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 1
Prove 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 = a2
5b2 = a2
a2/5 = b2 Hence, the number 5 is dividing a2
Here, according to the theorem if p is a prime number and it divides a2, 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 5b2 = a2
If we put a = 5c, then
5b2 = (5c)2
5b2 = 25c2
b2 = 1/5×25c2
b2 = 5c2, 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 2
Prove 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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