Class 10 Mathematics Notes Chapter 1 (Chapter 1) – Examplar Problems (English) Book

Alright class, let's get straight into Chapter 1: Real Numbers from your NCERT Exemplar. This chapter builds upon what you learned in Class 9 and is crucial for various government exams, as number systems form the bedrock of quantitative aptitude. We'll focus on the key concepts and problem types you'll encounter.

Chapter 1: Real Numbers - Key Concepts for Exam Preparation

1. 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.
• Significance: This is the cornerstone of many number theory concepts. It allows us to break down numbers into their fundamental building blocks (primes).
• Application:
  • Finding HCF (Highest Common Factor): Product of the smallest power of each common prime factor in the numbers.
  • Finding LCM (Least Common Multiple): Product of the greatest power of each prime factor involved in the numbers.
  • Example: Find HCF and LCM of 90 and 144.
    • 90 = 2 × 3² × 5¹
    • 144 = 2⁴ × 3²
    • HCF(90, 144) = 2¹ × 3² = 2 × 9 = 18 (Smallest power of common primes 2 and 3)
    • LCM(90, 144) = 2⁴ × 3² × 5¹ = 16 × 9 × 5 = 720 (Greatest power of all primes involved: 2, 3, 5)

2. Relationship between HCF and LCM

• Formula: For any two positive integers a and b,
  HCF(a, b) × LCM(a, b) = a × b
• Important Note: This formula is valid only for two numbers. It does not directly apply to three or more numbers.
• Application: If you know any three of the four values (HCF, LCM, a, b), you can find the fourth. This is a common type of problem.
  • Example: The HCF of two numbers is 18 and their product is 12960. Find their LCM.
    • LCM = (Product of numbers) / HCF = 12960 / 18 = 720

3. Revisiting Irrational Numbers

• Definition: A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Their decimal expansions are non-terminating and non-repeating.
• Key Examples: √2, √3, √5, π, 0.101101110...
• Properties:
  • The sum or difference of a rational and an irrational number is irrational. (e.g., 2 + √3, 5 - √2)
  • The product or quotient of a non-zero rational and an irrational number is irrational. (e.g., 3√7, √5 / 2)
  • The sum, difference, product, or quotient of two irrationals need not be irrational. (e.g., √2 + (-√2) = 0 (rational); √3 × √3 = 3 (rational))
• Proof Technique (for exams where subjective questions might appear, less common in MCQs but builds understanding): Proof by contradiction. Assume the number is rational (p/q), simplify, and arrive at a contradiction (like a common factor between p and q when they were assumed co-prime, or an even number equaling an odd number).

4. Revisiting Rational Numbers and Their Decimal Expansions

• Definition: A number is rational if it can be written in the form p/q, where p and q are integers and q ≠ 0.
• Decimal Expansions: Rational numbers have decimal expansions that are either:
  • Terminating: The decimal expansion ends after a finite number of digits.
  • Non-terminating Repeating (Recurring): The decimal expansion goes on forever, but a block of digits repeats indefinitely.
• The Key Condition (Theorem): Let x = p/q be a rational number, where p and q are co-prime (no common factors other than 1).
  • x has a terminating decimal expansion if and only if the prime factorization of the denominator q is of the form 2ⁿ5ᵐ, where n and m are non-negative integers (i.e., the only prime factors of q are 2s and/or 5s).
  • x has a non-terminating repeating decimal expansion if and only if the prime factorization of the denominator q has at least one prime factor other than 2 or 5.
• Application: Determining the nature of the decimal expansion without actual division.
  • Example 1: 17/8 = 17/2³. Denominator is 2³. Form 2ⁿ5ᵐ (n=3, m=0). So, terminating.
  • Example 2: 29/343 = 29/7³. Denominator has factor 7 (other than 2 or 5). So, non-terminating repeating.
  • Example 3: 77/210 = (7×11)/(2×3×5×7) = 11/(2×3×5). Denominator has factor 3. So, non-terminating repeating. (Crucial Step: Always simplify the fraction to co-prime p and q first!)
• Finding the number of decimal places (for terminating decimals): The number of decimal places is the maximum of the powers of 2 and 5 in the denominator's prime factorization (after simplification).
  • Example: 13 / (2² × 5³) = 13 / 500. Max power is 3 (from 5³). So, it terminates after 3 decimal places (0.026).

5. Euclid's Division Lemma (Contextual Understanding)

• Statement: Given positive integers a and b, there exist unique integers q (quotient) and r (remainder) satisfying a = bq + r, where 0 ≤ r < b.
• Note: While removed from the Class 10 board syllabus, the concept of division, remainders, and its application in finding HCF (Euclid's Division Algorithm) is fundamental. Some competitive exams might still reference the underlying principle. The algorithm repeatedly applies the lemma until the remainder is 0; the last non-zero remainder is the HCF.

Key Takeaways for Exams:

• Master prime factorization for HCF and LCM.
• Remember the HCF × LCM = Product formula (for two numbers).
• Understand the conditions for terminating and non-terminating repeating decimals based on the denominator's prime factors (2ⁿ5ᵐ rule). Always simplify the fraction first!
• Be clear about the properties of rational and irrational numbers (sums, products, etc.).

Now, let's test your understanding with some Multiple Choice Questions.

MCQs based on Chapter 1: Real Numbers (Exemplar Level)

1. If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is:
   (A) 4
   (B) 2
   (C) 1
   (D) 3

2. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is:
   (A) 13
   (B) 65
   (C) 875
   (D) 1750

3. If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then HCF(a, b) is:
   (A) xy
   (B) xy²
   (C) x³y³
   (D) x²y²

4. The decimal expansion of the rational number 14587 / 1250 will terminate after:
   (A) One decimal place
   (B) Two decimal places
   (C) Three decimal places
   (D) Four decimal places

5. Which of the following is NOT an irrational number?
   (A) 3 + √7
   (B) 3√7
   (C) (√2 - √3)(√2 + √3)
   (D) 5 / √2

6. The product of a non-zero rational and an irrational number is:
   (A) always irrational
   (B) always rational
   (C) rational or irrational
   (D) one

7. The LCM of two numbers is 1200. Which of the following cannot be their HCF?
   (A) 600
   (B) 500
   (C) 200
   (D) 400

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

9. The number (√5 + √2)² is:
   (A) Rational
   (B) Irrational
   (C) An integer
   (D) Cannot be determined

10. If n is a natural number, then 9^(2n) - 4^(2n) is always divisible by:
    (A) 5
    (B) 13
    (C) Both 5 and 13
    (D) Neither 5 nor 13

Answer Key:

1. (B) 2 (HCF(65, 117) = 13. 65m - 117 = 13 => 65m = 130 => m = 2)
2. (A) 13 (Required number is HCF(70-5, 125-8) = HCF(65, 117) = 13)
3. (B) xy² (HCF = Product of smallest powers of common primes: x¹ y²)
4. (D) Four decimal places (1250 = 2¹ × 5⁴. Max power is 4. So, 4 decimal places)
5. (C) (√2 - √3)(√2 + √3) (This simplifies to (√2)² - (√3)² = 2 - 3 = -1, which is rational)
6. (A) always irrational (Standard property)
7. (B) 500 (HCF must always be a factor of LCM. 500 is not a factor of 1200)
8. (C) 2q (Definition of an even integer)
9. (B) Irrational ((√5 + √2)² = (√5)² + (√2)² + 2(√5)(√2) = 5 + 2 + 2√10 = 7 + 2√10, which is irrational)
10. (C) Both 5 and 13 (Use a² - b² = (a-b)(a+b). 9^(2n) - 4^(2n) = (9ⁿ)² - (4ⁿ)² = (9ⁿ - 4ⁿ)(9ⁿ + 4ⁿ). For n=1, (9-4)(9+4) = 5 × 13. For n=2, (81-16)(81+16) = 65 × 97 = (5×13) × 97. It's always divisible by (9-4)=5 and (9+4)=13)

Study these notes carefully, practice problems from the Exemplar, and pay close attention to the conditions and theorems. Good luck with your preparation!