## Monday Math 83

Previously, I talked about expressing fractions as decimals. I showed how to determine if a fraction gives a terminating decimal, a pure repeating decimal, or a mixed repeating decimal. I also showed how to determine the number of digits in a terminating decimal from the denominator of the fraction.
Now, let us consider the number of digits that repeat in a pure repeating decimal.

Let us first consider a pure repeating decimal with a one-digit repeat, for example 0.55555…
Let our repeating decimal be x. If we multiply it by ten, we still have the same repeating sequence after the decimal point, but we also now have one occurence of it to the left of the decimal point:
If x=0.5555…, then 10x=5.55555…
Now, then we can subtract x from our 10x, thus removing everything to the right of the decimal point:
10xx=(5.555…)-(0.555…)=5
But this gives us 9x, and thus x is the repeated digit over 9:
0.55555…=5/9. Similarly, 0.6666…=6/9=2/3, and so on.

Now, for a repeat of two digits, one can multiply by 100, to obtain one repeat to the left of the decimal point, and subtract the original number to get an integer. In general, let us consider a pure repeating decimal which repeats n digits. Let a be the integer formed by the repeated digits; e.g. for 0.148148148…, n=3 and a=148. Then our number is
.
And, the series at the end is a geometric series, so we see
,
so that
.
For example, 0.148148148…=148/(103-1)=148/999=4/27.
Now, we have a way of quickly converting a pure repeating decimal into a fraction. However, it also tells us something about the reverse; namely that a fraction that converts to a pure repeating decimal with an n digit repeat must have a denominator which divides 10n-1. More specifically, the length of the repeat for a pure repeating decimal is the smallest n such that the denominator divides 10n-1. For example, 1/37: 37 does not divide 9 or 99, but does divide 999=27*37, so 1/37 has a three-digit repeat as a decimal (1/37=27/999=0.027027027…).