TITLE: Density of ring of algebraic integers in $\mathbb C$
QUESTION [5 upvotes]: Clearly, $\mathbb Z$ is not dense in $\mathbb Q$ (and not dense in $\mathbb C$).
But why is the ring of algebraic integers $\overline{\mathbb Z}$ dense in $\overline{\mathbb Q}$?
In particular, if this is so, it is dense in $\mathbb C$.

REPLY [6 votes]: Let $\alpha$ be a real number, $\alpha>1$.  If
$$k=\lfloor \alpha^n\rfloor$$
then it is not hard to prove that
$$0\le \alpha-k^{1/n}\le\frac1n\ .$$
Since $k^{1/n}$ is an algebraic integer, and since you can choose $n$ as large as you like, there is an algebraic integer as close as you like to $\alpha$.  By subtracting a rational integer you can do the same for any real $\alpha$.  And finally, by doing the same for a real number $\beta$ you can get an algebraic integer as close as you like to the complex number $\alpha+i\beta$.