Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 1

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Theorem

Let $p$ be a prime number.

Let $b \in \Z_{>0}$ such that $b$ and $p$ are coprime.

Let $a \in \Z$.

Then:

$\forall n \in \N: \exists r_n \in \Z : \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b} \in \set{0, 1, \ldots, p^{n + 1} - 1}$

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:

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

$p^{n + 1} \divides \paren{a c_n - A_n}$

By definition of divisor:

$\exists x_n \in \Z : x_n p^{n + 1} = a c_n - A_n$

Re-arranging terms:

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

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 $b$

\(\ds \leadsto \ \ \)

\(\ds A_n\)

\(=\)

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

rearranging terms

By definition of least positive residue:

$A_n \in \set{0, 1, \ldots, p^{n + 1} - 1}$

The result follows.

$\blacksquare$