NCERT Class 10th Math chapter 1 Exercise 1.3 Question Answer. NCERT Solution for Class 10 Maths Chapter 1 Exercise 1.3 of real numbers NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.
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NCERT Class 10 Math chapter 1 Real Numbers Exercise 1.3 Question Answer Solution.
Class 10 Math Chapter 1 Ex. 1.3 Question Answer
1. Prove that √5 is irrational.
Ans. Let us assume, to the contrary, that √5 is rational.
That is, we can find integers a and b(b≠0) such that √5 = a/b
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b√5 = a
Squaring on both sides, and rearranging, we get 5b2 = a2
Therefore, a2 is divisible by 5 and a is also divisible by 5.
So, we can write a = 5c for some integer c.
Substituting for a, we get 5b2 = 25c2 that is, b2 = 5c2 .
This means that b2 is divisible by 5, and so b is also divisible by 5.
Therefore, a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.
2. Prove that 3 + 2√5 is irrational.
Ans. Let us assume, to the contrary, that 3 + 2√5 is rational.
That is, we can find coprime a and b(b≠0) such that 3 + 2√5 = a/b
Rearranging, we get 2√5 = a/b – 3
= √5 = a/2b – 3/2
Since 2, a and b are integers, we get a/2b – 3/2 is rational, and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 is irrational.
3. Prove that the following are irrationals:
(i) 1/√2
Ans. Let us assume, to the contrary, that 1/√2 is rational.
That is, we can find integers a and b(b≠0) such that 1/√2 = a/b
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b = a√2
Squaring on both sides, and rearranging, we get b2 = 2a2
Therefore, b2 is divisible by 2 and b is also divisible by 2.
So, we can write b = 2c for some integer c.
Substituting for a, we get 2a2 = 4c2 that is, a2 = 2c2 .
This means that a2 is divisible by 2, and so b is also divisible by 2.
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.
So, we conclude that 1/√2 is irrational.
(ii) 7√5
Ans. Let us assume, to the contrary, that 7√5 is rational.
That is, we can find coprime a and b(b≠0) such that 7√5 = a/b
Rearranging, we get √5 = a/7b
Since 7, a and b are integers, a/7b is rational, and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So, we conclude that 7√5 is irrational.
(iii) 6 + √2
Ans. Let us assume, to the contrary, that 6 + √2 is rational.
That is, we can find coprime a and b(b≠0) such that 6 + √2 = a/b
Rearranging, we get √2 = a/b – 6
Since a and b are integers, we get a/b – 6 is rational, and so √2 is rational.
But this contradicts the fact that √2 is irrational.
So, we conclude that 6 + √2 is irrational.