Let $\Gamma$ be a totally ordered abelian group (written additively), and let $K$ be a field. A valuation of $K$ with values in $\Gamma$ is a mapping $v:K^* \to \Gamma$ such that
$1)$ $v(xy)=v(x)+v(y)$
$2)$ $v(x+y) \ge \min (v(x),v(y))$,
for all $x,y \in K^*$.
Show that the set $A=\{x\in K^*:v(x) \ge 0\} \cup \{0\}$ is a valuation ring of $K$.
I want to show that $-1 \in A$. From $v(1)=v(1)+v(1)$, $v(1)=0$ and $v(1)=v(-1)+v(-1)$ so that $2v(-1)=0$. Is it possible to conclude that $v(-1)=0$?
In general for any totally ordered abelian group, does $2x=0$ imply $x=0$?