It needs to be shown that $\nu_p$: :$(1): \quad \nu_p$ is a mapping into $\Z \cup \set {+\infty}$ :$(2): \quad \nu_p$ satisfies the valuation axioms $\text V 1$, $\text V 2$ and $\text V 3$ :$(3): \quad \nu_p$ extends $\nu_p^\Q$. Let $x, y \in \Q_p$. === $\nu_p$ is a mapping into $\Z \cup \set {+\infty}$ === If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition. Let $x \ne 0$. By P-adic Norm of p-adic Number is Power of p then: :$\exists v \in \Z: \norm x_p = p^{-v}$ Hence: {{begin-eqn}} {{eqn | l = \map {\nu_p} x       | r = -\log_p \norm {x}_p       | c = Since $x \ne 0$ }} {{eqn | r = -\log_p p^{-v}       | c = Definition of $v$ }} {{eqn | r = -\paren {-v}       | c = {{Defof|Real General Logarithm}} }} {{eqn | r = v }} {{eqn | o = \in       | r = \Z       | c = Definition of $v$ }} {{end-eqn}} {{qed|lemma}} === $\nu_p$ satisfies $(\text V 1)$ === If $x = 0$ then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {0 \cdot y}       | r = \map {\nu_p} 0 }} {{eqn | r = +\infty       | c = Definition of $\nu_p$ }} {{eqn | r = +\infty \cdot \map {\nu_p} y       | c = {{Defof|Extended Real Multiplication}} }} {{eqn | r = \map {\nu_p} 0 \cdot \map {\nu_p} y       | c = Definition of $\nu_p$ }} {{end-eqn}} Similarly, if $y = 0$ then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {x \cdot 0}       | r = \map {\nu_p} x \cdot \map {\nu_p} 0 }} {{end-eqn}} If $x \ne 0, y \ne 0$ then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {x y}       | r = -\log \norm {x y}_p       | c = $x y \ne 0$ }} {{eqn | r = -\log \norm x_p \norm y_p       | c = {{NormAxiom|2}} }} {{eqn | r = -\paren {\log \norm x_p + \log \norm y_p}       | c = Sum of General Logarithms }} {{eqn | r = \paren {-\log \norm x_p} + \paren {-\log \norm y_p}       | c =  }} {{eqn | r = \map {\nu_p} x + \map {\nu_p} y       | c = Definition of $\nu_p$ }} {{end-eqn}} {{qed|lemma}} === $\nu_p$ satisfies $(\text V 2)$ === If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition. If $x \ne 0$ then $\map {\nu_p} x \in \Z$ by the above. Hence: :$\map {\nu_p} x = +\infty \iff x = 0$  {{qed|lemma}} === $\nu_p$ satisfies $(\text V 3)$ === Suppose $x = 0$. Then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {0 + y}       | r = \map {\nu_p} y }} {{eqn | o = \ge       | r = \min \set {\map {\nu_p} 0, \map {\nu_p} y}       | c = {{Defof|Min Operation}} }} {{end-eqn}} Similarly, if $y = 0$ then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {x + 0}       | o = \ge       | r = \min \set {\map {\nu_p} x, \map {\nu_p} 0} }} {{end-eqn}} Suppose $x + y = 0$. Then: {{begin-eqn}} {{eqn | l = \map {\nu_p} {x + y}       | r = \map {\nu_p} 0 }} {{eqn | r = +\infty       | c = Definition of $\nu_p$ }} {{eqn | o = \ge       | r = \map {\nu_p} x       | c = {{Defof|Extended Real Number Line}} }} {{eqn | o = \ge       | r = \min \set {\map {\nu_p} x, \map {\nu_p} y}       | c = {{Defof|Min Operation}} }} {{end-eqn}} Suppose $x \ne 0, y \ne 0, x + y \ne 0$. Then: {{begin-eqn}} {{eqn | l = \norm {x + y}       | o = \le       | r = \max \set {\norm x_p, \norm y_p}       | c = {{NormAxiom|4}} }} {{eqn | o = }} {{eqn | ll = \leadsto       | l = \log \norm {x + y}       | o = \le       | r = \log \max \set {\norm x_p, \norm y_p}       | c = Logarithm is Strictly Increasing }} {{eqn | r = \max \set {\log \norm x_p, \log \norm y_p}       | c = Logarithm is Strictly Increasing }} {{eqn | o = }} {{eqn | ll = \leadsto       | l = -\log \norm {x + y}       | o = \ge       | r = -\max \set {\log \norm x_p, \log \norm y_p}       | c = Inversion Mapping Reverses Ordering in Ordered Group }} {{eqn | r = \min \set {-\log \norm x_p, -\log \norm y_p}       | c =  }} {{eqn | o = }} {{eqn | ll = \leadsto       | l = \map {\nu_p} {x + y}       | o = \ge       | r = \min \set {\map {\nu_p} x, \map {\nu_p} y}       | c = Definition of $\nu_p$ }} {{end-eqn}} {{qed|lemma}} === $\nu_p$ extends $\nu_p^\Q$ === Let $x \in \Q$. If $x = 0$ then $\map {\nu_p} 0 = +\infty = \map {\nu_p^\Q} 0$. Let $x \ne 0$. The $p$-adic norm $\norm{\,\cdot\,}_p$ on $p$-adic numbers $\Q_p$ is an extension of the $p$-adic norm $\norm{\,\cdot\,}_p$ on rational numbers $\Q$ by definition. Hence: {{begin-eqn}} {{eqn | l = \map {\nu_p} x       | r = -\log \norm x_p       | c = Definition of $\nu_p$ }} {{eqn | r = \map {\nu_p^\Q} x       | c = {{Defof|P-adic Norm|$p$-adic norm}} $\norm{\,\cdot\,}_p$ on rational numbers $\Q$ }} {{end-eqn}} {{qed}}