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Theorem: For an integer $n$ greater than or equal to $2$, the period length of the decimal expression for the rational number $1/n$ is at most $n-1$ and has lower bound $1$.

For the first part I found this page (Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$). But for the second part I find no proof.

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    $\begingroup$ Hem, what would be a period of length $0$ ? And there is an ambiguity in the question: period or period length ? $\endgroup$
    – user65203
    Dec 31, 2019 at 11:36
  • $\begingroup$ @YvesDaoust: The period length. Question edited. $\endgroup$
    – Safwane
    Dec 31, 2019 at 11:39
  • $\begingroup$ Doesn't 1/n have period length 1 for n=9? And actually period length 0 for n=2? $\endgroup$
    – Klaas van Aarsen
    Dec 31, 2019 at 11:48
  • $\begingroup$ Period length 0 would mean a non-periodic expansion, but expansions of rational numbers are either eventually periodic or terminate, i.e. have periodically repeating 0-s at the end, see Rational numbers and decimal expansions. $\endgroup$
    – Conifold
    Dec 31, 2019 at 11:59
  • $\begingroup$ @Conifold: So, the period length for the case where $n=2$ is not zero. $\endgroup$
    – Safwane
    Dec 31, 2019 at 14:48

1 Answer 1

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If by "lower bound" you mean the minimum, then yes, it is enough to show that for one $n$ you have the period $1$: $\frac{1}{3}=0.33333\ldots$ would suffice.

If by "lower bound" you mean it is the minimum of all accumulation points, i.e. if you need to prove that the period $1$ is reached for infinitely many $n$'s, then it can be done as well. For example:

$$\frac{1}{3}=0.333333\ldots$$ $$\frac{1}{30}=0.033333\ldots$$ $$\frac{1}{300}=0.003333\ldots$$ $$\frac{1}{3000}=0.000333\ldots$$

etc.

(That all excluding "trivial" cases which has a "tail" of zeros, e.g. $\frac{1}{2}=0.500000\ldots$. Not sure how those are regarded in this problem - if you include those, then you are getting an even easier example.)

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  • $\begingroup$ The question is for all $n$. I am wondering about the cases similar to $n=2$. $\endgroup$
    – Safwane
    Dec 31, 2019 at 14:51
  • $\begingroup$ Ah well ... $1/10=0.100000\ldots$, $1/100=0.010000\ldots$, $1/1000=0.001000\ldots$ etc. $\endgroup$
    – user700480
    Dec 31, 2019 at 16:21

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