# Leigh.Samphier/Sandbox/P-adic Expansion Representative of P-adic Number is Unique

## Contents

## 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.

That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where:

- $\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$
- $\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.

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

Let $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $\mathbf a$.

Then:

- $(1) \quad m = k$
- $(2) \quad \forall i \ge m : d_i = e_i$

That is, the $p$-adic expansions $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ are identical.

## Proof

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

By Rational Numbers are Dense Subfield of P-adic Numbers:

- the constant sequence $\tuple {p, p, p, \dots}$ represents $\mathbf p \in \Q_p$.

From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:

- $\mathbf p^n \mathbf a \in \Z^\times_p$
- $\norm {\mathbf a}_p = p^n$

where $\Z^\times_p$ is the set of $p$-adic units.

Let $l = -n$.

From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:

- $l = \min \set {i: i \ge m \land d_i \ne 0}$

and

- $l = \min \set {i: i \ge k \land e_i \ne 0}$

Now:

\(\displaystyle m\) | \(<\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle d_m\) | \(\ne\) | \(\displaystyle 0\) | Definition of $p$-adic Expansion | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m\) | \(=\) | \(\displaystyle l\) | Definition of $l$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle l\) | \(<\) | \(\displaystyle 0\) |

Similarly:

- $k < 0 \leadsto k = l$

### Proof of statement $(1)$

#### Case $l < 0$

Let $l < 0$.

Then:

\(\displaystyle l\) | \(<\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m\) | \(<\) | \(\displaystyle 0\) | as $m \le l$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m\) | \(=\) | \(\displaystyle l\) | from above |

Similarly:

- $l < 0 \leadsto k = l$

So:

- $m = l = k$.

$\Box$

#### Case $l \ge 0$

Let $l \ge 0$.

Then:

\(\displaystyle l\) | \(\ge\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m\) | \(\not <\) | \(\displaystyle 0\) | as $m < 0 \leadsto l < 0$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m\) | \(=\) | \(\displaystyle 0\) | Definition of $p$-adic Expansion |

Similarly:

- $l \ge 0 \leadsto k = 0$

So:

- $m = 0 = k$.

$\Box$

### Proof of statement $(2)$

By definition of $l$, for all $i$ such that $m \le i < l$:

- $d_i = 0$
- $e_i = 0$

So for all $m \le i < l$:

- $d_i = e_i$.

Consider the series:

- $\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$

From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number

- $\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$

is a representative of $\mathbf a$

From Multiple Rule for Cauchy Sequences in Normed Division Ring:

- $\displaystyle p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$

is a representative of $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.

By definition of a $p$-adic expansion:

- $\forall i \in N : 0 \le d_{i + 1} < p - 1$

Thus the series:

- $\displaystyle \sum_{i \mathop = 0}^\infty d_{i + l} p^i$

is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.

Similarly the series:

- $\displaystyle \sum_{i \mathop = 0}^\infty e_{i + l} p^i$

is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.

From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion it follows that:

- $\forall i \in N : d_{i + 1} = e_{i + 1}$

That is:

- $\forall i \ge l : d_i = e_i$

The result follows.

$\blacksquare$