Class 10 Mathematics Notes Chapter 1 (Real numbers) – Mathematics Book

In this revision note, let's focus on Chapter 1: Real Numbers. This is a foundational chapter, not just for your Class 10 boards but also crucial for many concepts you'll encounter in higher mathematics and various government exams. We need to understand the properties and relationships between different types of numbers.

Chapter 1: Real Numbers - Detailed Notes for Exam Preparation

1. Introduction to Real Numbers

• Real Numbers (R): The collection of all rational and irrational numbers. They can be represented on a number line.
• Rational Numbers (Q): Numbers that can be expressed in the form p/q, where p and q are integers, and q ≠ 0.
  • Examples: 1/2, -3/4, 7 (since 7 = 7/1), 0 (since 0 = 0/1), 0.333... (1/3), 0.25 (1/4).
  • Includes: Integers (Z), Whole Numbers (W), Natural Numbers (N).
• Irrational Numbers: Numbers that cannot be expressed in the form p/q, where p and q are integers, and q ≠ 0. Their decimal expansions are non-terminating and non-repeating.
  • Examples: √2, √3, √5, π (pi), 0.101101110...

2. Euclid's Division Lemma (Conceptual Basis) & Algorithm (Application)

• Euclid's Division Lemma (Statement - for understanding): Given positive integers 'a' and 'b', there exist unique integers 'q' (quotient) and 'r' (remainder) satisfying a = bq + r, where 0 ≤ r < b.
  • Think of basic division: Dividend = Divisor × Quotient + Remainder. The remainder is always non-negative and less than the divisor.
• Euclid's Division Algorithm (Key Application: Finding HCF): This is a technique to compute the Highest Common Factor (HCF) of two positive integers, say 'a' and 'b' (where a > b).

  • Step 1: Apply Euclid's division lemma to 'a' and 'b'. Find whole numbers 'q' and 'r' such that a = bq + r, 0 ≤ r < b.

  • Step 2: If r = 0, then 'b' is the HCF of 'a' and 'b'.

  • Step 3: If r ≠ 0, apply the division lemma to 'b' (the previous divisor) and 'r' (the remainder).

  • Step 4: Continue this process until the remainder becomes zero. The divisor at this stage will be the required HCF.

  • Example: Find HCF of 135 and 225.

    • 225 = 135 × 1 + 90 (Remainder is 90, not 0)
    • 135 = 90 × 1 + 45 (Remainder is 45, not 0)
    • 90 = 45 × 2 + 0 (Remainder is 0)
    • The last non-zero divisor is 45. So, HCF(225, 135) = 45.

3. The Fundamental Theorem of Arithmetic

• Statement: Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
  • Prime Number: A natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7, 11...).
  • Composite Number: A natural number greater than 1 that is not prime (e.g., 4, 6, 8, 9, 10...).
• Prime Factorization: The process of finding the prime numbers that multiply together to make the original number.
  • Example: Prime factorization of 360.
    • 360 = 2 × 180
    • = 2 × 2 × 90
    • = 2 × 2 × 2 × 45
    • = 2 × 2 × 2 × 3 × 15
    • = 2 × 2 × 2 × 3 × 3 × 5
    • = 2³ × 3² × 5¹
• Applications of the Fundamental Theorem:

  • Finding HCF: The HCF of two or more numbers is the product of the smallest powers of each common prime factor in the numbers.

  • Finding LCM: The LCM of two or more numbers is the product of the greatest powers of each prime factor involved in the numbers.

  • Example: Find HCF and LCM of 96 and 404.

    • 96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3¹
    • 404 = 2 × 202 = 2 × 2 × 101 = 2² × 101¹ (101 is prime)
    • HCF(96, 404): Common prime factor is 2. Smallest power is 2². HCF = 2² = 4.
    • LCM(96, 404): Prime factors involved are 2, 3, 101. Greatest power of 2 is 2⁵. Greatest power of 3 is 3¹. Greatest power of 101 is 101¹. LCM = 2⁵ × 3¹ × 101¹ = 32 × 3 × 101 = 96 × 101 = 9696.

  • Important Relationship (for TWO numbers 'a' and 'b'):
    HCF(a, b) × LCM(a, b) = a × b

    • Check for 96 and 404: HCF × LCM = 4 × 9696 = 38784. Product = 96 × 404 = 38784. (It matches!)
    • Caution: This formula does not generally hold for three or more numbers.

4. Revisiting Irrational Numbers

• Proof by Contradiction: This is the standard method to prove a number is irrational.
  • Assume the opposite (e.g., assume √2 is rational).
  • Express it in the form p/q, where p and q are co-prime integers (no common factors other than 1) and q ≠ 0.
  • Manipulate the equation algebraically to arrive at a contradiction (e.g., showing that p and q must have a common factor, contradicting the co-prime assumption).
  • Since the assumption leads to a contradiction, the assumption must be false. Therefore, the number is irrational.
  • Key Results (often assumed in exams unless proof is specifically asked): √2, √3, √5, √p (where p is prime) are irrational.
• Properties:
  • The sum or difference of a rational number and an irrational number is irrational. (e.g., 5 - √3 is irrational).
  • The product or quotient of a non-zero rational number and an irrational number is irrational. (e.g., 3√2 is irrational, √5 / 2 is irrational).

5. Revisiting Rational Numbers and Their Decimal Expansions

• Theorem 1: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form p/q, where p and q are co-prime, and the prime factorization of q is of the form 2ⁿ 5ᵐ, where n and m are non-negative integers.
  • In simple terms: If a fraction (in simplest form) has a denominator with only 2s and/or 5s as prime factors, its decimal expansion will terminate.
  • Example: 13/80. 80 = 16 × 5 = 2⁴ × 5¹. Since the denominator has only prime factors 2 and 5, 13/80 will have a terminating decimal expansion (0.1625).
• Theorem 2 (Converse of Theorem 1): Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5ᵐ, where n and m are non-negative integers. Then x has a decimal expansion which terminates.
• Theorem 3: Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2ⁿ 5ᵐ, where n and m are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).
  • In simple terms: If a fraction (in simplest form) has a prime factor in its denominator other than 2 or 5, its decimal expansion will be non-terminating and repeating.
  • Example: 1/7. The denominator is 7 (prime, not 2 or 5). So, 1/7 has a non-terminating repeating decimal expansion (0.142857...).
  • Example: 77/210. Simplify first: 77/210 = (7 × 11) / (7 × 30) = 11/30. Denominator 30 = 2 × 3 × 5. It contains the prime factor 3 (other than 2 and 5). Therefore, 77/210 has a non-terminating repeating decimal expansion.

Key Takeaways for Exams:

• Master finding HCF using Euclid's Algorithm.
• Master finding HCF and LCM using Prime Factorization.
• Remember the relationship: HCF(a, b) × LCM(a, b) = a × b (for two numbers only).
• Understand the conditions for terminating vs. non-terminating repeating decimal expansions based on the prime factors of the denominator (in the simplified fraction).
• Know that √p (p prime) is irrational, and combinations of rationals and irrationals behave predictably (sum/difference/product/quotient).

Multiple Choice Questions (MCQs)

1. The HCF of 84 and 144 using Euclid's Division Algorithm is:
   (A) 12
   (B) 24
   (C) 6
   (D) 8

2. The prime factorization of 156 is:
   (A) 2 × 3 × 13
   (B) 2² × 3 × 13
   (C) 2 × 3² × 13
   (D) 2² × 39

3. If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is:
   (A) 1800
   (B) 900
   (C) 150
   (D) 90

4. Which of the following numbers has a terminating decimal expansion?
   (A) 29/343
   (B) 64/455
   (C) 17/8
   (D) 77/210

5. Which of the following is an irrational number?
   (A) √16
   (B) √(12/3)
   (C) √12
   (D) √100

6. The number 3 + 2√5 is:
   (A) Rational
   (B) Irrational
   (C) An integer
   (D) A whole number

7. According to the Fundamental Theorem of Arithmetic, the prime factorization of a composite number is:
   (A) Not possible
   (B) Unique, including the order of factors
   (C) Unique, apart from the order of factors
   (D) Has only two factors

8. The decimal expansion of 131 / (2³ × 5²) will:
   (A) Terminate after 2 decimal places
   (B) Terminate after 3 decimal places
   (C) Terminate after 5 decimal places
   (D) Be non-terminating repeating

9. If p and q are two co-prime numbers, then their HCF is:
   (A) p
   (B) q
   (C) pq
   (D) 1

10. For some integer 'm', every even integer is of the form:
    (A) m
    (B) m + 1
    (C) 2m
    (D) 2m + 1

Answer Key:

1. (A)
2. (B)
3. (C)
4. (C)
5. (C)
6. (B)
7. (C)
8. (B) [Hint: Make powers of 2 and 5 equal in the denominator: 131 / (2³ × 5²) = (131 × 5) / (2³ × 5³) = 655 / 10³ = 0.655]
9. (D)
10. (C)

Study these notes thoroughly. Practice applying Euclid's algorithm and prime factorization. Understand the conditions for decimal expansions. Good luck!