My Math Forum Help Proving Whether a Certain Rational Exists - Tried and I'm Stumped

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 March 25th, 2018, 07:02 AM #1 Senior Member   Joined: Jun 2014 From: USA Posts: 568 Thanks: 43 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? $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$ Last edited by AplanisTophet; March 25th, 2018 at 07:40 AM.
 March 25th, 2018, 09:00 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,387 Thanks: 746 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
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Quote:
 Originally Posted by Maschke Do we know any other "alternative" posters who, failing to make their point in one thread, simply start another?
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
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 Originally Posted by AplanisTophet 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? $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$
The two statements had the $p$ and the $q$ or $q'$ backwards. So sorry. Corrected it reads:

Finally, one of the following statements must be true. The question is which?

$$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{q'-p : p \in B\} = \{q-p : p \in A\}$$

$$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{q'-p : p \in B\} = \{q-p : p \in A\}$$

 March 25th, 2018, 02:19 PM #5 Senior Member   Joined: Jun 2014 From: USA Posts: 568 Thanks: 43 I believe I got an answer. Statement 2 is true. Proof: $$\sup(\{q'-p : p\in B\})-\sup(\{q-p : p\in A\})=(q'-\inf B)-(q-\inf A)=(q'-r')-(q-r')=q'-q \neq 0$$ 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
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Quote:
 Originally Posted by AplanisTophet 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 $r'$
What does that even mean? Does the sequence $\frac{1}{n}$ approach $0$ to the "exact same distance" that the sequence $\frac{1}{2^n}$ does? It seems as if you're confusing yourself about convergence.

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
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 Originally Posted by Maschke What does that even mean? Does the sequence $\frac{1}{n}$ approach $0$ to the "exact same distance" that the sequence $\frac{1}{2^n}$ does? It seems as if you're confusing yourself about convergence. 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.
Not sure what it means. I know that both $A$ and $B$ exist within $r'$ and $r'+1$ and that both sets are essentially copies of each other, just one hits only irrationals while the other only rationals. For both of them to fit within $r'$ and $r'+1$, they either must align or one must be a copy that is shifted ever so slightly (as evidenced by the lack of alignment and differing suprema) so as to make sure both copies still fit. That shifting implies that one ought to get closer to $r'$ than the other which in turn gets closer to $r'+1$. You know I have no mathematical way to assert this though, so in a philosophical sense only, can you follow what I'm saying at least?

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: 568 Thanks: 43 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
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Quote:
 Originally Posted by AplanisTophet ... can you follow what I'm saying at least?
Planning to make a run at it.

Last edited by Maschke; March 25th, 2018 at 04:25 PM.

March 25th, 2018, 06:48 PM   #10
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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 you-know-who 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:
 Originally Posted by AplanisTophet This one is just for you. It summarizes everything up to this point.
I'm halfway through your OP in this thread. Is that sufficient for the moment? I'd rather complete working through it than start with this new one. For now. Is that ok?

[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 right-shift 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:
 Originally Posted by AplanisTophet With this, I think I’ve found all I care to know about math quite frankly. I view math as the study of statements that we can prove are true given the most basic of assertions that we take to be true, but I fear those basic assertions have left us at a point where we have both much we would like to know but cannot and much we might know to be true but cannot prove. I have hopes that discoveries in mathematics will still lead to results in physics, but delving further into the realm of pure mathematics involving the infinite seems quite fruitless.
Now look, THAT is cranky. The sane response to this should be, "Can you please help me understand what I'm missing?" That's exactly what I'm doing, and it's why I'm taking the trouble to unpack your exposition line by line and symbol by symbol.

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 symbol-pushing. Help me understand; and

b) My intuition doesn't match my symbol-pushing. 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.

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