P-adic Expansion is a Cauchy Sequence in P-adic Norm

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $p$ be a prime number.

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals numbers $\Q$.

Let $\ds \sum_{n \mathop = m}^\infty d_n p^n$ be a $p$-adic expansion.


Then the sequence of partial sums of the series:

$\ds \sum_{n \mathop = m}^\infty d_n p^n$

is a Cauchy sequence in the valued field $\struct{\Q, \norm{\,\cdot\,}_p}$.


Corollary 1

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic Numbers.


Then the sequence of partial sums of the series:

$\ds \sum_{n \mathop = m}^\infty d_n p^n$

represents a $p$-adic number of $\struct {\Q_p,\norm {\,\cdot\,}_p}$.


Corollary 2

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Then the sequence of partial sums of the series:

$\ds \sum_{n \mathop = m}^\infty d_n p^n$

converges to a $p$-adic number in $\struct{\Q_p, \norm{\,\cdot\,}_p}$.


Proof

Let $\sequence {s_N}$ be the sequence of partial sums defined by:

$\forall N \in \Z_{\ge m}: s_N = \ds \sum_{n \mathop = m}^N d_n p^n$


From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm:

the sequence $\sequence {s_N}$ is a Cauchy sequence if:
$\forall N \in \Z_{\ge m}: s_{N + 1} \equiv s_N \pmod {p^n}$


Now for all $N \in \Z_{\ge m}$:

\(\ds s_{N + 1} - s_N\) \(=\) \(\ds \sum_{n \mathop = m}^{N + 1} d_n p^ n - \sum_{n \mathop = m}^{N} d_n p^n\) Definition of Partial Sum
\(\ds \) \(=\) \(\ds d_{N + 1} p^{N + 1}\)
\(\ds \leadsto \ \ \) \(\ds s_{N + 1}\) \(\equiv\) \(\ds s_N \pmod {p^N}\) Definition of Congruence modulo $p^N$

The result follows.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=P-adic_Expansion_is_a_Cauchy_Sequence_in_P-adic_Norm&oldid=575727"