August 6th, 2014, 09:08 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  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 09:16 AM. 
August 6th, 2014, 09:49 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, 12:56 PM  #3  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  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 01:17 PM.  
August 7th, 2014, 04:45 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, 06:55 AM  #5  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  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, 07:08 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, 07:27 AM  #7  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  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, 07:57 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, 09:45 AM  #9  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  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, 10:32 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  

Tags 
dyadics, messing, reals, relating, sets 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Approach reals by 3smooth numbers  Maurice1969  Number Theory  11  October 22nd, 2013 11:24 PM 
Independence of Reals  mathbalarka  Number Theory  1  May 9th, 2013 05:51 AM 
Prove a set is a subring of the reals  HairOnABiscuit  Abstract Algebra  11  September 11th, 2012 11:37 AM 
Integration over the reals? finding R(dx)  cernlife  Real Analysis  5  May 30th, 2011 08:37 PM 
functions in reals such that inequality holds  maximus101  Real Analysis  5  February 14th, 2011 05:19 PM 