NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers
Introduction to Real Numbers
Chapter 1 of Class 10 Mathematics deals with Real Numbers, building upon concepts learned in earlier classes. This chapter is fundamental for understanding number systems and is important for board exams as well as competitive exams.
Key Concepts
Euclids Division Lemma
For any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b
Here, a is the dividend, b is the divisor, q is the quotient, and r is the remainder.
Euclids Division Algorithm
A technique to find HCF of two positive integers using Euclids Division Lemma repeatedly.
Steps:
- Apply Euclids Division Lemma to a and b where a > b
- If remainder r = 0, then b is the HCF
- If r is not 0, apply Euclids lemma to b and r
- Continue until remainder becomes 0
- The divisor at this stage is the HCF
Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of primes, and this factorization is unique (apart from the order of prime factors).
Example: 12 = 2 x 2 x 3 = 2^2 x 3
Finding LCM and HCF using Prime Factorization
- HCF: Product of smallest power of each common prime factor
- LCM: Product of greatest power of each prime factor
- Important: HCF x LCM = Product of two numbers
NCERT Exercise Solutions
Exercise 1.1
Q1. Use Euclids division algorithm to find HCF of:
(i) 135 and 225
Solution:
- 225 = 135 x 1 + 90
- 135 = 90 x 1 + 45
- 90 = 45 x 2 + 0
Since remainder is 0, HCF = 45
(ii) 196 and 38220
Solution:
- 38220 = 196 x 195 + 0
Since remainder is 0, HCF = 196
(iii) 867 and 255
Solution:
- 867 = 255 x 3 + 102
- 255 = 102 x 2 + 51
- 102 = 51 x 2 + 0
HCF = 51
Exercise 1.2
Q1. Express each number as product of its prime factors:
(i) 140
140 = 2 x 2 x 5 x 7 = 2^2 x 5 x 7
(ii) 156
156 = 2 x 2 x 3 x 13 = 2^2 x 3 x 13
(iii) 3825
3825 = 3 x 3 x 5 x 5 x 17 = 3^2 x 5^2 x 17
(iv) 5005
5005 = 5 x 7 x 11 x 13
(v) 7429
7429 = 17 x 19 x 23
Q2. Find LCM and HCF of following pairs using prime factorization:
(i) 26 and 91
26 = 2 x 13
91 = 7 x 13
HCF = 13 (common factor)
LCM = 2 x 7 x 13 = 182
(ii) 510 and 92
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23 = 2^2 x 23
HCF = 2
LCM = 2^2 x 3 x 5 x 17 x 23 = 23460
(iii) 336 and 54
336 = 2^4 x 3 x 7
54 = 2 x 3^3
HCF = 2 x 3 = 6
LCM = 2^4 x 3^3 x 7 = 3024
Exercise 1.3
Q1. Prove that square root of 5 is irrational.
Proof by contradiction:
Assume sqrt(5) is rational. Then sqrt(5) = a/b where a and b are coprime integers.
Squaring: 5 = a^2/b^2
So: a^2 = 5b^2
This means a^2 is divisible by 5, so a is divisible by 5.
Let a = 5c. Then 25c^2 = 5b^2, giving b^2 = 5c^2
This means b is also divisible by 5.
But this contradicts that a and b are coprime.
Therefore, sqrt(5) is irrational.
Q2. Prove that 3 + 2sqrt(5) is irrational.
Assume 3 + 2sqrt(5) is rational = a/b
Then sqrt(5) = (a/b – 3)/2 = (a – 3b)/(2b)
This would mean sqrt(5) is rational, which is false.
Therefore, 3 + 2sqrt(5) is irrational.
Exercise 1.4
Q1. Without actually dividing, determine which decimals are terminating:
(i) 13/3125
3125 = 5^5 (only prime factor is 5)
Since denominator has only 2 or 5 as prime factors, it is terminating.
(ii) 17/8
8 = 2^3 (only prime factor is 2)
Terminating decimal.
(iii) 64/455
455 = 5 x 7 x 13 (has factors other than 2 and 5)
Non-terminating, repeating decimal.
Important Formulas
- HCF(a, b) x LCM(a, b) = a x b
- A rational number p/q has terminating decimal if q = 2^n x 5^m
- For three numbers: HCF x LCM is not equal to product of numbers
Tips for Board Exams
- Practice Euclids algorithm thoroughly
- Learn proof methods for irrational numbers
- Remember the condition for terminating decimals
- Practice prime factorization of large numbers
Key Concepts: Real Numbers (Class 10 Maths Chapter 1)
Real Numbers is a crucial chapter for Class 10 CBSE Maths. It builds on number theory concepts and is a frequent source of questions in board exams — often carrying 6–8 marks.
Euclid’s Division Lemma
For any two positive integers a and b, there exist unique non-negative integers q (quotient) and r (remainder) such that:
a = bq + r, where 0 ≤ r < b
This is used to find the HCF (Highest Common Factor) of two numbers step by step.
Euclid’s Division Algorithm — Steps
- Step 1: Apply Euclid’s Division Lemma to the larger number (a) and smaller number (b): a = bq + r
- Step 2: If r = 0, then HCF = b
- Step 3: If r ≠ 0, apply the algorithm again with b and r: b = rq₁ + r₁
- Repeat until remainder = 0. The divisor at that step is the HCF.
Example: Find HCF of 870 and 225.
870 = 225 × 3 + 195
225 = 195 × 1 + 30
195 = 30 × 6 + 15
30 = 15 × 2 + 0
HCF = 15
Fundamental Theorem of Arithmetic
Every composite number can be expressed (factored) as a product of primes, and this factorisation is unique (apart from the order of prime factors).
This is the basis for finding LCM and HCF using prime factorisation:
- HCF = Product of the smallest powers of common prime factors
- LCM = Product of the greatest powers of all prime factors
- Key relationship: HCF × LCM = Product of the two numbers (for two numbers)
Rational and Irrational Numbers
A rational number can be expressed as p/q where p and q are integers and q ≠ 0. A rational number’s decimal expansion either terminates or is non-terminating repeating.
An irrational number cannot be expressed as p/q. Its decimal expansion is non-terminating and non-repeating. Examples: √2, √3, √5, π
Proving irrationality by contradiction (common board exam question type): Assume the number is rational, write it as p/q in lowest terms, then derive a contradiction to prove p and q share a common factor — contradicting our assumption.
Terminating and Non-Terminating Decimals
A rational number p/q (in lowest terms) has a terminating decimal expansion if and only if q can be written as 2ⁿ × 5ᵐ (i.e., q has no prime factor other than 2 and 5).
Examples:
3/8 = 3/(2³) → Terminating (0.375)
7/6 = 7/(2 × 3) → Non-terminating (3 is a prime factor of denominator)
Important Board Exam Questions
- Find HCF and LCM of 6, 72, and 120 using prime factorisation
- Prove that √2 is irrational
- Prove that √3 + √5 is irrational (given √15 is irrational)
- Without actually performing the long division, state whether 17/8 will have a terminating or non-terminating decimal expansion
- Show that any positive odd integer is of the form 4q + 1 or 4q + 3
- If HCF (a, b) = 4 and a × b = 180, find LCM (a, b)
Common Mistakes to Avoid
- Confusing HCF and LCM — remember HCF is always ≤ both numbers; LCM is always ≥ both numbers
- In HCF × LCM = a × b — this formula only works for exactly two numbers
- When proving irrationality, the contradiction must show p and q have a common factor ≥ 2, which contradicts p/q being in its lowest terms
- In the terminating decimal check, always simplify p/q to its lowest terms first before checking the denominator’s prime factors
