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

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

Then:


 * $\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.

That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof
Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.

Lemma 1
Let $x_1 \in \Z_{\gt 0}: p \nmid x_1, x_1 \gt \dfrac {p+1} 2$

Let $q$ be a prime such that $q \neq p$.

By the definition of a prime number, $p \nmid q$

Let $a = x_1^q + p$

Lemma 2
Let $f \paren{X} \in \Z [X]$ be the polynomial:
 * $X^q - a$

Lemma 3
Let $f' \paren{X} \in \Z [X]$ be the formal derivative of $f \paren{X}$.

Lemma 4
By Hensel's Lemma there exists a sequence of integers $\sequence {x_n}$ such that:
 * $(1) \quad \forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
 * $(2) \quad \forall n: x_{n+1} \equiv x_n \pmod {p^n}$

Lemma 5
By corollary of Characterisation of Cauchy Sequence in Non-Archimedean Norm then:
 * $\sequence{x_n}$ is a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$.

Aiming for a contradiction suppose $\sequence{x_n}$ converges in $\struct {\Q, \norm{\,\cdot\,}_p}$.

Suppose for some $c \in \Q$:
 * $\displaystyle \lim_{n \to \infty} x_n = c$

By product rule for convergent sequences then:
 * $\displaystyle \lim_{n \to \infty} x_n^q = c^q$

Hence:
 * $c^q = a$.

It follows that $c^q \in \Z$

By Nth Root of Integer is Integer or Irrational then:
 * $c \in \Z$

This contrdicts Lemma 2.

So the sequence $\sequence{x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.

The result follows.