How to construct a dense subset of $\mathbb R$ other than rationals.

Solution 1:

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.

Solution 2:

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 $\cup_{n \in \mathbb{Z}} (n + E)$.