Question

Give an example of two irrational numbers whose :
(i) difference is an irrational number
(ii) sum is an irrational number
(iii) product is an irrational number
(iv) division is an irrational number

Answer 1

Hint:In the question given to us, we have to consider two such irrational numbers whose difference, sum, product, and division are also irrational numbers. We must know that irrational numbers are those which can be expressed as ${\displaystyle \frac{p}{q}}$ , where p and q are integers and q is not equal to zero. So, we can consider $\pi ,\sqrt{2},\sqrt{3}....$ and then find the answers to each part.

Complete step by step answer:
To solves this question given to us, we need to first understand the concept of irrational numbers and their properties. In mathematics, those numbers are considered as irrational numbers, which are real numbers that cannot be expressed as the ratio of two integers. Or we can say that that they cannot be expressed in the form of ${\displaystyle \frac{p}{q}}$, where p and q are integers and q is not equal to zero. They also have one more characteristic, that is their decimal expansion does not repeat or terminate. So, let us consider each part of the question one by one and solve them.
(i) Difference is an irrational number : If we consider the two numbers as $\sqrt{3}$ and $\sqrt{2}$, then their difference will be given as, $\sqrt{3}-\sqrt{2}=\sqrt{1}$. We can see that their difference is also an irrational number.
(ii) Sum is an irrational number: If we consider the two numbers as $\pi$ and $\sqrt{2}$, then their sum would be, $\pi +\sqrt{2}$ , which is an irrational number.
(iii) Product is an irrational number : Let us consider the same numbers again as $\pi$ and $\sqrt{2}$, then their product would be, $\pi \times \sqrt{2}=\pi \sqrt{2}$ , which is an irrational number.
(iv) Division is an irrational number : In this case let us take the numbers as, $\sqrt{6}$ and $\sqrt{2}$, so when we divide them we will get, ${\displaystyle \frac{\sqrt{6}}{\sqrt{2}}}=\sqrt{{\displaystyle \frac{6}{2}}}=\sqrt{3}$, which is again an irrational number.
Therefore, we can say the answers are as following,
$\begin{array}{rl}& \left(i\right)\sqrt{3},\sqrt{2}\\ & \left(ii\right)\pi ,\sqrt{2}\\ & \left(iii\right)\pi ,\sqrt{2}\\ & \left(iv\right)\sqrt{6},\sqrt{2}\end{array}$

Note:
 In this question, we need to consider such numbers which are irrational keeping in mind that the requirement of the question must be fulfilled. There can be irrational numbers whose, difference, sum, product, or division may be rational numbers, so we have to make the specific choice from among the wide range of irrational numbers. For example, if we consider two irrational numbers $1-\sqrt{2}$ and $1+\sqrt{2}$, the sum would be 2, which is a rational number.