P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number
Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ be a $p$-adic expansion that represents $a$.
Let $l$ be the first index $i \ge m$ such that $d_i \ne 0$
Then the series:
- $\ds \sum_{i \mathop = l}^\infty d_i p^i$
also represents $a$.
Proof
Let $\sequence {\alpha_n}$ be the sequence of partial sums:
- $\ds \forall n \in \N: \alpha _n = \sum_{i \mathop = 0}^n d_{n + m} p^{n + m}$
Let $\sequence {\beta_n}$ be the sequence of partial sums:
- $\ds \forall n \in \N: \beta _n = \sum_{i \mathop = 0}^n d_{n + l} p^{n + l}$
Then:
\(\ds \beta_n\) | \(=\) | \(\ds \sum_{i \mathop = 0}^n d_{n + l} p^{n + l}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = l}^{n + l} d_n p^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = m}^{l - 1} d_n p^n + \sum_{i \mathop = l}^{n + l} d_n p^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = m}^{n + l} d_n p^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^{n + l - m} d_{n + m} p^{n + m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha_{n + l - m}\) |
By definition of $l$:
- $m \le l$
So:
- $\forall n \in \N : n + l - m \ge n$
Thus $\sequence {\beta_n}$ is a subsequence of $\sequence {\alpha_n}$ by definition.
From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
- $\ds \sum_{i \mathop = l}^\infty d_i p^i$ is a Cauchy Sequence in $\Q$.
From Subsequence is Equivalent to Cauchy Sequence:
- $\ds \lim_{n \mathop \to \infty} {\alpha_n - \beta_n} = 0$
That is, the sequence $\sequence {\alpha_n - \beta_n}$ is a null sequence.
By definition of $p$-adic number:
- $\sequence {\alpha_n}$ and $\sequence {\beta_n}$ represent the same $p$-adic number
Since $\sequence {\alpha_n}$ represents $a$, it follows that $\sequence {\beta_n}$ represents $a$.
$\blacksquare$