Equivalence of Definitions of P-adic Norms

Theorem
Let $p \in \N$ be a prime.

Let $\Q$ denote the rational numbers.

Proof
From Negative Powers of Group Elements, Definition 2 can be rewritten as:
 * $\forall r \in \Q: \norm r_p = \begin {cases} 0 & : r = 0 \\ p^{-k} & : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n \end {cases}$

Hence if follows that Definition 1 and Definition 2 are equivalent if it is shown:
 * $\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$

Let $r \in \Q_{\ne 0}$.

Let $r = \dfrac a b : a, b \in \Z_{\ne 0}$

We have:

Let:
 * $k_a := \map {\nu_p} a$
 * $k_b := \map {\nu_p} b$

Lemma 1
From Lemma 1:
 * $a = p^{k_a} m$
 * $b = p^{k_b} n$
 * $p \nmid m, n$

Hence:

Let:
 * $k = k_a - k_b$.

It follows that:
 * $\map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$

Since $r$ was arbitrary:
 * $\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$