distance between a real and R\Q Please how to prove that $d(x, R\setminus Q)=0, d(x,Q)=0$ for all $x\in 
\mathbb{R}$ ?
I know that $d(x,R\setminus Q)=\inf_{a\in R\setminus Q} d(x, a)$ but how to continue ? 
Can i say that As $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}$ and $d(x, \overline{A})=d(x,A)$ then $d(x,\mathbb{R}\setminus \mathbb{Q})=d(x,\mathbb{R}) =0$
Thank you
 A: Both $\mathbb{Q}$ and $\mathbb{R-Q}$ are dense in $\mathbb{R}$. Thus for all $x\in\mathbb{R},\epsilon>0$, there exists an $a\in\mathbb{Q}$ and a $b\in\mathbb{R-Q}$ such that $d(x,a)<\epsilon$ and $d(x,b)<\epsilon$. The result follows.
A: To show that $d (x, \mathbb{Q})=0$ for every $x \in \mathbb{R}$ what you need to show is that for every $\epsilon >0$ there is a rational number such that $|x-y|< \epsilon$. I assume you are aware of the fact that every real number is the limit of a sequence of rational numbers. This is basically the same thing. 
To show that $d (x, \mathbb{R} \setminus \mathbb{Q})=0$ for every $x \in \mathbb{R}$ what you need to show is that for every $\epsilon >0$ there is an irrational number such that $|x-y|< \epsilon$.
This is more or less the same as above, but I might set up the proof differently. For $x$ is irrational the assertion is clear (take $x=y$), and 
for rational $x$ consider $x + \sqrt{2}/n$ for growing integral $n$ (or take whatever irrational instead of $\sqrt{2}$). 
