# Show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N} \}$ is dense in $\mathbb {R}$ [duplicate]

Possible Duplicate:
Density of a Set on $\mathbb{R}$?

I have to show that show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N}\}$ is dense in $\mathbb {R}$.

A set A is dense in $\mathbb {R}$ if $\overline A=\mathbb {R}$.

But also $Y$ is a subset of $X$, we say that $Y$ is dense in $X$, if for every $x\in X$ , there is $y \in Y$ that is arbitary close to $x$.

So ,I have to prove that for every $x \in \mathbb {R}$ ,there is a number $\frac{m}{2^n}$ arbitrarily close to $x$.So $\forall \epsilon,x ,\exists y$ such that $|y-x|<\epsilon$.

I got a little stuck at this point...Could anyone give me a hint?Thanks a lot!

-

## marked as duplicate by Ross Millikan, Leonid Kovalev, Zhen Lin, Chris Eagle, t.b.Aug 17 '12 at 10:44

Show that for $x$ in $\Bbb R$ and $n$ fixed, there is an interval $[m/2^n, (m+1)/2^n)$ to which $x$ belongs. Note you can force these intervals to have as small a diameter as you wish. –  David Mitra Aug 16 '12 at 22:35
More generally, an additive subgroup $G$ of $\mathbb R$ is either discrete or dense and this is decided by whether $\inf \{ x \in G : x>0 \}>0$. –  lhf Aug 16 '12 at 22:43
@QuinCulver Thanks!I was trying to find it.I knew that there could be this question,how I didn´t find it,I posted.Thanks for letting me know! –  HipsterMathematician Aug 17 '12 at 0:17

In order to prove that these numbers - called the dyadic rationals, I believe - are dense in $\mathbb R$, it suffices to show that any real number is a limit of a sequence of such numbers. For a given $x \in \mathbb R$, consider the sequence $\big( \frac{\lfloor 2^n x \rfloor}{2^n} \big)$, where $\lfloor \cdot \rfloor$ is the usual "largest-integer-less-than-or-equal-to" function.

-

Do not worry about the downvote. It is an attack on my answers.

I'll sketch the proof for you. If $x$ is a real number then one can find $a_1$ and $a_2 \in A$ such that, $$a_1 < x < a_2$$ Choosing $a_1 = \frac{m-1}{2^n}$ and $a_1 = \frac{m+1}{2^n}$ gives

$$\frac{m-1}{2^n} < x < \frac{m+1}{2^n} \Rightarrow - \frac{1}{2^n}<x-\frac{m}{2^n} < \frac{1}{2^n} \Rightarrow \left|x-\frac{m}{2^n}\right|<\frac{1}{2^n}=\epsilon$$

-
@Serial-downvoters: This is a perfect proof for this problem!! –  Mhenni Benghorbal Jan 22 '14 at 19:54
I did not downvote, but by assuming that $x$ must be in an interval of the form $[(m-1)/2^n, (m+1)/2^n]$ for each $n$, you are essentially assuming what has to be shown. Also, it is strangely written. You never made any assumptions about $\epsilon$ and $n$ in the first place, the reader has to come up with them themselves. –  Michael Greinecker Jan 25 '14 at 9:28
@MichaelGreinecker: This is just to give the OP a start (it is an idead). I have not given him a full proof. Whoever is interested can fill the gaps. So, what's wrong with this? –  Mhenni Benghorbal Jan 25 '14 at 19:25
@MichaelGreinecker: They down voted this answer many times. It is the first answer to be posted. –  Mhenni Benghorbal Jan 25 '14 at 19:43
I would have started with let $\epsilon>0$ be given, then there is an $n\in\Bbb N$ such that $1/2^n<\epsilon$. Then consider the set $\{k/2^n: k\in\Bbb Z\}$. Then there is a least $k$ such that $x<k/2^n$. This $k$ is your $m+1$. But great proof otherwise; I don't know why anyone would downvote it. –  Bryan Jun 16 '14 at 23:25

$$\frac{\lfloor 2^nx\rfloor}{2^n}\leqslant x\lt \frac{\lfloor 2^nx\rfloor+1}{2^n}$$

-

A thought: Given $\epsilon > 0$ and $x\in\mathbb{R}$, there is some rational, say $\frac{p}{q}$ that is within $\epsilon/2$ of $x$. Then can you find $m,n\in\mathbb{Z}$ such that $|\frac{p}{q} - \frac{m}{2^n}| < \epsilon/2$?

If so, you could then apply the triangle inequality.

-