Canonical P-adic Expansion of Rational is Eventually Periodic

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Theorem

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.

Let $x \in \Q_p$.


Then:

$x$ is a rational number if and only if the canonical expansion of $x$ is eventually periodic.


Proof

Necessary Condition

Let $x$ be a rational number.

Let $\ldots d_n \ldots d_2 d_1 d_0 . d_{-1} d_{-2} \ldots d_{-m}$ be the canonical expansion of $x$.

It is sufficient to show that the canonical expansion $\ldots d_n \ldots d_2 d_1 d_0$ is eventually periodic.


Let $y$ be the $p$-adic number with canonical expansion:

$\ldots d_n \ldots d_2 d_1 d_0$

We have:

$y = x - \ds \sum_{i \mathop = -m}^{-1} d_i p^i$

So:

$y$ is a rational number


By definition of $p$-adic integer:

$y$ is a $p$-adic integer


Let:

$y = \dfrac a b : a \in \Z, b \in Z_{> 0}$ are coprime

From Characterization of Rational P-adic Integer:

$p \nmid b$

From Prime not Divisor implies Coprime:

$b, p$ are coprime

Lemma 1

$\forall n \in \N: \exists r_n \in \Z : \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b} \in \set{0, 1, \ldots, p^{n + 1} - 1}$

$\Box$

Lemma 2

$\exists n_0 \in \N : \forall n \ge n_0 : -b \le r_n \le 0$

$\Box$


For all $n \in \N$, let:

$A_n = \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b}$

Then:

$\dfrac a b = A_n + \paren{p^{n + 1} \dfrac {r_n} b}$

From Lemma 1, for all $n \in \N$:

$A_n \in \N$
$0 \le A_n \le p^{n+1} - 1$

Lemma 3

$\ds \lim_{n \mathop \to \infty} A_n = \dfrac a b$

$\Box$


Lemma 4

$\forall n \in \N: r_n = d_{n + 1} b + p r_{n + 1}$

$\Box$


Lemma 5

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

$\Box$


It follows that $\ldots d_n \ldots d_2 d_1 d_0$ is eventually periodic.

$\Box$


Sufficient Condition

Let the canonical expansion of $x$ be eventually periodic.


Lemma 6

$\exists r \in \Q, n \in \Z, y \in \Z_p$:
$(1) \quad x = r + p^n y$
$(2) \quad$ the canonical expansion of $y$ is periodic.

$\Box$


To show that $x$ is a rational number it is sufficient to show that $y$ is a rational number.


Let:

$\dots d_{k - 1} \ldots d_1 d_0 d_{k - 1} \ldots d_1 d_0 d_{k - 1} \ldots d_1 d_0$

be the periodic canonical expansion of $y$.

By definition of a canonical expansion:

$y = d_0 + d_1 p + \cdots + d_{k - 1} p^{k - 1} + d_0 p^k + d_1 p^{k + 1} + \cdots + d_{k - 1} p^{2 k - 1} + d_0 p^{2 k} + \cdots$


Let $a = d_0 + d_1 p + \cdots + d_{k - 1} p^{k - 1}$.

Then:

$y = a \paren {1 + p^k + p^{2 k} + \cdots}$


Lemma 7

$1 + p^k + p^{2 k} + p^{3 k} + \cdots = \dfrac 1 {1 - p^k}$

$\Box$


Then:

$y = \dfrac a {1 - p^k}$

Hence:

$y$ is a rational number


It follows that $x$ is a rational number.

$\blacksquare$


Sources

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