Recall that the $p$-adic norm is defined as: :$\forall q \in \Q: \norm r_p := \begin{cases}   0 & : r = 0 \\   p^{- k} & : r \ne 0 \end{cases}$ where: :$\displaystyle r = p^k \frac m n$ and: :$k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$ where $\nmid$ stands for "does not divide". We must show the following hold for all $r_1$, $r_2 \in \Q$: {{begin-axiom}} {{axiom | n = \text N 1         | q = \forall r \in \Q         | ml= \norm r_p = 0         | mo= \iff         | mr= x = 0 }} {{axiom | n = \text N 2         | q = \forall r_1, r_2 \in \Q         | ml= \norm {r_1 r_2}         | mo= =         | mr= \norm {r_1}_p \times \norm {r_2}_p }} {{axiom | n = \text N 3         | q = \forall r_1, r_2 \in \Q         | ml= \norm {r_1 + r_2}_p         | mo= \le         | mr= \norm {r_1}_p + \norm {r_2}_p }} {{end-axiom}} === Norm Axiom $(\text N 1)$ === Let $r \in \Q : r \ne 0$. Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$. Suppose $r = 0$. By definition: :$\norm {r}_p = 0$ Suppose $\displaystyle r = p^k \frac m n \ne 0$ By definition: :$\displaystyle \norm {r}_p = \frac 1 {p^k} > 0$ Suppose $\norm r_p = 0$. By definition: :$r = 0$ {{qed|lemma}} === Norm Axiom $(\text N 2)$ === Suppose $r_1 = 0$ or $r_2 = 0$. From axiom $(\text N 1)$, $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$. Suppose $r_1 \ne 0 \ne r_2$. Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$ Let $\displaystyle r_1 = p^{k_1} \frac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$ Then: :$\displaystyle r_1 r_2 = p^{k_1 + k_2} \frac {m_1 m_2}{n_1 n_2}$ We have that $p \nmid m_1$, $p \nmid m_2$. Since $p$ is prime: :$p \nmid m_1 m_2$. Similarly: :$p \nmid n_1 n_2$. Therefore: {{begin-eqn}} {{eqn| l = \norm {r_1 r_2}_p       | r = p^{- \paren {k_1 + k_2} } }} {{eqn| r = p^{-k_1} p^{-k_2}  }} {{eqn| r = \norm {r_1}_p \norm {r_2}_p }} {{end-eqn}} {{qed|lemma}} === Norm Axiom $(\text N 3)$ === Suppose one of the following is true: :$r_1 = 0$ :$r_2 = 0$ :$r_1 + r_2 = 0$ Then the result is straightforward. Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$. Let $\displaystyle r_1 = p^{k_1} \frac {m_1}{n_1}, r_2 = p^{k_2} \frac{m_2}{n_2}$ where: :$\displaystyle k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$ Then: {{begin-eqn}} {{eqn | l = r_1 + r_2       | r = \frac {p^{k_1} m_1 n_2 + p^{k_2} m_2 n_1} {n_1 n_2} }} {{eqn | r = p^{\map \min {k_1, k_2} } \frac {p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2}{n_1 n_2}       | c = {{Defof|Min Operation}} }} {{eqn | r = p^{\map \min {k_1, k_2} } \frac {\tilde m}{n_1 n_2}       | c = $\displaystyle \tilde m := p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2$ }} {{end-eqn}} By Fundamental Theorem of Arithmetic: :$\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$ Obviously, $p \nmid n_1 n_2$ Hence: {{begin-eqn}} {{eqn | l = \norm {r_1 + r_2}_p       | r = \frac 1 {p^{\tilde k + \map \min {k_1, k_2} } } }} {{eqn | o = \le       | r = \frac 1 {p^{\map \min {k_1, k_2} } } }} {{eqn | r = \map \max {p^{-k_1}, p^{-k_2} }       | c = {{Defof|Max Operation}} }} {{eqn | r = \map \max {\norm {r_1}_p, \norm {r_1}_p} }} {{eqn | o = \le       | r = \map \max {\norm {r_1}_p, \norm {r_2}_p} + \map \min {\norm {r_1}_p, \norm {r_2}_p} }} {{eqn | r = \norm {r_1}_p + \norm {r_2}_p }} {{end-eqn}} {{qed|lemma}} All norm axioms are seen to be satisfied. Hence the result. {{qed}}