Density of ring of algebraic integers in $\mathbb C$

Question

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 $\mathbb C$.

Answer 1

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 $\alpha+i\beta$.