# Repeating decimals linked to reciprocals of primes

Now this question is base dependent, I will assume base 10 but feel free to generalize.

I was noticing that for small primes that are not factors of the base ($2$ and $5$ terminate) the reciprocal of the prime (read $\frac 1p$) was a repeating decimal.

Will the reciprocal of any prime other than the two mentioned form a repeating decimal?

Is there anything you can say about the order of the repeating decimal if you know the prime that generated it?

• Indeed, there is an interesting connection to Fermat's Little Theorem within this question. Oct 28, 2015 at 17:34
• This is a very interesting topic, and if you want to learn more about it, you might start by looking here or here. Oct 28, 2015 at 17:39
• If $10^k/p$ is an integer, for some $k$, then the prime $p$ divides $10$ - is that what you are asking? (At least for the first part of the quesiton) Oct 28, 2015 at 17:42

In fact, if $p$ is a prime greater than $5$ (this excludes both primes $2$ and $5$ and also $3$), the decimal expansion of $1/p$ is periodic with period length equals $e=\operatorname{ord}_p(10)$ the order of $10$ modulo $p$, i.e., $e$ is the smallest positive integer such that $$10^e\equiv 1 \pmod p$$ This result can be generalizated in any number base $B>1$ and extended to any fraction $x/n$ where the numerator $x\in \mathbb{U}_n$, the set of positive integers less than and relatively primes to $n$ and $\gcd(B,n)=1$. With this hypothesis, the decimal expansion of $x/n$ in the base $B$ is periodic and its period length is the order of $B$ modulo $n$ (which exists because of the relatively prime condition between $B$ and $n$). Note that the period length of $x/n$ don't depend of $x$, so each one of the fraction $x/n$ always have the same period length.

• That is exactly what I was looking for, linking the prime $p$ to the period length. Oct 28, 2015 at 18:27

Given a prime $p \neq 2,5$, it is always the case that $1/p$ has a repeating decimal expansion.

Lemma: Every prime $p \neq 2, 5$ divides a repunit.

Proof of Lemma:

Fix a prime $p \neq 2,5$. Let $\textbf{A}$ be the set of repunits, so

$$\textbf{A} = \left\{\displaystyle\sum\limits_{k=1}^{n} 10^{k-1} \, \mid \, n \in \mathbb{N} \right\} = \left\{\frac{10^n -1}{9} \, \mid \, n \in \mathbb{N} \right\}$$

Consider the repunits, modulo $p$. Since $\mathbb{N}$ is not a finite set, neither is $\textbf{A}$. There are a finite number of remainders modulo $p$ (specifically, $p$ possible remainders).

There are (infinitely) more repunits than remainders modulo $p$. Thus, there must exist two distinct repunits with the same residue modulo $p$. So $$\exists \, a, b \in \textbf{A} \,\, \text{s.t.} \,\,\,\,\,\, a \equiv b \pmod{p}, \,\, a \neq b$$

Without loss of generality, assume $a > b$.

Since $a, b \in \textbf{A}$, $\exists \, x, y \in \mathbb{N}$ with $x > y$ such that

$$a = \frac{10^x - 1}{9}$$

$$b = \frac{10^y - 1}{9}$$

We can substitute in to $a \equiv b \pmod{p}$ to get:

$$\frac{10^x - 1}{9} \equiv \frac{10^y - 1}{9} \pmod{p}$$

$$\frac{\left(10^x - 1\right)-\left( 10^y - 1\right)}{9}\equiv 0 \pmod{p}$$

$$\frac{10^x-10^y}{9} \equiv 0 \pmod{p}$$

$$\frac{\left(10^y\right)\left(10^{x-y}-1 \right)}{9}\equiv 0 \pmod{p}$$

We know that $p \nmid 10^y$, because $p$ is not $2$ or $5$. Since $\mathbb{Z}/p\,\mathbb{Z}$, the ring of integers modulo $p$, has no zero divisors (because $p$ is prime),

$$\frac{10^{x-y}-1}{9}\equiv 0 \pmod{p}$$

This is a repunit.

Since our choice of $p \neq 2, 5$ was arbitrary, we have proved that every prime that is not $2$ or $5$ divides a repunit. It follows that every prime that is not $2$ or $5$ divides nine times a repunit (a positive integer whose digits are all nines).

This means that for every $p$ that is not $2$ or $5$, there exist positive integers $m$ and $n$ such that such that

$$\frac{1}{p} = \frac{m}{10^{n}-1}$$

Since $m$ is not equal to $10^n -1$, this implies that $1/p$ has a repeating decimal representation, and further that the period of repetition divides $n$.

• Note: I borrowed from myself here. If we are allowed to assume without proof that every rational number has a decimal representation that either terminates or eventually repeats, then this proof can be done in one line (as someone in the comments pointed out): assume that $1/p$ terminates. Then the following is an integer for some $k$: $$\frac{10^k}{p}$$ Therefore $p=2,5$. Oct 28, 2015 at 17:59