# Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 5

This article needs proofreading.Please check it for mathematical errors.If you believe there are none, please remove `{{Proofread}}` from the code.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Proofread}}` from the code. |

## Theorem

Let $p$ be a prime.

Let $b \in Z_{> 0}$ such that $b, p$ are coprime.

Let $\sequence{d_n}$ be a sequence of $p$-adic digits.

Let $\sequence{r_n}$ be a sequence of integers:

- $(\text a) \quad \forall n \in \N: r_n = d_{n+1} b + p r_{n+1}$
- $(\text b) \quad \exists n_0 \in \N : \forall n \ge n_0 : -b \le r_n \le 0$

Then:

- $\exists \mathop m, l \in \N : \forall n \ge m: r_n = r_{n + l}$ and $d_n = d_{n + l}$

## Proof

By hypothesis the set $\set{r_n : n \in \N}$ of values of the sequence $\sequence{r_n}$ is a subset of:

- $\set{r_0, r_1, \ldots, r_{n_0}} \cup \set{-b, -b + 1, -b + 2, \ldots, 2, 1, 0}$

It follows that the set $\set{r_n : n \in \N}$ takes only finitely many values.

Hence:

- $\exists m_0, l \in \N : l > 0 : r_{m_0} = r_{ {m_0} + l}$

### Lemma 10

Let:

- $n, k \in \N : k > 0 : r_n = r_{n + k}$

Then:

- $d_{n+1} = d_{n + k + 1}$
- $r_{n+1} = r_{n + k + 1}$

$\Box$

Let $m = m_0 + 1$

The proof now proceeds by induction.

For all $n \ge m$, let $\map P {n}$ be the proposition:

- $r_n = r_{n+l}$ and $d_n = d_{n + l}$

### Basis for the Induction

$\map P {m}$ is the proposition:

- $r_m = r_{m+l}$ and $d_m = d_{m + l}$

It has already been shown that:

- $r_{m_0} = r_{ {m_0}+l}$

From lemma 10:

- $d_m = d_{m+l}$
- $r_m = r_{m+l}$

This proves proposition $\map P {m}$.

$\Box$

### Induction Hypothesis

Let $n \ge m$.

The induction hypothesis is the proposition $\map P {n}$:

- $r_n = r_{n+l}$ and $d_n = d_{n + l}$

It has to be shown that the proposition $\map P {n+1}$ is true:

- $r_{n + 1} = r_{n + l + 1}$ and $d_{n + 1} = d_{n + l + 1}$

### Induction Step

From the induction hypothesis:

- $r_n = r_{n+l}$

From lemma 10:

- $d_{n + 1} = d_{n + l + 1}$
- $r_{n + 1} = r_{n + l + 1}$

$\Box$

Hence:

- $\forall n \ge m: r_n = r_{n + l}$ and $d_n = d_{n + l}$

The result follows.

$\blacksquare$