Mathematics · Class 9-10 · CBSE

Real Numbers: Prime Factorisation and Irrationality Proofs

Master the Fundamental Theorem of Arithmetic, prime factorisation method for HCF and LCM, proving the irrationality of √2, √3, and √5 by contradiction, and decimal expansion conditions for rational numbers.

1Fundamental Theorem of Arithmetic and HCF-LCM Relations

TheFundamental Theorem of Arithmeticstates that every composite number can be expressed (factorised) uniquely as a product of prime numbers, apart from the order in which the prime factors occur.

Product Property for Two Positive Integers

HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

Prime Factorisation Definitions: HCF vs LCM

  • HCF (Highest Common Factor)

    Prime Factor Definition
    Product of the SMALLEST power of each common prime factor involved in the numbers
    Calculation Procedure
    36 = 2² × 3²; 84 = 2² × 3 ×
    7. Common primes: 2 (min power 2), 3 (min power 1)
    Example for 36 and 84
    HCF = 2² × 3 = 12
  • LCM (Lowest Common Multiple)

    Prime Factor Definition
    Product of the GREATEST power of each prime factor involved in the numbers
    Calculation Procedure
    Highest powers: 2², 3², 7¹
    Example for 36 and 84
    LCM = 2² × 3² × 7 = 4 × 9 × 7 = 252

2Proof of Irrationality by Contradiction

A number isIrrationalif it cannot be written in the form p/q where p, q are integers and q ≠
0. We prove numbers like √2, √3, √5 are irrational using the technique of
Proof by Contradiction.

Step-by-Step Proof that √2 is Irrational

  1. 1

    1. Assumption: Assume to the contrary that √2 is rational. Hence √2 = a/b, where a and b are co-prime integers (HCF = 1) and b ≠ 0.

  2. 2

    2. Squaring both sides: 2 = a²/b² ⇒ 2b² = a². Therefore, 2 divides a². By Theorem, if a prime p divides a², then p divides a. Hence 2 divides a.

  3. 3

    3. Substitute a = 2c: 2b² = (2c)² = 4c² ⇒ b² = 2c². This implies 2 divides b²; hence 2 divides b.

  4. 4

    4. Contradiction: 2 divides both a and b. This contradicts our initial premise that a and b are co-prime (have no common factor other than 1).

  5. 5

    5. Conclusion: Our assumption that √2 is rational was false. Therefore, √2 is strictly irrational.

3Decimal Expansions of Rational Numbers

Terminating vs Non-Terminating Recurring Decimals

  • Terminating Decimal

    Condition on Denominator q (in simplest p/q form)
    Prime factorisation of q is strictly of the form 2ⁿ · 5ᵐ (where n, m are non-negative integers)
    Decimal Behavior
    Ends after finite digits
    Example
    13/80 = 13/(2⁴ × 5¹) = 0.1625
  • Non-Terminating Repeating

    Condition on Denominator q (in simplest p/q form)
    Prime factorisation of q contains prime factors other than 2 and 5 (e.g. 3, 7, 11)
    Decimal Behavior
    Infinite repeating block of digits
    Example
    17/30 = 17/(2 × 3 × 5) = 0.5666...

4Exam Corner: Tips, Mistakes & Board Questions

Critical Board Exam Takeaways

  • The relation HCF(a, b) × LCM(a, b) = a × b holds true ONLY for two numbers; it is FALSE for three or more numbers.
  • The sum, difference, product, or quotient of a non-zero rational number and an irrational number is always irrational (e.g. 3 + 2√5 is irrational).
  • To find whether p/q terminates, always reduce the fraction to simplest lowest terms first so that p and q are co-prime.

Exam question · CBSE Class 10 Board

If HCF(306, 657) = 9, what is LCM(306, 657)?
  1. A.22338
  2. B.21456
  3. C.23450
  4. D.20198

Remember

  • Every composite number has a uniqueprime factorisation.
  • For any two positive integers: ==HCF(a, b) × LCM(a, b) = a × b==.
  • Irrationality of 2,3,5\sqrt{2}, \sqrt{3}, \sqrt{5} is established viacontradiction.
  • A rational fraction p/qp/q terminates if and only if ==q=2n⋅5mq = 2^n \cdot 5^m==.

Check yourself

  1. Question 1

    After how many decimal places will the decimal expansion of 23 / (2⁴ × 5³) terminate?

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