Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 3

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Theorem

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

Let:

$a \in \Z, b \in Z_{> 0}$

Let:

$\forall n \in \N: \exists A_n, r_n \in \Z$:
$(\text a) \quad \dfrac a b = A_n + p^{n+1} \dfrac {r_n} b$
$(\text b) \quad \exists n_0 \in \N : \forall n \ge n_0 : -b \le r_n \le 0$


Then:

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


Proof

Let $\epsilon \in \R_{> 0}$.

Let $M = \max \set {\norm{r_0}_p, \norm{r_1}_p, \ldots, \norm{r_{n_0}}_p, \norm{-1}_p, \norm{-2}_p, \ldots, \norm{-b}_p}$

From Power Function is Unbounded Above:

$\exists N \in \N: p^{N+1} > \dfrac M {\epsilon \norm b_p}$


We have:

\(\ds \forall n \in \N: n \ge N: \, \) \(\ds \norm{\dfrac a b - A_n}_p\) \(=\) \(\ds \norm{p^{n+1} \dfrac {r_n} b} _p\) By hypothesis
\(\ds \) \(=\) \(\ds \norm p_p^{n+1} \dfrac {\norm{r_n}_p} {\norm b_p}\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(\le\) \(\ds \norm p_p^{n+1} \dfrac M {\norm b_p}\) Definition of Max Operation
\(\ds \) \(\le\) \(\ds \dfrac 1 {p^{n+1} } \dfrac M {\norm b_p}\) Definition of P-adic Norm
\(\ds \) \(\le\) \(\ds \dfrac 1 {p^{N+1} } \dfrac M {\norm b_p}\) Power Function on Base between Zero and One is Strictly Decreasing
\(\ds \) \(<\) \(\ds \dfrac {\epsilon \norm b_p } M \dfrac M {\norm b_p}\) Choice of $N$
\(\ds \) \(=\) \(\ds \epsilon\) Cancelling terms

By definition of convergence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$:

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

$\blacksquare$