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Number Theory 1: Fermat's Dream asks the reader to verify the following. They then use this to extend the definition of Ord$_p$ to $\mathbb{Q}_p$.

Let $p$ be prime and $a\neq 0$.

If $(x_n)_{n\geq 1}$, $n \in \mathbb{N}$ is a p-adic Cauchy sequence of rationals whose class is $a \in \mathbb{Q}_p$, show that ord$_p(x_n)$ is constant for sufficiently large $n$.

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I may be missing something, but I think you require $a\neq 0$ for this to be true. Maybe the problem really says $\mathbb Q_p^\times$? – Thomas Andrews Jan 31 '13 at 20:03
Correct. I edited that. Thanks. – user43666 Jan 31 '13 at 20:13
When $a\neq0$, the distance from $x_n$ to $a$ will eventually fall below $|a|_p$. What does the non-archimedean triangle inequality say about $|x_n|_p=|(x_n-a)+a|_p$ after that point? – Jyrki Lahtonen Jan 31 '13 at 21:23

If $(x_n)$ is a sequence of rational numbers that doesn't converge to zero, then the sequence of valuations, $|x_n|_p$ is constant for sufficiently large $n$.

Proof: $(x_n)$ does not tend to zero, so there exists a natural number $N$ and rational $\varepsilon > 0$ such that $n \ge N$ implies $|x_n| > \varepsilon$. $(x_n)$ fulfills the Cauchy condition for convergence, so there exists another natural number $N'$ for which the following statement is true: if $n, m \ge N'$ then $|x_n - x_m| < \varepsilon$.

Now set $M$ to be the larger of these two numbers, $N$ and $N'$ and notice, that from $n, m \ge M$ it follows that $|x_n - x_m| < \max(|x_n|, |x_m|)$. Finally the non-archimedean property gives $|x_n| = |x_m|$.

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