# Math Help - Are algebraic integers dense in the reals?

1. ## Are algebraic integers dense in the reals?

Title says it all. Let $\mathbb{A}$ be the ring of algebraic integers; is $\mathbb{A} \cap \mathbb{R}$ dense in $\mathbb{R}$?
I bet "yes" is the answer.

Perhaps my question is trivial but I can't think of an easy answer.

P.S. I hesitated a bit between this forum and Analysis but I figured here was a better place.

2. Originally Posted by Bruno J.
Title says it all. Let $\mathbb{A}$ be the ring of algebraic integers; is $\mathbb{A} \cap \mathbb{R}$ dense in $\mathbb{R}$?
I bet "yes" is the answer.

Perhaps my question is trivial but I can't think of an easy answer.

P.S. I hesitated a bit between this forum and Analysis but I figured here was a better place.
I have just thought of a short proof. You should agree that, since $\mathbb{A}$ is a subgroup of $(\mathbb{R},+)$, it suffices to show that it contains elements arbitrarily close to 0.

Note that, for any $n,m\in\mathbb{N}$, $\sqrt{n}-m\in\mathbb{A}$ (it is a root of $X^2+2mX-n+m^2$). And that $\sqrt{n^2+1}-n=\frac{1}{\sqrt{n^2+1}+n}\to_n 0$. That's it.

3. I’d also say yes is the answer. If $\alpha\in\mathbb R,$ then for any real interval $I_\epsilon=(\alpha-\epsilon,\,\alpha+\epsilon),$ there is a rational number $\beta$ such that $\alpha<\beta<\epsilon;$ any rational number is algebraic, therefore $\beta\in\mathbb A\cap\mathbb R$ and $\beta\in I_\epsilon\setminus\{\alpha\}.$ So every open neighbourhood of any number $\alpha\in\mathbb R$ contains a point in $\mathbb A\cap\mathbb R$ other than $\alpha,$ i.e. $\mathbb A\cap\mathbb R$ is dense in $\mathbb R.$

4. Originally Posted by TheAbstractionist
I’d also say yes is the answer. If $\alpha\in\mathbb R,$ then for any real interval $I_\epsilon=(\alpha-\epsilon,\,\alpha+\epsilon),$ there is a rational number $\beta$ such that $\alpha<\beta<\epsilon;$ any rational number is algebraic, therefore $\beta\in\mathbb A\cap\mathbb R$ and $\beta\in I_\epsilon\setminus\{\alpha\}.$ So every open neighbourhood of any number $\alpha\in\mathbb R$ contains a point in $\mathbb A\cap\mathbb R$ other than $\alpha,$ i.e. $\mathbb A\cap\mathbb R$ is dense in $\mathbb R.$
Actually, the question is about "algebraic integers", i.e. roots of monic polynomials with integer coefficients. Notably, the only rational numbers in $\mathbb{A}$ are integers... But I had the same reaction as you at first

5. Originally Posted by Laurent
Actually, the question is about "algebraic integers", i.e. roots of monic polynomials with integer coefficients. I had the same reaction as you at first
Oops, I didn’t see that.

6. Very good Laurent! Indeed that is pretty short.