Integers with Metric Induced by P-adic Valuation

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Theorem

Let $p \in \N$ be a prime.

Let $d: \Z^2 \to \R_{\ge 0}$ be the mapping defined as:

$\forall x, y \in \Z: \map d {x, y} = \begin {cases} 0 & : x = y \\ \dfrac 1 r & : x - y = p^{r - 1} k: r \in \N_{>0}, k \in \Z, p \nmid k \end {cases}$

Then $d$ is a metric on $\Z$.


Proof

From Characterization of P-adic Valuation on Integers:

$d$ is well-defined.


We prove the metric space axioms.


Metric Space Axiom $\text M 1$

This follows immediately from the definition of $d$.

$\Box$


Metric Space Axiom $\text M 2$

Let $x, y, z \in \Z$.

From Characterization of P-adic Valuation on Integers:

$\exists r, s, t \in \N_{>0}, k, l, m \in \Z, p \nmid k, l , m:$
$x - y = p^{r - 1} k$
$y - z = p^{s - 1} l$
$x - z = p^{t - 1} m$


Let $t' = \min \set{r, s}$.

Then:

\(\ds x - z\) \(=\) \(\ds \paren{x - y} + \paren{y - z}\) Add $-y + y$
\(\ds \) \(=\) \(\ds p^{r - 1} k + p^{s - 1} l\)
\(\ds \) \(=\) \(\ds p^{t' - 1} \paren{p^{r - t'} k + p^{s - t'} l}\)

Hence:

$p^{t' - 1} \divides p^{t - 1} m$

Then:

$t' \le t$


Hence:

\(\ds \map d {x, z}\) \(=\) \(\ds \dfrac 1 t\)
\(\ds \) \(\le\) \(\ds \dfrac 1 {t'}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\min \set{r, s} }\)
\(\ds \) \(=\) \(\ds \max \set{ \dfrac 1 r, \dfrac 1 s}\)
\(\ds \) \(\le\) \(\ds \dfrac 1 r + \dfrac 1 s\)
\(\ds \) \(=\) \(\ds \map d {x, y} + \map d {y, z}\)

$\Box$


Metric Space Axiom $\text M 3$

Let $x, y \in \Z$.


Case 1: $x = y$

Let $x = y$.

By definition of $d$:

$\map d {x, y} = 0 = \map d {y, x}$

$\Box$


Case: $x \ne y$

Let $x \neq y$.

Then

\(\ds x - y\) \(=\) \(\ds p^{r - 1} k:\) \(\ds : r \in \N_{>0}, k \in \Z, p \nmid k\) Characterization of P-adic Valuation on Integers
\(\ds \) \(=\) \(\ds p^{r - 1} \paren{-k'}\) \(\ds : r \in \N_{>0}, k' \in \Z, p \nmid k'\) Replacing $k$ with $k' = -k$
\(\ds \leadstoandfrom \ \ \) \(\ds y - x\) \(=\) \(\ds p^{r - 1} k'\) \(\ds : r \in \N_{>0}, k' \in \Z, p \nmid k'\)

Hence:

$\map d {x, y} = \dfrac 1 r = \map d {y, x}$

$\Box$


Metric Space Axiom $\text M 4$

Let $x, y \in \Z$ such that:

$x \ne y$

Let $x - y = p^{r - 1} k: r \in \N_{>0}, k \in \Z, p \nmid k$.

We have:

\(\ds \map d {x, y}\) \(=\) \(\ds \dfrac 1 r\) Definition of $d$
\(\ds \) \(>\) \(\ds 0\) as $r > 0$

$\blacksquare$


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