Show that this set $B=${$\frac{p} {q}$, $p \in \mathbb{Z}, q \in \mathbb{N}$, $q=2^n, n \in \mathbb{N}$} is dense in $\mathbb{R}$

Question

I know a set $B$ is dense in $\mathbb{R}$ if an element of $B$ can be found between any two real numbers $a,b$ $s.t.$ $a<b$. I have an inkling that, as $n \rightarrow \infty$, $q=2^n \rightarrow \infty$, and we can play with the values of $\dfrac {p} {q}$ such that for any $a,b \in \mathbb{R}$, $0<b-a< \dfrac {p} {q}$.

Exactly how is a good way to show this? I can start with an interval $(0,1)$ and find a $\dfrac {p} {q}=1/2 $, which is in $(0,1)$. I can shrink the interval and find a $q$ sufficiently large such that $0<b-a< \dfrac {p} {q}$.

I don't think this constitutes a proof but perhaps I'm on the right track? Any suggestions? Thanks!

Possible duplicate of How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals? — Viktor Vaughn, Feb 04 '16 at 06:43
@SpamIAm, I always search first, for some reason this one did not come up in a standard search. I was also looking for feedback on my own, unique thoughts. Thanks. — Jabernet, Feb 04 '16 at 06:46
Fair enough. The term I used was "dyadic number" while searching. Here is another thread. — Viktor Vaughn, Feb 04 '16 at 06:49
@SpamIAm,I have not encountered the word "dyadic" and the answers provided in the linked threads (thanks, by the way, I do appreciate it) confuse me more than help me. I am relatively new to some of these concepts and am struggling a bit. Leaving this thread up for a bit will be beneficial to me, anyway. — Jabernet, Feb 04 '16 at 06:54

score 2 · Accepted Answer · answered Feb 04 '16 at 06:51

Let $a,b\in\mathbb{R}$ such that $a<b$. There is $n\in\mathbb{N}$ such that $0<\frac{1}{2^n}<\frac{b-a}{3}$.

Moreover, $2^na<[2^na]+1=[2^na+1]\le 2^na+1$. Thus

$$a<\frac{[2^na+1]}{2^n}\le a+\frac{1}{2^n}<a+\frac{b-a}{3}<b$$

Then, every open interval $(a,b)$ has a number of the form $\frac{[2^na+1]}{2^n}$.

Please, note that the only property we use was $1<2$ for found $n$

score 1 · Answer 2 · answered Feb 04 '16 at 06:40

1

Hint: The numbers of the form $\frac{p}{2^n}$ where $p,n$ are integers are regularly spaced like the markings of a ruler. If you want to guarantee that one of the markings is between $a,b$, how fine divisions do you need?

answered Feb 04 '16 at 06:40

user21820

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@user21312: Yes you can do it with any integer $q > 1$, since nothing I stated relies on the $2$. – user21820 Jun 16 '18 at 02:40

Show that this set $B=${$\frac{p} {q}$, $p \in \mathbb{Z}, q \in \mathbb{N}$, $q=2^n, n \in \mathbb{N}$} is dense in $\mathbb{R}$

2 Answers2