Period of Reciprocal of Prime

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Theorem

Consider the decimal expansion of the reciprocal $\dfrac 1 p$ of a prime $p$.

If $p \nmid a$, the decimal expansion of $\dfrac 1 p$ is periodic in base $a$ and its period of recurrence is the order of $a$ modulo $p$.

If $p \divides a$, the decimal expansion of $\dfrac 1 p$ in base $a$ terminates.


Proof

Case $1$: $p \divides a$

Let $q = \dfrac a p$.

Then $\dfrac 1 p = \dfrac q a$.

So the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.q$ and terminates.

$\Box$


Case $2$: $p \nmid a$

Let $d$ be the order of $a$ modulo $p$.

By definition, $d$ is the smallest integer such that:

$a^d \equiv 1 \pmod p$

By definition of modulo arithmetic:

$\exists x \in \Z: a^d - 1 = xp$


We can rearrange the terms to achieve the following expression:

\(\displaystyle \frac 1 p\) \(=\) \(\displaystyle \frac x {a^d - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac x {a^d} \frac 1 {1 - \frac 1 {a^d} }\) factorizing
\(\displaystyle \) \(=\) \(\displaystyle \frac x {a^d} \paren {1 + \paren {\frac 1 {a^d} } + \paren {\frac 1 {a^d} }^2 + \paren {\frac 1 {a^d} }^3 + \cdots}\) From Power Series Expansion of Reciprocal of 1 + x, noting that $-1 < -\dfrac 1 {a^d} < 1$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac 1 {a^d} } x + \paren {\frac 1 {a^d} }^2 x + \paren {\frac 1 {a^d} }^3 x + \cdots\) rerranging

so the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.xxx...$ and is periodic of length $d$, which is the order of $a$ modulo $p$.

$\blacksquare$