# The period length of the decimal expression for a rational number $1/n$ has lower bound $1$.

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.

• Hem, what would be a period of length $0$ ? And there is an ambiguity in the question: period or period length ?
– user65203
Dec 31, 2019 at 11:36
• @YvesDaoust: The period length. Question edited. Dec 31, 2019 at 11:39
• Doesn't 1/n have period length 1 for n=9? And actually period length 0 for n=2? Dec 31, 2019 at 11:48
• 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. Dec 31, 2019 at 11:59
• @Conifold: So, the period length for the case where $n=2$ is not zero. Dec 31, 2019 at 14:48

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.)

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