Question

What is the product of two irrational numbers?
${\text{A}}{\text{. }}$Always irrational
${\text{B}}{\text{. }}$Always rational
${\text{C}}{\text{. }}$Can be both rational and irrational
${\text{D}}{\text{. }}$Always an integer

Answer 1

Hint- Here, we will proceed by taking different examples of the irrational numbers and then we will be multiplying these irrational numbers to see whether the corresponding product comes out to be rational or irrational for the various considered cases.

Complete step-by-step solution -
Let us take two irrational numbers as p and q.
Case-I: Let $p = \sqrt 3 $ and $q = \sqrt 3 $ are the values of the two considered irrational numbers.
Here, the product of these two irrational numbers i.e., $p = \sqrt 3 $ and $q = \sqrt 3 $ is given by
$pq = \left( {\sqrt 3 } \right)\left( {\sqrt 3 } \right) = {\left( {\sqrt 3 } \right)^2} = 3$
Since, 3 is a rational number.
Therefore, for this case the product of the two irrational numbers is coming rational.
Case-II: Let $p = \sqrt 3 $ and $q = \sqrt 2 $ are the values of the two considered irrational numbers.
Here, the product of these two irrational numbers i.e., $p = \sqrt 3 $ and $q = \sqrt 2 $ is given by
$pq = \left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right) = \left( {\sqrt {3 \times 2} } \right) = \sqrt 6 $
Since, $\sqrt 6 $ is a rational number.
Therefore, for this case the product of the two irrational numbers is coming irrational.
Therefore, by considering both the above cases we can say that the product of two irrational numbers can be both rational and irrational.
Hence, option C is correct.

Note- A rational number is a number that can be expressed as a fraction (or ratio) in the form $\dfrac{a}{b}$ where the numbers a and b are integers and the value of b can never be equal to zero because then the fraction $\dfrac{a}{b}$ will become not defined. An irrational number is a number that cannot be expressed as a fraction with integer values in the numerator and denominator.