Infinitely Small?

[Melllvar]

Melllvar. For many things in life 99 percent sure is good enough. In crossing the street, better than 99 percent is recommended. In mathematics only 100 percent is acceptable.

I started reading your post. I read this part:
You started with the epsilon. We don't care what that epsilon is. You said the rational is inside that neighborhood. Then you said it must be greater than the epsilon. NOT IF IT IS INSIDE. When you said that, I stopped.

Notice that cardinality is not mentioned in these statements. Cardinality is irrelevant.

Hold on. You said, "Proof by contradiction." In order to have proof by contradiction, you must have a clearly stated premise. I don't see that. The only stated premise I see is "pick an epsilon > 0. A proof must have clearly stated steps. You won't find them though because what you are after is false ... but just in case I'm wrong, you're welcome to try.

Pi, there is no way in hell you have read any of my past few posts. I've given you multiple contradictions for your supposed proof (not the last one, the "finding a smaller rational than the irrational" one) and explained how your thinking is off here at length. As I said, all your proof showed was that you could create two irrational numbers that had a rational number between them. On the other hand I already offered a direct proof that there must be at least two irrationals that do not have rationals between them - among other things, the cardinality of the set of irrationals is greater than the rationals.

Proof by contradiction is exactly what you did in your proof of a smaller number than N, what you call reductio ad absurdum. Assuming a premise is true and then show that it leads to a contradiction, thereby proving it false. My assumption was to assume that you were correct and any infinitesimally small neighborhood of numbers close to zero would contain a rational number, if that neighborhood contained any non-zero numbers at all.

Sorry, but in the end you are refusing to accept this because it conflicts with basic intuition about numbers (that's developed solely because of the way they're written), when if you look at the basic properties of the rational/irrational numbers it obviously has to be true. Irrational numbers have infinite decimal expansions, they occupy continuous regions along the number line (unlike rationals), they are uncountable while the rationals are merely countable. It doesn't matter to me, I've had fun researching it and coming to understand it. The 99% was referring to my explanation, as I've made a lot of stupid mistakes already in this discussion, but I have no doubt left about the theorem itself.

 

[GYX_Kid]

Sometimes when I'm lying in bed almost asleep, I'll be made to imagine something infinitely big. I think it's more likely to happen if my tongue is positioned against the roof of my mouth in a certain way.

 

[Melllvar]

Ok, I can think of no simpler or more undeniable proof than this:

Assumptions (these can be verified):
The real numbers form a continuous set (i.e. the number line).
The irrationals do not form a continuous set, but do form continuous subsets.
The set of rationals is not continuous.
The real numbers in any region of the number line are just the union of the rationals and the irrationals in that region.

1. Create a set S that is the union of the irrationals and zero. This gives us a number line which is continuous everywhere except non-zero rational numbers.
2. In an arbitrarily small region surrounding zero, we now have a continuous subset of the real numbers.
3. The region cannot be empty of non-zero numbers because that would imply zero was discontinuous with the rest of the real numbers, but there are no rationals in that region surrounding zero, so we can conclude that every non-zero number in that continuous region must be an irrational.
4. Because that region is continuous with zero, those irrationals much also have absolute values less than all numbers which are not in that region.
5. Hence those irrationals all have absolute values lower than every rational number.
7. Therefore there must be irrational numbers which are smaller than all rational numbers.

 

[BigApplePi]

I have no doubt left about the theorem itself.

I have nothing to say myself at the moment except I think you've tried very hard to state your case. Sometimes mathematics can get very difficult and errors are hard to find.

I will let the below speak for me since I don't know what else to say. It makes a general statement about the theorem. Skip the stuff at the beginning and scroll to the highlighted two blue boxes. They contain the theorem in question called "Lemma 1."

http://www.cut-the-knot.org/do_you_know/numbers.shtml

 

[Otherside]

It's true that you can divide something infinitely, but what law says that you cant stop at some point and remove the last speck of it from some area?

 

[Otherside]

Doesn't matter what your denominator is if your numerator is zero.

 

[BigApplePi]

It's true that you can divide something infinitely, but what law says that you cant stop at some point and remove the last speck of it from some area?

Are you asking that mathematically or physically in the real world?

 

[Melllvar]

I have nothing to say myself at the moment except I think you've tried very hard to state your case. Sometimes mathematics can get very difficult and errors are hard to find.

I will let the below speak for me since I don't know what else to say. It makes a general statement about the theorem. Skip the stuff at the beginning and scroll to the highlighted two blue boxes. They contain the theorem in question called "Lemma 1."

http://www.cut-the-knot.org/do_you_know/numbers.shtml

*facepalm*

Actually it turns out you may be right, I did some research into the concept of dense sets and dense orders which seems to contradict the assertion that there must be a neighborhood of irrationals with no rationals present. But again I don't see how continuity of the real numbers arises if that is the case, and it also would (to me) seem to conflict with the fact that you can define the rationals in terms of this. So I am not convinced either way now, but I don't have any more arguments for either side at the moment either.

Nevertheless this discussion has become incredibly frustrating when initially I thought it was one of the most interesting I'd ever had here. I mean to be honest you blindly accept the most simple and obvious conclusion, refuse to question the assumptions on which it is based, cease trying to understand the points being made at all once you *think* you have found a flaw, assume you simply must be correct with such conviction that you dismiss opposing arguments without even finding a flaw in them, refuse to respond to the alleged flaws pointed out in your own supposed solutions, then resort to merely referencing some website claiming the same thing you are despite the fact that it has no proof given for it's assertion either (that I saw). The only reason to accept it is because it seems equivalent to saying that the rationals are dense in the reals, and I can easily find proofs of that.

Sorry, but even if I'm wrong I've probably expanded my knowledge of number theory by an order of magnitude over the course of this discussion. I had never even heard of the continuum hypothesis or dense sets or summations over arbitrary indices before this. Meanwhile you're happy to unquestioningly assume the truth of your beliefs rather than dig for the actual answer. I really do not feel that I am the one who is losing out on anything here, despite the fact that the dense-ness (density?) of the rationals does heavily argue in your favor.

No offense intended, just a personal POV.

The irrationals form continuous subsets.

2. In an arbitrarily small region surrounding zero, we now have a continuous subset of the real numbers.
3. The region cannot be empty of non-zero numbers because that would imply zero was discontinuous with the rest of the real numbers...

Also I've been trying to figure out where I went wrong, assuming I did, and it seems one of these.

 

[BigApplePi]

*facepalm*

Actually it turns out you may be right, I did some research into the concept of dense sets and dense orders which seems to contradict the assertion that there must be a neighborhood of irrationals with no rationals present. But again I don't see how continuity of the real numbers arises if that is the case, and it also would (to me) seem to conflict with the fact that you can define the rationals in terms of this. So I am not convinced either way now, but I don't have any more arguments for either side at the moment either.

Nevertheless this discussion has become incredibly frustrating when initially I thought it was one of the most interesting I'd ever had here. I mean to be honest you blindly accept the most simple and obvious conclusion, refuse to question the assumptions on which it is based, cease trying to understand the points being made at all once you *think* you have found a flaw, assume you simply must be correct with such conviction that you dismiss opposing arguments without even finding a flaw in them, refuse to respond to the alleged flaws pointed out in your own supposed solutions, then resort to merely referencing some website claiming the same thing you are despite the fact that it has no proof given for it's assertion either (that I saw). The only reason to accept it is because it seems equivalent to saying that the rationals are dense in the reals, and I can easily find proofs of that.

Sorry, but even if I'm wrong I've probably expanded my knowledge of number theory by an order of magnitude over the course of this discussion. I had never even heard of the continuum hypothesis or dense sets or summations over arbitrary indices before this. Meanwhile you're happy to unquestioningly assume the truth of your beliefs rather than dig for the actual answer. I really do not feel that I am the one who is losing out on anything here, despite the fact that the dense-ness (density?) of the rationals does heavily argue in your favor.

No offense intended, just a personal POV.



Also I've been trying to figure out where I went wrong, assuming I did, and it seems one of these.

Umm. Lots of things being said here. I believe many of the things you've brought up here had been worthy of pursuit. You must forgive me (or not) if I've dropped the ball in places and not addressed everything you've brought up. There are so many loose threads here how can they possibly be tied all up? Tying one if not perfect loosens another leading to endless (open) situations resulting in frustration. The only way to relieve this, I suppose, is to agree on closing some areas. But who can agree on what areas? Etc etc etc.

Added: Doing a lot of things right and getting credit for that or not is not enough. Where you went wrong and where I went wrong IS a worthy enterprise and that's how we learn. Allow me to look over some of this thread and I'll see if I can dig up any loose threads unexplained. The only thing I dread is if I go over old ground too much and you say, "I already covered that" ... the idea is to avoid going in circles.

 

[BigApplePi]

I know what you're saying, and it doesn't work. You claim that there is at least one irrational between every two rationals and at least one rational between every two irrationals? If there were true we could put the rationals and irrationals in a one-to-one correspondence, one rational for each irrational. But we can't, the irrationals are uncountable, so there have to be more irrationals than rationals. This means there is at least one pair of irrationals with no rational number between them.

Melllvar I went back and see I failed to point this out. I make no such claim this would lead to a 1-1 correspondence. Maybe that could be one source of misunderstanding.

BTW I agree with you (if I have this right) that the rationals are countable while the irrationals are not.

 

[BigApplePi]

Ok, I can think of no simpler or more undeniable proof than this:

Assumptions (these can be verified):
The real numbers form a continuous set (i.e. the number line).
The irrationals do not form a continuous set, but do form continuous subsets.
The set of rationals is not continuous.
The real numbers in any region of the number line are just the union of the rationals and the irrationals in that region.

1. Create a set S that is the union of the irrationals and zero. This gives us a number line which is continuous everywhere except non-zero rational numbers.
2. In an arbitrarily small region surrounding zero, we now have a continuous subset of the real numbers.
3. The region cannot be empty of non-zero numbers because that would imply zero was discontinuous with the rest of the real numbers, but there are no rationals in that region surrounding zero, so we can conclude that every non-zero number in that continuous region must be an irrational.
4. Because that region is continuous with zero, those irrationals much also have absolute values less than all numbers which are not in that region.
5. Hence those irrationals all have absolute values lower than every rational number.
7. Therefore there must be irrational numbers which are smaller than all rational numbers.

Line 1 under assumptions. Yes true. The real numbers form a continuous set.

Line 2 under assumptions: "The irrationals do not form a continuous set, but do form continuous subsets." I agree they do not form a continuous set. But they do not form continuous subsets. That would be jumping to a conclusion without proof.

This renders 3. to question. No longer can it be concluded, "there are no rationals in that region surrounding zero."

 

[Melllvar]

The only thing I dread is if I go over old ground too much and you say, "I already covered that" ... the idea is to avoid going in circles.

Ha, fair enough, and I don't mean to be like that. Actually I didn't mean to request any responses from you at all, it's more just something that often gets on my nerves with people. You have an advantage here in that so far I can't find any way to show you aren't correct, and indeed the latest info supports the idea that the original theorem was false, but I go through this same thing with people trying to explain things that are less questionable. E.g. the constancy of the speed of light relative to all observers - most people *just know* it can't be true, or even if they've come to blindly accept that, try explaining the relativity of the ordering of events. No amount of explanation will ever get them to consider other, less obvious, possibilities.

So forgive me if I saw a slight parallel here.

Melllvar I went back and see I failed to point this out. I make no such claim this would lead to a 1-1 correspondence. Maybe that could be one source of misunderstanding.

BTW I agree with you (if I have this right) that the rationals are countable while the irrationals are not.

Well, you don't have to explicitly claim it, if it's implied by what you're saying. I thought a 1-1 correspondence was, but now I'm not so sure. Among other things, the stuff on dense sets said that it was possible for a dense subset to have lower cardinality than it's parent set. This doesn't seem obvious (or possible) to me, but it's apparently true and would explain how their can be "more irrationals" and still have a rational existing between every irrational.

Again though it's more that I'm not really satisfied to just say, "Ok I guess Pi was right," when even that answer just implies there are more things I'm misunderstanding about the structure of the numbers. How can you always find another element of type A between elements of type B if there are more elements of type B in the set? (Obviously has to do with the uncountable-ness, but I still want to think about it more. Superficial understanding is kind of worthless.)

Line 2 under assumptions: "The irrationals do not form a continuous set, but do form continuous subsets." I agree they do not form a continuous set. But they do not form continuous subsets. That would be jumping to a conclusion without proof.

Yeah, that may have been flawed too (I am not sure). I can think of no less than four different reasons I made that assertion, but I won't bother explaining. In any case I haven't been able to find anything to verify or falsify the statement, so far.

Nevertheless I am still forced to wonder whether the theorem itself is true, and just "wicked to prove" as Vrecknidj originally said, or if it's simply been mis-stated and hence isn't true. One can think they've proven it false, but even then I'm wary that a mistake hasn't just been made, or some implicit assumption ignored.

Side note: Ugh. This entire thread reminds me why I dropped out of college.