Difference of Consecutive terms of Coherent Sequence
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Theorem
Let $p$ be a prime number.
Let $\sequence {\alpha_n}$ be a coherent sequence.
Then:
- for all $n \in \N_{>0}$ there exists $c_n \in \N$ such that:
- $0 \le c_n < p$
- $\alpha_n - \alpha_{n - 1} = c_n p^n$
Proof
By definition of a coherent sequence:
- $\forall n \in \N_{>0}: \alpha_n \equiv \alpha_{n - 1} \pmod {p^n}$
That is:
- $\forall n \in \N_{>0}: \exists c_n \in \Z : \alpha_n - \alpha_{n - 1} = c_n p^n$
So it remains to show that:
- $\forall n \in \N_{>0} : 0 \le c_n < p$
Aiming for a contradiction, suppose for some $N \in \N_{>0}$:
- $c_N \ge p$
Then:
- $c_N p^N \ge p^{N + 1}$
By definition of a coherent sequence:
- $\alpha_{N - 1} \ge 0$
Then:
- $\alpha_N = c_N p^N + \alpha_{N - 1} \ge p^{N + 1}$
This contradicts the coherent sequence condition:
- $\alpha_N \le p^{N + 1} - 1$
So:
- $\forall n \in \N_{>0} : c_n < p$
Aiming for a contradiction, suppose for some $M \in \N_{>0}$:
- $c_M < 0$
Then:
- $-c_M \ge 1$
Now:
- $c_M p^M + \alpha_{M - 1} = \alpha_M \ge 0$
So:
- $\alpha_{M - 1} \ge -c_M p^M \ge p^M$
This contradicts the coherent sequence condition:
- $\alpha_{M - 1} < p^M$
So:
- $\forall n \in \N_{>0} : c_n \ge 0$
$\blacksquare$