P-adic Norm is Norm/Proof 1

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 $x$, $y \in \Q$:

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

By definition of the $p$-adic norm it follows that:
 * $\forall x \in \Q: \norm x_p = 0 \iff x = 0$

Thus the $p$-adic norm fulfils axiom $(\text N 1)$.

Norm Axiom $(\text N 2)$
Let $x = 0$ or $y = 0$.

Then $\norm x_p = 0$ or $\norm y_p = 0$ from axiom $(\text N 1)$, and:

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

Then:

Thus the $p$-adic norm fulfils axiom $(\text N 2)$.

Norm Axiom $(\text N 3)$
Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.

Let $x = 0$.

Then:

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

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:


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

Then:

Thus the $p$-adic norm fulfils axiom $(\text N 3)$.

All norm axioms are seen to be satisfied.

Hence the result.