Converting a Bicimal to a Fraction (Series Method)

I’ve shown you two ways to convert a bicimal to a fraction: the subtraction method and the direct method. In this article, I will show you a third method — a common method I call the series method — that uses the formula for infinite geometric series to create the fraction.

Viewing Repeating Decimals as Infinite Series

First, let’s take a look at repeating decimals. The repeating portion of the decimal can be modeled as an infinite geometric series. For example, consider the (pure) repeating decimal 0.39:

0.39 = 0.393939…

= 0.39 + 0.0039 + 0.000039 + …

= 39/100 + 39/10000 + 39/1000000 + …

= 39(1/100 + 1/10000 + 1/1000000 + … )

= 39(1/102 + 1/104 + 1/106 + …)

= 39(1/(102)1 + 1/(102)2 + 1/(102)3 + …)

= 39((1/102)1 + (1/102)2 + (1/102)3 + …)

\mbox{\footnotesize{\displaystyle = 39\sum_{i=1}^\infty \left(\frac{1}{10^{2}}\right)^{i}}}

The summation notation (∑) is just a concise way to represent an infinite sum. We can compute that sum — and use it in place of the summation — by using a variation on a well-known formula. This will allow us to express the decimal as a fraction.

Infinite Series Formula

The formula that represents an infinite geometric series (with a first term of ‘1’) is usually stated as such:

\mbox{\footnotesize{\displaystyle\sum_{i=0}^\infty r^{i} = \frac{1}{1-r}}} ,

where |r| < 1.

I think a repeating decimal is more naturally modeled as a series starting at i = 1, as my derivation in the above example shows. Let’s adjust the starting point and derive a new formula to apply in this case. This is done simply, by subtracting r0 = 1 from the right hand side:

\mbox{\footnotesize{\displaystyle \frac{1}{1-r} \; $-$ \; 1}}}

\mbox{\footnotesize{\displaystyle = \frac{1}{1-r} \; $-$ \;}}} \mbox{\footnotesize{\displaystyle\frac{1-r}{1-r}}}}

\mbox{\footnotesize{\displaystyle = \frac{1-(1-r)}{1-r}}}}

\mbox{\footnotesize{\displaystyle = \frac{r}{1-r}}}}

So the formula for starting at i = 1 is

\mbox{\footnotesize{\displaystyle\sum_{i=1}^\infty r^{i} = \frac{r}{1-r}}}}

Completing the Example

Let’s go back to our example; this is where we left it:

0.39 =

\mbox{\footnotesize{\displaystyle 39\sum_{i=1}^\infty \left(\frac{1}{10^{2}}\right)^{i}}}

Let’s remove the summation by applying the formula with r = 1/102:

\mbox{\footnotesize{\displaystyle \frac{r}{1-r}}}}

\mbox{\footnotesize{\displaystyle = \frac{\frac{1}{10^{2}}}{1-\frac{1}{10^{2}}}}}}

\mbox{\footnotesize{\displaystyle = \frac{\frac{1}{10^{2}}}{\frac{10^{2}-1}{10^{2}}}}}}

\mbox{\footnotesize{\displaystyle = \frac{1}{10^{2}-1}}}}

\mbox{\footnotesize{\displaystyle = \frac{1}{99}}}}

Now multiply this by 39 to get our answer: 0.39 = 39/99.

Generalizing the Formula

The above analysis applies to any pure repeating decimal. The integer made up of the repeating digits is multiplied by the infinite series that uses r = 1/10n, where n is the length of the repeating cycle. By the formula, the value of the series is 1/(10n – 1), a unit fraction with a denominator of n 9s.

Mixed Repeating Decimals

The infinite series method only works for pure repeating decimals. If you want to use it on a mixed repeating decimal, you’ll have to shift the non-repeating part to the left of the decimal point, and then apply the method to the remaining (pure repeating) fractional part. You then have to add in the whole part and divide by the power of ten used to shift the decimal.

For example, take 0.135954. Shift it left three places (multiply by 103) to get 135.954. Apply the series method to 0.954, getting 954/999. Add 135 = 134865/999 to it, getting 135819/999. Finally, divide it by 103 to undo the shifting, getting 135819/999000 = 15091/111000.

The Series Method for Bicimals

The infinite series formula works for bicimals as well; you just work in powers of two instead of powers of ten. For example, let’s convert 0.101 to a fraction:

0.101 = 0.101101101…

= 0.101 + 0.000101 + 0.000000101+ …

= 101/1000 + 101/1000000 + 101/1000000000 + …

= 5/8 + 5/64 + 5/512 + … (converting to decimal numerals makes it easier)

= 5(1/8 + 1/64 + 1/512 + … )

= 5(1/23 + 1/26 + 1/29 + …)

= 5(1/(23)1 + 1/(23)2 + 1/(23)3 + …)

= 5((1/23)1 + (1/23)2 + (1/23)3 + …)

\mbox{\footnotesize{\displaystyle = 5\sum_{i=1}^\infty \left(\frac{1}{2^{3}}\right)^{i}}}

Now remove the summation by applying the formula with r = 1/23:

\mbox{\footnotesize{\displaystyle \frac{r}{1-r}}}}

\mbox{\footnotesize{\displaystyle = \frac{\frac{1}{2^{3}}}{1-\frac{1}{2^{3}}}}}}

\mbox{\footnotesize{\displaystyle = \frac{\frac{1}{2^{3}}}{\frac{2^{3}-1}{2^{3}}}}}}

\mbox{\footnotesize{\displaystyle = \frac{1}{2^{3}-1}}}}

\mbox{\footnotesize{\displaystyle = \frac{1}{7}}}}

Multiply this by 5 to get our answer: 0.101 as a fraction is 5/7, or in binary, 101/111.

This method also generalizes to any pure repeating bicimal with cycle n. By the formula, the value of the series is 1/(2n – 1), a unit fraction with a denominator of n 1s.

(Mixed repeating bicimals can be handled just as mixed repeating decimals, except the shifting is in powers of two and not powers of ten.)

Comparing the Three Methods

Of the three methods, I think the subtraction method is the best. It handles all numbers uniformly, whether pure repeating or mixed repeating or whether there is a whole number part or not. The direct method is the quickest, but you have to handle three cases (pure repeating, mixed repeating, and numbers with whole parts). The series method is a little “scary” in that it uses infinite sums, and also requires handling of separate cases.

The direct method is nice because you know the denominator patterns directly: decimals have fractions with denominators of all 9s or 9s followed by 0s, and bicimals have fractions with denominators of all 1s or 1s followed by 0s. In a sense, the subtraction and series methods serve as proofs of the direct method; they explain the patterns the direct method produces.

Dingbat
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