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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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    $\begingroup$ As written here all dyadic rationals are positive. They can't be dense in $\mathbb R$. $\endgroup$ – Asaf Karagila Dec 21 '12 at 9:48
  • $\begingroup$ I have edited the OP's formulation of the problem. $\endgroup$ – Haskell Curry Dec 21 '12 at 10:34
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So, basically, we have to show that the set of dyadic rationals are dense in $\mathbb{R}$. For this, if we show that the set $\left\lbrace \frac{m}{2^n}:m\in \mathbb{N}\cup\lbrace 0 \rbrace,n\in\mathbb{N}\text{ and }0\leq m \leq 2^n \right\rbrace$ is dense in $[0,1]$, then we are done. From this, one can easily extend $\textit{denseness}$ of the above numbers from $[0,1]$ into whole $\mathbb{R}$.

For the above purpose, let $\epsilon >0$, and let $x\in[0,1]$. From the Archimedean property, there is $n\in\mathbb{N}$ such that $\frac{1}{2^n}<\epsilon$. Let $m=\lfloor x\cdot2^n \rfloor$ where $\lfloor\cdot\rfloor$ represents Floor function. Then it is clear that $0\leq m \leq 2^n$. From the usual properties of Floor function, we have $$m=\lfloor x\cdot2^n \rfloor \leq x\cdot2^n \leq m+1=\lfloor x\cdot2^n \rfloor+1 $$, i.e., $$ \frac{m}{2^n}\leq x\leq \frac{m+1}{2^n} $$, i.e., $$ 0\leq x-\frac{m}{2^n}\leq \frac{1}{2^n} <\epsilon $$ which is exactly the statement we want to prove. Thus the set $\left\lbrace \frac{m}{2^n}:m\in \mathbb{N}\cup\lbrace 0 \rbrace,n\in\mathbb{N}\text{ and }0\leq m \leq 2^n \right\rbrace$ is dense in $[0,1]$. Now just extend this to whole $\mathbb{R}$.

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