## Saturday, 5 July 2014

### 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.

## 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.