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It is well known that every decimal of a fraction becomes periodic at one point or the other, e.g: 1/3 starts repeating at period 1, 27/91 at period 6 and 3923/6173 at period 3086.

Now, as there are an infinite number of non-zero values that we may place in the numerator or denominator of a fraction, the following question arises: does there exist a fraction whose repeating decimal has a period of some highest number or can the number of this period take infinitely many values?

In other words, does there exist some fraction whose repeating decimal has a period of 577291485 or even 5*10^444? And furthermore, is it true that there are an infinite number of fractions whose decimal number has the period 5?

It should be kept in mind however that the numerator/denominator of the fractions in question may contain numbers not repeating thrice/twice, e.g: 586192/247591, or 1/43817 and 1/39820, so it's not just 1/9 or 1/999.

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  • $\begingroup$ Have you thought about how to write a repeating decimal fraction with period (say) one million? $\endgroup$
    – hardmath
    Nov 10, 2017 at 17:22
  • $\begingroup$ "does there exist a fraction whose repeating decimal has a period of some highest number or can the number of this period take infinitely many values? " Neither. A period can be any finite length and there is no highest. An infinite period however is a nonsensical string of words that make no sense and is as logically inconsistent as "four-sided triangle". $\endgroup$
    – fleablood
    Nov 10, 2017 at 22:28
  • $\begingroup$ "is it true that there are an infinite number of fractions whose decimal number has the period 5?" There are each digit of a string of 5 characters can have any of $10$ values there are $10^5$ different possible strings. Last time I check $10^5$ was a finite number. But we can precede the fraction by anything so .... yes. But a reduced fraction $\frac 1 n$ there are only $10^5$ possible with period of $5$. $\endgroup$
    – fleablood
    Nov 10, 2017 at 22:35
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    $\begingroup$ @fleablood: You seem to have misunderstood the gist of the question. The OP is asking not whether the period is in itself infinite, he is asking whether the period can take infinitely many values, whether there may exist fractions whose repeating decimal has a period of any number, even 5867473710 or numbers ranging around 10^400, he illustrated this with various examples. There is nothing wrong with this and it has already been given a satisfactory answer. $\endgroup$
    – Fine
    Nov 10, 2017 at 22:55
  • $\begingroup$ He/she however imposed an important restriction which would deter us from giving this an arbitrary solution by considering that 1/9, 1/99, 1/999... and so the period of these fractions will increase infinitely along with the continual addition of 9 in the denominator. Dealing with this restriction adequately, I think, would require combinatorial notions. $\endgroup$
    – Fine
    Nov 10, 2017 at 23:05

2 Answers 2

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Consider $\cfrac 19, \cfrac 1{99}, \cfrac 1{999} \dots$

For the second part, what happens to the period of the fraction when you multiply the denominator by $10$?

In general a fraction which repeats will have the form $$f=\frac p{10^r}+\frac q{10^{r}}\left(\frac 1{10^n}+\frac 1{10^{2n}}+\frac 1{10^{3n}}+\dots\right)$$ Where $p$ is an integer with up to $r$ digits ($r$ and $p$ may be zero) which represents the non-recurring part of the fraction, and $q$ is the repeating part, with up to $n$ digits. Now we see that $$10^rf-p=qs$$ where $s$ is the sum of the series. Multiply by $10^n$ and obtain $$10^{r+n}f-10^np=10^nqs=q+qs=q+10^rf-p$$ and $$f=\frac p{10^r}+\frac q{10^r(10^n-1)}$$

If we take $p=r=0$ for the moment so that $f=\cfrac q{10^n-1}$ is the form of a fraction which recurs without a non-recurring part, we see that a fraction which has a denominator which is a factor of $10^n-1$ will recur after $n$ places. It may recur after fewer places than that, but not after more.

Every odd prime $k$ is a factor of $10^n-1$ for some minimum value of $n$, and $\frac 1k$ will then recur with period $n$.

There are more answers on this site which go into more detail than this, and there is a full treatment in Number Theory book by Hardy and Wright. If you explore it a bit yourself you may find some interesting things.

The reason denominators $9, 99, 999$ etc come into it (they correspond to $q=1$) is that they are the values of $10^r-1$.

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  • $\begingroup$ I have edited my question. Can you please take a look and modify your own answer if needed? $\endgroup$
    – Jasmus
    Nov 10, 2017 at 17:26
  • $\begingroup$ @Jasmus I have added some things. Try computing the period of the reciprocals of small primes and factoring the numbers $9, 99, 999, 9999$ etc to get a feel for what is going on. $\endgroup$
    – Mark Bennet
    Nov 10, 2017 at 17:48
  • $\begingroup$ Thank you very much for editing your answer, it answered my inquiries nicely. However, I'd greatly appreciate it if you could tell me in what chapter of Hardy & Wright's book these concepts and this topic is discussed. $\endgroup$
    – Jasmus
    Nov 10, 2017 at 18:23
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    $\begingroup$ @Jasmus Chapter IX The representation of Numbers by Decimals $\endgroup$
    – Mark Bennet
    Nov 10, 2017 at 18:27
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$\frac{1}{10^p - 1}$ will have period $p$ for any positive integer $p$.

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