Question

Prove that the sum of two irrational numbers given by $3+\sqrt{2}\text{ and 3}-\sqrt{2}$ is a rational number.

Answer 1

Hint: We will have to know about the rational numbers and irrational numbers. The rational numbers are numbers which can be expressed as a fraction and also as positive numbers, negative numbers and zero. It can be written as p/q, where q is not equal to zero whereas the irrational numbers can not be written as the ratio of two integers and having endless non-repeating digits after the decimal point. Here, we will add the given number and express the sum as a rational number.

Complete step-by-step answer:
We have been given two irrational number $3+\sqrt{2}\text{ and 3}-\sqrt{2}$ whose sum is as follows:
\[\Rightarrow \left( 3+\sqrt{2} \right)+\left( 3-\sqrt{2} \right)=3+{\sqrt{2}}+3-{\sqrt{2}}=\dfrac{6}{1}\]
So, the sum of the given two irrational numbers is equal to 6 which is a rational number in the form of p/q where p=6 and q=1 both are integers.
Therefore, it is proved that the sum of the two given irrational numbers is a rational number.

Note: Just remember that fact about rational numbers and irrational numbers that the rational numbers includes perfect squares and irrational numbers includes surds.
Also the decimal expansion of a rational number executes finite or recurring decimals whereas the decimal expansion of irrational numbers executes non-finite and non-recurring decimals.