Question

Prove that: $4 - 5\sqrt 3 $ is an irrational number.

Answer 1

Hint: In order to prove such types of questions we will start with the assumption that a given number is rational then we will proceed further by definition of rational number which is defined as a number that can be expressed as the quotient or fraction of two integers.

Complete step-by-step answer:

Let us assume that $4 - 5\sqrt 3 $ is a rational number.
$
   \Rightarrow 4 - 5\sqrt 3 = \dfrac{a}{b}{\text{ where a,b}} \in {\text{Z}} \\
   \Rightarrow {\text{ - 5}}\sqrt 3 = \dfrac{a}{b} - 4 \\
   \Rightarrow \sqrt 3 = \dfrac{4}{5} - \dfrac{a}{b} \\
 $

LHS= irrational number as ($\sqrt 3 $ is an irrational number)

RHS= rational number

So, $RHS \ne LHS$

Hence, our assumption is wrong and $4 - 5\sqrt 3 $ is an irrational number.

Note: In order to solve these types of questions remember the definitions of all the numbers such as rational, natural, whole and irrational numbers. Irrational numbers are those numbers which cannot be expressed in the $\dfrac{p}{q}$ form. Also remember all natural numbers are rational numbers.