NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

NCERT Solutions Class 10 Maths
Chapter – 1 (Real Numbers)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 1 Real Numbers  Exercise 1.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 1 Real Numbers

• NCERT Class 10 Maths Solution Ex – 1.1
• NCERT Class 10 Maths Solution Ex – 1.2
• NCERT Class 10 Maths Solution Ex – 1.4

Exercise – 1.3

1. Prove that √5 is irrational.
Solutions – Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 =
Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
a = √5b
a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k)² = 5b²  This means that b² is divisible by 5 and hence, b is divisible by 5.
b² = 5k²  This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence,√5 cannot be expressed as or it can be said that √5 is irrational.

2. Prove that 3 + 2√5 + is irrational.
Solutions – Let 3 + 2√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
3 + 2√5 =
2√5 =  – 3

√5 =
Since a and b are integers, will also be rational and therefore,√5 is rational.
This contradicts the fact that √5 is irrational. Hence, our assumption that 3 + 2√5 is rational is false.
Therefore, 3 + 2√5  is irrational.

3. Prove that the following are irrationals:
(i)
(ii) 7√5
(iii) 6 + √2

Solutions –
(i)
Let   is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that

=

√2 =

is rational as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and is irrational.

(ii) 7√5
Let 7√5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that

7√5  =  for some integers a and b

∴ √5 =

is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational.
Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.

(iii) 6 + √2
Let 6 + √2 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that
6 + √2 =
√2 = – 6
Since a and b are integers, – 6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational.
Therefore, our assumption is false and hence, 6 + √2 is irrational.

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