# Valuation of congruent elements in a local division ring

Let $$K$$ be a complete local division ring (note $$v$$ its valuation). For $$x,y\in K$$ ($$y\ne0$$), one puts $$x^y=yxy^{-1}$$. Let $$r\in\mathbb N$$. Consider $$x,y\in K$$ and $$a,b\in K^*$$ such that $$v(x-y)\ge r$$ and $$v(a-b)\ge r$$. Do we have $$v(x^a-y^b)\ge r$$? In the commutative case, it is obvious but in the non-commutative case, I can not see the answer.

$$\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$$ Not necessarily.
Take $$K = \Q_3 + \Q_3 i + \Q_3 j + \Q_3 ij$$ with $$i^2=-1$$ and $$j^2=3$$, $$r=2$$, $$x=y=j$$, $$a=3i$$ and $$b=3(1+i)$$. Then $$v(j)=1$$, and the maximal order of $$K$$ is $$\Z_3+\Z_3 i+\Z_3 j + \Z_3 ij$$. We have $$a\equiv b \bmod 3$$ so that $$v(a-b) = 2$$, but $$x^a = -j$$ and $$y^b=ij$$, which are not congruent modulo $$3$$.
Basically the problem is that $$v(a-b)\ge r$$ gives you no information since from arbitrary $$a,b$$ you can always ensure that this condition holds by multiplying by a high power of the uniformiser of the center of $$K$$, which does not change $$x^a$$ and $$y^b$$.