In Part A of this lesson, we saw how to convert a terminating decimal number to a fraction. If a decimal is repeating (in other words, after a while, some pattern of digits repeats over and over), then we have to use a different strategy.

Let $N$ be the number, and let $n$ be the number of digits in the repeating block. For example,

$0.01\overline{83}$

has $2$ digits in the repeating block, whereas

$0.01\overline{83}$

has only $1$.

Setting $N$ equal to its value, multiply both sides of the equation by ${10}^{n}$. Then subtract $N$ from both sides. You will end up with an equation which you can solve for $N$ as a fractional value.

**Example:**

Convert the decimal

$0.01\overline{83}$

into a fraction.

Start with the equation:

$N=0.01\overline{83}$

There are two digits in the repeating block, so multiply both sides by ${10}^{2}=100$.

$100N=1.\overline{83}$

Now subtract $N$ from both sides. Notice that the repeating part cancels out.

$\begin{array}{l}100N=1.\overline{83}\\ \underset{\_}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}N=0.01\overline{83}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}99N=1.82\end{array}$

Divide both sides by $99$, multiply numerator and denominator by a power of $10$ to get rid of the decimal point, and simplify.

$N=\frac{1.82}{99}=\frac{182}{9900}=\frac{91}{4950}$