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📑 In This Chapter Guide (Table of Contents)
1. The Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorized) as the product of powers of primes, and this prime factorization is unique, apart from the order in which the prime factors occur.
Composite Number x = p₁ᵃ¹ · p₂ᵃ² · p₃ᵃ³ ... · pₙᵃⁿ
Calculating HCF and LCM via Prime Factorization:
- HCF (Highest Common Factor): Product of the smallest power of each common prime factor involved in the numbers.
- LCM (Lowest Common Multiple): Product of the greatest power of each prime factor involved in the numbers.
- Crucial Formula (For any two positive integers a and b):
HCF(a, b) × LCM(a, b) = a × b
⚠️ Note: This relationship holds ONLY for two numbers, not for three numbers!
2. Step-by-Step Proof: Proving √2 and √5 are Irrational
These proofs rely on the classical method of Contradiction, assuming the contrary and demonstrating an absurd logical impossibility.
Theorem: Prove that √5 is an irrational number.
1. Let us assume, on the contrary, that √5 is rational.
2. Then there exist coprime integers a and b (where b ≠ 0 and HCF(a, b) = 1) such that:
√5 = a / b ⟹ a = b√5
3. Squaring both sides: a² = 5b² ... (Equation 1)
4. Since 5 divides 5b², it follows that 5 divides a². By fundamental arithmetic theorem, if a prime p divides a², then 5 divides a.
5. Therefore, we can write a = 5c for some integer c.
6. Substituting a = 5c into Equation 1:
(5c)² = 5b² ⟹ 25c² = 5b² ⟹ b² = 5c²
7. This implies that 5 divides b², and consequently, 5 divides b.
8. From steps 4 and 7, both a and b share at least 5 as a common factor.
9. But this contradicts our foundational fact that a and b are coprime (HCF = 1).
10. This contradiction arises because of our incorrect assumption that √5 is rational. Hence, √5 is irrational. (Hence Proved)
3. Decimal Expansions of Rational Numbers
Let x = p / q be a rational number in simplest form (where p and q are coprime):
- Terminating Decimal: The prime factorization of denominator q is strictly of the form
2ⁿ · 5ᵐ, where n and m are non-negative integers. E.g.,13 / 125 = 13 / 5³ = 13 × 2³ / (5³ × 2³) = 104 / 1000 = 0.104. - Non-Terminating Repeating (Recurring) Decimal: The prime factorization of denominator q contains any prime factor other than 2 or 5 (such as 3, 7, 11). E.g.,
1 / 7,5 / 6.
💡 Frequently Asked Questions (FAQ)
❓ Can the number 6ⁿ end with the digit 0 for any natural number n?
For a number to end with the digit 0, its prime factorization must contain both primes 2 and 5 (since 10 = 2 × 5). The prime factorization of 6ⁿ is (2 × 3)ⁿ = 2ⁿ × 3ⁿ. By the uniqueness of the Fundamental Theorem of Arithmetic, there are no prime factors of 5. Hence, 6ⁿ can never end with the digit zero for any natural number n.
❓ If HCF(306, 657) = 9, find LCM(306, 657).
Using the formula HCF(a, b) × LCM(a, b) = a × b: LCM(306, 657) = (306 × 657) / 9 = 34 × 657 = 22,338.
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