Question

Give an example of two irrational numbers, whose difference is a rational number.
(a) $4+\sqrt{2},2+\sqrt{2}$
(b) $4-\sqrt{2},2+\sqrt{2}$
(c) $4+\sqrt{2},2-\sqrt{2}$
(d) $4+\sqrt{2},-2-\sqrt{2}$

Answer 1

Hint: Try to recall the definition of rational and irrational numbers. Use the property that the sum or difference of a rational and an irrational number is always an irrational number. So, all the numbers given in the options are irrational, so just check each and every option to see which option satisfies the condition given in the question to get the answer.

Complete step-by-step solution -
Before moving to the options, let us talk about the definitions of rational numbers followed by irrational numbers.
So, rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers and $q\ne 0$ . In other words, we can say that the numbers which are either terminating or recurring when converted to decimal form are called rational numbers. All the integers fall under this category.
Now, moving to irrational numbers.
Those real numbers which are non-terminating and non-recurring are termed as irrational numbers.
The roots of the numbers which are not perfect squares fall under the category of irrational numbers. $\pi \text{ and }e$ are also the standard examples of irrational numbers.
Now moving to the solution.
We know that the sum or difference of a rational and an irrational number is always an irrational number. Also, as 2 is not a perfect square, $\sqrt{2}$ is irrational. So, all the numbers in the options are irrational. Now we will have to check the options one by one to get the option which satisfies the condition given in the question to reach the answer.
Let us start with (a) $4+\sqrt{2},2+\sqrt{2}$ . If we subtract the second number from the first, we get
$4+\sqrt{2}-\left( 2+\sqrt{2} \right)$
$=2$
As 2 is rational, this is the correct answer.
We have reached a conclusion, however, for a better understanding we will check option (b) as well. Let us subtract the second number given in option (b) from the first number. On doing so, we get
$4-\sqrt{2}-\left( 2+\sqrt{2} \right)$
$=2-2\sqrt{2}$
As $2\sqrt{2}$ is irrational, so $2-2\sqrt{2}$ is irrational so option (b) is not correct. Similarly the results for option (c) and option (d) are $2+2\sqrt{2}$ and $6+2\sqrt{2}$ , which are also irrational.
Therefore, we can conclude that the answer to the above question is option (a).

Note: Always remember that operations like addition, subtraction, multiplication and division between a rational and an irrational number always yields an irrational number, rational numbers involved in division and multiplication are not zero. Also, you can solve the above question based on your observation. See, all the numbers in the options have $\sqrt{2}$ , which makes them irrational so, when we subtract the two numbers the $\sqrt{2}$ terms needs to be eliminated and while we subtract the sign of the $\sqrt{2}$ term of the subtracted number gets changed. So, for the two numbers from the given options to yield a rational number when subtracted must have $\sqrt{2}$ term with the same sign, so that they are eliminated when subtracted, which is only in case of option (a).