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 :
⋅
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 :
Therefore,
Or
Since a and b are integers, we get :
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 :
Solution (i) Let us assume, to the contrary, that
That is, we can find coprime a and b (b ≠ 0) such that
Or
Rearranging, we get :
Since a and b are integers,
But this contradicts the fact that √ 2 is irrational.
So, we conclude that -
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 :
⋅
Rearranging, we get:
⋅
Since p and q are integers,
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 :
Therefore,
Since p and q are integers, we get :
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.
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 |
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 |
| 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 |
⋅ 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 |
| 1 | a | |
| = | ||
| √ 2 | b |
| 1×√ 2 | a | |
| = | ||
| √ 2×√ 2 | b |
Or
| √ 2 | a | |
| = | ||
| 2 | b |
Rearranging, we get :
| 2a | ||
| √ 2 | = | |
| b |
| 2a | |
| is rational, So √ 2 is rational | |
| b |
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 |
| p | ||
| √5 | = | |
| 7q |
| p | |
| is rational, So √5 is rational | |
| 7q |
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 |
| p | ||||
| √2 | = | − | 6 | |
| q |
Since p and q are integers, we get :
| p | ||
| − 6 | is rational | |
| q |
⋅ 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.
