Characterisation of Terminal P-adic Expansion/Necessary Condition

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Theorem

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

Let $x \in \Q_p$.

Let the $p$-adic expansion of $x$ terminate.


Then:

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

Proof

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.

$\blacksquare$