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This is the second of our many quizzes on classifying real numbers. Click on this to access more such competitive quizzes on real numbers.

Classify Real Numbers Quiz 2

mathematics

 $\dfrac{{{2^4} \times {3^4}}}{{{5^2} \times {7^2}}}$ is which type of number?

 The equation is in the form of $\dfrac{P}{Q}$. 

  $\dfrac{{{2^4} \times {3^4}}}{{{5^2} \times {7^2}}} \\ \Rightarrow \dfrac{{16 \times 81}}{{25 \times 49}} \\ \Rightarrow \dfrac{{81}}{{25}} \\ \text{Hence, it is a rational number.}$

 If ${2, 3, 4}$ are natural numbers then $\dfrac{{{2^2} \times {4^3}}}{{{3^0}}}$ will be which type of number?

 ${a^0} = 1$, with help of this solve the expression, and determine the type of number.

  $\dfrac{{{2^2} \times {4^3}}}{{{3^0}}} \\ \Rightarrow \dfrac{{4 \times 64}}{1} \\ \Rightarrow 256\text{ or }\dfrac{{256}}{1} \\ \text{ So, }\dfrac{{{2^2} \times {4^3}}}{{{3^0}}}\text{ is both rational as well a natural number.}$

 If $\sqrt 2 $ is irrational number, then $\dfrac{{\sqrt 2  \times {{\left( {\sqrt 2 } \right)}^2}}}{{\dfrac{{{{\left( {2\sqrt 2 } \right)}^2}}}{{\sqrt 2 }}}}$ will be which type of number?

 Rational number are that type of numbers which can be expressed in term of $\dfrac{P}{Q},\left( {{\text{where, Q}} \ne {\text{0}}} \right)$.

 $\text{Given equation can be written as, } \dfrac{{\sqrt 2  \times 2 \times \sqrt 2 }}{{\left( {2\sqrt 2 } \right)\left( {2\sqrt 2 } \right)}} \\ \Rightarrow \dfrac{{4}}{8} = \dfrac{1}{2}\\ \text{ Hence, The given equation will be a rational number.}$

 Which of the following statements is correct?

 Rational number are that type of numbers which can be expressed in term of $\dfrac{P}{Q},\left( {{\text{where, Q}} \ne {\text{0}}} \right)$

 $2\sqrt 2  \in I \\ {\left( {2\sqrt 2 } \right)^2} \Rightarrow 4 \times 2 = 8 \\ \left( {2\sqrt 2 } \right)^2,\text{ is a natural / rational number} \\  \Rightarrow {\left( {2\sqrt 2 } \right)^2} \notin I \\ \Rightarrow 2 + \sqrt 2\text{ As }\sqrt{2}\text{ is a irrational number} \\ \Rightarrow 2 + \sqrt 2  \in I $ 

 $\text{Assertion: }0,\dfrac{1}{2}, - 0.33,\sqrt 4\text{are rational numbers.} \\ \text{Reason: All fractions are rational numbers.}$

 0, cannot be written in form of $\dfrac{P}{Q},\left( {{\text{where, Q}} \ne {\text{0}}} \right)$, so it will not be a rational number.

 Assertion: $ \\ 0,\dfrac{1}{2}, - 0.33,\sqrt 4  \\ \Rightarrow \dfrac{0}{0},\dfrac{1}{2},\dfrac{1}{3},2 \\ $ As, rational number can be converted in form of $ \dfrac{P}{Q},\left( {{\text{where, Q}} \ne {\text{0}}} \right).\\ $ 0 will not be a rational number. $ \\ $ So, Assertion is incorrect. $\\ $ Reason: $ \\ $ All fractions are rational number $ \\ $ As, fractions can be converted in form of $ \\ \dfrac{{\text{P}}}{{\text{Q}}}{\text{ (where Q}} \ne {\text{0)}}\\ $ So, all fractions are rational numbers. Hence, reason is correct, but reason is not a correct explanation of Assertion.