How to construct a dense subset of $\mathbb R$ other than rationals. How to construct a dense subset say $A$, of the real numbers other than rationals? By dense I mean that there should be an element of $A$ between any two real numbers.
 A: Take any irrational number $\alpha$ and consider the set $E = \{n\alpha \bmod 1\ : n \in \mathbb{N}\}$.  By the equidistribution theorem this set is uniformly distributed (and thus must be dense) on $[0,1]$.  For a set dense on all of $\mathbb{R}$ take $\cup_{n \in \mathbb{Z}} (n + E)$.
A: Let's construct dense sets in $[0,1]$. We can then get a dense set in $\Bbb R$ by taking the union of dense sets for the intervals $[n, n+1]$ (a dense set for $[n,n+1]$ can be obtained from a dense set of $[0,1]$ by shifting).  To make things more interesting, we will find dense sets  for which, given any two distinct elements in the set there is a number between them that is not in the set.
For a dense set in $[0,1]$, you can take:
The irrationals in $[0,1]$. 
Or: take $[0,1]$. Take its midpoint $1/2$ to be an element in the, to be constructed, dense set. Then take the midpoints of $(0,1/2)$ and $(1/2,1)$ to be elements in the sense set.  Then take the midpoints of the four sets obtained by splitting the two prior sets in two... 
Or: use any similar, carefully done,   construction similar to the preceding example. For instance, you could successively split $[0,1]$ as above (always splitting the previous sets in   half), but choose  irrationals in each piece.
A: Let $a_n$ be a positive sequence going to zero then $p a_n$ with $p\in \mathbb{Z}$ is dense in $\mathbb{R}$.
A: The Liouville numbers are dense in $\mathbb{R}$.
A: I guess you have a countable dense set $A$ in mind, since otherwise you could just put $A:={\mathbb R}$. I don't know what your intuition of the real numbers is, but I assume that you are happy with the idea that a real number is an infinite decimal, like $34.5210071856\ldots\ $.
The set 
$$A\ :=\ \bigcup_{r=0}^\infty \left\{{k\over 10^r}\ \bigm| k\in{\mathbb Z}\right\}$$
of finite decimal fractions is a union of countable sets, therefore it is countable. Given any two  real numbers
$$\alpha:=a_0.a_1\, a_2\, a_3\,\ldots,\qquad\ a_0\in{\mathbb Z},\quad a_k\in\{0,1,\ldots,9\}\ \  (k\geq 1)$$
and $$\beta:=b_0.b_1\, b_2\, b_3\,\ldots,\qquad\ b_0\in{\mathbb Z},\quad b_k\in\{0,1,\ldots,9\}\ \  (k\geq 1)$$
with $\alpha<\beta$ there is a minimal $k\geq0$, call it $k$, such that $a_k<b_k$. Since we have assumed $\alpha<\beta$ the case 
$$(a_k, a_{k+1}, a_{k+2}\ldots)=(a_k,9,9,9,\ldots)\quad\wedge\quad (b_k , b_{k+1}, b_{k+2}\ldots)=(a_k+1,0,0,0,\ldots)$$
is excluded. There are a few cases to be distinguished, but all in all it is easy to produce a finite decimal expansion
$$\xi=x_0.x_1\, x_2\, x_3\,\ldots x_{k-1}\, x_k\ \in A$$
such that $\alpha<\xi<\beta$.
A: Take every real number with infinite number of occurences of the string '56787773' in the decimal expansion.
A: First note that you can "cheat" by taking any subset and take a union with the rationals.
Second note that you can always cheat by taking some real number $x$ and considering the set $\{x+q\mid q\in\mathbb Q\}$. If $x$ is irrational then the set is not the rationals.
Now more seriously, you can note that the irrationals ($\mathbb R\setminus\mathbb Q$) are dense, as well all the irrational algebraic numbers ($\sqrt2$ and such). More interestingly the set $\{\sin n\mid n\in\mathbb N\}$ is dense in $[0,1]$ so it can be stretched (or multiplied) into a dense set of $\mathbb R$.
However an important fact is that every countable dense linear order is isomorphic to the rationals, so if your dense set is countable it will not differ too much from the rationals.
A: For any fixed $s>0$,
$$\Bigg\{\sum_{n=1}^\infty\frac{a_n}{n^s}: \{a_n\}_{n=1}^\infty \text{ is a periodic sequence of integers.} \Bigg\}$$
Is a countable set which is dense in the reals. To demonstrate this we can apply checkmath's answer. 
A: Use the axiom of choice to choose one element from each open interval.  (I suppose that with this method there's no guarantee you won't get the rationals.)
