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
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