# Equivalence Class in P-adic Numbers Contains Unique P-adic Expansion

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## Theorem

Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.

Let $\mathbf a$ be an equivalence class in $\Q_p$.

Then $\mathbf a$ has exactly one representative that is a $p$-adic expansion.

## Proof

### Case 1

Let $\norm{\mathbf a}_p \le 1$.

From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion, $\mathbf a$ has exactly one representative that is a $p$-adic expansion of the form:

$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$
$\displaystyle \sum_{i \mathop = 0}^\infty e_i p^i$ is the only $p$-adic expansion that represents $\mathbf a$

$\Box$

### Case 2

Let $\norm{\mathbf a}_p > 1$.

Let $\textbf p$ denote the equivalence class that is identified with the prime number $p$.

By definition of the $p$-adic numbers as a quotient of Cauchy sequences:

the constant sequence $\tuple {p, p, p, \dotsc}$ represents $\mathbf p \in \Q_p$.
$\exists m \in \Z: \mathbf p^m \textbf a \in \Z_p^\times$

where $\Z_p^\times$ denotes the $p$-adic units.

By definition a $p$-adic unit is a $p$-adic integer.

$\mathbf p^m \textbf a$ has exactly one representative that is a $p$-adic expansion of the form:
$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$
$\norm {\mathbf p^m \textbf a}_p = 1 = p^0$
$d_0 \neq 0$

Let $\sequence{x_n}$ be a Cauchy sequence of rational numbers that represents $\mathbf a \in \Q_p$.

By definition of the $p$-adic numbers as a quotient of Cauchy sequences:

the constant sequence $\tuple {p^m, p^m, p^m, \dotsc}$ represents $\mathbf p^m \in \Q_p$.

By definition of the product in a quotient ring:

the Cauchy sequence $\sequence{p^m x_n}$ is a representative of $\mathbf p^m \mathbf a \in \Q_p$.
$\sequence {p^m x_n - \displaystyle \sum_{i \mathop = 0}^n d_i p^i}$ is a null sequence
$\sequence {x_n - \displaystyle p^{-m} \sum_{i \mathop = 0}^n d_i p^i}$ is a null sequence
$\sequence {\displaystyle p^{-m} \sum_{i \mathop = 0}^\infty d_i p^i}$ is a representative of $\mathbf a$.

That is:

$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a representative of $\mathbf a$

where:

$\forall i \ge -m: e_i = d_{i + m}$

By definition of a $p$-adic expansion;

$\forall i \ge 0: 0 \le d_i < p$

Then:

$\forall i \ge -m: 0 \le e_i = d_{i + m} < p$

Since $d_0 \neq 0$, then $e_{-m} = d_0 \neq 0$.

By definition of a $p$-adic expansion;

$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a $p$-adic expansion that represents $\mathbf a$
$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is the only $p$-adic expansion that represents $\mathbf a$

$\blacksquare$