Question

Which of the following is an irrational number?
$
  {\text{a}}{\text{. }}\sqrt {41616} \\
  {\text{b}}{\text{. 23}}{\text{.232323}}........... \\
  {\text{c}}{\text{. }}\dfrac{{{{\left( {1 + \sqrt 3 } \right)}^3} - {{\left( {1 - \sqrt 3 } \right)}^3}}}{{\sqrt 3 }} \\
  {\text{d}}{\text{. 23}}{\text{.10100100010000}}........ \\
$

Answer 1

Hint: - An irrational number cannot be written in the form of $\dfrac{p}{q},q \ne 0$and it has non-repeating or non-terminating sequence after decimal place.
Check Option (a)
$ \Rightarrow \sqrt {41616} = \sqrt {{{\left( {204} \right)}^2}} = 204$
So, 204 is a rational number.
$ \Rightarrow \sqrt {41616} $ Is a rational number.
Now check Option (b)
$
   \Rightarrow 23.232323........ \\
  {\text{Let, }}y = 23.232323.......{\text{ }}..................\left( 1 \right) \\
$
Multiply by 100 in both sides
$ \Rightarrow 100y = 2323.2323.......{\text{ }}..................\left( 2 \right)$
Subtract equation (2) from (1)
$
   \Rightarrow 99y = 2323.2323...... - 23.232323....... \\
   \Rightarrow 99y = 2300 \Rightarrow y = \dfrac{{2300}}{{99}} \\
$
So, 23.2323……….. Is also a rational number.
Now check Option (c)
As we know${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b,{\text{ }}{\left( {a - b} \right)^3} = {a^3} - {b^3} + 3a{b^2} - 3{a^2}b$, so use these properties we have
$\dfrac{{{{\left( {1 + \sqrt 3 } \right)}^3} - {{\left( {1 - \sqrt 3 } \right)}^3}}}{{\sqrt 3 }} \Rightarrow \dfrac{{1 + 3\sqrt 3 + 3\sqrt 3 + 9 - \left( {1 - 3\sqrt 3 - 3\sqrt 3 + 9} \right)}}{{\sqrt 3 }} = \dfrac{{12\sqrt 3 }}{{\sqrt 3 }} = 12$

So, $\dfrac{{{{\left( {1 + \sqrt 3 } \right)}^3} - {{\left( {1 - \sqrt 3 } \right)}^3}}}{{\sqrt 3 }}$ is also a rational number.
Now check Option (d)
${\text{23}}{\text{.10100100010000}}.............$
Since this number is non-repeating and non-terminating and cannot be expressed as a fraction.
$ \Rightarrow $It is an irrational number.
Hence option (d) is correct.

Note: - A rational number is a number that can be expressed as the quotient or fraction $\dfrac{p}{q},q \ne 0$of two integers, a numerator p and a non-zero denominator q. since q may be equal to 1, every integer is a rational number. An irrational number is a number that cannot be expressed as a fraction for any integers, so one by one check all options then we will get the required answer.