Introduction to Class 10 Maths Chapter 1
Class 10 Maths Chapter 1 on Real Numbers is a fundamental chapter that introduces students to the Euclidean Division Algorithm and the Fundamental Theorem of Arithmetic. This chapter builds upon the concepts learned in Class IX and provides the mathematical foundation for understanding the structure of real numbers.
The chapter covers three main areas: the Fundamental Theorem of Arithmetic, revisiting irrational numbers, and exploring the decimal expansion of rational numbers. Understanding these concepts is crucial for success in higher mathematics.
The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorization is unique apart from the order of prime factors. This theorem forms the backbone of number theory and has numerous applications in mathematics.
Key Applications of the Fundamental Theorem:
- Finding HCF and LCM using prime factorization
- Proving irrationality of numbers
- Understanding decimal expansions of rational numbers
- Solving problems related to divisibility
NCERT Solutions for Class 10 Maths Chapter 1
Exercise 1.1: Fundamental Theorem of Arithmetic Applications
Class 10 Maths Chapter 1 Exercise 1.1 Solutions
Solution 1: Express each number as a product of its prime factors
(i) 140 Prime factorization of 140: 140 = 4 × 35 = 4 × 5 × 7 = 2² × 5 × 7
(ii) 156 Prime factorization of 156: 156 = 4 × 39 = 4 × 3 × 13 = 2² × 3 × 13
(iii) 3825 Prime factorization of 3825: 3825 = 25 × 153 = 25 × 9 × 17 = 5² × 3² × 17
(iv) 5005 Prime factorization of 5005: 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
(v) 7429 Prime factorization of 7429: 7429 = 17 × 437 = 17 × 19 × 23
Solution 2: Find LCM and HCF and verify LCM × HCF = product of numbers
(i) 26 and 91
- 26 = 2 × 13
- 91 = 7 × 13
HCF(26, 91) = 13 LCM(26, 91) = 2 × 7 × 13 = 182
Verification: LCM × HCF = 182 × 13 = 2366 Product of numbers = 26 × 91 = 2366 ✓
(ii) 510 and 92
- 510 = 2 × 3 × 5 × 17
- 92 = 4 × 23 = 2² × 23
HCF(510, 92) = 2 LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
Verification: LCM × HCF = 23460 × 2 = 46920 Product of numbers = 510 × 92 = 46920 ✓
(iii) 336 and 54
- 336 = 16 × 21 = 2⁴ × 3 × 7
- 54 = 2 × 27 = 2 × 3³
HCF(336, 54) = 2 × 3 = 6 LCM(336, 54) = 2⁴ × 3³ × 7 = 3024
Verification: LCF × HCF = 3024 × 6 = 18144 Product of numbers = 336 × 54 = 18144 ✓
Solution 3: Find LCM and HCF using prime factorization method
(i) 12, 15, and 21
- 12 = 2² × 3
- 15 = 3 × 5
- 21 = 3 × 7
HCF(12, 15, 21) = 3 (common factor) LCM(12, 15, 21) = 2² × 3 × 5 × 7 = 420
(ii) 17, 23, and 29 All three numbers are prime. HCF(17, 23, 29) = 1 LCM(17, 23, 29) = 17 × 23 × 29 = 11,339
(iii) 8, 9, and 25
- 8 = 2³
- 9 = 3²
- 25 = 5²
HCF(8, 9, 25) = 1 (no common factors) LCM(8, 9, 25) = 2³ × 3² × 5² = 1800
Solution 4: Given HCF(306, 657) = 9, find LCM(306, 657)
Using the formula: HCF × LCM = Product of numbers LCM(306, 657) = (306 × 657) ÷ 9 = 201042 ÷ 9 = 22338
Solution 5: Check whether 6ⁿ can end with digit 0
For a number to end with 0, it must be divisible by 10 = 2 × 5. 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factorization of 6ⁿ contains only powers of 2 and 3. Since 5 is not a factor, 6ⁿ cannot be divisible by 10. Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.
Solution 6: Explain why given expressions are composite
(a) 7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 13 × 78 Since this can be expressed as a product of integers greater than 1, it is composite.
(b) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5(1008 + 1) = 5 × 1009 Since this can be expressed as a product of integers greater than 1, it is composite.
Solution 7: Circular path problem
Sonia takes 18 minutes, Ravi takes 12 minutes. They will meet again at the starting point after LCM(18, 12) minutes.
Prime factorization:
- 18 = 2 × 3²
- 12 = 2² × 3
LCM(18, 12) = 2² × 3² = 36 minutes
Exercise 1.2: Irrational Numbers
Class 10 Maths Chapter 1 Exercise 1.2 Solutions
Solution 1: Prove that √5 is irrational
Proof by contradiction: Assume √5 is rational, so √5 = a/b where a and b are coprime integers.
Squaring both sides: 5 = a²/b² Therefore: 5b² = a²
This means 5 divides a², so by the fundamental theorem, 5 divides a. Let a = 5c for some integer c.
Substituting: 5b² = (5c)² = 25c² Therefore: b² = 5c²
This means 5 divides b² and hence 5 divides b.
Since both a and b are divisible by 5, they have a common factor, contradicting our assumption that they are coprime.
Therefore, √5 is irrational.
Solution 2: Prove that 3 + 2√5 is irrational
Proof by contradiction: Assume 3 + 2√5 is rational = p/q where p, q are integers and q ≠ 0.
Then: 2√5 = p/q – 3 = (p – 3q)/q Therefore: √5 = (p – 3q)/(2q)
Since p, q are integers, (p – 3q)/(2q) is rational. But this contradicts the fact that √5 is irrational (proved above).
Therefore, 3 + 2√5 is irrational.
Solution 3: Prove the following are irrational
(i) 1/√2 Assume 1/√2 is rational = a/b where a, b are coprime integers. Then √2 = b/a, which would make √2 rational. This contradicts the known fact that √2 is irrational. Therefore, 1/√2 is irrational.
(ii) 7√5 Assume 7√5 is rational = p/q where p, q are coprime integers. Then √5 = p/(7q), which would make √5 rational. This contradicts the fact that √5 is irrational. Therefore, 7√5 is irrational.
(iii) 6 + √2 Assume 6 + √2 is rational = r/s where r, s are coprime integers. Then √2 = r/s – 6 = (r – 6s)/s, which would make √2 rational. This contradicts the fact that √2 is irrational. Therefore, 6 + √2 is irrational.
Key Problems and Solutions
Determining Nature of Decimal Expansions
For fractions like 13/3125:
- 3125 = 5⁵
- Since denominator has only factor 5, decimal expansion terminates
For fractions like 17/8:
- 8 = 2³
- Since denominator has only factor 2, decimal expansion terminates
For fractions like 15/1600:
- 1600 = 2⁶ × 5²
- Since denominator has only factors 2 and 5, decimal expansion terminates
Advanced Problem-Solving
Problems involving prime factorization and divisibility
Real-world applications of HCF and LCM
Integration of rational and irrational number concepts
Class 10 Maths Chapter 1 Solutions: Study Strategy
Effective Learning Approach:
- Master the Fundamental Theorem: Understand that every composite number has unique prime factorization
- Practice Prime Factorization: Regular practice with factor trees and division method
- Understand Proof Techniques: Learn proof by contradiction for irrational numbers
- Apply HCF-LCM Relationships: Use the formula HCF × LCM = Product of numbers
- Analyze Decimal Patterns: Predict terminating vs. non-terminating expansions
Common Mistakes to Avoid:
- Incorrect prime factorization
- Forgetting to check if fractions are in lowest terms
- Incomplete proofs for irrational numbers
- Misapplying the HCF × LCM formula for three or more numbers
- Confusing terminating and non-terminating decimal conditions
Examination Tips for Class 10 Maths Chapter 1
Important Points for Board Exams:
- Show all steps in prime factorization
- State theorems clearly before applying them
- Use proper mathematical notation in proofs
- Verify answers using alternative methods where possible
- Practice word problems involving HCF and LCM applications
Time Management:
- Allocate sufficient time for proof-based questions
- Practice quick prime factorization techniques
- Memorize squares and cubes of numbers up to 25
- Learn to identify composite numbers quickly
Real-World Applications
Practical Uses of Chapter 1 Concepts:
- Cryptography: RSA encryption uses properties of prime numbers
- Computer Science: Algorithms for finding GCD and LCM
- Engineering: Gear ratios and periodic phenomena
- Music Theory: Mathematical relationships in musical scales
- Architecture: Proportional relationships in design
Advanced Concepts and Extensions
Connections to Higher Mathematics:
The concepts in Class 10 Maths Chapter 1 form the foundation for:
- Abstract algebra and group theory
- Advanced number theory
- Cryptographic algorithms
- Mathematical analysis and real number system
- Discrete mathematics and combinatorics
Conclusion
Class 10 Maths Chapter 1 provides essential tools for understanding the structure of numbers. Through systematic study of the Fundamental Theorem of Arithmetic, properties of irrational numbers, and decimal expansions, students develop critical mathematical reasoning skills.
The NCERT solutions for Class 10 Maths Chapter 1 presented here offer comprehensive explanations for all exercises, ensuring students can master these fundamental concepts. Regular practice of these Class 10 Maths Chapter 1 exercise solutions will build confidence and prepare students for both board examinations and advanced mathematical studies.
Success in this chapter requires understanding the underlying principles rather than memorizing procedures. Focus on the logical structure of proofs, the systematic approach to prime factorization, and the connections between different mathematical concepts to achieve mastery in Class 10 Maths Chapter 1.