**Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 – Real Numbers**

**1. Prove that √5 is irrational.**

**Proof:** Let √5 is not irrational, i.e., rational number. Then there exist two integers a and b, where (b ≠ 0) such that √5 = .

Let a and b have a common factor other than 1, then dividing a and b by that common factor we get √5 = , where a_{1 }and b_{1 }are co prime.

√5 = gives a_{1 }= √5 b_{1}

Squaring both side,

(a_{1})^{2}= 5 (b_{1})^{2 }…..(1).

Therefore (a_{1})^{2 }is divisible by 5 and hence a_{1} is divisible by 5

So we can write a_{1 }= 5k, for some integer k. substituting this in (1) we get

25k^{2}= 5 (b_{1})^{2}

(b_{1})^{2} = 5k

(b_{1})^{2} is also divisible by 5 and hence b_{1 }is divisible by 5.

Therefore a_{1 }and b_{1 }have least common factor 5, which is a contradiction to the fact that a_{1 }and b_{1 }are co prime. So, our assumption is wrong that √5 is rational.

**2. Prove that 3+ 2√5 is irrational.**

**Proof:** Let 3+2√5 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

3+2√5 =

⇒ – 3 = 2√5

⇒ = √5

Since a and b are integers then is rational and therefore √5 is rational, which is a contradiction to the fact that √5 is irrational.

Hence, our assumption is wrong that 3+ 2√5 is rational.

Therefore, 3+ 2√5 is irrational.

**3. Prove that the following are irrational.**

**(i) ****(ii) 7√5 ****(iii) 6+√2**

**(i)** Let is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

=

⇒ = √2.

Since, a and b are integers then is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

Hence, our assumption is wrong and is irrational.

**(ii)** Let 7√5 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

7√5 =

⇒ = √5.

Since, a and b are integers, then is rational

Therefore, √5 is rational, which is a contradiction to the fact that √5 is irrational.

Hence, our assumption is wrong and 7√5 is irrational.

**(iii)** Let 6+√2 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

6+√2 =

⇒ – 6 = √2

⇒ = √2

Since, a and b are integers, then is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

Hence, our assumption is wrong and 6+√2 is irrational.

### NCERT Solutions For Class 10 Maths Chapter 1

**NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1****NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2****NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4**

**Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 – Real Numbers**

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