P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2

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 $\struct {\Q, \norm {\,\cdot\,}_p }$.

Case: $p \gt 3$
Suppose $p \gt 3$, then there is $a \in \Z: 1 \lt a < p-1$.

Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 \lt a < p-1$.

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

That is:
 * $\displaystyle \lim_{n \to \infty} \norm {x_{n+1} - x_n } = 0$

By Characterisation of Cauchy Sequence in Non-Archimedean Norm


 * $\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.

$\sequence {x_n}$ converges to some $x \in \Q$.

That is:
 * $x = \displaystyle \lim_{n \mathop \to \infty} x_n$

By Modulus of Limit on a Normed Division Ring:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$

Since $\forall n, p \nmid a^{p^n} = x_n$, then:
 * $ \norm {x_n }_p = 1$

So:
 * $\norm {x }_p = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = 1$

By Axiom (N1) of a norm on a division ring then:
 * $x \ne 0$.

Since:

and $x \ne 0$ then:


 * $x^{p-1} = 1$.

So:
 * $x = 1$ or $x = -1$

and so $a-x$ is an integer:
 * $0 \lt a-x \lt p$

It follows that:
 * $p \nmid \paren{a-x}$

and so:
 * $\norm {x-a}_p = 1$

Since $x_n \to x$ as $n \to \infty$ then:
 * $\exists N: \forall n \gt N: \norm {x_n - x}_p \lt \norm {x - a}_p$

That is:
 * $\exists N: \forall n \gt N: \norm {a^{p^n} - x}_p \lt \norm {x - a}_p$

Let $n \gt N$:

Since $\norm {x - a^{p^n}}_p \lt \norm {x - a}_p$ then:

This contradicts the earlier assertion that $\norm {x-a}_p = 1$.

In conclusion:
 * $\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.