# Integers with Metric Induced by P-adic Valuation

## 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

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

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

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 $r > 0$

$\blacksquare$