Let $R\hspace{.005 in}$ be a division ring. $\;\;$ Let $\:\leq\:$ be a total order on $R\hspace{.005 in}$ such that for all elements $x,y,z$ of $R$ :

if $\: x\leq y \:$ then $\:\: x+z\:\leq\:y+z \:\:$

if $\;\;\; 0\leq x \:$ and $\: 0\leq y \;\;\;$ then $\:\: 0\:\leq\:x\cdot y \:\:$

Define $\mathcal{T}\hspace{.05 in}$ to be the order topology on $R\hspace{.005 in}$.

Define $\;\; f \: : \: (R-\{0\}) \: \to \: (R-\{0\}) \;\;$ by $\;\; f(x)\cdot x \: = \: 1 \: = \: x\cdot f(x) \;\;$.

Does it follow that $f\hspace{.02 in}$ is continuous with respect to the subspace topology from $\: \langle R\hspace{.01 in},\mathcal{T}\hspace{.06 in}\rangle \:$ ?