Canonical P-adic Expansion of Rational is Eventually Periodic/Sufficient 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 canonical expansion of $x$ be eventually periodic.
Then:
- $x$ be a rational number
Proof
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.
$\blacksquare$