NCERT Solutions for Class 10 Maths Ex 1.2

# NCERT Solutions for Class 10 Maths Ex 1.2

## NCERT Solutions for Class 10 Maths Ex 1.2

Q1. Prove that √5 is irrational.

Solution:  Let us assume that √5 is a rational number.
Let a and b are two co-prime numbers and b is not equal to 0 such that √5 = a/b
Multiply by b on both sides, we get
b√5 = a
To remove root, squaring on both sides, we get
5b2 = a2     … (i)

Therefore, 5 divides a2 and according to theorem of rational number (Theorem 1.3), for any prime number p which is divides a2, it will divide a also.
That means 5 divides a. So we can write
a = 5c
a2 = (5c)2

Putting value of a2 in equation (i), we get
5b2 = (5c)2
5b2 = 25c2
Dividing by 5, we get
b2 = 5c2
Therefore, 5 divides b2 and according to theorem of rational number (Theorem 1.3), 5 divides a.
We have already find that a is divide by 5.
This means a and b both are divisible by 5.

So it contradicts our assumption that a and b are co-prime numbers.
Hence, √5 is not a rational number, it is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Solution: Let us assume that 3 + 2√5 is a rational number.
So, we can write this number as
3 + 2√5 = a/     . . . (i)
Where a and b are two co-prime numbers and b is not equal to 0.
From equation (i), we get
2√5 = a/b – 3
2√5 = (– 3b)/b
√5 = (– 3b)/2b    . . . (ii)
In eq (ii), a and b are integers, so (– 3b)/2b is a rational number.

Therefor, √5 should be a rational number.

But √5 is an irrational number.

Hence, 3 + 2√5 is irrational.

Q3.  Prove that the following are irrationals:
(i) 1/√2                                     (ii) 7√5                                     (iii) 6 + √2

Solution:  (i) Let us assume that 1/√2 is a rational number.
So, we can write this number as
1/√2 = a/b
Where a and b are two co-prime numbers and b is not equal to 0.
In above equation, by cross-multiplication, we get
b = (a√2)
Dividing by a on both sides, we get
b/a = √2
Here, a and b are integers, so b/a is a rational number.

Therefore, √2 should be a rational number.

But √2 is an irrational number.

Hence, 1/√2 is an irrational number.

(ii) Let us assume that 7√5 is a rational number.
So, we can write this number as
7√5 = a/b   . . . (i)
Where a and b are two co-prime numbers and b is not equal to 0.
From eq (i), we get
√5 = a/(7b)
Here, a and b are integers, so a/7b is a rational number.

Therefore, √5 should be a rational number.

But √5 is an irrational number.

Hence, 7√5 is an irrational number.

(iii) Let us assume that 6 + √2 is a rational number.
So, we can write this number as
6 + √2 = a/b    . . . (i)
Where a and b are two co-prime numbers and b is not equal to 0.
From eq (i), we get
√2 = a/b – 6
√2 = (a – 6b)/b
Here, a and b are integers, so (a – 6b)/b is a rational number.

Therefore, √2 should be a rational number.
But √2 is an irrational number.