Express in $\dfrac{p}{q}$ form $0.\overline 6 $

Hint: First, we need to know about the concept of rational numbers. From the question, we have $0.\overline 6 $ which is equal to $0.6666666$ . Take any variable as the given number, here we have taken $x$, so we get $x = 0.\overline 6 $ and solve them by multiplying with $10$ to get the value of the variable $x$ . Hence, we can express it in the $\dfrac{p}{q}$ form.

Complete step by step answer:
Since from given that $0.\overline 6 $ which means $0.\overline 6 = 0.6666666$
The repeated numbers in the decimal format denoted as $0.\overline 1 = 0.11111$
Thus, take the given numbers into any variable so that we can solve the given problem easily with the multiplication of the number ten.
Assume the given as $x$ then we get $x = 0.\overline 6 $
Now multiply both sides with the number $10$ then we get $10 \times x = 10 \times 0.\overline 6 $
Since we know that $0.\overline 6 = 0.6666666$ and then we have $10 \times x = 10 \times 0.\overline 6 \Rightarrow 10 \times x = 10 \times 0.666666$
Using the multiplication operation, we get $ \Rightarrow 10x = 6.6666666$
Again, using the same method $0.\overline 6 = 0.6666666$ then we have $ \Rightarrow 10x = 6.6666666 \Rightarrow 10x = 6.\overline 6 $
Now using the subtraction operation of the two founded values $x = 0.\overline 6 $ and $10x = 6.\overline 6 $ we get $10x - x = 6.\overline 6 - 0.\overline 6 $
Thus, canceling the common terms, we have $10x - x = 6.\overline 6 - 0.\overline 6 \Rightarrow 9x = 6$
Using the division operation, we get $9x = 6 \Rightarrow x = \dfrac{6}{9}$
Canceling with the number $3$ on numerator and denominator we get $x = \dfrac{6}{9} \Rightarrow x = \dfrac{2}{3}$
Hence $0.\overline 6 $ can be written as $\dfrac{p}{q}$ form of $\dfrac{2}{3}$

Note:
Note that every whole number has number one as its denominator like $2 = \dfrac{2}{1}$ and hence which is also the rational number, we can say that every natural number are the rational number (without the number zero)
Also, there are infinitely rational numbers between any two numbers.
Using the basic operations like subtraction and division we solved the given problem.
Subtraction operation, minus of given two or more than two numbers, but here comes with the condition that in subtraction the greater number sign represented in the number will stay constant example $2 - 3 = - 1$
The process of the inverse of the multiplication method is called division. Like $x \times y = z$ is multiplication thus the division sees as $x = \dfrac{z}{y}$. Like $\dfrac{4}{2} = 2$