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Math Help - Are algebraic integers dense in the reals?

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    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.
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    Quote Originally Posted by Bruno J. View Post
    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.
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    Senior Member TheAbstractionist's Avatar
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    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.
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    Quote Originally Posted by TheAbstractionist View Post
    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
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Laurent View Post
    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 didnít see that.
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    MHF Contributor Bruno J.'s Avatar
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    Very good Laurent! Indeed that is pretty short.
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