Class 10 Maths Real Numbers – Complete Chapter Guide with Exercise Solutions

Chapter Overview

Real Numbers form the foundation of mathematics and represent one of the most crucial chapters in Class 10 Maths curriculum. This comprehensive chapter introduces students to the classification of numbers, including natural numbers, whole numbers, integers, rational and irrational numbers. Students learn about Euclid’s division lemma, the fundamental theorem of arithmetic, and methods to find HCF and LCM of large numbers. The chapter also covers decimal expansions of rational numbers and their terminating or non-terminating nature. Understanding these concepts is essential for advanced mathematical topics and forms approximately 6-8% of board examination marks. Mastering this chapter strengthens logical reasoning and problem-solving abilities required throughout mathematics.

NCERT Solutions – Exercise-wise Solutions

Exercise 1.1 Solutions

Question 1: Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution: Let ‘a’ be any positive integer. By Euclid’s division lemma: a = 3q + r, where 0 ≤ r < 3 So r = 0, 1, or 2

Case 1: When r = 0 a = 3q a² = (3q)² = 9q² = 3(3q²) = 3m, where m = 3q²

Case 2: When r = 1 a = 3q + 1 a² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q

Case 3: When r = 2 a = 3q + 2 a² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1

Therefore, the square of any positive integer is either 3m or 3m + 1.

Question 2: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Solution: Let ‘a’ be any positive integer. By Euclid’s division lemma: a = 3q + r, where r = 0, 1, or 2

Case 1: When a = 3q a³ = (3q)³ = 27q³ = 9(3q³) = 9m, where m = 3q³

Case 2: When a = 3q + 1 a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1, where m = 3q³ + 3q² + q

Case 3: When a = 3q + 2 a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8, where m = 3q³ + 6q² + 4q

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Exercise 1.2 Solutions

Question 1: Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solutions:

(i) 140 = 2² × 5 × 7 Method: 140 = 4 × 35 = 4 × 5 × 7 = 2² × 5 × 7

(ii) 156 = 2² × 3 × 13 Method: 156 = 4 × 39 = 4 × 3 × 13 = 2² × 3 × 13

(iii) 3825 = 3² × 5² × 17 Method: 3825 = 25 × 153 = 25 × 9 × 17 = 5² × 3² × 17

(iv) 5005 = 5 × 7 × 11 × 13 Method: 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23 Method: First check divisibility by small primes systematically

Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Verification: 26 × 91 = 2366; 13 × 182 = 2366 ✓

(ii) 510 and 92 510 = 2 × 3 × 5 × 17 92 = 2² × 23 HCF = 2 LCM = 2² × 3 × 5 × 17 × 23 = 23460 Verification: 510 × 92 = 46920; 2 × 23460 = 46920 ✓

Exercise 1.3 Solutions

Question 1: Prove that √5 is irrational.

Solution: Let us assume that √5 is rational. Then √5 = a/b, where a and b are integers with no common factors and b ≠ 0.

Squaring both sides: 5 = a²/b² Therefore: 5b² = a²

This means a² is divisible by 5, so a is divisible by 5. Let a = 5c for some integer c.

Substituting: 5b² = (5c)² = 25c² Therefore: b² = 5c²

This means b² is divisible by 5, so b is divisible by 5.

But if both a and b are divisible by 5, they have a common factor 5, which contradicts our assumption that a and b have no common factors.

Therefore, our assumption is wrong, and √5 is irrational.

Question 2: Prove that 3 + 2√5 is irrational.

Solution: Let us assume that 3 + 2√5 is rational. Then 3 + 2√5 = r, where r is rational.

Rearranging: 2√5 = r – 3 Therefore: √5 = (r – 3)/2

Since r is rational and 3 is rational, (r – 3) is rational. Since 2 is rational, (r – 3)/2 is rational.

This means √5 is rational, which contradicts the fact that √5 is irrational.

Therefore, our assumption is wrong, and 3 + 2√5 is irrational.

Exercise 1.4 Solutions

Question 1: Without actually performing the long division, state whether the following rational numbers will have a terminating or non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343

Solutions:

(i) 13/3125 3125 = 5⁵ Since denominator has only 5 as prime factor → Terminating

(ii) 17/8 8 = 2³ Since denominator has only 2 as prime factor → Terminating

(iii) 64/455 455 = 5 × 7 × 13 Since denominator has prime factors other than 2 and 5 → Non-terminating

(iv) 15/1600 1600 = 2⁶ × 5² Since denominator has only 2 and 5 as prime factors → Terminating

(v) 29/343 343 = 7³ Since denominator has prime factors other than 2 and 5 → Non-terminating

Question 2: Write down the decimal expansions of those rational numbers which have terminating decimal expansions.

Solutions:

(i) 13/3125 = 13/5⁵ = (13 × 2⁵)/(5⁵ × 2⁵) = 416/100000 = 0.00416

(ii) 17/8 = 17/2³ = (17 × 5³)/(2³ × 5³) = 2125/1000 = 2.125

(iv) 15/1600 = 15/(2⁶ × 5²) = (15 × 2⁴)/(2¹⁰ × 5²) = 240/25600 = 0.009375

Important Formulas and Concepts

Key Definitions

Real Numbers Classification:

• Natural Numbers (N): {1, 2, 3, 4, …}
• Whole Numbers (W): {0, 1, 2, 3, 4, …}
• Integers (Z): {…, -2, -1, 0, 1, 2, …}
• Rational Numbers (Q): Numbers that can be expressed as p/q where p, q ∈ Z and q ≠ 0
• Irrational Numbers: Numbers that cannot be expressed as p/q
• Real Numbers (R): Union of rational and irrational numbers

Euclid’s Division Lemma

For any two positive integers a and b: a = bq + r, where 0 ≤ r < b

Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes, and this factorization is unique (except for the order of factors).

HCF and LCM Formulas

• For two numbers a and b: HCF × LCM = a × b
• HCF by Prime Factorization: Product of lowest powers of common prime factors
• LCM by Prime Factorization: Product of highest powers of all prime factors

Decimal Expansion Rules

A rational number p/q (in lowest terms) has:

• Terminating decimal: if q = 2ᵐ × 5ⁿ for some non-negative integers m, n
• Non-terminating repeating decimal: if q has prime factors other than 2 and 5

Irrational Number Properties

• √p is irrational if p is not a perfect square
• Sum/difference of rational and irrational = irrational
• Product of non-zero rational and irrational = irrational

Tips and Tricks Section

Quick Problem-Solving Strategies

1. Euclid’s Division Algorithm Speed Tip: Always divide the larger number by the smaller one. Continue until remainder becomes 0. The last non-zero remainder is the HCF.

2. Prime Factorization Shortcut: Start with smallest primes (2, 3, 5, 7, 11…). Use divisibility tests:

• Divisible by 2: Last digit even
• Divisible by 3: Sum of digits divisible by 3
• Divisible by 5: Last digit 0 or 5

3. Proving Irrationality: Always use contradiction method. Assume the number is rational, then show this leads to a contradiction.

4. Decimal Expansion Quick Check: For p/q in lowest terms:

• Check prime factors of q
• Only 2 and 5 → Terminating
• Any other prime → Non-terminating

5. Memory Techniques:

• RICE: Rational, Irrational numbers Complete the rEal numbers
• “2 and 5 Terminate”: Remember only denominators with factors 2 and 5 give terminating decimals

Exam Strategy Tips

Time Management:

• Exercise 1.1: 8-10 minutes per question
• Exercise 1.2: 5-7 minutes per question
• Exercise 1.3: 10-12 minutes for proof questions
• Exercise 1.4: 3-5 minutes per question

Common Mistakes to Avoid:

1. Not simplifying fractions to lowest terms before checking decimal expansion
2. Forgetting to verify HCF × LCM = Product of numbers
3. Incomplete prime factorization
4. Not stating the contradiction clearly in irrationality proofs

Scoring Tips:

• Always show step-by-step working
• State theorems/lemmas before using them
• Double-check calculations in factorization
• Write clear conclusions for proof questions

This comprehensive guide covers all aspects of Class 10 Maths Chapter 1 Exercise solutions with detailed explanations, formulas, and practical tips for examination success.