Partial Sums of P-adic Expansion forms Coherent Sequence
Theorem
Let $p$ be a prime number.
Let $\ds \sum_{n \mathop = 0}^\infty d_n p^n$ be a $p$-adic expansion.
Let $\sequence {\alpha_n}$ be the sequence of partial sums; that is:
- $\forall n \in \N :\alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$.
Then $\sequence {\alpha_n}$ is a coherent sequence.
Proof
From the definition of a coherent sequence, it needs to be shown that $\sequence {\alpha_n}$ is a sequence of integers such that:
- $(1): \quad \forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
- $(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
That the sequence $\sequence {\alpha_n}$ is a sequence of integers follows immediately from the assumption that the series begins at $n = 0$ and so the terms of each summation are integers.
First note:
| \(\ds \forall n \in \N: \, \) | \(\ds \alpha_{n + 1}\) | \(=\) | \(\ds \ds \sum_{i \mathop = 0}^{n + 1} d_i p^i\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \ds \sum_{i \mathop = 0}^n d_i p^i + d_{n + 1} p^{n + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \alpha_n + d_{n + 1} p^{n + 1}\) |
Partial Sums satisfy (1)
This is proved by induction on $n$:
Basis for the Induction
$n = 0$
By definition: $\alpha_0 = d_0$
By definition of a $p$-adic expansion:
- $0 \le d_0 < p$
So shown for basis for the induction.
Induction Hypothesis
This is our induction hypothesis:
- $0 \le \alpha_k < p^{k + 1}$
Now we need to show true for $n=k+1$:
- $0 \le \alpha_{k + 1} < p^{k + 2}$
Induction Step
This is our induction step:
| \(\ds 0\) | \(\le\) | \(\ds \alpha_k + d_{k + 1} p^{k + 1}\) | as $\alpha_k, d_{k + 1}, p \ge 0$ | |||||||||||
| \(\ds \) | \(<\) | \(\ds p^{k + 1} + d_{k + 1} p^{k + 1}\) | induction hypothesis | |||||||||||
| \(\ds \) | \(\le\) | \(\ds p^{k + 1} + \paren{p - 1} p^{k + 1}\) | Definition of $p$-adic expansion means $0 \le d_{k + 1} < p$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds p \cdot p^{k + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds p^{k + 2}\) |
Since:
- $\alpha_{k + 1} = \alpha_k + d_{k + 1} p^{k + 1}$
then:
- $0 \le \alpha_{k + 1} < p^{k + 2}$
By induction:
- $\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
$\Box$
Partial Sums satisfy (2)
Let $n \in \N$.
Since:
- $\alpha_{n + 1} - \alpha_n = d_{n + 1}p^{n + 1}$
Then:
- $p^{n + 1} \divides \alpha_{n+1} - \alpha_n$
So:
- $\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$
The result follows.
$\blacksquare$
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The theorem is stated for the special set of $p$-adic expansions where the series index begins at $0$ and not the more general $m \in \Z_{\le 0}$.
From P-adic Integer has Unique P-adic Expansion Representative, it is seen that this set of $p$-adic expansions is indeed the $p$-adic integers.