Equivalence of Definitions of P-adic Norms/Lemma 1

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

Let $\nu_p: \Z \to \N \cup \set {+\infty}$ be the $p$-adic valuation on the integers.

Then:
 * $\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$

Proof
Let $x \in \Z_{\ne 0}$.

By definition of the $p$-adic valuation:
 * $\map {\nu_p} x = \sup \set {v \in \N: p^v \divides x}$

Let $\map {\nu_p} x = k$.

Then:
 * $p^k \nmid x$

By definition of a divisor:
 * $\exists y \in Z : x = p^k y$


 * $p \divides y$
 * $p \divides y$

By definition of a divisor:
 * $\exists y' \in Z : y = p y'$

Hence:

This contradicts:
 * $k = \sup \set {v \in \N: p^v \divides x}$

Hence:
 * $p \nmid y$

Since the choice of $x \in \Z_{\ne 0}$ was arbitrary:
 * $\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$