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

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Theorem

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p > 3$.


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

Let $p > 3$.

Then there exists $a \in \Z: 1 < a < p-1$.

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

Let $n \in \N$.

Then:

$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$

From Euler's Theorem (Number Theory): Corollary $1$:

$a^{p^n \paren {p - 1} } - 1 \equiv 0 \pmod {p^n}$

so:

$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$


That is:

$\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$

By Characterisation of Cauchy Sequence in Non-Archimedean Norm

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


Aiming for a contradiction, suppose $\sequence {x_n}$ converges to some $x \in \Q$.

That is:

$x = \ds \lim_{n \mathop \to \infty} x_n$

By Modulus of Limit on a Normed Division Ring:

$\ds \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$

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

$ \norm {x_n }_p = 1$

So:

$\norm x_p = \ds \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$

By Axiom (N1) of a norm on a division ring then:

$x \ne 0$.


Since:

\(\ds x\) \(=\) \(\ds \lim_{n \mathop \to \infty} x_n\) Definition of $x$
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} x_{n + 1}\) Limit of Subsequence equals Limit of Sequence
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {x_n}^p\) Definition of $x_n$
\(\ds \) \(=\) \(\ds \paren {\lim_{n \mathop \to \infty} x_n}^p\) Product Rule for Sequences in Normed Division Ring
\(\ds \) \(=\) \(\ds x^p\) Definition of $x$

and $x \ne 0$ then:

$x^{p-1} = 1$

So:

$x = 1$ or $x = -1$

and so $a-x$ is an integer:

$0 < a - x < p$

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:

$\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$

That is:

$\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$

Let $n > N$:

\(\ds \norm {x - a}_p\) \(=\) \(\ds \norm {x - a^{p^n} + a^{p^n} - a}_p\)
\(\ds \) \(\le\) \(\ds \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}\) P-adic Norm on Rational Numbers is Non-Archimedean Norm


As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:

\(\ds \norm {x - a}_p\) \(=\) \(\ds \norm {a^{p^n} - a}_p\) Three Points in Ultrametric Space have Two Equal Distances
\(\ds \) \(=\) \(\ds \norm a_p \norm {a^{p^n - 1} - 1}_p\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(=\) \(\ds \norm {a^{p^n - 1} - 1}_p\) as $\norm a_p = 1$
\(\ds \) \(<\) \(\ds 1\) Fermat's Little Theorem: Corollary $4$

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

$\blacksquare$

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