Theorem: P-adic Norm is Norm

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 $x$, $y \in \Q$:
{{begin-axiom}}
{{axiom | n = \text N 1
        | q = \forall x \in \Q
        | ml= \norm x_p = 0
        | mo= \iff
        | mr= x = 0
}}
{{axiom | n = \text N 2
        | q = \forall x, y \in \Q
        | ml= \norm {x y}_p
        | mo= =
        | mr= \norm x_p \times \norm y_p
}}
{{axiom | n = \text N 3
        | q = \forall x, y \in \Q
        | ml= \norm {x + y}_p
        | mo= \le
        | mr= \norm x_p + \norm y_p
}}
{{end-axiom}}
=== {{Norm-axiom-mult|1|nolink}} ===
By Power of Positive Real Number is Positive:
:$\ds \forall s \in \R: \frac 1 {p^s} > 0$
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-mult|1}}.
{{qed|lemma}}
=== {{Norm-axiom-mult|2|nolink}} ===
Let $x = 0$ or $y = 0$.
Then $\norm x_p = 0$ or $\norm y_p = 0$ from {{Norm-axiom-mult|1}}, and:
{{begin-eqn}}
{{eqn | l = x y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm {x y}_p
      | r = 0
      | c = {{Norm-axiom-mult|1}}
}}
{{eqn | r = \norm x_p \times \norm y_p
      | c = 
}}
{{end-eqn}}
Let $x, y\in \Q_{\ne 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {x y}_p
      | r = \frac 1 {p^{\map {\nu_p} {x y} } }
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | r = \frac 1 {p^{\map {\nu_p} x + \map {\nu_p} y} }
      | c = $p$-adic Valuation is Valuation
}}
{{eqn | r = \frac 1 {p^{\map {\nu_p} x} p^{\map {\nu_p} y} }
      | c = Product of Powers
}}
{{eqn | r = \norm x_p \times \norm y_p
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{end-eqn}}
Thus the $p$-adic norm fulfils {{Norm-axiom-mult|2}}.
{{qed|lemma}}
=== {{Norm-axiom-mult|3|nolink}} ===
Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.
Let $x = 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm x_p
      | r = 0
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | ll= \leadsto
      | l = \norm x_p + \norm y_p
      | r = \norm y_p
      | c = 
}}
{{eqn | r = \norm {x + y}_p
      | c = 
}}
{{end-eqn}}
and so $\norm {x + y}_p \le \norm x_p + \norm y_p$
The same argument holds for $y = 0$.
Let $x + y = 0$.
{{begin-eqn}}
{{eqn | l = x + y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm {x + y}_p
      | r = 0
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = \le
      | r = \norm x_p + \norm y_p
      | c = as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from {{Norm-axiom-mult|1}}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \norm {x + y}_p
      | r = p^{-\map {\nu_p} {x + y} }
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = \le
      | r = \max \set {p^{-\map {\nu_p} x}, p^{-\map {\nu_p} y} }
}}
{{eqn | r = \max \set {\norm x_p, \norm y_p}
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = \le
      | r = \norm x_p + \norm y_p
}}
{{end-eqn}}
Thus the $p$-adic norm fulfils {{Norm-axiom-mult|3}}.
{{qed|lemma}}
All norm axioms are seen to be satisfied.
Hence the result.
{{qed}}