# Prove that the following are irrationals:

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

**Solution:**

Irrational numbers are the subset of real numbers that cannot be expressed in the form of a fraction, p/q where p and q are integers. The denominator q is not equal to zero (q ≠ 0).

(i) 1/√2

Let us assume that 1/√2 is a rational number.

Then, 1/√2 = a/b, where a and b have no common factors other than 1.

√2 × a = b

√2 = b/a

Since b and a are integers, b/a is a rational number and so, √2 is rational.

But we know that √2 is irrational.

So, our assumption was wrong. Therefore, 1/√2 is an irrational number.

(ii) 7√5

Let us assume that 7√5 is a rational number.

Then, 7√5 = a/b, where a and b have no common factors other than 1.

(7√5) b = a

√5 = a/7b

Since, a, 7, and b are integers, so, a/7b is a rational number. This means √5 is rational. But this contradicts the fact that √5 is irrational.

So, our assumption was wrong. Therefore, 7√5 is an irrational number.

(iii) 6 + √2

Let us assume that 6 + √2 is rational.

Then, 6 + √2 = a/b, where a and b have no common factors other than 1.

√2 = (a/b) - 6

Since, a, b, and 6 are integers, so, a/b - 6 is a rational number. This means √2 is also a rational number.

But this contradicts the fact that √2 is irrational. So, our assumption was wrong.

Therefore, 6 + √2 is an irrational number.

**Video Solution:**

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

### NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.3 Question 3 - Chapter 1 Exercise 1.3 Question 3:

**Summary:**

We have proved that (i) 1/√2 (ii) 7√5 (iii) 6 + √2 are irrationals using the contradiction method.