# Exercise 1.2(Real numbers)

### Question 1: Prove that √5 is irrational.

Solutions: Let us assume, that 5 is rational number.where, x and y are co-primes and y ≠ 0.

i.e. 5 = x/y.

⇒ y5= x

Squaring both the sides, we get,

(y5)2 = x2

⇒5y2 = x2……………………………….. (1)

Thus, x2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5c, for some value of c and substituting the value of x in equation (1), we get,

5y2 = (5c)2

⇒y2 = 5c2

is divisible by 5 it means y is divisible by 5.

thus, x and y are not co-primes. Thus, our assumption about 5 is rational is incorrect.

Hence, √5 is an irrational number.

### Question 2: Prove that 3 + 2√5  is irrational.

Solutions: Let us assume 3 + 25 is rational.

Then we can find co-prime x and y (y ≠ 0) such that

3 + 2√5 = x/y

⇒ 2√5 = x/y – 3  Since, x and y are integers, thus, is a rational number.

Therefore, 5 is also a rational number. But this contradicts the fact that 5 is irrational.

Hence, 3 + 2√5 is irrational.

(i) 1/√2

(ii) 7√5

(iii) 6 + 2

### Solutions:

(i) 1/2

Let us assume 1/√2 is rational. Then we can find co-prime x and y (y ≠ 0) such that

1/√2 = x/y

⇒ √2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, 1/√2 is irrational.

(ii) 75

Let us assume 7√5 is a rational number.Then we can find co-prime x and y (y ≠ 0) such that

7√5 = x/y

⇒  √5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, 7√5 is irrational.

(iii) 6 +2

Let us assume 6 +√2 is a rational number. Then we can find co-primes x and y (y ≠ 0) such that

6 +√2 = x/y⋅

⇒  √2 = (x/y) – 6 Since, x and y are integers, thus is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, 6 +√2 is irrational.

## Chapter 1 Real Number Class 10 Ncert maths

### Some Case based question   gmath.in