Results 1 to 6 of 6

Thread: Are algebraic integers dense in the reals?

  1. #1
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1

    Are algebraic integers dense in the reals?

    Title says it all. Let $\displaystyle \mathbb{A}$ be the ring of algebraic integers; is $\displaystyle \mathbb{A} \cap \mathbb{R}$ dense in $\displaystyle \mathbb{R}$?
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Bruno J. View Post
    Title says it all. Let $\displaystyle \mathbb{A}$ be the ring of algebraic integers; is $\displaystyle \mathbb{A} \cap \mathbb{R}$ dense in $\displaystyle \mathbb{R}$?
    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 $\displaystyle \mathbb{A}$ is a subgroup of $\displaystyle (\mathbb{R},+)$, it suffices to show that it contains elements arbitrarily close to 0.

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

  3. #3
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    I’d also say yes is the answer. If $\displaystyle \alpha\in\mathbb R,$ then for any real interval $\displaystyle I_\epsilon=(\alpha-\epsilon,\,\alpha+\epsilon),$ there is a rational number $\displaystyle \beta$ such that $\displaystyle \alpha<\beta<\epsilon;$ any rational number is algebraic, therefore $\displaystyle \beta\in\mathbb A\cap\mathbb R$ and $\displaystyle \beta\in I_\epsilon\setminus\{\alpha\}.$ So every open neighbourhood of any number $\displaystyle \alpha\in\mathbb R$ contains a point in $\displaystyle \mathbb A\cap\mathbb R$ other than $\displaystyle \alpha,$ i.e. $\displaystyle \mathbb A\cap\mathbb R$ is dense in $\displaystyle \mathbb R.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by TheAbstractionist View Post
    I’d also say yes is the answer. If $\displaystyle \alpha\in\mathbb R,$ then for any real interval $\displaystyle I_\epsilon=(\alpha-\epsilon,\,\alpha+\epsilon),$ there is a rational number $\displaystyle \beta$ such that $\displaystyle \alpha<\beta<\epsilon;$ any rational number is algebraic, therefore $\displaystyle \beta\in\mathbb A\cap\mathbb R$ and $\displaystyle \beta\in I_\epsilon\setminus\{\alpha\}.$ So every open neighbourhood of any number $\displaystyle \alpha\in\mathbb R$ contains a point in $\displaystyle \mathbb A\cap\mathbb R$ other than $\displaystyle \alpha,$ i.e. $\displaystyle \mathbb A\cap\mathbb R$ is dense in $\displaystyle \mathbb R.$
    Actually, the question is about "algebraic integers", i.e. roots of monic polynomials with integer coefficients. Notably, the only rational numbers in $\displaystyle \mathbb{A}$ are integers... But I had the same reaction as you at first
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Very good Laurent! Indeed that is pretty short.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Aug 23rd 2010, 08:23 AM
  2. Replies: 1
    Last Post: May 14th 2010, 01:53 AM
  3. Reals & Integers
    Posted in the Math Challenge Problems Forum
    Replies: 6
    Last Post: Jun 27th 2009, 12:49 AM
  4. Algebraic Integers trouble...
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: May 8th 2009, 10:02 AM
  5. Algebraic integers
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: Apr 11th 2009, 02:35 PM

Search Tags


/mathhelpforum @mathhelpforum