 QUESTION

# There is a rational number and an irrational number. What will be the product of these two numbers.

Hint: In this question to prove that the product is an irrational number, first we take an example of a product of a rational and an irrational number. Then we assume that the given number is a rational number and later we will contradict our assumption to prove that the given number is an irrational number.

In the given question, we have to find the product of a rational and an irrational number.
First of all, take a number $2\sqrt 3$.
Now suppose that the given number is a rational number of the form $\dfrac{{\text{a}}}{{\text{b}}}$ , where a and b have no common factor other than 1 and b$\ne$ 0.
$\therefore$ .$2\sqrt 3$ =$\dfrac{{\text{a}}}{{\text{b}}}$
On squaring both sides, we get:
${\left( {2\sqrt 3 } \right)^2} = \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^2}}} \\ \Rightarrow 12 = \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^2}}} \\ \Rightarrow 12{{\text{b}}^2} = {{\text{a}}^2} \\$
$\because$ 12 divides ${{\text{b}}^2}$.
$\therefore$ It will also divide ${{\text{a}}^2}$and hence it will divide ‘a’.
Now, let a =12k
Putting the value of ‘a’ in above equation, we get:
$12{{\text{b}}^2} = {\left( {12{\text{k}}} \right)^2}$
On further solving, we get:
$12{{\text{b}}^2} = 144{{\text{k}}^2} \\ {{\text{b}}^2} = 12{{\text{k}}^2} \\$
$\because$ 12 divides ${{\text{k}}^2}$.
$\therefore$ It will also divide ${{\text{b}}^2}$and hence it will divide ‘b’.

Therefore, we can say that 12 is the common factor of both ‘a’ and ‘b’.

This contradicts our assumption that ‘a’ and ‘b’ have no common factor other than 1
Hence, the number $2\sqrt 3$ is an irrational number.

So, we can say that the product of a rational and an irrational number is always an irrational number.

Note: This type of question is always solved by a contradiction method. In this method you have to contradict what you have assumed. In the same way can also prove that $\sqrt a$ is an irrational number, where ‘a’ is greater than 1 and a non perfect square number. You should also remember the result that we have derived above that the product of a rational and an irrational number is an irrational number.