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 $2$.
Answer
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
$$1/7 = 0.(142857)(142857)\ldots$$
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