Error in or another way of calculating $\frac{1}{2} \in \mathbb{Q}_3$ I want to find the $p$-adic expansion of $\frac{1}{2} \in \mathbb{Q}_3$. I begin with noting that $- \frac{1}{2} = \frac{1}{1-3} = \sum_{n=0}^{\infty} 3^n$. Therefore, $\frac{1}{2} = 1 - \frac{1}{2} = 2 + 3 + 3^2 + \ldots$. So far so good. However; Note that $$\frac{1}{2}= -4 \cdot \frac{1}{1-3^2} = -4 \sum_{n=0}^{\infty} 3^{2n} = -4(1 + 3^2 + 3^4 + \cdots ).$$ I'm relatively new at the $p$-adic theory, but as I understand it, the coefficients all have to be considered modulo $p$, so in this case $-4 = 2$. So where did I went wrong? 
As always, any help would be greatly appreciated. 
EDIT As pointed out, it's a really big error just to set $-4=2$. However, if one should choose to continue with the reasoning that $$1/2 = -4\sum_{n=0}^{\infty} 3^{2n},$$ how could one then find a $p$-adic expansion of $1/2$?
 A: This "gotcha" illustrates a feature that has deceived many people: yes, a $p$-adic integer has what looks like a sort of power series expansion $\sum_{n\ge 0} c_n\cdot p^n$ with $0\le c_n<p$, but these expansions do not behave stably under addition or multiplication. That is, regrouping or rearranging is required to put the result of such operations back in this form. Also, there are many other (convergent!) expansions of $p$-adic numbers (as this example illustrates), much like there are many other expansions of real numbers beside decimal expansions $\sum c_n/10^n$ with $0\le c_n\le 9$.
Altogether, the situation argues for not thinking of $p$-adic numbers as being "exactly" those special expansions, although that does give a transitional viewpoint. Instead, a $p$-adic integer is "represented" by any sequence of ordinary integers $a_n$, such that $a_{n+1}=a_n \mod p^n$. Correctly-enough, this makes things look a bit different from the real numbers.
A: If $a_k=b_k$ for all $k\ne n$ and $a_n=b_n+c$ then $\sum a_k3^k=3^nc+\sum b_k3^k$ so that $\sum a_k3^k\ne\sum b_k3^k$ if $c\ne0$, even if $3\mid c$. Thus you cannot use $-4=2$ just like that here.
