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$

{\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
Verified

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, 23.2323â€¦â€¦â€¦.. 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.

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, 23.2323â€¦â€¦â€¦.. 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.

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