How do i show that $\{m/2^n:m,n\in\mathbb{R}\}$ is dense in $\mathbb{R}$? [closed]

Solution 1:

As pointed out in the comments the question is trivial as posed now, but if you want to show (what I think you meant) that $$ \left\{\frac{m}{2^n} \Bigm| m,n \in \mathbb{Z}\right\} $$ is dense in $\mathbb{R}$ then let $a,b \in \mathbb{R}$ where $a < b$. The idea here is to first expand the interval $[a,b]$ so be at least $1$ unit of length long and the collapsing that interval back down. We know that $\exists n \in \mathbb{Z}$ so that $$ (b-a)2^n > 1 \qquad (1) $$ thus we know that there is an $m \in \mathbb{Z}$ so that $m \in [a 2^n, b 2^n]$ (since the interval is of length greater than $1$). Now we have $$ a2^n < m < b2^n \implies a < \frac{m}{2^n} < b \qquad (2) $$ If you want specific values of $m,n$ in terms of $a,b$ then consider $$ n = \left\lfloor \log_2 \left( \frac{1}{b-a} \right) \right\rfloor + 1 $$ so that we have $(1)$ (since $x < \lfloor x \rfloor + 1$). Then put $$ m = \lfloor b 2^n \rfloor $$ so that with $(1)$ (and the fact that $x - 1< \lfloor x \rfloor$) we have $$ a 2^n < m \le b2^n $$ and of course if $b = \frac{m}{2^n}$ we're done, otherwise we indeed have $(2)$ thus we're done as well.

Solution 2:

I just give you the idea.

First: As noted in the comments above I assume that you are considering the set $$ \left\{\frac{m}{2^n} : n,m \in \mathbb{Z}\right\}. $$

Let $\epsilon >0$ be given. Let $x$ (reduced expression) be a rational number.

Now first pick $n$ such that

$$\frac{1}{2^n} < \epsilon/2.$$

That mans that for all $x\in \mathbb{Q}$ there is a $y\in \mathbb{Z}\frac{1}{2^n} \subseteq \mathbb{Q}$ such that $\lvert x - y\rvert < \epsilon$.

Hence your set is dense in $\mathbb{Q}$. And $\mathbb{Q}$ is ...

Solution 3:

I think that it should be $\{ \frac{m}{2^n} :m \in \mathbb{Z}, n \in \mathbb{Z} \}$.

Let $(a,b) \subset \mathbb{R}$ where $a<b$ be an open segment in $\mathbb{R}$. Then exists $n \in \mathbb{N}$ such that $|b-a|<\frac{1}{2^n}$. Divide $\mathbb{R}$ to subintervals $\mathbb{R}=\cup_{m=-\infty}^{\infty}[\frac{m}{2^n},\frac{m+1}{2^n})$. Then exists segment from this sum that its endpoint lay in $(a,b)$, because $(a,b)$ is shorter than each.