repeating (e.g., √2, π, e).
• **Real Numbers (R)**: The union of all rational and irrational numbers. Every point on the number line represents a real number.
• **Terminating Decimals**: Rational numbers whose denominators (in simplest form) have only 2 and/or 5 as prime factors. Example: 7/8 = 0.875 terminates.
• **Non-Terminating Repeating Decimals**: Rational numbers whose denominators contain prime factors other than 2 or 5. Example: 1/3 = 0.333... repeats.
• **Laws of Exponents for Real Numbers**: For any positive real numbers a, b and rational exponents m, n: a^m × a^n = a^(m+n), (a^m)^n = a^(mn), a^m / a^n = a^(m−n), (ab)^n = a^n b^n, a^0 = 1, a^(−n) = 1/a^n, a^(1/n) = nth root of a.
• **Rationalization**: Multiplying numerator and denominator by the conjugate to eliminate radicals from the denominator. For 1/(√a + √b), multiply by (√a − √b)/(√a − √b).

Formulas / Key Facts

• **a^m × a^n = a^(m+n)** — When multiplying same bases, add exponents.
• **(a^m)^n = a^(mn)** — Power of a power means multiply exponents.
• **a^m / a^n = a^(m−n)** — When dividing same bases, subtract exponents.
• **(ab)^n = a^n b^n** — Exponent distributes over multiplication.
• **(a/b)^n = a^n / b^n** — Exponent distributes over division.
• **a^0 = 1** (for any non-zero a) — Zero exponent always equals one.
• **a^(−n) = 1 / a^n** — Negative exponent means reciprocal.
• **a^(1/n) = ⁿ√a** — Fractional exponent represents roots.
• **Decimal test**: p/q (in lowest terms) terminates if and only if q = 2^m × 5^n for non-negative integers m, n.
• **Conjugate of (a + √b) is (a − √b)** — Product eliminates the radical: (a + √b)(a − √b) = a² − b.
• **√2, √3, √5, π, e are standard irrationals** — Their decimal forms never terminate or repeat.
• **Sum/product of rational and irrational**: Rational + Irrational = Irrational; Rational × Irrational = Irrational (if rational ≠ 0).

Worked Examples

**Example 1: Decimal Classification** *Determine whether 13/125 is terminating or repeating.*

**Solution:** Step 1: Simplify 13/125. Already in simplest form (GCD = 1). Step 2: Factor denominator: 125 = 5³. Step 3: Denominator has only prime factor 5 (no other primes). Step 4: By decimal test, 13/125 terminates. Verification: 13/125 = 13/(5³) = (13 × 8)/(5³ × 8) = 104/1000 = 0.104 ✓

**Example 2: Simplifying Exponents** *Simplify: (2⁵ × 3² × 2³) / (2⁴ × 3³)*

**Solution:** Step 1: Combine same bases in numerator: 2⁵ × 2³ = 2⁸. Step 2: Rewrite: (2⁸ × 3²) / (2⁴ × 3³). Step 3: Apply division law: 2⁸/2⁴ = 2⁴ and 3²/3³ = 3⁻¹. Step 4: Result: 2⁴ × 3⁻¹ = 16 × (1/3) = 16/3.

**Example 3: Rationalization** *Rationalize: 1 / (√7 − √3)*

**Solution:** Step 1: Identify conjugate: (√7 + √3). Step 2: Multiply numerator and denominator: [1 × (√7 + √3)] / [(√7 − √3)(√7 + √3)]. Step 3: Expand denominator using (a−b)(a+b) = a² − b²: (√7)² − (√3)² = 7 − 3 = 4. Step 4: Simplified form: (√7 + √3) / 4.

**Example 4: Proving Irrationality** *Show that √5 is irrational using contradiction.*

**Solution:** Assume √5 is rational, so √5 = a/b where a, b are coprime integers. Then 5 = a²/b², so a² = 5b². This means a² is divisible by 5, hence a is divisible by 5 (since 5 is prime). Let a = 5k. Then (5k)² = 5b², so 25k² = 5b², giving b² = 5k². Now b² is divisible by 5, so b is divisible by 5. Both a and b are divisible by 5, contradicting that they are coprime. Therefore √5 is irrational. ✓

Common Mistakes

• **Confusing terminating condition**: Students think 7/12 terminates because it's a "nice" fraction. Wrong — 12 = 2² × 3 has prime 3, so 7/12 = 0.58333... repeats. Fix: Always factor denominator and check for primes other than 2 or 5.

• **Adding exponents when multiplying different bases**: Writing 2³ × 3² = 6⁵ is incorrect. Fix: You can only add exponents for the same base. Keep different bases separate: 2³ × 3² = 8 × 9 = 72.

• **Forgetting negative sign in a^(−n)**: Writing 2⁻³ = −8 instead of 1/8. Fix: Negative exponent means reciprocal, not negative number. 2⁻³ = 1/(2³) = 1/8.

• **Incomplete rationalization**: Rationalizing 1/(2 + √3) by multiplying by √3 only, leaving a radical in denominator. Fix: Use the full conjugate (2 − √3) to eliminate both terms.

• **Assuming all roots are irrational**: Claiming √16 is irrational. Wrong — √16 = 4, a rational number. Fix: Only non-perfect roots (√2, √3, etc.) are irrational. Check if the number is a perfect square first.

Quick Reference

• **Terminating decimal ↔ denominator = 2^m × 5^n only** (in simplest form).
• **a^m × a^n = a^(m+n)** and **(a^m)^n = a^(mn)** — know the difference.
• **a^(−n) = 1/a^n** — negative exponent = reciprocal, not negative.
• **Rationalize using conjugate**: (a + √b) × (a − √b) = a² − b.
• **Proof by contradiction**: Assume rational, derive coprime factors both divisible by same prime → contradiction.
• **√(prime) is always irrational** — use this in classification problems.