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So, the definition I have for dyadic numbers is $$\mathbb D_2=\bigcup_{n\in\mathbb N}\{\frac{m}{2^n}\mid m\in\mathbb Z,n\in\mathbb N\}.$$

Q1) Do we have that $\mathbb D_2$ is the set of numbers written in basis $2$ ? i.e. $$\left\{\sum_{-n<i<m}a_i2^i\mid n,m\in\mathbb N\right\} \ \ ?$$

Q2) Is this $\mathbb Q_2$ ? (the field of $2-$adic number). The construction of $\mathbb Q_p$ is a bit weird for me).

Q3) In What $\mathbb D_2$ are particularly interesting ? ($\mathbb Q$ is also dense in $\mathbb R$ and countable, but several proof in probability use $\mathbb D_2$ instead of $\mathbb Q$, so it should have an advantage that $\mathbb Q$ has not).

Q4) I want to show that $\mathbb D_2$ is dense in $\mathbb R$. Can I do as follow ?

Let $a<b$. Let $p,q\in\mathbb Z$ s.t. $a<\frac{p}{q}<b$. Set $x_n^m=\lfloor a\rfloor+\frac{m}{2^n}(\lfloor b \rfloor+1-\lfloor a\rfloor)$. Now, I would like to show that there are $n$ and $m$ s.t. $x_n^m\in (a,b)$, but I don't really see how to do.

user659895
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  • $\Bbb{Z} = { \pm \sum_{i=0}^m a_i 2^i, a_i \in {0,1}}, \Bbb{Z}[1/2] = { \pm \sum_{i=-n}^m a_i 2^i, a_i \in {0,1}}$. It is interesting for many reason, for example the unique factorization in prime numbers still holds, except in $\Bbb{Z} $ the units are $\pm 1$ and $2$ is prime, in $\Bbb{Z} [1/2]$ the units are $\pm 2^n$ so $2$ isn't a prime anymore. – reuns Apr 20 '19 at 13:04
  • Q4: see this question. – Dietrich Burde Apr 20 '19 at 13:37
  • Q2: No. $\Bbb D_2$ is the subset of $\Bbb Q_2$ consisting of numbers with binary expansions of finite length, just like it is also the subset of $\Bbb R$ consisting of numbers with binary expansions of finite length. The only difference is that the other elements of $\Bbb R$ have expansions extending infinitely to the right, whereas the other elements of $\Bbb Q_2$ have expansions extending infinitely to the left. – Paul Sinclair Apr 20 '19 at 20:49
  • Q3: Mostly when $\Bbb D_2$ is used, it is simply because it is more straightforward to define some function on $\Bbb D_2$ than on all of $\Bbb Q$. And if you can get everything you need for an example or proof with $\Bbb D_2$, why bother with a more complicated definition for all of $\Bbb Q$? – Paul Sinclair Apr 20 '19 at 20:53

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