The p-adic expansion of 1/2 for odd p

Question

I just started self-studying p-adic analysis, using Alain M. Robert's book A Course in p-adic Analysis. So, I wanted to make sure that I got this correct before continuing and developing bad habits (for some reason something seems off to me):

$$\begin{array}{l}\frac{1}{2} = {a_0} + {a_1}p + {a_2}{p^2} + ...\\\\\frac{1}{2} \equiv {a_0}(\bmod p)\\\Rightarrow{a_0} = \frac{{p + 1}}{2}\\\\\frac{1}{2} \equiv {a_0} + {a_1}p(\bmod {p^2}) = \frac{{p + 1}}{2} + {a_1}p(\bmod {p^2}) \\\Rightarrow - \frac{p}{2} \equiv {a_1}p(\bmod {p^2}) \\\Rightarrow - \frac{1}{2} \equiv {a_1}(\bmod p)\\ \Rightarrow {a_1} = \frac{{p - 1}}{2}\\\\\frac{1}{2} \equiv {a_0} + {a_1}p + {a_2}{p^2}(\bmod {p^3}) = \frac{{p + 1}}{2} + \frac{{p - 1}}{2}p + {a_2}{p^2}(\bmod {p^3}) \\\Rightarrow - \frac{p}{2} \equiv \frac{{p - 1}}{2}p + {a_2}{p^2}(\bmod {p^3})\\ \Rightarrow - \frac{1}{2} \equiv \frac{{p - 1}}{2} + {a_2}p(\bmod {p^2}) \\\Rightarrow - \frac{p}{2} \equiv {a_2}p(\bmod {p^2}) \\\Rightarrow - \frac{1}{2} \equiv {a_2}(\bmod p)\\ \Rightarrow {a_2} = {a_1} = \frac{{p - 1}}{2}\\\\ \Rightarrow \frac{{p - 1}}{2} = {a_1} = {a_2} = {a_3} = ...\\\\\therefore\frac{1}{2} = \frac{{p + 1}}{2} + \frac{{p - 1}}{2}p + \frac{{p - 1}}{2}{p^2} + ...\end{array} $$

EDIT 1: Oh wait... I just realized that if I simply regroup the terms on the RHS (it is a telescoping sum) everything but 1/2 cancels. NOW I understand this stuff.

But now I have a new question... Is this expansion unique? It seems like one could find similar expansions... Am I wrong?

EDIT 2: The bit that I was asking about uniqueness was pertaining to the fact that there are other models of the p-adics depending on where we choose to place our "origin" (for lack of a better word) in the ordered group Z/pZ, or radix, for instance, there is the symmetric construction where we take coefficients in {-(p-1)/2,...,(p-1)/2}. Clearly there are (p-1) of these representations for a given element in the p-adic integers, but then by symmetry (p-1)/2 of these representations are distinct. Now, if we assume that the element is written in simplest terms, then all the models are equivalent, independently of our choice in "origin". This last part is what I was failing to realize. In other words, the sum is still telescoping, the coefficients are simply shifted.

Answer 1

Checking your calculations...

The first meaningful representation is $$-1= (p-1) + (p-1)p + (p-1) p^2 + \cdots $$ (think "odometer"),

so if $p$ is odd $$-\frac{1}{2}= \frac{(p-1)}{2} + \frac{(p-1)}{2}p + \frac{(p-1)}{2} p^2 + \cdots $$ now add $1$ to both sides.... seems OK.

Answer 2

It's a theorem that any element $x\in\mathbb{Q}_p$, has a unique representation as a finite-tailed Laurent series

$$\sum_{i\ge n}^\infty a_i p^i$$

Here $n$ can be negative, and the $a_i$'s are all integers between $0$ and $p-1$.

If you really want to understand this you should probably try to understand the statement: $\mathbb{Q}_p$ is the fraction field of the completion of $\mathbb{Z}$ at the ideal $(p)$. It's pretty clear from the definition of completions that you can write elements of $\mathbb{Z}_p$ as power series, and from general manipulations with power series, it's clear that inverses of power series are finite-tailed laurent series.