# P-adic Norm is Non-Archimedean Norm

## Theorem

The $p$-adic norm forms a non-Archimedean norm on the rational numbers $\Q$.

## Proof

First we note that the $p$-adic norm is a norm.

Let $\nu_p$ denote the $p$-adic valuation on the rational numbers.

Recall the definition of the $p$-adic norm:

$\forall q \in \Q: \norm q_p := \begin{cases} 0 & : q = 0 \\ p^{-\map {\nu_p} q} & : q \ne 0 \end{cases}$

We must show the following holds for all $x, y \in \Q$:

$\norm {x + y}_p \le \max \set {\norm x_p, \norm y_p}$

If $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial, as follows:

Let $x = 0$.

Then:

 $\displaystyle x$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \norm x_p$ $=$ $\displaystyle 0$ Definition of $p$-adic Norm $\displaystyle \leadsto \ \$ $\displaystyle \max \set {\norm x_p, \norm y_p}$ $=$ $\displaystyle \norm y_p$ as $\norm y_p \ge 0 = \norm x_p$ from Norm Axioms: Axiom $(\text N 1)$ $\displaystyle$ $=$ $\displaystyle \norm {x + y}_p$

and so $\norm {x + y}_p \le \max \left( \norm x_p, \norm y_p \right)$

The same argument holds for $y = 0$.

Let $x + y = 0$.

 $\displaystyle x + y$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \norm {x + y}_p$ $=$ $\displaystyle 0$ Definition of $p$-adic Norm $\displaystyle$ $\le$ $\displaystyle \max \set {\norm x_p, \norm y_p}$ as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from Norm Axioms: Axiom $(\text N 1)$

Let $x, y, x + y \in \Q_{\ne 0}$.

$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$

Then:

 $\displaystyle \norm {x + y}_p$ $=$ $\displaystyle p^{-\map {\nu_p} {x + y} }$ Definition of $p$-adic Norm $\displaystyle$ $\le$ $\displaystyle \max \set {p^{- \map {\nu_p} x}, p^{-\map {\nu_p} y} }$ $\displaystyle$ $=$ $\displaystyle \max \set {\norm x_p, \norm y_p}$ Definition of $p$-adic Norm

$\blacksquare$