Class 10 - Chapter 1: Real Numbers

Exercise 1.2

Q1. Prove that √5 is irrational.

We will use the method of contradiction.

Assume that √5 is rational.

Then it can be written in the form

√5 = a/b

where a and b are integers having no common factor and b ≠ 0.

Squaring both sides,

5 = a²/b²

a² = 5b²

Therefore a² is divisible by 5. Hence a is also divisible by 5.

Let a = 5k.

Substituting,

(5k)² = 5b²

25k² = 5b²

5k² = b²

Thus b² is divisible by 5, so b is also divisible by 5.

Therefore both a and b are divisible by 5.

This contradicts the assumption that a and b have no common factor.

Hence our assumption is wrong.

Therefore, √5 is an irrational number.

Q2. Prove that 3 + 2√5 is irrational.

Assume that 3 + 2√5 is rational.

Then

3 + 2√5 = r

where r is a rational number.

Subtracting 3 from both sides,

2√5 = r − 3

Since the difference of two rational numbers is rational, (r − 3) is rational.

Dividing by 2,

√5 = $\frac{\mathrm{r\; -\; 3}}{2}$

This implies √5 is rational.

But from Question 1, √5 is irrational.

This is a contradiction.

Hence our assumption is false.

Therefore, 3 + 2√5 is irrational.

Q3. Prove that the following are irrational.

(i) 1/√2

Assume that

1/√2

is rational.

Multiplying both sides by √2,

√2 = 1 ÷ (1/√2)

Since the quotient of two rational numbers is rational, √2 would be rational.

But √2 is irrational.

This is a contradiction.

Therefore, 1/√2 is irrational.

(ii) 7√5

Assume that 7√5 is rational.

Then

7√5 = r

where r is rational.

Dividing both sides by 7,

√5 = r/7

Since r/7 is rational, √5 becomes rational.

But √5 is irrational.

This is a contradiction.

Therefore, 7√5 is irrational.

(iii) 6 + √2

Assume that 6 + √2 is rational.

Then

6 + √2 = r

where r is rational.

Subtracting 6 from both sides,

√2 = r − 6

Since the difference of two rational numbers is rational, r − 6 is rational.

This means √2 is rational.

But √2 is irrational.

This is a contradiction.

Hence our assumption is false.

Therefore, 6 + √2 is irrational.