# P-adic Norm is Norm/Proof 1

## Theorem

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

## Proof

Let $v_p$ be the $p$-adic valuation on the rational numbers.

Recall that the $p$-adic norm is defined as:

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

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

 $(\text N 1)$ $:$ $\ds \forall x \in \Q:$ $\ds \norm x_p = 0$ $\ds \iff$ $\ds x = 0$ $(\text N 2)$ $:$ $\ds \forall x, y \in \Q:$ $\ds \norm {x y}_p$ $\ds =$ $\ds \norm x_p \times \norm y_p$ $(\text N 3)$ $:$ $\ds \forall x, y \in \Q:$ $\ds \norm {x + y}_p$ $\ds \le$ $\ds \norm x_p + \norm y_p$

### Norm Axiom $\text N 1$: Positive Definiteness

$\ds \forall s \in \R: \frac 1 {p^s} > 0$

By definition of the $p$-adic norm it follows that:

$\forall x \in \Q: \norm x_p = 0 \iff x = 0$

Thus the $p$-adic norm fulfils Norm Axiom $\text N 1$: Positive Definiteness.

$\Box$

### Norm Axiom $\text N 2$: Positive Homogeneity

Let $x = 0$ or $y = 0$.

Then $\norm x_p = 0$ or $\norm y_p = 0$ from Norm Axiom $\text N 1$: Positive Definiteness, and:

 $\ds x y$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \norm {x y}_p$ $=$ $\ds 0$ Norm Axiom $\text N 1$: Positive Definiteness $\ds$ $=$ $\ds \norm x_p \times \norm y_p$

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

Then:

 $\ds \norm {x y}_p$ $=$ $\ds \frac 1 {p^{\map {\nu_p} {x y} } }$ Definition of $p$-adic Norm $\ds$ $=$ $\ds \frac 1 {p^{\map {\nu_p} x + \map {\nu_p} y} }$ $p$-adic Valuation is Valuation $\ds$ $=$ $\ds \frac 1 {p^{\map {\nu_p} x} p^{\map {\nu_p} y} }$ Product of Powers $\ds$ $=$ $\ds \norm x_p \times \norm y_p$ Definition of $p$-adic Norm

Thus the $p$-adic norm fulfils Norm Axiom $\text N 2$: Positive Homogeneity.

$\Box$

### Norm Axiom $\text N 3$: Triangle Inequality

Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.

Let $x = 0$.

Then:

 $\ds x$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \norm x_p$ $=$ $\ds 0$ Definition of $p$-adic Norm $\ds \leadsto \ \$ $\ds \norm x_p + \norm y_p$ $=$ $\ds \norm y_p$ $\ds$ $=$ $\ds \norm {x + y}_p$

and so $\norm {x + y}_p \le \norm x_p + \norm y_p$

The same argument holds for $y = 0$.

Let $x + y = 0$.

 $\ds x + y$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \norm {x + y}_p$ $=$ $\ds 0$ Definition of $p$-adic Norm $\ds$ $\le$ $\ds \norm x_p + \norm y_p$ as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from Norm Axiom $\text N 1$: Positive Definiteness

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:

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

Thus the $p$-adic norm fulfils Norm Axiom $\text N 3$: Triangle Inequality.

$\Box$

All norm axioms are seen to be satisfied.

Hence the result.

$\blacksquare$