Coherent Sequence is Partial Sum of P-adic Expansion/Informal Proof

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Theorem

Let $p$ be a prime number.

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

Let $\sequence{\alpha_n}$ be a coherent sequence.


Then there exists a $p$-adic expansion of the form:

$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

such that:

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

Informal Proof

Consider the $\displaystyle \alpha_n$ written in base $p$.

To reduce an integer modulo $p^n$, it is simply a matter of stripping off all but the last $n$ digits.

So the coherence condition:

$\alpha_{n + 1} \equiv \alpha_n \mod{p^{n+1}}$

means that the last $n + 1$ digits of both integers are the same.

So the sequence $\sequence{\alpha_n}$ can be written:

\(\displaystyle \alpha_0\) \(=\) \(\displaystyle d_0\) $0 \le d_0 \le p - 1$
\(\displaystyle \alpha_1\) \(=\) \(\displaystyle d_0 + d_1 p\) $0 \le d_1 \le p - 1$
\(\displaystyle \alpha_2\) \(=\) \(\displaystyle d_0 + d_1 p + d_2 p^2\) $0 \le d_2 \le p - 1$
\(\displaystyle \alpha_3\) \(=\) \(\displaystyle d_0 + d_1 p + d_2 p^2 + d_3 p^3\) $0 \le d_3 \le p - 1$
\(\displaystyle \ldots\) \(\) \(\displaystyle \) $0 \le d_n \le p - 1$

Putting this together, we get the $p$-adic expansion:

$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

such that:

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

$\blacksquare$

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