P-adic Norm is Non-Archimedean Norm

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

From $p$-adic Valuation is Valuation:

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

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

$\blacksquare$


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