P-adic Norm not Complete on Rational Numbers

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

Let $d_p$ be the p-adic metric; that is, the metric induced by $\norm {\,\cdot\,}_p$.

Then:


 * $\struct {\Q, d_p}$ is not a complete metric space.

Informally, the valued field $\struct {\Q, \norm {\,\cdot\,}_p }$ does not define a complete metric space.

Proof
By definition of the p-adic metric:


 * $\forall x, y \in \Q: d_p \paren {x,y} = \norm {x-y}_p$

To show that $\struct {\Q, d_p}$ is not complete we need to show there exists a Cauchy sequence in $\Q$ which does not converge in $\struct {\Q, d_p}$.

We note that convergence in the metric space $\struct {\Q, d_p}$ is equivalent to convergence in the normed division ring $\struct {\Q, \norm {\,\cdot\,}_p }$.

Consider the sequence $\sequence {a^{p^n} }_{n \in \N} \subseteq \Q$ where $1 \le a \lt p$ is an integer.

Let $n \in \N$.

Then:


 * $\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \left({p - 1}\right)} - 1) }_p$

From the corollary to Euler's Theorem:
 * $a^{p^n \left({p - 1}\right)} - 1 \equiv 0 \pmod {p^n}$

so:
 * $\norm {a^{p^n} \left({a^{p^n \left({p - 1}\right)} - 1}\right)}_p \le p^{-n} \xrightarrow {n \to \infty} 0$

Hence $\sequence {a^{p^n} }_{n \in \N}$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.

On the other hand, let $x_n := a^{p^n}$.

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

Since:

and:

From:
 * $\norm {x - a}_p = \dfrac 1 {p^{\nu_p \paren {x - a} } } < 1$

it follows that:
 * $p \divides \left({x - a}\right)$

Hence from $x^p = x$ it follows that, if $a \ne 1$ and $a \ne p - 1$, then $x$ must be a non-trivial $p-1$-th root of unity which is also in $\Q$.

This is a contradiction.

In conclusion:
 * $x \notin \Q$