Trying to prove:

Let $p > 5$ be a prime and let $k$ be any positive integer $< p$. Show that the decimal expansion of $\frac{k}{p}$ consists of (p-1) repeating decimal digits. (hint: use Fermat's Little Theorem and Geometric Series).

I was trying to understand the theorem using an example but I am not very sure about why (p-1) repeating decimal digits.

E.g suppose $p = 11$, k = $4$, $5$, $7$.

  • $\frac{4}{11} = 0.36363636363636\ldots$
  • $\frac{5}{11} = 0.454545454545454545\ldots$
  • $\frac{7}{11} = 0.63636363636363636\ldots$

So where does $p-1 = 10$ come from? It seems the repeating length is always $2$.

  • $\begingroup$ Maybe $p-1$ does not need to be the minimal period, so since $2\mid10$ that is sufficient. $\endgroup$ – YiFan Oct 17 '19 at 23:29
  • $\begingroup$ Consider the consequences of $10^{p-1}\equiv1\pmod{p}$ and how one computes $1/p$ in base ten. (Look at the remainders.) $\endgroup$ – robjohn Oct 17 '19 at 23:35
  • 1
    $\begingroup$ @robjohn but still why (p-1) in the example? It seems not right based on the example I gave. $\endgroup$ – Jerry_Ge Oct 17 '19 at 23:38
  • $\begingroup$ @Jerry_Ge: Did you read YiFan's comment? $0.\overline{6363636363}$ has a period of $10$, but also a period of $2$. $\endgroup$ – robjohn Oct 18 '19 at 1:51
  • $\begingroup$ This might help: math.stackexchange.com/q/3295196/597411 $\endgroup$ – C. Melton Oct 18 '19 at 3:44

If $10^{p-1} = 1 + mp$, where $1 < m < 10^{p-1}$, that says $$ \frac{1}{p} = \frac{m}{10^{p-1}-1} = \frac{m}{10^{p-1}(1-10^{-p+1})} = \sum_{j=1}^\infty \frac{m}{10^{j(p-1)}} $$ The decimal representation of this sum consists of "$0.$" followed by the digits of $m$ (padded at the front with zeros if necessary to length $p-1$) repeated: each term of the sum represents one block of $p-1$ digits.

Nobody said that $p-1$ has to be the smallest period. You can consider $4/11$ as $0.(36)(36)(36)\ldots$ with period $2$, but you could also write it as $0.(3636363636)(3636363636)\ldots$ with period $10=11-1$. An example where $p-1$ is the smallest period is $$1/7 = 0.(142857)(142857)\ldots$$

  • $\begingroup$ Thanks for the clarification! 1. Curious how you get from m/(10^(p-1)(1-10^{-p+1}) to the summation using geometric series? [Seems not intuitive for me] 2. why p <= 5 will not work? $\endgroup$ – Jerry_Ge Oct 18 '19 at 16:13
  • $\begingroup$ @Jerry_Ge: $p=3$ works. But $p=2$ and $p=5$ don't work, because the decimal expansions of $k/2$ and $k/5$ terminate. $\endgroup$ – TonyK Nov 23 '19 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.