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 $x_1 \in \Z: b \le x_1 \lt p$

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

Let $a = {x_1}^2 + p$

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

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

From above:
 * $p \nmid x_1$
 * $p \nmid 2$

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

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:
 * $\norm{x_n^2 - a}_p \le \dfrac 1 {p^n}$