# Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 4

## Theorem

Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.

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

Let $\sequence{\alpha_j}$ be a coherent sequence that represents $\mathbf a$.

Then:

$\sequence{\alpha_j}$ is the only coherent sequence that represents $\mathbf a$.

## Proof

Let $\sequence{\alpha'_j}$ be a coherent sequence not equal to $\sequence{\alpha_j}$.

From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence{\alpha_j - \alpha'_j}$ is not a null sequence.

Since $\sequence{\alpha'_j} \neq \sequence{\alpha_j}$ then:

$\exists i_0 \in \N : \alpha'_{i_0} \neq \alpha_{i_0}$

By definition of coherent sequences:

$0 \le \alpha_{i_0}, \alpha'_{i_0} < p^{i_0 + 1}$
$\alpha_{i_0} \not \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1}}$

By definition of a coherent sequence, for all $i > i_0$:

$\alpha_i \equiv \alpha_{i_0} \pmod {p^{i_0 + 1}}$
$\alpha'_i \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1}}$

Then:

$\forall i > i_0: \alpha_i \equiv \alpha_{i_0} \not \equiv \alpha'_{i_0} \equiv \alpha'_i \pmod {p^{i_0 + 1}}$

That is:

$\forall i > i_0: p^{i_0 + 1} \nmid \alpha_i - \alpha'_i$

By definition of the $p$-adic norm on integers:

$\forall i > i_0: \norm{\alpha_i - \alpha'_i} > \dfrac 1 {p^{i_0 + 1}}$

By definition of convergence, $\sequence{\alpha_j - \alpha'_j}$ is not a null sequence.

The result follows.

$\blacksquare$