P-adic Valuation of Rational Number is Well Defined

Theorem

The $p$-adic valuation:

$\nu_p: \Q \to \Z \cup \set {+\infty}$

is well defined.

Proof

Let $\dfrac a b = \dfrac c d \in \Q$.

Thus:

$a d = b c \in \Z$

By Definition of Rational Number:

$b, d \ne 0$

By Definition of Restricted $p$-adic Valuation:

$\map {\nu_p^\Z} b, \map {\nu_p^\Z} d < +\infty$

Case 1 : $a \ne 0$

Let $a \ne 0$.

It follows:

$c \ne 0$

By Definition of Restricted $p$-adic Valuation:

$\map {\nu_p^\Z} a, \map {\nu_p^\Z} c < +\infty$

Then:

\(\ds \map {\nu_p^\Z} a + \map {\nu_p^\Z} d\)

\(=\)

\(\ds \map {\nu_p^\Z} {a d}\)

Restricted $p$-adic Valuation is Valuation: Axiom $\text V 1$

\(\ds \)

\(=\)

\(\ds \map {\nu_p^\Z} {b c}\)

\(\ds \)

\(=\)

\(\ds \map {\nu_p^\Z} c + \map {\nu_p^\Z} b\)

Restricted $p$-adic Valuation is Valuation: Axiom $\text V 1$

\(\ds \leadsto \ \ \)

\(\ds \map {\nu_p^\Z} a - \map {\nu_p^\Z} b\)

\(=\)

\(\ds \map {\nu_p^\Z} c - \map {\nu_p^\Z} d\)

$\Box$

Case 2 : $a = 0$

Let $a = 0$.

It follows:

$c = 0$

By Definition of Restricted $p$-adic Valuation:

$\map {\nu_p^\Z} a = \map {\nu_p^\Z} c = +\infty$

Then:

\(\ds \map {\nu_p^\Z} a - \map {\nu_p^\Z} b\)

\(=\)

\(\ds +\infty - \map {\nu_p^\Z} b\)

\(\ds \)

\(=\)

\(\ds +\infty\)

as $\map {\nu_p^\Z} b < +\infty$

\(\ds \)

\(=\)

\(\ds +\infty - \map {\nu_p^\Z} d\)

as $\map {\nu_p^\Z} d < +\infty$

\(\ds \)

\(=\)

\(\ds \map {\nu_p^\Z} c - \map {\nu_p^\Z} d\)

$\Box$

In either case:

\(\text {(1)}: \quad\)

\(\ds \map {\nu_p^\Z} a - \map {\nu_p^\Z} b\)

\(=\)

\(\ds \map {\nu_p^\Z} c - \map {\nu_p^\Z} d\)

So:

\(\ds \map {\nu_p^\Q} {\frac a b}\)

\(=\)

\(\ds \map {\nu_p^\Z} a - \map {\nu_p^\Z} b\)

Definition of $p$-adic Valuation

\(\ds \)

\(=\)

\(\ds \map {\nu_p^\Z} c - \map {\nu_p^\Z} d\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \map {\nu_p^\Q} {\dfrac c d}\)

Definition of $p$-adic Valuation

Thus, by definition, $\nu_p: \Q \to \Z \cup \set {+\infty}$ is well defined.

$\blacksquare$