# Lower bound for the length period of the decimal expression

We know (Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$) that for an integer $$n\geq 2$$, the period length of the decimal expression for the rational number $$\frac{1}{n}$$ is at most $$n-1$$.

I am asking if there is a lower bound if $$n>m$$ where $$m$$ is given integer.

For $$n$$ relatively prime to $$10$$, $$\dfrac1n$$ repeats after $$m$$ digits where $$n\mid10^m-1$$.

Then $$n<10^m$$, so the period length of the decimal expression for $$\dfrac1n$$ is at least $$\lceil\log_{10}n\rceil$$.

• I wondering if there a simple proof for this or at least a link to a proof. Dec 26, 2019 at 14:40
• think of calculating $\dfrac1n$ by long division; when the remainder is $1$ (i.e., $n\mid10^m-1$), then the process repeats itself Dec 26, 2019 at 14:49
• Ok. Thank you very much. Dec 26, 2019 at 14:51

Consider that $$\frac 1{100}=0.01$$, which has period 0 or 1 depending on your definition of the period.

In other words, the lower bound of the period length is always 0 or 1.

Or if you want a non-zero repetition, we can always find a number with period 1. We can pick for instance something like $$0.001111... = \frac 19\cdot \frac 1{100}=\frac 1{900}$$.