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So we got $ \exists k , l \in \mathbb{Z} $ such that $ kp+lq=1 $. So $ kp$ and $lq $ are relatively prime (Bezout ) .


$$ (ab)^{lq}=a^{1-kp}b^{lq}=a a^{-kp}=a=x^{lq}\\
(ab)^{kp}=a^{kp}b^{1-ql}=b=x^{pk} $$

Let $ A \subseteq \cup H_{i \in I } $ with $ H_{i} $ be open sets . So there existis some $ k $ such that $ x \in H_{k} $ so there exists some $ c>0 $ such that $$ B(x,c) \subseteq H_{k} $$ Since $ x_{n} $ converges to $ x $ there exists some $ N(c) $ such that if $ n \geq N(c) $ $$ x_{n} \in B(x,c)...

suppose there is one $$ p \in R-Q $$ such that : $$ f(p) \neq p $$ so for every $$ c>0 $$ the neiborhood $$ (p-c,p+c) $$ contains at least two rational numbers one of them $$ r_{1} \in (p-c,p) $$ and the other one $$ r_{2} \in (p,p+c) $$ we can see that $$ f(r_{1})<f(p)<f(r_{2}) \leftrightarrow r_{1...