P-adic Norm is Norm

Theorem
The p-adic measure of distance forms a norm on the $p$-adic numbers $\Q_p$.

Proof
Let $v_p$ be the valuation of the $p$-adic numbers.

Recall that the $p$-adic metric is defined by


 * $\displaystyle |x|_p = \begin{cases} \frac{1}{p^{v_p(x)}}, & x \neq 0, \\ 0, & x = 0\end{cases}$

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


 * $(1): \quad \left\vert {x} \right\vert = 0 \iff x = 0$
 * $(2): \quad \left\vert {x y} \right\vert = \left\vert{x}\right\vert \cdot \left\vert{y}\right\vert$
 * $(3): \quad \left\vert {x + y}\right\vert \leq \left\vert{x}\right\vert + \left\vert{y}\right\vert$

$(1): \quad$ This follows directly from the definition of the p-adic function $|\cdot|_p$ and the fact that $\displaystyle \frac 1{p^s} > 0$ for all $s \in \R$.

$(2): \quad$ If $x=0$ or $y=0$ the result is trivial by part 1.

Suppose that $x,y\in \Q_p$, $x,y \neq 0$.

Then $v_p (xy) = v_p (x) + v_p (y)$.

Therefore,

$(3): \quad$ If $x=0$, $y=0$, or $x+y=0$, the result is trivial.

Suppose now that $x,y, x+y \in \Q_p$ are all non-zero.

First, assume $x,y \in \Q$.

Let $\displaystyle x=\frac ab, y=\frac cd$ be the canonical form of rationals.

Then


 * $\displaystyle x+y = \frac{ad+bc}{bd}$

and


 * $v_p(x+y)=v_p(ad+bc)-v_p(b)-v_p(d)$.

The highest power of $p$ dividing the sum of two numbers is at least the minimum of the highest power dividing the first and the highest power dividing the second. Therefore,

Therefore,