Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 4

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $y $ be a rational $p$-adic integer.

Let $\ldots d_n \ldots d_2 d_1 d_0$ be the canonical expansion of $y$.

Let:

$y = \dfrac a b : a \in \Z, b \in Z_{> 0}$

Let:

$\forall n \in \N: \exists A_n, r_n \in \Z$:
$(\text a) \quad \dfrac a b = A_n + p^{n + 1} \dfrac {r_n} b$
$(\text b) \quad 0 \le A_n \le p^{n + 1} - 1$
$(\text c) \quad \ds \lim_{n \mathop \to \infty} A_n = \dfrac a b$


Then:

$\forall n \in \N: r_n = d_{n + 1} b + p r_{n + 1}$


Proof

We have:

\(\ds \forall n \in \N: \, \) \(\ds \dfrac a b\) \(=\) \(\ds A_n + p^{n + 1} \dfrac {r_n} b\) by hypothesis
\(\ds \) \(=\) \(\ds A_{n + 1} + p^{n + 2} \dfrac {r_{n + 1} } b\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds A_{n + 1} + p^{n + 2} \dfrac {r_{n + 1} } b\) \(=\) \(\ds A_n + p^{n + 1} \dfrac {r_n} b\)
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds A_{n + 1} - A_n\) \(=\) \(\ds p^{n + 1} \paren {\dfrac {r_n - p r_{n + 1} } b}\) rearranging terms
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds b \paren {A_{n + 1} - A_n}\) \(=\) \(\ds p^{n + 1} \paren {r_n - p r_{n + 1} }\) rearranging terms
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds b\) \(\divides\) \(\ds p^{n + 1} \paren {r_n - p r_{n + 1} }\)


From Characterization of Rational P-adic Integer:

$p \nmid b$

From Prime not Divisor implies Coprime:

$b, p$ are coprime

From Integer Coprime to all Factors is Coprime to Whole:

$b, p^{n+1}$ are coprime

Hence: $b \nmid p^{n + 1}$


From Euclid's Lemma:

$b \divides \paren {r_n - p r_{n + 1} }$

Then:

$\dfrac {r_n - p r_{n + 1} } b \in \Z$

Hence:

$A_{n + 1} \equiv A_n \pmod {p^{n + 1} }$


By definition of coherent sequence:

the sequence $\sequence{A_n}$ is a coherent sequence

From Coherent Sequence is Partial Sum of P-adic Expansion:

the sequence $\sequence{A_n}$ is the sequence of partial sums of a $p$-adic expansion $\ds \sum_{i \mathop = 0}^n e_i p^i$

Hence:

$y = \dfrac a b = \ds \sum_{n \mathop = 0}^\infty e_n p^n$

From P-adic Number is Limit of Unique P-adic Expansion:

$\forall n \in \N: e_n = d_n$

By definition of partial sums:

$\forall n \in \N: A_{n + 1} = A_n + d_{n + 1} p^{n + 1}$


Hence:

\(\ds \forall n \in \N: \, \) \(\ds d_{n + 1} p^{n + 1}\) \(=\) \(\ds A_{n + 1} - A_n\)
\(\ds \) \(=\) \(\ds p^{n + 1} \paren {\dfrac {r_n - p r_{n + 1} } b}\) a priori
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds d_{n + 1}\) \(=\) \(\ds \paren {\dfrac {r_n - p r_{n + 1} } b}\) dividing both sides by $p^{n + 1}$
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N: \, \) \(\ds r_n\) \(=\) \(\ds d_{n + 1} b + p r_{n + 1}\) rearranging terms

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Canonical_P-adic_Expansion_of_Rational_is_Eventually_Periodic/Lemma_4&oldid=657707"