Restricted P-adic Valuation is Valuation

Theorem
Let $\nu_p^\Z: \Z \to \Z \cup \left\{ {+\infty}\right\}$ be the $p$-adic valuation restricted to the integers.

Then $\nu_p^\Z$ is a valuation.

Proof
To prove that $\nu_p^\Z$ is a valuation it is necessary to demonstrate:

Axiom $(V1)$
Let $m, n \in \Z$.

Let $m = 0$ or $n = 0$.

Then:
 * $\nu_p^\Z \left({m}\right) = +\infty$

or:
 * $\nu_p^\Z \left({n}\right) = +\infty$

Also:
 * $n m = 0$

and hence:

Let $n m \ne 0$.

Then by definition of the restricted $p$-adic valuation:
 * $p^{\nu_p^\Z \left({n}\right)} \mathop \backslash n$
 * $p^{\nu_p^\Z \left({n}\right) + 1} \nmid n$

Also:
 * $p^{\nu_p^\Z \left({m}\right)} \mathop \backslash m$
 * $p^{\nu_p^\Z \left({m}\right) + 1} \nmid m$

Hence:
 * $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)} \mathop \backslash n m$
 * $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right) + 1} \nmid n m$

So:
 * $\nu_p^\Z \left({n m}\right) = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$

Axiom $(V2)$
By definition of the restricted $p$-adic valuation:


 * $\forall n \in \Z: \nu_p^\Z \left({n}\right) = +\infty \iff n = 0$

Axiom $(V3)$
Let $m, n \in \Z$.

WLOG let:
 * $p^\alpha \mathop \backslash n$
 * $p^\beta \mathop \backslash m$

where $\alpha \ge \beta$.

Then $\exists t \in \Z, k \in \Z$ such that:

Thus:
 * $p^\beta \mathop \backslash \left({m + n}\right)$

Hence by the definition of $\nu_p^\Z$:
 * $\nu_p^\Z \left({m + n}\right) \ge \min \left\{{\nu_p^\Z \left({m}\right), \nu_p^\Z \left({n}\right)}\right\}$