August 6th, 2014, 10:08 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 422 Thanks: 26  Messing Around  Relating Sets of Dyadics to Reals
I haven't seen any new threads here for a while. As always, I promise this is Fields Medal material , but perhaps someone will find it interesting anyways. To start, in my hopeless quest to find a discrete uniform distribution over $\mathbb{N}$, I was thinking in terms of base 2 on the unit interval and came across the following: Let $D = \{d \in (0,1]: d \text{ is a dyadic rational } \}$ Let $\alpha(x) = \{ d \in D : d < x \}$ Conjecture: There exists $x,y \in (0,1]$ where $x \not= y$ and $\alpha(x) = \alpha(y)$. Can anyone give an example where this is true? I'm thinking no, though it better be true if $\mathbb{R}_{(0,1]} > D$, so the lack of a realworld example isn't going to deter me here. Assuming the conjecture is true, we can create a unique interval on (0,1] for each dyadic in $D$: Let $\beta(x) = d \rightarrow (d \in D \wedge \alpha(x) = \alpha(d))$. Each dyadic $d \in D$ then defines its own interval equal to $\{ x \in (0,1] : \beta(x) = d \}$. These intervals are interesting. Their measure would have to be 0, because the sum of the measures would equal $\infty$ otherwise, but at the same time we must assume that the set of elements in each interval has a cardinality equal to $2^{\aleph_0}$ and is continuous. Is this possible? (clearly, it is, but ) Let $z$ be a randomly selected element of (0,1] given a uniform distribution. As CRGreathouse pointed out in an earlier thread, we can do this by randomly choosing each bit $x_i$ of $0.x_1x_2x_3...$ (temporarily ignoring the patchable problem of each dyadic having two binary representations, except 1, and all other elements of (0,1] having only one representation). This is the paragraph where CRGreathouse will have to come save the day (ie... potential crankery warning)... That said, we are guaranteed to have $\beta(z)$ equal to a dyadic in (0,1]. All dyadics in $D$ had an equal probability of being equal to $\beta(z)$, having selected $z$ uniformly at random, so we must conclude that our dyadic $\beta(z)$ is selected uniformly at random as well. ... Might as well finish it. Let $\psi$ be a bijection from $D$ onto $\mathbb{N}$. Then we have a unique positive integer, $\psi(\beta(z))$, where again all positive integers had an equal chance of being equal to $\psi(\beta(z))$, so I conclude that $\psi(\beta(z))$ is also selected uniformly at random. Last edited by AplanisTophet; August 6th, 2014 at 10:16 AM. 
August 6th, 2014, 10:49 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
For those of you following along: "dyadic rational" is the name for rational numbers with a denominator (in lowest terms) which is a power of 2. Quote:
I have no idea why you think these two are related. Quote:
I skipped this part since it isn't.  
August 6th, 2014, 01:56 PM  #3  
Senior Member Joined: Jun 2014 From: USA Posts: 422 Thanks: 26  Quote:
The set $\gamma(x)$ will then contain at least one element if the conjecture is false (and $x \leq 1$). Further, the collection $\{ \gamma(x) : x \in (0,1] \}$ must be pairwise disjoint. Let $\delta(d) = x \rightarrow d \in \gamma(x)$ Then, $\delta$ is a function from (a subset of) $D$ onto $\mathbb{R}_{(0,1]}$, and the cardinality of $D$ must be equal to or greater than the cardinality of $\mathbb{R}_{(0,1]}$. Last edited by AplanisTophet; August 6th, 2014 at 02:17 PM.  
August 7th, 2014, 05:45 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
$$ \gamma(x)=\left[(0,x)\setminus\bigcup_{y<x}(0,y)\right]\cap D=\{z:\ z<x\wedge\forall y<x,\ z>y\}\cap D=\emptyset\cap D=\emptyset $$ No, the conjecture is false and $\gamma(x)$ is uniformly empty.  
August 7th, 2014, 07:55 AM  #5  
Senior Member Joined: Jun 2014 From: USA Posts: 422 Thanks: 26  Quote:
Specifically, if the conjecture is false, we have $y < x \rightarrow \alpha(y) \subset \alpha(x)$. This should hold true for every possible value of $y < x$. It doesn't though, so at what point do we hit a $y$ where it doesn't?  
August 7th, 2014, 08:08 AM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Quote:
Expanding, this is $0<y<x<1\rightarrow(0,y)\cap D\subsetneq(0,x)\cap D.$ In other words, $(y,x)\cap D\ne\emptyset$ when $0<y<x<1$ ("there is some dyadic rational between y and x when y < x"). This is pretty obvious, yes? It's a standard undergraduate exercise to prove this for $\mathbb{Q}$ in place of D, and this version is no harder.  
August 7th, 2014, 08:27 AM  #7  
Senior Member Joined: Jun 2014 From: USA Posts: 422 Thanks: 26  Quote:
If $\gamma(x)$ is empty and $y$ never equals $x$, then there must exist a $y$ and $x$ where $\alpha(y) = \alpha(x)$. I don't see any way around this.  
August 7th, 2014, 08:57 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
That I may properly understand you: you're claiming that there are real numbers $y<x$ such that there are no dyadic rationals in $(y,x)$. Correct?

August 7th, 2014, 10:45 AM  #9  
Senior Member Joined: Jun 2014 From: USA Posts: 422 Thanks: 26  Quote:
We know that as $x$ increases, $\alpha(x)$ changes (grows, I'll say). Let $dx$ be an infinitesimal change in $x$. Does $\alpha(x) = \alpha(x + dx)$? If $x \in D$, the answer is no, it doesn't. If $x \not\in D$, then I believe the answer is yes. Now what say you?  
August 7th, 2014, 11:32 AM  #10 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  

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dyadics, messing, reals, relating, sets 
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