P-adic Integer has Unique Coherent Sequence Representative/P-adic Expansion

Theorem
Let $p$ be a prime number.

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

Let $\mathbf a$ be an equivalence class in $\Q_p$ such that $\norm{\mathbf a}_p \le 1$.

Then $\mathbf a$ has exactly one representative that is a $p$-adic expansion of the form:
 * $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

Proof
From Equivalence Class in P-adic Integers Contains Unique Coherent Sequence, $a$ has exactly one representative coherent sequence.

Let $\sequence{\alpha_n}$ be the coherent sequence such that:
 * $\sequence{\alpha_n}$ is a representative of $\mathbf a$.

From Coherent Sequence is Partial Sum of P-adic Expansion, there exists a $p$-adic expansion of the form:
 * $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

such that:
 * $\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$

Then $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ is a representative of $\mathbf a$.

Let $\displaystyle \sum_{n \mathop = 0}^\infty d'_n p^n$ be a $p$-adic expansion that is a representative of $\mathbf a$.

From Partial Sums of P-adic Expansion forms Coherent Sequence, the partial sums:
 * $\forall n \in \N: \alpha'_n = \displaystyle \sum_{i \mathop = 0}^n d'_i p^i$

are a coherent sequence that is a representative of $\mathbf a$.

Since $\sequence{\alpha_n}$ is the only coherent sequence that is a representative of $a$, then:
 * $\sequence{\alpha'_n} = \sequence{\alpha_n}$

It follows that:
 * $\forall n \in \N : d'_n = d_n$

Then $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ is the only $p$-adic expansion that is a representative of $\mathbf a$.

Also see

 * P-adic Integer is Limit of Unique P-adic Expansion


 * Equivalence Class in P-adic Integers Contains Unique Coherent Sequence


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