Additive inverse of elements in the p-adic numbers $\mathbb Q_p$ I am trying to find out if 
$-a = -1\cdot a$ 
where $-a$ is the additive inverse of $a=a_{-m}\cdot p^{-m}+\cdots + a_{-1}p^{-1} + a_0 + a_1\cdot p +\cdots$ in $\mathbb Q_p$ and $-1= (p-1)+(p-1)\cdot p + \cdots \in \mathbb Q_p$ 
but I am struggling to do $-1\cdot a$ as a general formal sum. Is there a nicer way to do this?
Thanks
 A: I think, from our exchange of comments, that I understand your question. Here’s the situation: we have the well-known rings $\Bbb Z\subset\Bbb Q$, and we know (from elementary-school math) that the inclusion is a ring homomorphism. That means, that whether “$\star$” is taken to be “$+$”, “$-$”, or multiplication, for integers $m$ and $n$, $m\star n$ as integers is the same as $m\star n$ as rational numbers. Now the question is, if $m$ and $n$ are integers, whether the integer $m\star n$ is the same as the $p$-adic number $m\star n$.
Your question is a serious one, and needs to be addressed whenever we design a completion like $\Bbb Q_p$ out of $\Bbb Q$. One addresses it by a formal proof, in which I guess the important ingredient is the Theorem that if the limits exist, then $\lim_n(a_n\star b_n)=\lim_na_n\star\lim_nb_n$, where again the star can represent addition, subtraction, multiplication, or (with obvious restrictions) division.
What makes things difficult for you is that you’ve been tied to what might be called the “standard representation” of $p$-adic numbers, rather than to a more abstract definition. It’s like insisting that a real number really is its decimal representation, and that to prove things about real numbers, you have to think of them as infinite decimals, and worse, deal with them as infinite decimals.
Let me give an example, dependent on a special version of the Binomial Series. The binomial expansion of $(1+t)^{1/2}$ turns out to start out this way:
$$
1 + \frac12t - \frac18t^2 + \frac1{16}t^3 - \frac5{128}t^4 + \frac7{256}t^5 - \frac{21}{1024} +\frac{33}{2048}t^7 - \frac{429}{32768}t^8 +\cdots\,,
$$
in which you notice (and can prove) that the denominators are all powers of $2$. This means that for any odd prime $p$,  $\sqrt{1+p\,}$ is an element of $\Bbb Q_p$, in fact of the $p$-adic integers $\Bbb Z_p$. The reason is that when you replace $t$ in the series by $p$, you get a $p$-adically convergent series of rational numbers, whose limit is, by the definition of completion, in $\Bbb Q_p$. But if you wanted to use this series and deal with the separate parts only by way of their standard representation, you’d be in a huge tub of hot water. It’s just not the way to think about $\Bbb Q_p$ and $\Bbb Z_p$.
