How to Convert a Repeating Decimal to a Fraction
To convert a repeating decimal into a fraction, I use the following formula:$$ \text{Fraction} = \frac{N - A}{D} $$ where:
- \(N\) is the number written without the decimal point, including all digits from both the repeating and non-repeating parts.
- \(A\) is the non-repeating portion written without the decimal point.
- \(D\) is the denominator, formed by writing as many \(9\)s as there are digits in the repeating part, followed by as many \(0\)s as there are digits in the non-repeating part.
A worked example
Consider the repeating decimal \(13.2\overline{31}\). Its fractional form is:
$$ \frac{13231 - 132}{990} = \frac{13099}{990} $$
Here, \(N = 13231\), \(A = 132\), and \(D = 990\). The repeating part has two digits, which gives two 9s, while the non-repeating part has one digit, which contributes a single 0.
Therefore,
$$ 13.2\overline{31} = \frac{13099}{990} $$
This procedure is completely general and applies to any repeating decimal.
An alternative method
This approach relies on setting up and solving a simple linear equation.
To convert a repeating decimal into a fraction:
- Introduce a variable to represent the number. For example, \(x = 0.777...\).
- Multiply by an appropriate power of 10 so that the repeating block shifts to the left of the decimal point. For instance, \(10x = 7.777...\).
- Subtract the original equation from the new one to eliminate the repeating part. For example, \(10x - x = 7\).
- Solve the resulting equation for \(x\), yielding \(x = \frac{7}{9}\).
Why does this work? This method works because every repeating decimal can be expressed as an infinite geometric series. By subtracting a suitably shifted copy of the number, the repeating portion cancels out, reducing the problem to a simple linear equation.
Example
Convert the repeating decimal \(0.777...\) into a fraction.
Let
$$ x = 0.777... $$
Multiply both sides by 10:
$$ 10 \cdot x = 10 \cdot 0.777... $$
$$ 10x = 7.777... $$
Subtract the first equation from the second:
$$ 10x - x = 7.777... - 0.777... $$
$$ 9x = 7 $$
Solve for \(x\):
$$ x = \frac{7}{9} $$
Thus, \(0.777...\) is equal to \(\frac{7}{9}\).
Example 2
Now consider the number \(0.16 \overline{23}\).
Let
$$ x = 0.16232323... $$
Multiply both sides by \(100\) to eliminate the non-repeating part:
$$ 100 \cdot x = 100 \cdot 0.16232323... $$
$$ 100x = 16.232323... $$
Multiply again by \(100\) to align the repeating block:
$$ 100 \cdot 100x = 100 \cdot 16.232323... $$
$$ 10000x = 1623.232323... $$
Subtract the previous equation from this one:
$$ 10000x - 100x = 1623.232323... - 16.232323... $$
$$ 9900x = 1607 $$
Solve for \(x\):
$$ x = \frac{1607}{9900} $$
Hence, \(0.16 \overline{23}\) is equal to \(\frac{1607}{9900}\).
The same reasoning applies in general.
