Characterisation of Terminal P-adic Expansion

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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:

the $p$-adic expansion of $x$ terminates

if and only if

$\exists a \in \N : \exists k \in \Z : x = \dfrac a {p^k}$


Proof

Necessary Condition

Let the $p$-adic expansion of $x$ be:

$x = \ds \sum_{n \mathop = m}^\infty d_n p^n$

where:

$m \in \Z_{\le 0}$
$\forall n \in \Z_{\ge m}: d_n$ is a $p$-adic digit
$m < 0 \implies d_m \ne 0$


By the definition of terminates:

$\exists n_0 \in \N : n_0 \ge m : \forall n \ge n_0 : d_n = 0$


We have:

\(\ds x\) \(=\) \(\ds \sum_{n \mathop = m}^{n_0} d_n p^n\) All trailing terms are $0$
\(\ds \) \(=\) \(\ds p^m \paren{\sum_{n \mathop = m}^{n_0} d_n p^{n - m} }\) Extract common $p^m$ factor from each term
\(\ds \) \(=\) \(\ds p^m \paren{\sum_{n \mathop = 0}^{n_0 - m} d_{n + m} p^n }\) Re-indexing the terms


Let:

$k = -m$

Let:

$a = \ds \sum_{n \mathop = 0}^{n_0 - m} d_{n + m} p^n $

Then:

$x = \dfrac a {p^k}$


The result follows.

$\Box$

Sufficient Condition

From Basis Representation Theorem, $a$ can be expressed uniquely in the form:

$\ds a = \sum_{j \mathop = 0}^n d_j p^j$

where:

$n$ is such that $p^n \le a < p^{n + 1}$
all the $d_j$ are such that $0 \le d_j < p$.


We have:

\(\ds x\) \(=\) \(\ds \dfrac a {p^k}\) Hypothesis
\(\ds \) \(=\) \(\ds \dfrac {\ds \sum_{j \mathop = 0}^n d_j p^j} {p^k}\) Replacing $a$ with base $p$ expression
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^n d_j p^{j - k}\) Dividing each term by $p^k$
\(\ds \) \(=\) \(\ds \sum_{i \mathop = -k}^{n-k} d_{i+ k} p^i\) Re-indexing with $i = j - k$


Let:

$m = \begin{cases}

-k & : -k \le 0\\ 0 & : 0 < -k \end{cases}$


For each $i : m \le i \le n-k$, let:

$e_i = \begin{cases}

d_{i + k} & : -k \le i \le n-k\\ 0 & : m \le i < -k \end{cases}$


For each $i > n-k$, let:

$e_i = 0$


Then:

$x = \ds \sum_{i \mathop = m}^\infty e_i p^i$

where:

$\forall i \ge m: 0 \le e_i < p$
$\forall i > n-k: e_i = 0$


Hence $\ds \sum_{i \mathop = m}^\infty e_i p^i$ is a terminal $p$-adic expansion by definition.

From P-adic Expansion Representative of P-adic Number is Unique, the $p$-adic expansion of $x$ is:

$x = \ds \sum_{i \mathop = m}^\infty e_i p^i$

$\blacksquare$

Sources

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