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

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## Theorem

Let $p$ be a prime number.

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

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 $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ 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 $p$.

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 $l = -n$.

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

and

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

Now:

 $\ds m$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds d_m$ $\ne$ $\ds 0$ Definition of $p$-adic Expansion $\ds \leadsto \ \$ $\ds m$ $=$ $\ds l$ Definition of $l$ $\ds \leadsto \ \$ $\ds l$ $<$ $\ds 0$

Similarly:

$k < 0 \leadsto k = l$

### Proof of statement $(1)$

#### Case $l < 0$

Let $l < 0$.

Then:

 $\ds l$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds m$ $<$ $\ds 0$ as $m \le l$ $\ds \leadsto \ \$ $\ds m$ $=$ $\ds l$ from above

Similarly:

$l < 0 \leadsto k = l$

So:

$m = l = k$.

$\Box$

#### Case $l \ge 0$

Let $l \ge 0$.

Then:

 $\ds l$ $\ge$ $\ds 0$ $\ds \leadsto \ \$ $\ds m$ $\not <$ $\ds 0$ as $m < 0 \leadsto l < 0$ $\ds \leadsto \ \$ $\ds m$ $=$ $\ds 0$ 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:

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

is a representative of $\mathbf a$

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

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:

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

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

Similarly the series:

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

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.

$\blacksquare$