# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

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 Book Name Class 10 Mathematics NCERT Textbook Chapter Chapter 1 Real Numbers Exercise Ex 1.3

1. Prove that √5 is irrational.

Solution

Let √5 be a rational number.

∴ We have to find two integers a and b (where, b ≠ 0 and a and b are coprime) such that

a/b = √5

⇒ a = √5.b      ... (1)

Squaring both sides, we have

a2 = 5b2

∴ 5 divides a2

⇒ 5 divides a   ...(2)

[∵ a prime number ‘p’ divides a2 then ‘p’ divides ‘a’, where ‘a’ is positive integer.]

∴ a  =  5c,  where c is an integer.

∴ Putting a = 5c in (1), we have

5c = √5. b

or (5c)2 =  5b2

⇒ 25c2 = 5b2

⇒ 5c2 = b2

⇒ 5 divides b2

⇒ 5 divides b         ...(3)

From (2) and (3)

a and b have at least 5 as a common factor.

i.e., a and b are not coprime.

∴ Our supposition that √5 is rational is wrong.

Hence, √5 is irrational.

2. Prove that 3+2√5 is irrational.

Solution

Let 3+2√5 is rational.

∴ We can find two coprime integers ‘a’ and ‘b’ such that

[3+2√5] =  a/b,  where  b ≠ 0

⇒ (1) is a rational

⇒ √5 is a rational

But this contradicts the fact that √5 is irrational.

∴ Our supposition is wrong.

3+2√5  is an irrational.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6+√2

Solution

(i) We have

since, the division of two integers is rational.

∴ 2a/b is a rational.

From (1), √2 is a rational number which contradicts the fact that √2 is irrational.

∴ Our assumption is wrong.

Thus, 1√2 is irrational.

(ii) Let us suppose that 7√5 is rational.

Let there be two coprime integers ‘a’ and ‘b’.

such that 7√5 =  a/b , where b ≠ 0

Now, = 7√5 = a/b

⇒ √5 is a rational

This contradicts the fact that √5 is irrational.

∴ We conclude that 7√5 is irrational.

(iii) Let us suppose that 6+√2 is rational.

∴ We can find two coprime integers ‘a’ and ‘b’ (b ≠ 0), such that

6+√2  = a/b

[∵ subtraction of integers is also an integer]

[∵ Division of two integers is a rational number]

⇒ a-6b/b is a rational number.

From (1), √2 is a rational number, which contradicts the fact that √2 is an irrational number.

∴ Our supposition is wrong.

⇒ 6+ √2 is an irrational number.

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