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Since rational numbers are real algebraic numbers, clearly the real algebraic numbers must be dense in $\mathbb{R}$. So it seems natural to ask: are the (complex) algebraic numbers dense in $\mathbb{C}$? If not, is the (topological) closure of the algebraic numbers known?

user829347
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    Here's a proof: the algebraic complex numbers are closed under addition, and for every algebraic real $\alpha$ both $\alpha+0i$ and $0+\alpha i$ are algebraic complex numbers. So $\alpha+\beta i$ is an algebraic complex number whenever $\alpha,\beta$ are algebraic real numbers, and so the density of the algebraic complex numbers in $\mathbb{C}$ follows from the density of the algebraic real numbers in $\mathbb{R}$. – Noah Schweber Jun 15 '21 at 18:20
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    (Note that this proof isn't quite as immediate as it may first appear since the fact that algebraicity is closed under addition used in the first sentence is surprisingly nontrivial - see e.g. here.) – Noah Schweber Jun 15 '21 at 18:20

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