Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}$, is $S$ a dense set on $\mathbb{R}$?

share|cite|improve this question
Magic words are "dyadic rationals." – Arturo Magidin May 23 '12 at 3:29
Didn't a similar question with 10 as the denominator surface a couple of days ago? – copper.hat May 23 '12 at 3:49
@copper.hat: I believe that that one asked for a little more, namely, an explicit sequence converging to any given real. – Brian M. Scott May 23 '12 at 3:53
@BrianM.Scott: You're right:… – copper.hat May 23 '12 at 3:56
Choose $x \in \mathbb{R}$, and let $x_n = \frac{\lfloor 2^n x \rfloor}{2^n}$. Then $x_n \in S$, $\forall n$, and $|x-x_n| < \frac{1}{2^n}$. Hence $x_n \rightarrow x$, so $S$ is dense. – Riemann May 23 '12 at 4:09
up vote 8 down vote accepted

Hint: Every real number has a binary expansion.

share|cite|improve this answer

Yes, is it, given open interval $(a,b)$ (suppose $a$ and $b$ positives) you can find $n\in\mathbb{N}$ such that $1/2^n<|b-a|$. Then consider the set:

$$X=\{k\in \mathbb{N}; k/2^n > b\}$$

This is a subset of $\mathbb{N}$, for well ordering principe $X$ has a least element $k_0$ then is enought taking $(k_0-1)/2^n\in(a,b)$.

The same is if $a$, $b$ or both are negatives (because $(a,b)$ is bounded).

share|cite|improve this answer

This set looks really close to the rationals. Maybe you could use the density of the rationals and see if you can put an element of this set between any two rationals and go from there.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.