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.
$\blacksquare$