Convert \[4.6\] (\[6\] being repeated) to a fraction.

Hint: A fraction represents a part of a whole. In this question we are provided with a recurring decimal number. So we first multiply it by \[10\], subtract the old number from the new, get an integer value and then divide this by \[9\].

Complete step by step solution:
In the given decimal the number \[6\] keeps on repeating, that is it is a recurring decimal.
We know, \[4.6666....\]can be represented as \[4.\bar 6\].
First multiply \[4.\bar 6\] with \[10\]:
\[4.\bar 6 \times 10 = 46.\bar 6\]..(1)
Subtract \[4.\bar 6\] from \[46.\bar 6\]:
\[46.\bar 6 - 4.\bar 6\] \[ = \] \[42\]..(2)
Substitute the value of \[46.\bar 6\] from equation (1) into equation (2):
\[\left( {4.\bar 6 \times 10} \right)\] \[ - \] \[4.\bar 6\] \[ = \] \[42\]
Take \[4.\bar 6\] common in left hand side of the equation:
\[ \Rightarrow \] \[4.\bar 6\left( {10 - 1} \right) = 42\]
\[ \Rightarrow \] \[4.\bar 6 \times 9 = 42\]
Dividing both sides of the equation by \[9\]:
\[ \Rightarrow \] \[4.\bar 6 = \dfrac{{42}}{9}\]
\[ \Rightarrow \] \[4.\bar 6 = \dfrac{{14 \times 3}}{{3 \times 3}}\]
Remove common factor of \[3\]:
\[ \Rightarrow \] \[4.\bar 6 = \dfrac{{14}}{3}\]

Hence \[4.6666....\] \[ = \] \[\dfrac{{14}}{3}\] \[ = 4\dfrac{2}{3}\].

Additional information:
A recurring decimal or repeating decimal, is the decimal representation of a number whose digits are repeated periodically and the infinitely repeated portion is not \[0\]. They are represented by putting a bar or a dot above the digit or digits that are periodically repeated.
Examples: \[0.\dot 7\] , \[0.\mathop {82}\limits^{\_\_} \]

Note:
Avoid converting any recurring decimal to a rounded off form and then converting it to a fraction. In that case the result obtained is less accurate than the result obtained by the described method. For example in this case if the decimal is rounded off it becomes \[4.67\] and the fraction becomes \[\dfrac{{467}}{{100}}\] , observe that though this fraction is close to the fraction of our answer that is \[\dfrac{6}{9}\], the latter is much more accurate.