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March 25th, 2018, 10:59 AM   #51
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 Originally Posted by Maschke When it comes to your sliding idea, I have already paid my dues. If I'm reluctant to dive in again, that's why. And when I asked you earlier if this latest version is essentially the same as your idea from 2016, you said yes. Did you not?
I already told you that the last time I pondered this I had zero knowledge of real analysis. My thread here stems from a similar set up, but what you taught me last time I acknowledged right away:

Quote:
 Originally Posted by AplanisTophet Despite our ability to slide the uppermost piece of paper with God-like precision, it is seemingly still impossible to win the game. The length that we would have to slide the upper piece of paper to the right must be a rational number because sliding an irrational length to the right would imply that none of the rational numbers align (ie, a rational plus an irrational will always be irrational). However, if we slide a rational length to the right, then $r$ and $r’$ could not be vertically aligned. In that case, the Archimedean property of the real numbers would ensure that an infinite number of rational numbers on each line segment could also not be vertically aligned.1 References: 1 The proof that there is a rational between any two irrationals stems from the Archimedean property. See, e.g., Theorems 1.1.4 and 1.1.6 on pages 5 and 6 of the following: Trench, W. (2013). Introduction to Real Analysis. Available at http://ramanujan.math.trinity.edu/wt...L_ANALYSIS.PDF
I am going beyond what we talked about before as opposed to rehashing it. I told you what we discussed in the past still doesn't sit well with my intuition. I have not asserted that we can 'win the game' and that the point of this was to more closely examine this stuff so as to gain a better understanding. Note the bolded part in particular:

Quote:
 Originally Posted by AplanisTophet We'll see how much farther I can push the line between being a crank and trying to make valid mathematical and philosophical suggestions. I of course view my job here in challenging some of math's most basic principles as that of being a devil's advocate who tries to nit pick all of the standard rules that I expect everyone here to adhere to in general. I'm having fun, hope you are too. Will try to keep it interesting.
This isn't about trying to break math, assert I'm right, or anything like that. I have no point to make.

Zylo is out to disprove a theorem and become the greatest mathematician that ever lived, so quit comparing us. We agreed to disagree, but I had genuine concerns about his well being, which I was up front with zylo about, and then chose to refrain from playing the 'disprove Cantor' game with him. If you think I might be suffering too say so, otherwise quit comparing us. I have been doing this for fun because I'm up to my eyeballs in work and for me this is a nice way to take my mind off of my career at the moment.

March 25th, 2018, 11:47 AM   #52
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Quote:
 Originally Posted by AplanisTophet To make $q'$ well defined, I would define it as follows: For any given $q \in A = \{r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$, let $q’$ be the $$q’ \in \, (r’, r’+1\, ) \cap \mathbb{Q} \text{ such that } \{p-q’ : p \in \,(r’,r’+1\, ) \cap \mathbb{Q} \} = \{p-q : p \in A \}$$
I had the $p$ and each of $q$ and $q'$ backwards. I'm sorry! Corrected:

For any given $q \in A = \{r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$, let $q’$ be the
$$q’ \in \, (r’, r’+1\, ) \cap \mathbb{Q} \text{ such that } \{q'-p : p \in \,(r’,r’+1\, ) \cap \mathbb{Q} \} = \{q-p : p \in A \}$$

March 25th, 2018, 12:09 PM   #53
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Quote:
 Originally Posted by AplanisTophet I already told you that the last time I pondered this I had zero knowledge of real analysis. My thread here stems from a similar set up, but what you taught me last time I acknowledged right away:
You're responding to things I didn't say.

You said that I didn't CARE to read your latest. I explained that it's not due to a lack of care, but rather to an EXCESS of care the last time we went around on your sliding idea. I already cared way too much.

It was only your particular phrasing that I objected to. No point in rehashing subjects we've agreed to disagree on.

Regarding the much more interesting point of why the hyperreals are not a counterexample to the density of the rationals in the reals depending on the Archimedean principle, it's a fact that in the hyperreals, if $\epsilon$ is infinitesimal, then there's clearly no standard rational between $\epsilon$ and $2 \epsilon$. So density fails in that particular sense

BUT ... there's something strange going on. Clearly (in the hyperreals) the standard rationals are dense in the standard reals. I need to understand this point better. There are some subtleties going on in terms of what statements of the reals can be transferred to the hyperreals, and with what modifications.

Last edited by Maschke; March 25th, 2018 at 12:12 PM.

 March 25th, 2018, 01:38 PM #54 Senior Member   Joined: Aug 2012 Posts: 2,342 Thanks: 731 ps -- The answer is that everyone uses the Archimedean property as a shortcut to prove the density of the rationals in the reals; but the hyperreals show that the Archimedean property is not required. If you construct the reals as Dedekind cuts, you can prove the density of the rationals in the reals without the use of the Archimedean property. So my original point holds. Forget Archimedes, he has nothing to do with this.
March 25th, 2018, 01:44 PM   #55
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Quote:
 Originally Posted by Maschke You're responding to things I didn't say. You said that I didn't CARE to read your latest. I explained that it's not due to a lack of care, but rather to an EXCESS of care the last time we went around on your sliding idea. I already cared way too much.
As are you. I'm really not offended if you don't want to take a trip to idiotville with me so it's not like I would expect you to care. I'm the one who finds this goes against my intuition and am digging for something I know probably isn't there. But I'm having fun, so what's the harm?

What annoys me is for you to compare me to zylo and call me a crank. Zylo isn't a crank on things outside of Cantor's argument (as far as I know) and I am not someone who cares enough to be a crank in the first place. You're (presumably) a retired math professor or something, but I'm just a guy doing this for your amusement as much as mine (don't even think of denying that you love being the 'big man on campus' here, so you're welcome), and I'll eventually go on with my life outside of math either way.

I don't spend much time here, but you always seem to be here, so if I were to teach you something for a change, I would notice that you sitting around a math forum calling people cranks is pretty sad. There's a big beautiful world out there, so if being here is making you feel cranky, you should get out and enjoy it. My apologies if you can't for some reason.

March 25th, 2018, 01:45 PM   #56
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Quote:
 Originally Posted by Maschke ps -- The answer is that everyone uses the Archimedean property as a shortcut to prove the density of the rationals in the reals; but the hyperreals show that the Archimedean property is not required. If you construct the reals as Dedekind cuts, you can prove the density of the rationals in the reals without the use of the Archimedean property. So my original point holds. Forget Archimedes, he has nothing to do with this.
The variable $x$ in my post would contradict the Archimedean property. You keep ignoring that.

March 25th, 2018, 01:58 PM   #57
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Quote:
 Originally Posted by AplanisTophet The variable $x$ in my post would contradict the Archimedean property. You keep ignoring that.
I'm not ignoring that. I made the limited technical point that the density of the reals in the rationals does not depend on the Archimedean property, despite the fact that the Archimedean property can be used to get a short proof of density and is used that way by just about everyone.

If you have developed some argument that the Archimedean property is false in the reals, then your argument is flawed. If the existence of some number $x$ disproves the Archimedean property, then there is no such $x$.

Since I already spent a HUMONGOUS amount of time and energy working through your sliding argument in 2016, I didn't feel like doing it again. That's not ignoring you, it's simply declining to spend any more time and energy than I already have.

It's a symptom of your own narcissism that you think I'm ignoring you. Anyone who clicks on any of the many conversations I've had with you over the past two years knows that I have spent tremendous amounts of time and energy working with your ideas. Far more than anyone here or anywhere else on the Internet.

Your recent ungrateful and snide comments make me disinclined to toss any more energy down that unproductive rat hole. I really think you're playing this all wrong. You're alienating the person who's spent the most time working with you. Every time I start to get ready to work through your latest screed you toss out another gratuitous insult. You've made it impossible for me to respond to you.

Last edited by Maschke; March 25th, 2018 at 02:09 PM.

March 25th, 2018, 02:20 PM   #58
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Quote:
 Originally Posted by AplanisTophet What annoys me is for you to compare me to zylo and call me a crank.

Leave it alone dude.

I tried to explain to you that although your real analysis text uses the Archimedean property to prove the density of the rationals in the reals, that assumption is not actually necessary. I found that interesting and can't understand why you simply want to argue about stuff I stopped arguing with you about days ago, and in fact apologized for.

Last edited by Maschke; March 25th, 2018 at 02:24 PM.

March 25th, 2018, 03:05 PM   #59
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Quote:
 Originally Posted by Maschke I haven't done this for days. Why are you still going on about this? Leave it alone dude.

Quote:
 Originally Posted by Maschke Do we know any other "alternative" posters who, failing to make their point in one thread, simply start another?
Look, you are knowledgeable, you enjoy the math forum, and you can be very interesting. It's great you recognized that the density of the rationals in the reals might be proven without the Archimedean property.

I know you aren't inclined to go into this sort of discussion because I'm questioning the Archimedean property, which I told you early on, and we both know you hate philosophical discussions. I never got around to thinking you were ignoring me (it's your own narcissism that leads you to imply that I think you're ignoring when I didn't care either way).

I'm not trying to alienate you. I don't know much about math overall and I don't spend a lot of time with it. Every time I sign off the forum, there's a good chance I might not ever sign on again. If I do and you want to be helpful, great. I've always thanked you sincerely when you were. If not, I won't think anything less of you either. Maybe it seems like an insult to you, but telling you to stop and smell the roses as opposed to worrying about me and zylo was good advice. Us non-math folks would see it that way anyways, but you can take it as a gratuitous insult if you want.

For the last time, I'm not arguing about anything here. I am trying to pinpoint why we can't get the rationals to align when they all fit within $r'$ and $r'+1$. If the only reason is that we can't calculate the infinitesimal distance $x$ that we must shift by so that they align (and none lie outside of $\, (r',r'+1\, )$), then I'm inclined to think there is something wrong with what we're saying. If I can figure it out, great. If not, great.

March 25th, 2018, 03:55 PM   #60
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Beneath me. Couldn't resist. At least 60 Minutes isn't doing an exposé of my sex life. Could be worse, right?

Quote:
 Originally Posted by AplanisTophet I know you aren't inclined to go into this sort of discussion because I'm questioning the Archimedean property ...
No no no, not at all. I keep telling you why I can't dive in but you don't believe me. It's because I burned myself out on the 2016 thread. I am actually hoping to make a run at your latest.

Quote:
 Originally Posted by AplanisTophet , which I told you early on, and we both know you hate philosophical discussions.
I love philosophical discussions!

Quote:
 Originally Posted by AplanisTophet I never got around to thinking you were ignoring me (it's your own narcissism that leads you to imply that I think you're ignoring when I didn't care either way).
You explicitly said I was ignoring your point and now you deny it. You are ignoring yourself!!

Quote:
 Originally Posted by AplanisTophet I'm not trying to alienate you.
I took exception a while back regarding your feelings about other alternative posters. I thought you were lacking in self-awareness. I've said my piece. I won't mention anymore. Unless I do.

Quote:
 Originally Posted by AplanisTophet I don't know much about math overall and I don't spend a lot of time with it. Every time I sign off the forum, there's a good chance I might not ever sign on again.
Wouldn't be the same without you. Or the other alternative posters. I'd rather see alternative posters than upper-division students asking for answers without showing any work, which is a popular pastime around here.

Quote:
 Originally Posted by AplanisTophet If I do and you want to be helpful, great. I've always thanked you sincerely when you were. If not, I won't think anything less of you either. Maybe it seems like an insult to you, but telling you to stop and smell the roses as opposed to worrying about me and zylo was good advice. Us non-math folks would see it that way anyways, but you can take it as a gratuitous insult if you want.
LOL. Ok. Really. I'm actually trying to muster the energy to dive into your latest sliding idea.

Quote:
 Originally Posted by AplanisTophet For the last time, I'm not arguing about anything here. I am trying to pinpoint why we can't get the rationals to align when they all fit within $r'$ and $r'+1$. If the only reason is that we can't calculate the infinitesimal distance $x$ that we must shift by so that they align (and none lie outside of $\, (r',r'+1\, )$), then I'm inclined to think there is something wrong with what we're saying. If I can figure it out, great. If not, great.
I'll make a run at it. One preliminary thought is that when you slide an interval by an irrational distance, all the rationals go to irrationals. And almost all of the irrationals also go to irrationals, except for a handful of exceptional irrationals that happen to map to rationals. If the original rational and the slide length are in the same equivalence class mod $\mathbb Q$ that will happen. I wonder if that's part of your idea. If not, that's ok. Just a thought I had.

ps -- If I did make a run at this, which version should I start with? I definitely haven't got the heart to read through everything that's been written. As best I understand it, you're shifting an interval by an irrational number and then something funny allegedly happens. I do agree that some unlucky irrationals will match up to rationals. Other than that, I can't see what might go wrong.

Last edited by Maschke; March 25th, 2018 at 04:07 PM.

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