Question

Which of the following is an irrational number?
\[
  A.{\text{ }}\sqrt {\dfrac{8}{{18}}} \\
  B.{\text{ }}\sqrt {\dfrac{{12}}{3}} \\
  C.{\text{ }}\sqrt {\dfrac{{28}}{8}} \\
  D.{\text{ }}\sqrt {81} \\
 \]

Answer 1

Hint: In order to solve the given problem use the basic definition of rational number and try to simplify all the options given above and bring it in the general form of rational number. The number which cannot be converted to a general form of rational number will be an irrational number.

Complete step by step answer:

As we know that any number which can be converted in the form of $\dfrac{p}{q}$ , where both p and q are integers are called rational numbers and any number which does not fulfil this criteria is an irrational number.

Since we have 4 options let us check for all the options.
Let us start with option A
The number is $\sqrt {\dfrac{8}{{18}}} $
Let us try to simplify the number in the lowest form and we will try to remove the root.
\[
   \Rightarrow \sqrt {\dfrac{8}{{18}}} = \sqrt {\dfrac{{2 \times 2 \times 2}}{{2 \times 3 \times 3}}} \\
   = \sqrt {\dfrac{{2 \times 2}}{{3 \times 3}}} \\
   = \dfrac{2}{3} \\
 \]
Since the number has been converted into $\dfrac{p}{q}$ form where p = 2 and q = 3 and both are integers so option A is a rational number.
Let us proceed to option B
The number is $\sqrt {\dfrac{{12}}{3}} $
Let us try to simplify the number in the lowest form and we will try to remove the root.
\[
   \Rightarrow \sqrt {\dfrac{{12}}{3}} = \sqrt {\dfrac{{2 \times 2 \times 3}}{3}} \\
   = \sqrt {\dfrac{{2 \times 2}}{1}} \\
   = \dfrac{2}{1} = 2 \\
 \]
Since the number has been converted into $\dfrac{p}{q}$ form where p = 2 and q = 1 and both are integers so option B is also a rational number.
Let us move to option C
The number is $\sqrt {\dfrac{{28}}{8}} $
Let us try to simplify the number in the lowest form and we will try to remove the root.
\[
   \Rightarrow \sqrt {\dfrac{{28}}{8}} = \sqrt {\dfrac{{2 \times 2 \times 7}}{{2 \times 2 \times 2}}} \\
   = \sqrt {\dfrac{7}{2}} \\
 \]
Since the number could be simplified only till $\sqrt {\dfrac{7}{2}} $ and could not be converted into $\dfrac{p}{q}$ form with both as integer so option C is an irrational number.
Let us move to option D
The number is $\sqrt {81} $
Let us try to simplify the number in the lowest form and we will try to remove the root.
\[
   \Rightarrow \sqrt {81} = \sqrt {3 \times 3 \times 3 \times 3} \\
   = 3 \times 3 = 9 \\
   = \dfrac{9}{1} \\
 \]
Since the number has been converted into $\dfrac{p}{q}$ form where p = 9 and q = 1 and both are integers so option D is also rational number
Hence, $\sqrt {\dfrac{7}{2}} $ is an irrational number.

So, option C is the correct option.

Note: A rational number is a number that can be expressed as the quotient or fraction p/q 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. But every rational number is not an integer.