Here we are giving you the solution of NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.3 . You can also download Free NCERT Solutions for Class 10 Maths Chapter 1 PDF on our website . NCERT Maths class 10 chapter 1 exercise 1.3 solutions This post has been prepared by the experienced teachers of Educationforindia.com . Detailed answers of all the questions in Class 10th Maths Chapter 1

## Class 10 Maths Chapter 1 Real Numbers Ex 1.3

**Exercise 1.3 class 10th Question 1Prove that \sqrt\mathbf{5} is irrational.Solutions :**Here, we have to prove that \sqrt\mathbf{5} is an irrational, to solve the question let’s consider opposite of this first that is \sqrt\mathbf{5} is a rational number and not an irrational number.

If is it true than we can write \sqrt\mathbf{5} in the form \ \frac{a}{b}

Then a and b will be co-primes and b ≠ 0.

Therefore, = \sqrt\mathbf{5} = \ \frac{a}{b}

\sqrt\mathbf{5b} = a, now by squaring both sides we will get,

(\sqrt\mathbf{5b})

^{2}= a

^{2}

5b

**= a**

^{2}

^{2}a

^{2}/5 = b

^{2}Hence, the number 5 is dividing a

^{2}

Here, according to the theorem if p is a prime number and it divides a

^{2}, p also divides a, where a is a positive number.

That is why 5 should also divide a

So, a/5 = c where c is some integer, then a = 5c

Now, that’s known to us that 5b

^{2}= a

^{2}

If we put a = 5c, then

5b

^{2}= (5c)

^{2}

5b

^{2}= 25c

^{2}

b

^{2}= 1/5×25c

^{2}

b

^{2}= 5c

^{2}, it is clear that 5 divides b also.

As 5 divides both a and b so 5 is a factor of both a and b,

So, they are not co-prime and what we have assumed is wrong. \sqrt\mathbf{5} is irrational- it is proved.

**2. Exercise 1.3 class 10th Question 2Prove that 3+2\sqrt\mathbf{5} is irrational.**

**Solutions :**

Let’s take, 3+2\sqrt\mathbf{5} is rational number, then

we can write it as ⇒3+2\sqrt\mathbf{5} = b a

a and b will be two co-prime numbers and b ≠ 0

2\sqrt\mathbf{5} = a/b – 3

2\sqrt\mathbf{5} = a-3b/b

Therefore, a-3b/2b

As a and b are integers so a-3b/2b is a rational number, that’s why \sqrt\mathbf{5} should also be a rational number but \sqrt\mathbf{5} is irrational

Ans. 3+2\sqrt\mathbf{5} is irrational and it is proved.

**2. Exercise 1.3 class 10th Question 3****Prove that the following are irrationals:****(i) \frac{\mathbf{1}}{\sqrt\mathbf{2}} (ii) 7\sqrt5 (iii) 6+\sqrt\mathbf{2}**

**(i) \frac{\mathbf{1}}{\sqrt\mathbf{2}}**

**Solutions :**

Let’s assume \frac{\mathbf{1}}{\sqrt\mathbf{2}}

**as rational, now let’s write this number as \frac{\mathbf{1}}{\sqrt\mathbf{2}} = \frac{{a}}{{b}} ; so that a and b are two co-prime numbers and b ≠ 0. By multiplying \sqrt2 on both sides we get,**

1 = \frac{a\sqrt2}{b} , then divide by b,

b = a \sqrt2 or \frac{{b}}{{a}} = \sqrt2

in this case a and b are integers so, \frac{{b}}{{a}} will be consider as a rational number,

as \sqrt2 must be a rational number but it is not. It is an irrational number so it is contradictory

**Ans.**Hence it is proved that \frac{\mathbf{1}}{\sqrt\mathbf{2}}

**is an irrational number.**

**(ii) 7\sqrt5****Solutions :**

Let’s consider 7\sqrt5 ** **as rational number, and let’s write it in the form

7\sqrt5 = \frac{a}{b}

so that a and b are two co-prime numbers and b ≠ 0, now let’s divide the equation by 7 on both sides, we get.

\sqrt5 = \frac{a}{7b} , a and b are integers so \frac{a}{7b} is a rational number, according to that \sqrt5 should also be a rational number but it is an irrational number so it is contradictory.**Ans. **it is proved that 7\sqrt5 is an irrational number.

**(iii) 6+\sqrt\mathbf{2}****Solutions :**

Let’s assume 6+\sqrt\mathbf{2} ** **is rational

It can be written in the form

6+\sqrt\mathbf{2}** =** ** **\frac{a}{b}

so that a and b are two co-prime numbers and b ≠ 0. As we proceed to simplify the above equation we will subtract 6 on both sides and we will get,

\sqrt2 **=** \frac{a}{b}** **– 6

\sqrt5** =** ** **\frac{{a}-{6}{b}}{{b}} as a and b are integers so, \frac{{a}-{6}{b}}{{b}} should be a rational number, that is why \sqrt2 ** **should also be a rational number but it is not so it is contradictory.**Ans. **It is proved that **6 +** **\sqrt2 **is an irrational number.

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