A proposition about valuation ring 
Q1
$x \in m\Rightarrow x~\text{is a element of an ideal}\Rightarrow ax~\text{is a element of an ideal}\Rightarrow ax~\text{is a non-unit}$
What's the mean of $(ax)^{-1} \in B$ ?
Q2
$x,y \in m\subseteq B,x\ne0,y\ne0\Rightarrow x^{-1},y^{-1}\notin B\Rightarrow xy^{-1} \text{and}~x^{-1}y \notin B $
Either $xy^{-1} \in B $ or $x^{-1}y \in B$?Why?
 A: Here is a question I asked myself that might help: Valuation rings and total order.
Q1) If $ax\notin\mathfrak{m}$, then $ax$ is not a nonunit in $B$. Thus $ax$ is a unit which means that its inverse $(ax)^{-1}\in B$.
Q2) This is closely related to my own question. Note that by definition of $B$ either $x$ or $x^{-1}$ is in $B$. Since $(x^{-1}y)^{-1}= y^{-1}x$, either one of those elements has to lie in $B$.
Let me know if you need more help;)
A: For Q1: you have to remember that all of this is taking place inside the field of fractions $K$ for $B$. Every nonzero element (in $B$ and outside of $B$) is invertible there.
So $(ax)^{-1}$ certainly exists (in $K$), but of course it does not have to be inside of $B$. For $ax\in m$, the existence of an inverse for $ax$ in $K\setminus B$ doesn't contradict the  fact that $ax$ isn't a unit of $B$. An element is a unit relative to a particular ring, just as $2$ is a unit in $\Bbb Q$ but not a unit in $\Bbb Z$, despite the fact that $\Bbb Z\subseteq \Bbb Q$.
For Q2: This is just observing that $(xy^{-1})^{-1}=(yx^{-1})$, and applying the definition of valuation rings that says at least one of $xy^{-1}$ and $(xy^{-1})^{-1}$ must lie in $B$.
