P-adic Norm is Norm

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: \left\Vert{q}\right\Vert_p := \begin{cases}

0 & : q = 0 \\ p^{- \nu_p \left({q}\right)} & : q \ne 0 \end{cases}$

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

Norm Axiom $(N1)$
By Power to Real Number is Positive:
 * $\displaystyle \forall s \in \R: \frac 1 {p^s} > 0$

By definition of the $p$-adic norm it follows that: $:\forall x \in \Q: \left\Vert{x}\right\Vert_p = 0$

Thus the $p$-adic norm fulfils axiom $(N1)$.

Norm Axiom $(N2)$
Let $x = 0$ or $y = 0$.

Then $\left\Vert{x}\right\Vert_p = 0$ or $\left\Vert{y}\right\Vert_p = 0$ from axiom $(N1)$, and:

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

Then:

Thus the $p$-adic norm fulfils axiom $(N2)$.

Norm Axiom $(N3)$
Let $x = 0$ or $y = 0$, or $x+y=0$, the result is trivial.

Let $x = 0$.

Then:

and so $\left\Vert{x + y}\right\Vert_p \le \left\Vert{x}\right\Vert_p + \left\Vert{y}\right\Vert_p$

The same argument holds for $y = 0$.

Let $x + y = 0$.

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

From P-adic Valuation is Valuation:
 * $\nu_p \left({x + y}\right) \ge \min \left\{ {\nu_p \left({x}\right), \nu_p \left({y}\right)}\right\}$

Then: