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 $p \ge 3$.

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.

Let $b = \dfrac {p + 1} 2$

Since $p$ is odd then $p + 1$ is even and $b \in \Z$

Then:

and

Hence:
 * $0 \lt b \lt p$

Let $x_1 \in \Z_{\gt 0}: b \le x_1 \lt p$

Let $a = x_1^2 + p$

Then $a \in \Z_{\gt 0}$

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

Then:
 * $\map f {x_1} = x_1^2 - \paren {x_1^2 + p} = -p \equiv 0 \pmod p$

By Corollary to Absolute Value of Integer is not less than Divisors then:
 * $p \nmid x_1$
 * $p \nmid 2$

By Euclid's Lemma for Prime Divisors then:
 * $p \nmid 2x_1 = \map {f'} {x_1}$

Hence:
 * $2x_1 \not \equiv 0 \mod p$

The formal derivative $f' \paren{X} \in \Z [X]$ is by definition:
 * $2X$

Then:
 * $\map {f’} {x_1} = 2x_1 \not \equiv 0 \pmod p$

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

From (1) it follows that:
 * $p^n \divides \paren{x_n^2 - a}$

Hence the $p$-adic norm of $x_n^2 - a$ is by definition:
 * $\norm{x_n^2 - a}_p \le \dfrac 1 {p^n}$

By Sequence of Powers of Number less than One then:
 * $\displaystyle \lim_{n \to \infty} \dfrac 1 {p^n} = 0$

By Inequality Rule for Real Sequences then:
 * $\displaystyle \lim_{n \to \infty} \norm{x_n^2 - a}_p = 0$.

By the definition of convergence in $\struct {\Q, \norm{\,\cdot\,}_p}$ then:
 * $\displaystyle \lim_{n \to \infty} x_n^2 = a$

From (2) it follows that:
 * $p^n \divides \paren{x_{n+1} - x_n}$

Hence the $p$-adic norm of $x_n^2 - a$ is by definition:
 * $\norm{x_{n+1} - x_n}_p \le \dfrac 1 {p^n}$

By Inequality Rule for Real Sequences then:
 * $\displaystyle \lim_{n \to \infty} \norm{x_{n+1} - x_n}_p = 0$.

By 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:
 * $\displaystyle \lim_{n \to \infty} x_n^2 = c^2$

Hence:
 * $c^2 = a$.

It follows that $c^2 \in \Z$

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

Since $a \neq 0$ then $c \neq 0$

Let $d = \size c$

Then $d \in \Z_{\gt 0}$

From above it follows that $d^2 = c^2 = x_1^2 + p$

Hence:

Also $d - x_1 \in \Z$ and $d + x_1 \in \Z$

So $d - x_1$ and $d + x_1$ are factors of $p$

The factors of $p$ by definition are:
 * $\pm 1$ and $\pm p$

Since $d, x_1 \in \Z_{\gt 0}$ then $d + x_1 \ge 2$

Hence $d + x_1 = p$

It follows that $d - x_1 = 1$

That is, $d = x_1 + 1$

Then

This contrdicts the previous conclusion that $c + x_1 = p$

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