Question

The rational number of decimal numbers $0.\bar 6$ is:
A) $\dfrac{{33}}{{50}}$
B) $\dfrac{2}{3}$
C) $\dfrac{{167}}{{111}}$
D) $\dfrac{1}{3}$

Answer 1

Hint:In this question, we will assume x = $0.\bar 6$ and this would be the first equation. In the next step we will multiply both sides of the first equation with 10. In this way we will get two equations and after subtracting the first equation from the second equation we will get the value of x.

Complete step-by-step answer:
Real numbers is a set of rational numbers and irrational numbers. Imaginary numbers are unreal numbers, which cannot be expressed on the number line and commonly used to represent a complex number. Rational numbers are represented in the $\dfrac{p}{q}$ form where q is not equal to zero. Any fraction with non-zero denominators is a rational number. Zero is also a rational number. Here $\dfrac{0}{1},\dfrac{0}{2},\dfrac{0}{3}$, etc. are also a rational number but $\dfrac{1}{0},\dfrac{2}{0},\dfrac{3}{0}$, etc. are not rational, as they give infinite value.
In given question
Let $x = 0.66666…. (1)$
(Multiplying both side by 10)
$10x = 6.66666…. (2)$
Now Subtracting (1) from (2), we get;
$10x – x = 6.66666… - 0.66666…$
         $9x = 6$
$\therefore x = \dfrac{2}{3}$
Thus, the rational number of decimal numbers $0.\bar 6$ is $\dfrac{2}{3}$

So, the correct answer is “Option B”.

Note:Irrational numbers are real numbers that cannot be expressed as a simple fraction. It cannot be expressed in the form of $\dfrac{p}{q}$. An irrational number is a real number that cannot be expressed as a ratio of integers, for example $\sqrt 3 $ is an irrational number. The decimal expansion of an irrational number is neither terminating nor recurring. The value of that is $\dfrac{{22}}{7}$ an irrational number because it is non-terminating.