P-adic Expansion Representative of P-adic Number is Unique

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 $\Q_p$.

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
From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:
 * $\mathbf p^n \mathbf a \in \Z^\times_p$
 * $\norm a_p = p^n$

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:
 * $d_l$ is the first index $i \ge m$ such that $d_i \neq 0$
 * $e_l$ is the first index $i \ge k$ such that $e_i \neq 0$

Now:

Similarly:

Case $l < 0$
Let $l < 0$.

Then:

Similarly:

So:
 * $m = l = k$.

Case $l \ge 0$
Let $l \ge 0$.

Then:

Similarly:

So:
 * $m = 0 = k$.

Proof of statement $(2)$
By definition of $l$, for all $i$ such that $m \le i < l$ then
 * $d_i = 0$
 * $e_i = 0$

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 Leigh.Samphier/Sandbox/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:
 * $\sum_{i \mathop = 0}^\infty d_{i+l} p^i$

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

Similarly the series:
 * $\sum_{i \mathop = 0}^\infty e_{i+l} p^i$

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.