P-adic norm, valuation, and expansion

[MostlyHarmless]

In this book I'm working from (P-adic numbers: An introduction by Fernando Gouvea), he gives an explanation of the p-adic expansion of a rational number which I'm pretty sure I understand this, but a bit later he talks about the p-adic absolute value, which from what I understand is the same as the norm. He defines the valuation like this:

...Define $v_p: ℤ-0\rightarrowℝ$ then for each integer $n≠0$ let $v_p(n)$ be the unique positive integer satisfying $n=p^{v_p(n)}n'$ with p not dividing n'...

Which I can understand that, my interpretation is that this is basically another way of decomposing an integer that emphasizes the prime factor that we care about. So $v_p(n)$ is just the exponent on p in the the prime factorization of n.

Then he discusses the p-adic absolute value, and in short he defines it as
$$
|x|_p=\left\{
\begin{array}{l}
|x|_p=p^{-\alpha}\text{ if } p^{\alpha}|x\\
|x|_p=1 \text{ if }p\not|\text{ } x
\end{array}
\right.
$$

I'm a little foggy as to how these 3 things are related. I'm pretty sure he defines the valuation the way he does for the proof that $|x|_p$ is a metric. But I don't see how the expansions are related.

 

[MostlyHarmless]

I've resolved this question. When we find the p-adic expansion we can directly read off the value of the p-adic norm. Numbers whose p-adic norms are equal to 1 correspond to the numbers who p-adic expansions have a non-zero constant term.