P-adic Norm is Norm

Theorem
The p-adic norm forms a norm on the rational numbers $\Q$ and hence a metric.

Proof
Let $v_p$ be the valuation of the rational numbers.

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


 * $\displaystyle |x|_p = \frac{1}{p^{v_p(x)}}$

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


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

$(1): \quad$ 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 $x,y\in \Q$, $x,y \neq 0$.

From P-adic Valuation is Valuation we get that $v_p (xy) = v_p (x) + v_p (y)$.

Therefore,

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

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

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

and from P-adic Valuation is Valuation we get that $v_p(x+y)\ge \min\{v_p(x),v_p(y)\}$

Therefore,