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}         | 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 $(\text N 1)$ === 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 $(\text N 1)$. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = x y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \norm {x y}_p       | r = 0       | c = Norm Axioms: Axiom $(\text N 1)$ }} {{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 = Exponent Combination Laws: 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 axiom $(\text N 2)$. {{qed|lemma}} === Norm Axiom $(\text N 3)$ === 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 Axioms: Axiom $(N1)$ }} {{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 axiom $(\text N 3)$. {{qed|lemma}} All norm axioms are seen to be satisfied. Hence the result. {{qed}}