if $x\in Q^p$, then there exists $-x\in Q^p$

Question

Show that if $x\in Q^p$, then there exists $-x\in Q^p$ where $$Q^p=\{a_{-l}p^{-l}+a_{-l+1}p^{-l+1}+...|l\in Z,a_i\in\{0,1,...,p-1\}\}$$ and p is a prime number.

Actually I don't quite understand p-adic numbers and how addition and multiplication work in this number system. For this question, I think I need to find a $y\in Q^p$ such that x+y=0 but I don't know how to start with this question.

Answer 1

Let $x\in\Bbb Q_p$. By definition, there exists a sequence of rationals $x_n\in \Bbb Q$ that converge to $x$ in the $p$-adic sense. What can you say about the convergence of the sequence $-x_n$?

Answer 2

Think "carries to the infinity" and try $$ y=\color{red}{(p-a_{-\ell})}p^{-\ell}+(p-1-a_{-\ell+1})p^{-\ell+1}+(p-1-a_{-\ell+2})p^{-\ell+2}+\cdots $$ Not unlike the usual addition of base ten integers (with carries) such as $$ 142857142857+85714285714\color{red}{3}=1000000000000. $$ Only this time we "push that leading $1$ out of bounds" (remember that in $p$-adics $p^n\to0$ when $n\to\infty$).