Leigh.Samphier/Sandbox/Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 1

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Theorem

Let $p$ be a prime number.

Let $b \in \Z_{> 0}$:

$b, p$ are coprime

Let $a \in \Z$.


Then:

$\forall n \in \N: \exists A_n, r_n \in \Z$ :
$(1) \quad \dfrac a b = A_n + p^{n+1} \dfrac {r_n} b$
$(2) \quad 0 \le A_n \le p^{n+1} - 1$
$(3) \quad \dfrac {a - \paren{p^{n+1} - 1} b } {p^{n+1}} \le r_n \le \dfrac a {p^{n+1}}$


Proof

Let $n \in \N$.

From Integer Coprime to all Factors is Coprime to Whole:

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

From Integer Combination of Coprime Integers:

$\exists c_n, d_n \in \Z : c_n b + d_n p^{n+1} = 1$

Multiplying both sides by $a$:

$a = a c_n b + a d_n p^{n+1}$

Let $A_n$ be the least positive residue of $a c_n \pmod {p^{n+1}}$.

By definition of least positive residue:

$0 \le A_n \le p^{n+1} - 1$


Let $x_n$ be the multiple of $p^n$:

$a c_n = x_n p^{n+1} + A_n$


We have:

\(\ds a\) \(=\) \(\ds \paren{x_n p^{n+1} + A_n} b + a d_n p^{n+1}\)
\(\ds \) \(=\) \(\ds A_n b + \paren{b x_n + a d_n} p^{n+1}\) Rearranging terms


Let $r_n = b x_n + a d_n$.

Then:

\(\ds a\) \(=\) \(\ds A_n b + r_n p^{n+1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac a b\) \(=\) \(\ds A_n + p^{n+1} \dfrac {r_n} b\) Dividing both sides by $b$
\(\ds \leadsto \ \ \) \(\ds r_n\) \(=\) \(\ds \dfrac {a - A_n b} {p^{n+1} }\) Rearranging terms


We have:

\(\ds \) \(\) \(\ds 0 \le A_n \le p^{n+1} - 1\)
\(\ds \) \(\leadsto\) \(\ds - \paren{p^{n+1} - 1} \le - A_n \le 0\)
\(\ds \) \(\leadsto\) \(\ds - \paren{p^{n+1} - 1} b \le - A_n b \le 0\)
\(\ds \) \(\leadsto\) \(\ds a - \paren{p^{n+1} - 1} b \le a - A_n b \le a\)
\(\ds \) \(\leadsto\) \(\ds \dfrac {a - \paren{p^{n+1} - 1} b} {p^{n+1} } \le \dfrac {a - A_n b} {p^{n+1} } \le \dfrac a {p^n}\)
\(\ds \) \(\leadsto\) \(\ds \dfrac {a - \paren{p^{n+1} - 1} b} {p^{n+1} } \le r_n \le \dfrac a {p^{n+1} }\)

The result follows.

$\blacksquare$