March 25th, 2018, 07:02 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26  Help Proving Whether a Certain Rational Exists  Tried and I'm Stumped
Trying to solve this: Let $r \in \, (0,1\, )$ and $r \notin \mathbb{Q}$. Let $r’ \in \, (0,1\, )$, $r’>r$, $r’ \notin \mathbb{Q}$, and $r’r \notin \mathbb{Q}$. Let $q \in A = \{ r’r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$ Finally, one of the following statements must be true. The question is which? $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{pq’ : p \in B\} = \{pq : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{pq’ : p \in B\} = \{pq : p \in A\}$$ Last edited by AplanisTophet; March 25th, 2018 at 07:40 AM. 
March 25th, 2018, 09:00 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574 
Do we know any other "alternative" posters who, failing to make their point in one thread, simply start another?

March 25th, 2018, 09:06 AM  #3 
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26  What point is that? I know I haven’t made it a point to assert anything other than the fact that that my intuition seems flawed, so I’m sure you’re not talking about me. If you have an issue with me asking questions and trying to gain a deeper understanding then go $&@$ yourself.

March 25th, 2018, 11:40 AM  #4  
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26  Quote:
Finally, one of the following statements must be true. The question is which? $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{q'p : p \in B\} = \{qp : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{q'p : p \in B\} = \{qp : p \in A\}$$  
March 25th, 2018, 02:19 PM  #5 
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26 
I believe I got an answer. Statement 2 is true. Proof: $$\sup(\{q'p : p\in B\})\sup(\{qp : p\in A\})=(q'\inf B)(q\inf A)=(q'r')(qr')=q'q \neq 0$$ If the two sets have different suprema, then they cannot be the same sets. The proof relies on asserting that $\inf B = \inf A = r'$. In so doing, we are essentially saying that the irrationals of $A$ can approach $r'$ to the EXACT same distance that the rationals of $B$ can approach $r'$. I'm not sure I buy that personally because there would be an 'infinitesimal difference,' but that's my common sense talking and not the math. 
March 25th, 2018, 02:31 PM  #6  
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  Quote:
Can you express "exact same distance" more precisely? You know, it's not that I'm ignoring your question. It's just that it's so very similar to the conversation we had in 2016 that I'm having a hard time diving back into it. A personal problem on my part I'm sure. It took all the concentration I had back then to get to the bottom of the issue, and I'm finding myself psychologically blocked from diving in again. Last edited by Maschke; March 25th, 2018 at 02:44 PM.  
March 25th, 2018, 03:17 PM  #7  
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26  Quote:
Last edited by AplanisTophet; March 25th, 2018 at 03:21 PM.  
March 25th, 2018, 03:24 PM  #8 
Senior Member Joined: Jun 2014 From: USA Posts: 366 Thanks: 26 
Getting tired and making typos. Time for a walk and some fresh air. I think the immediately preceding post is ok now though.

March 25th, 2018, 03:45 PM  #9 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  Planning to make a run at it. ps  Can you please tell me which version to start with? I can't read everything that's already been written. Last edited by Maschke; March 25th, 2018 at 04:25 PM. 
March 25th, 2018, 06:48 PM  #10  
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574 
Reposting this from the other thread. This is exactly the problem with starting a separate thread. It's hard to follow them both. You post actively to them both. Can you see how confusing this is for people trying to work with you? It's annoying when youknowwho does it and it's annoying when you do it. Surely that's within the bounds of discussion. It's always bad form to start a new thread on the same topic as another active thread, then post back and forth to both of them to confuse the heck out of the people trying to have a conversation. Quote:
[Note  I am referring to the "just for me" version you posted on the OTHER thread. So confusing]. I did make one notational change. $r$ is some irrational and $s \in \mathbb R \setminus \mathbb Q, s > 0$, the shift. I think that makes the geometry more clear. You start with an open interval of length $1$ whose left endpoint is irrational; and you rightshift by some irrational distance $s$. Presumably $r$ and $r  s$ are not in the same equivalence class mod $\mathbb Q$, else the irrational $r$ would get shifted to a rational and we'd all get confused. [I don't think I notated that right but you know what I mean. I just don't want to shift $r$ onto a rational by accident]. Is that ok so far? Just yes or no, I haven't got bandwidth for any new info. I'm just working line by line through your OP. Drawing pictures. Do you draw pictures? I find them very helpful. ps  I'm going to ignore your most recent exposition. I read up till where you introduced the dyadic rationals. That's one complication over the line. I'm staying with your exposition in the OP here. ps  WHAT???? I skimmed the rest of your latest post and found this: Quote:
But for you to think you found some incomprehensible mystery about the real numbers  that my friend is a delusion. You can see this, right? You did not break math. You're misunderstanding something about the reals. I'm diving in to try to help you sort it out. You go back and forth between: a) My intuition doesn't match my symbolpushing. Help me understand; and b) My intuition doesn't match my symbolpushing. Math is wrong. When you float over to (b) you sell yourself short. You are trying to understand the real numbers, people are trying to help. The fault I assure you is with your understanding of the real numbers. Not with the real numbers themselves. Last edited by Maschke; March 25th, 2018 at 06:59 PM.  

Tags 
exists, proving, rational, stumped 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proving that root 4 is rational by contradiction  YES q THE zU19  Applied Math  7  April 11th, 2016 06:07 AM 
Proving P exists in complex equation  jb8992  Complex Analysis  2  March 5th, 2014 12:49 PM 
Proving that a limit of a function of two variables exists  Akcope  Real Analysis  6  February 19th, 2014 06:10 PM 
Proving that a function exists  eddybob123  Algebra  5  March 14th, 2013 11:56 PM 
Proving that x is rational  Cissek  Elementary Math  1  September 27th, 2011 07:10 AM 