Restricted P-adic Valuation is Valuation

Theorem
Consider $\nu_p^\Z:\Z\to\N\cup\{+\infty\}$, the restricted $p$-adic valuation.

Then the following holds:


 * $(1):\quad \nu_p^\Z \left({n m}\right) = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$.
 * $(2):\quad \nu_p^\Z \left({n + m}\right) \geq \min \left\{{\nu_p^\Z \left({n}\right), \nu_p^\Z \left({m}\right)}\right\}$

$(1)$
Let $n, m \in \Z$.

Suppose $m = 0$ or $n = 0$

Then $\nu_p^\Z \left({n}\right) = +\infty$ or $\nu_p^\Z \left({m}\right) = +\infty$.

Now $n m = 0$, and hence $\nu_p^\Z \left({n m}\right) = +\infty = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$.

Next, suppose $n m \ne 0$.

Then
 * $p^{\nu_p^\Z \left({n}\right)} \mid n$, and
 * $p^{\nu_p^\Z \left({n}\right) + 1} \nmid n$;

also
 * $p^{\nu_p^\Z \left({m}\right)} \mid m$, but
 * $p^{\nu_p^\Z \left({m}\right) + 1} \nmid m$.

It follows immediately that
 * $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)} \mid n m$, and
 * $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right) + 1} \nmid n m$.

Hence $\nu_p^\Z \left({n m}\right) = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$.

$(2)$
Consider $n,m \in \Z$.

Assume that $p^\alpha\mid n$ y $p^\beta\mid m$; where $\alpha\geq\beta$.

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

In conclusion $p^\beta\mid (n+m)$.

Hence $\nu_p^\Z(n+m)\geq\min\{\nu_p^\Z(n),\nu_p^\Z(m)\}$ by the definition of $\nu_p^\Z$.