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March 18th, 2018, 06:47 PM   #21
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Quote:
 Originally Posted by AplanisTophet If I had instead stated that $r’-r \in \mathbb{Q}$
But it's not. I was talking about the situation in hand, where we would need the first rational greater than $r’-r \in \mathbb R \setminus \mathbb{Q}$.

March 18th, 2018, 07:14 PM   #22
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Quote:
 Originally Posted by v8archie But it's not. I was talking about the situation in hand, where we would need the first rational greater than $r’-r \in \mathbb R \setminus \mathbb{Q}$.
So you’re asserting if we moved the paper a distance of $r’-r$, we would then need to ‘nudge’ to the right? Why not the left? There is no first rational as you stated though, and we don’t need one (which is evident when $r’-r \in \mathbb{Q}$, as it makes no difference in this case either).

March 18th, 2018, 08:35 PM   #23
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 Originally Posted by AplanisTophet That is very interesting and I thank you for posting it. My one question for you is whether you can offer proof of the above statement or if it is merely hand-waivy?
It isn't true that EVERY pair of linearly independent irrational numbers is algebraically independent. However, what I meant by my comment is that without some additional constraints (which you haven't required in your problem statement), 2 irrational numbers are algebraically indepdent with probability 1.

The proof of this is not hard. Let $a$ be an arbitrary irrational number. Then $a \in \mathbb{Q}[a]$ and the set of real numbers which are algebraic over $\mathbb{Q}[a]$ is countable (since this is a field extension of degree 2). Since the irrational reals are uncountable, it follows that if $b$ is an irrational chosen arbitrary, then it is algebraic over this field with probability zero or in other words, $a,b$ are algebraically independent with probability 1.

March 18th, 2018, 08:50 PM   #24
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Quote:
 Originally Posted by AplanisTophet So you’re asserting if we moved the paper a distance of $r’-r$, we would then need to ‘nudge’ to the right? Why not the left?
No reason whatever, this is a symmetrical situation.

March 18th, 2018, 09:01 PM   #25
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Quote:
 Originally Posted by SDK It isn't true that EVERY pair of linearly independent irrational numbers is algebraically independent. However, what I meant by my comment is that without some additional constraints (which you haven't required in your problem statement), 2 irrational numbers are algebraically indepdent with probability 1. The proof of this is not hard. Let $a$ be an arbitrary irrational number. Then $a \in \mathbb{Q}[a]$ and the set of real numbers which are algebraic over $\mathbb{Q}[a]$ is countable (since this is a field extension of degree 2). Since the irrational reals are uncountable, it follows that if $b$ is an irrational chosen arbitrary, then it is algebraic over this field with probability zero or in other words, $a,b$ are algebraically independent with probability 1.
The question wasn't whether two linearly independent irrationals are algebraically independent, but whether it would imply that the rationals can't be aligned in my game if they were. You've now offered proof that the irrationals used in my game will most likely be linearly independent irrational numbers that are algebraically independent, which is great, but I still fail to see how that implies the rationals cannot be aligned when moving the paper to the right as instructed in my game. Proving what I asked is, I believe, quite impossible using this method. It's actually quite redundant given that I already specified $r'-r \notin \mathbb{Q}$, is it not?

March 18th, 2018, 09:23 PM   #26
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Quote:
 Originally Posted by AplanisTophet The question wasn't whether two linearly independent irrationals are algebraically independent, but whether it would imply that the rationals can't be aligned in my game if they were. You've now offered proof that the irrationals used in my game will most likely be linearly independent irrational numbers that are algebraically independent, which is great, but I still fail to see how that implies the rationals cannot be aligned when moving the paper to the right as instructed in my game. Proving what I asked is, I believe, quite impossible using this method. It's actually quite redundant given that I already specified $r'-r \notin \mathbb{Q}$, is it not?
The answer to this question was the subject of my initial post. Namely, diophantine conditions are precisely how you prove that the rationals won't line up. However, as I alluded to earlier, this topic is broad and there is no way to provide a self contained proof on a forum like this.

If you are particularly interested in this one result, I would guess that the result I mentioned before (classification of homeomorphisms of the circle) can be adapted to prove exactly the question you are asking about. A quick google search turned up the following:

http://math.bu.edu/people/cew/preprints/introkam.pdf

The first 11 pages contain a complete proof of the classification result for diffeomorphisms of the circle and also seems to contain essentially all of the important technical tools required to address your problem.

March 19th, 2018, 07:01 AM   #27
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Quote:
 Originally Posted by SDK The answer to this question was the subject of my initial post. Namely, diophantine conditions are precisely how you prove that the rationals won't line up. However, as I alluded to earlier, this topic is broad and there is no way to provide a self contained proof on a forum like this. If you are particularly interested in this one result, I would guess that the result I mentioned before (classification of homeomorphisms of the circle) can be adapted to prove exactly the question you are asking about. A quick google search turned up the following: http://math.bu.edu/people/cew/preprints/introkam.pdf The first 11 pages contain a complete proof of the classification result for diffeomorphisms of the circle and also seems to contain essentially all of the important technical tools required to address your problem.
This approach is no different than noticing that when $r$ and $r'$ are vertically aligned, none of the rationals will be. It's like using the death star to blow up a peanut, imho.

Ask yourself, what fills the space between the unaligned rationals when $r$ and $r'$ are aligned? That is more or less the starting point for the problem. My longer exposition to a8archie then follows, as he went right to that point (which is what anyone should do...).

March 19th, 2018, 07:48 AM   #28
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Quote:
 Originally Posted by AplanisTophet Ask yourself, what fills the space between the unaligned rationals when $r$ and $r'$ are aligned?
More unaligned rationals and also irrationals. It's turtles all the way down.

The question you seem to be trying to ask is "what is there between consecutive rationals?", forgetting that such beasties don't exist.

Last edited by v8archie; March 19th, 2018 at 07:51 AM.

March 19th, 2018, 09:09 AM   #29
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Quote:
 Originally Posted by v8archie More unaligned rationals and also irrationals. It's turtles all the way down. The question you seem to be trying to ask is "what is there between consecutive rationals?", forgetting that such beasties don't exist.
I asked the question in #1 of my philosophical notes, but did not incorporate it into the mathematics of the game because we agree there aren’t consecutive rationals.

On the flip side, you (we) question what is between $r$ and $r’$ when the rationals are aligned in the game. That distance would be the same distance that separates the philosophically unaligned rationals when $r$ and $r’$ are aligned, right?

 March 19th, 2018, 10:44 AM #30 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra Not necessarily. I see it as just a distance, not any sort of minimum possible distance. To me it has no philosophical meaning because no such "minimum" exists. All these questions boil down to asking about the bottom turtle.

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