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If you divide 1 by 7, you'll of course have a repeating decimal sequence of 142857. This of course repeats forever. Are there any scenarios where a quotient seems to repeat forever, but then changes at some point? Is this even possible? I'm not a mathematician (obviously), but rather a programmer. The question came up when I was thinking of a programmatic way to determine whether a division result is repeating.

Example (not real of course):


P.S. I can't even find an appropriate tag for this question, so feel free to edit my tags.

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closed as off-topic by Ricardo Andrade, darij grinberg, Karl Schwede, Eric Wofsey, Andrey Rekalo Sep 24 '13 at 7:34

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I don't follow. It is impossible to determine whether a number is rational by looking at any finite sequence of its digits. – Qiaochu Yuan Oct 2 '10 at 8:03
@Qiaochu Yuan, I agree with you, as you'll see in my answer below. If, however, Orokusaki is generating his sequence as the result of dividing $a \div b$, where $a,b \in \mathbb{Z}$, then obviously $a \div b$ is rational and will ultimately repeat. ... though it may take $b-1$ digits before it does so... – sleepless in beantown Oct 2 '10 at 10:08
up vote -1 down vote accepted

If you want to check if the result of the division is repeating you can make use of this (from

the decimal representation of any rational number will either terminate, or eventually become periodic. (As a bonus challenge, can you figure out how to tell the difference between rational numbers whose decimal expansion terminates, and those whose expansion repeats?) Note that we also know something else: since there are only (q-1) possible non-zero remainders when dividing by q, the repeating portion of the decimal expansion of a rational number with a denominator of q can be at most (q-1) digits long. It could be shorter, but it can’t be any longer. You can also look at this in reverse; for example, if you see a periodic decimal whose repeating portion is ten digits long, you know that the rational number it represents must have a denominator of at least 11.

So in your example I think you would only need to check up to the 7th digit and for sleepless's example, you would only need to check up the 13th digit.

I'm also a programmer and not a mathematician so I probably have this all wrong :)

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@mattjames - awesome, thanks for getting that for me. – orokusaki Oct 2 '10 at 4:07
@mattjames, note that the repeating fraction $1/7$ (which has $q$=7) repeats the six digits 142857. Six is $q-1$, even though the numeral $7$ is one digit long. If you didn't know that $a/(10^{n}-1)$ was going to repeat after $n$ digits, you'd have to calculate it out to $10^n-1$ digits to check. For example, take $1 \div 103$. It requires checking 102 digits of the decimal expansion to check for the repeat length. Or for $a/(10^{n}-3)$, where $a,n \in \mathbb{Z}$, all $10^{n}-3$ digits would have to assessed to determine repeat length. That's a lot of digits, increasing exponentially in n – sleepless in beantown Oct 2 '10 at 9:36
@mattjames, technically, you'd have to check the $121212121213 \div 999999999999$ example out to $999999999998$ digits to make sure of the repeat length if it weren't for the facts described below. I can't predict how many digits are in the repeat of $121212121213 \div 999999999997$ beyond saying that it will repeat in less than $999999999997$ digits. That's a very large number of digits to expand out to and check for repeats. – sleepless in beantown Oct 2 '10 at 10:07

There is no such thing as a non-repeating division sequence if the quotient is formed by the division of two integers or by the divsion of two rational numbers.

Orokusaki, any rational sequence of any length (length=$n$) at all can be constructed as a fraction of the form

$$ \frac{\sum_{i=1}^{i=n} d_i 10^{i-1}}{10^n -1} $$

where $d_i$ is the $i$-th digit counting from the right (the lowermost powers of 10).

For example, the fractional sequence repeating 12 as 0.121212... can be constructed as $12/99$.

The repeating decimal 0.121212121213 which continually repeats 12 five times followed by 13 once is the fraction


So the fact that you have $m$ repeats of a $n$-digit length sequence is not a guarantee that the repeats will continue ad infinitum.

Sequences which repeat a sequence a fixed number of times followed by infinite repeats of another sequence can also be constructed in a similar fashion.

If, however, you are generating this digit sequence as the quotient of two integer values (call them $a,b, a\in \mathbb{Z}, b \in \mathbb{Z}$), then you are guaranteed that the sequence will repeat. This is because all rational fractions of integers can be represented in the fraction format $c/({10^n-1})$ with $c \in \mathbb{Z}$ as described above, though it is not always easy to determine what length of repeat may occur.

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Coo, I can read that ∑ thing just like any other word :| – orokusaki Oct 2 '10 at 3:51

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