Find countable dense subset of $D$, with $D\subset \mathbb{R}\setminus\mathbb{Q}$ If $D\subset \mathbb{R}\setminus\mathbb{Q}$. How can we find a countable dense subset of $D$?
 A: How about $\left\{q+\sqrt{2}\,\big|\,q\in\mathbb Q\right\}$?
A: Suppose $\emptyset\ne D\subseteq\mathbb R$. Choose $d\in D$. Let $\langle I_n:n\in\mathbb N\rangle$ be an enumeration of the open intervals with rational endpoints. Use the axiom of choice to get a sequence $\langle s_n:n\in\mathbb N\rangle$ such that $s_n\in D\cap I_n$ whenever $D\cap I_n\ne\emptyset$, otherwise $s_n=d$; and define $S=\{s_n:n\in\mathbb N\}$. Then $S$ is a countable dense subset of $D$.
A: Let’s see whether I can do this fairly efficiently, without getting egg all over me. Without loss of generality, we may assume that $D\subset[0,1]$. Now, for each $n\ge1$, consider the $2^n$ subintervals $I_{n,j}$ of length $1/2^n$, namely $[j/2^n,(j+1)/2^n]$, with $0\le j<2^n$. For $j,n$ such that $D\cap I_{n,j}\ne\emptyset$, let $x_{n,j}$ be a point of $D$ in this intersection. The set of all such $x_{n,j}$ is certainly countable, and seems to me to be dense in $D$.
A: Fix an irrational number $\alpha$. Denoting by a bar the fractional part of a real number, take the set $\{ m + \overline{n\alpha} \mid m,n\in \mathbf{Z}, n\neq0 \}$.
EDIT:
After clarification from Matt, I am revising my answer: it depends on the choice of $D$. Take $D$ to be all positive integral multiples of $\sqrt2$, This consists of irrationals , and  is a discrete set.
