## Theorem

Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $x \in \Q_p: x \ne 0$.

Then:

$\exists v \in \Z: \norm x_p = p^{-v}$

## Lemma

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.

Let $\sequence {x_n}$ be a Cauchy sequence in $\struct{\Q, \norm {\,\cdot\,}_p}$ such that $\sequence {x_n}$ does not converge to $0$.

Then:

$\exists v \in \Z: \displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$

## Proof

By definition of the $p$-adic numbers $\Q_p$, the rational numbers $\Q$ are dense in $\Q_p$.

By the definition of a dense subset then $\map \cl \Q = \Q_p$.

there exists a sequence $\sequence {x_n} \subseteq \Q$ that converges to $x$.

That is:

$\displaystyle \lim_{n \mathop \to \infty} x_n = x$

From Modulus of Limit:

$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = \norm x_p$

By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.

From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$.

From Lemma:

$\exists v \in \Z: \norm x_p = \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$

$\blacksquare$