Question

Prove that $ 5 - \sqrt 3 $ is an irrational number.

Answer 1

Hint: Before attempting this question, one should have prior knowledge about the irrational Number. According to the definition, irrational numbers are all the real numbers which are not rational numbers. So we can say that, irrational numbers can’t be expressed as the ratio of two integers

Complete step-by-step answer:
Let suppose $ 5 - \sqrt 3 $ as a rational number
As we know that $ 5 - \sqrt 3 $ is an rational number then $ 5 - \sqrt 3 $ = $ \dfrac{p}{q} $ where p and q are coprime numbers and q is not equal to zero here Coprime numbers can be defined as the number or integers which have only ‘1’ as the highest common factor
So, we have $ 5 - \sqrt 3 $ = $ \dfrac{p}{q} $
Now Rearranging terms in the above equation i.e. $ 5 - \sqrt 3 = \dfrac{p}{q} $ , we get
 $ - \sqrt 3 = \dfrac{p}{q} - 5 $
 $ - \sqrt 3 = \dfrac{{p - 5q}}{q} $
 $ \sqrt 3 = \dfrac{{5q - p}}{q} $
Now, $ \dfrac{{5q - p}}{q} $ is clearly a rational number as both p and q are integers.
So, by the above statement we can say that $ \sqrt 3 $ is a rational number
Thus, our assumption is incorrect
Therefore, the number $ 5 - \sqrt 3 $ is irrational.

Note: In this above question to identify the given number is rational number or irrational number we separated the rational term and irrational terms to conclude whether the whole expression is either rational or irrational. Also, one should remember that an irrational number can’t be expressed as a simple fraction. E.g. $ \pi $ , $ \sqrt 2 $ ,etc.