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This question already has an answer here:

Let $A=\{a+b\sqrt{2}\mid a,b\in\mathbb Z\}$ and let $Y\subset A$ with $x\in Y$ iff $x\in [0,1]$.

I'm trying to prove that $A$ is dense in $\mathbb R$ and already noticed that it is enough to show that $Y$ is dense in $[0,1]$. Could any one point me in a direction?

I have already looked at a proof of the density of $\mathbb Q$ in $\mathbb R$, where the core of the proof consisted of showing that between any two reals there is a rational. This was done by multiplying the reals with a natural number until the difference between them became bigger than one and noticing that you can find a rational between the two multiplied reals and divide that by the amount the reals got multiplied with and you get a rational between the two reals.

This method does not work identically in this case, because dividing an element in $A$ by a natural number won't always yield another element in $A$.

I will use this, when I have a proof, to prove that the quotient $\mathbb R/{\sim}$ with $r\sim s$ if $r-s=a+b\sqrt{2},\;\exists a,b \in \mathbb Z$ is not hausdorff. The density of every eq. clas in this quotient implies that the quotient topology must be the trivial one, hence the quotient cannot be hausdorff.

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marked as duplicate by Parcly Taxel, GNUSupporter 8964民主女神 地下教會, Xam, Semsem, A. Salguero-Alarcón Feb 23 '18 at 19:43

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Note that $0\lt \sqrt{2}-1\lt 1$, and that any integer power of $\sqrt{2}-1$ is of the form $a+b\sqrt{2}$ where $a$ and $b$ are integers.

Given any $\epsilon\gt 0$, we can find an integer $q$ such that $0\lt (\sqrt{2}-1)^q\lt \epsilon$. Now consider the numbers $n(\sqrt{2}-1)^q$, as $n$ ranges over the integers, positive, negative, and $0$. It is not hard to show that for any real number $x$, there is an integer $n$ such that the distance from $n(\sqrt{2}-1)^q$ to $x$ is $\lt \epsilon$.

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  • $\begingroup$ Thank you for your response. Just the little direction I needed. $\endgroup$ – bbnkttp Mar 23 '13 at 4:45
  • $\begingroup$ You are welcome. One can prove a stronger result. Let $\gamma$ be any irrational number. Then the set of numbers of the form $a+b\gamma$, where $a$ and $b$ range over the integers, is dense in the reals. $\endgroup$ – André Nicolas Mar 23 '13 at 4:48
  • $\begingroup$ But of course, that is the beauty in your response, you don't use $\sqrt{2}$ explicitly, therefore it can be replaced by any irrational $\endgroup$ – bbnkttp Mar 23 '13 at 5:08
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Another approach is to use this application of the pigeonhole principle to show that the numbers $b\sqrt2-\lfloor b\sqrt2\rfloor$ for $b\in\Bbb Z$ are dense in $[0,1]$: clearly $\{b\sqrt2-\lfloor b\sqrt2\rfloor:b\in\Bbb Z\}\subseteq A$. The same argument works with any irrational in place of $\sqrt2$.

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  • $\begingroup$ Looking 'mod 1' is how I intuitively convinced myself that the set is indeed dense, but did not know how to make it explicit, thanks! Too bad I cannot accept 2 answers and upvote both. $\endgroup$ – bbnkttp Mar 23 '13 at 5:06
  • $\begingroup$ @Sadar: Glad to help! $\endgroup$ – Brian M. Scott Mar 23 '13 at 5:06
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You can approximate $\sqrt{2}$ as closely as you want with rational numbers $-a/b$. If $\dfrac{-a}{b}$ is close to $\sqrt{2}$, then $a+b\sqrt{2}$ is close to $0$. If you can make $a+b\sqrt{2}$ as close as you want to $0$ be suitable choice of $a,b\in\mathbb Z$, then there is no neighborhood of $0$ that excludes all nonzero numbers of the form $a+b\sqrt{2}$. If the set in question were discrete, then it would have only finitely many members in every bounded interval.

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  • $\begingroup$ You are going directly to the hausdorffness? Then you should not forget that 0 is in the same eq. class as $a+b\sqrt{2}$ for any $a,b\in\mathbb Z$. How can this be rephrased for any real in $[0,1]$? $\endgroup$ – bbnkttp Mar 23 '13 at 4:36
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    $\begingroup$ @Sadar: I think the point is more along the lines that if you can make arbitrarily small numbers, then you can use them to get arbitrarily close to any number. $\endgroup$ – Hurkyl Mar 23 '13 at 4:45

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