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.

Thanks in advance

Asymptotic formula for number of square-free numbers congruent to $1$ modulo $p$


Knowing that $ \sum_{n\leq x}\mu^2(n)=\frac{6}{\pi^2}x+O(\sqrt{x})$ , prove that:

$ $ \sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)=\frac{6}{\pi^2(p-1)}x+O(\sqrt{x})$ $

Using Dirichlet charaters, we have: \begin{align*} \sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\sum_{\chi(\text{mod }p)}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{\chi(\text{mod }p)}\sum_{n\leq x}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)+\underbrace{\frac{1}{\varphi(p)}\sum_{\chi\neq \chi_0}\sum_{n\leq x}\mu^2(n)\chi(n)}_{=:\Delta(x)} \end{align*} Where $ \chi_0$ is the principal character. I was able to prove that $ \Delta(x)=O(\sqrt{x})$ basically by using the fact that $ \mu^2(n)=\sum_{d|n}\mu(d)$ , and that $ \left|\sum_{n\leq x}\chi(n)\right|\leq \varphi(p)$ . My problem is the main term, which doesn’t match my calculation.

By definition, $ \chi_0(n)=1$ if $ p\not| n$ and $ \chi_0(n)=0$ if $ p|n$ . Therefore: \begin{align*} \frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq x,\,\,p|n}\mu^2(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq \frac{x}{p}}\mu^2(n)\ &=\frac{1}{\varphi(p)}\left(\frac{6}{\pi^2}x-\frac{6}{\pi^2}\frac{x}{p}\right)+O(\sqrt{x})\ &=\frac{6x}{\pi^2\varphi(p)}\left(1-\frac{1}{p}\right)+O(\sqrt{x})\ &=\frac{6}{\pi^2p}x+O(\sqrt{x}) \end{align*} What am I missing?