P-adic Norm not Complete on Rational Numbers

Theorem
The normed vector space $\left({\Q, \left\vert{\,\cdot\,}\right\vert_p}\right)$ does not define a complete metric space.

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

Let $n \in \N$.

Then:


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

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

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

Hence $\left({a^{p^n} }\right)_{n \in \N}$ is a cauchy sequence in $\left({\Q, \left\vert{\,\cdot\,}\right\vert_p}\right)$.

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

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

Since:

and:

From:
 * $\left\vert{x - a}\right\vert_p = \dfrac 1 {p^{\nu_p \left({x - a}\right)} } < 1$

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

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 in $\Q$.

This is a contradiction.

In conclusion:
 * $x \notin \Q$