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Numbers of the form $ \dfrac{m}{2^{n}} $, where $ m $ is an integer and $ n $ is a non-negative integer, are called dyadic rational numbers.

How can one show that the dyadic rationals are dense in $ \mathbb{R} $?

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As written here all dyadic rationals are positive. They can't be dense in $\mathbb R$. –  Asaf Karagila Dec 21 '12 at 9:48
    
I have edited the OP's formulation of the problem. –  Haskell Curry Dec 21 '12 at 10:34

4 Answers 4

up vote 3 down vote accepted

Hint: show that for any non-empty open interval $\,(a,b)\,$ there exist $\,m,n\in\Bbb Z\,$ s.t. $\,\displaystyle{\frac{m}{2^n}\in(a,b)}\,$ (why is this enough?)

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Hint: if I walk from left to right on the real line taking steps of fixed length $\delta > 0$ and start to the left of some interval $I = (a,b)$, then if $\delta < b-a$, I must set foot in the interval $I$.

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+1 This is as simple and nice a hint as I can think of, though it is almost the whole solution. –  DonAntonio Dec 21 '12 at 9:56

HINT: Let $x,y\in\Bbb R$ with $x<y$. There is an $n\in\Bbb N$ such that $2^{-n}<y-x$. Show that $(x,y)$ must contain an integer multiple of $2^{-n}$ and therefore some dyadic rational with denominator $2^m$ for some $m\le n$.

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Show that the smallest closed set containing the set of all such dyadic numbers is $\mathbb{R}$. Specifically: any arbitrary open neighborhood $U$ of a real number $q$ not dyadic must intersect nontrivially with the set of all dyadic numbers. So $q$ is a limit point. Can you show why the intersection contains something other than $q$?

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