Exercise 1.3 Real Numbers | CBSE Class 10th Mathematics | Solutions

Exercise 1.3 Real Numbers | CBSE Class 10th Mathematics | Solutions

Question 1. Prove that √5 is irrational.

Solution : Let us assume, to the contrary, that √5 is rational.
That is, we can find integers a and b (≠ 0) such that :

		a
√5	=
		b

⋅

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b √5 = a⋅



Squaring on both sides, and rearranging, we get 5b² = a²

[Note: As we know that if p is a prime number and If p divides a2, then p also divides a, where a is a positive integer.]
Therefore, a² is divisible by 5
Then it a is also divisible by 5.
So, we can write a = 5c for some integer c.
Substituting for a, we get 5b² = 25c², that is, b² = 5c²
This means that b² is divisible by 5, and so b is also divisible by 5
Therefore, a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.

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

Solution : Let us assume, to the contrary, that 3 + 2 √5 is rational.
That is, we can find coprime a and b (b ≠ 0) such that : 

		a
3 + 2 √5	=
		b

Therefore, 

		a
2 √5	=		−	3
		b

Or 

		a		3
√5	=		−
		2b		2

Since a and b are integers, we get : 

a		3
	−		is rational
2b		2

And therefore, √5 is rational.

⋅ But this contradicts the fact that √5 is irrational.
This contradiction has arisen because of our incorrect assumption that 3 + 2 √5 is rational.
So, 3 + 2 √5 is an irrational number.

Question 3. Prove that the following are irrationals :

	1
(i)		(ii)	7√ 5	(iii)	6 + √2
	√ 2

Solution (i) Let us assume, to the contrary, that 

1
	is rational
√ 2

That is, we can find coprime a and b (b ≠ 0) such that 

1		a
	=
√ 2		b

1×√ 2		a
	=
√ 2×√ 2		b

Or 

√ 2		a
	=
2		b

Rearranging, we get : 

		2a
√ 2	=
		b

Since a and b are integers, 

2a
	is rational, So √ 2 is rational
b

But this contradicts the fact that √ 2 is irrational.

So, we conclude that - 

1
	is irrational
√ 2

Solution (ii) Let us assume, to the contrary, that 7√5 is rational
That is, we can find coprime p and q (q ≠ 0) such that : 

		p
7√5	=
		q

⋅

Rearranging, we get: 

		p
√5	=
		7q

⋅

Since p and q are integers, 

p
	is rational, So √5 is rational
7q

But this contradicts the fact that √ 5 is irrational.

So, we conclude that 7√5 is rational


Solution (iii) Let us assume, to the contrary, that 6 + √2 is rational.
That is, we can find coprime p and q (q ≠ 0) such that : 

		p
6 + √2	=
		q

Therefore, 

		p
√2	=		−	6
		q

Since p and q are integers, we get : 

p
	− 6	is rational
q

And therefore, √2 is rational.

⋅ But this contradicts the fact that √2 is irrational.
Our  assumption that 6 + √2 is rational, is incorrect.
So, 6 + √2 is an irrational number. 

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Points to remember

• A number ‘s’ is called irrational if it cannot be written in the form p/q , where p and q are integers and q ≠ 0. Examples : √2, √3, √15, π etc.
• For p being a prime number , If p divides a2, then p also divides a, where a is a positive integer.
• The sum or difference of a rational and an irrational number is irrational and
• The product and quotient of a non-zero rational and irrational number is irrational.