# P-adic Valuation of Rational Number is Well Defined

## Theorem

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

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.

$\blacksquare$