So, remember back in December, I wrote a post about a Cantor crank who had a Knol page supposedly refuting Cantor's diagonalization?
This week, I foolishly let myself get drawn into an extended conversation with him in comments. Since it's a comment thread on an old post that had been inactive for close to two months before this started, I assume most people haven't followed it. In an attempt to salvage something from the time I wasted with him, I'm going to share the discussion with you in this new post. It's entertaining, in a pathetic sort of way; and it's enlightening, in that it's one of the most perfect demonstrations of the behavior of a crank that I've yet encountered. Enjoy!
I'm going to edit for formatting purposes, and I'll interject a few comments, but the text of the messages is absolutely untouched - which you can verify, if you want, by checking the comment thread on the original post. The actual discussion starts with this comment, although there's a bit of content-free back and forth in the dozen or so comments before that.
Comment 176 (John Gabriel @ 2/2/2010, 2:02pm)
You hurled the first insult by calling me a crank. Let's drop the insults and start again. I shall prove to you step by step that my argument is correct. However, in order to do this, I need you to respond to each of my questions. Let me begin with the first question:Do you agree that every real number in the interval (0,1) can be represented in decimal?
YES or NO
I do not want you to say anything else. Just YES or NO. I am waiting for your response before I continue my proof. I will ask you a few more questions and then I will show you my proof. Deal?
This was fallowed immediately by:
Comment 177 (John Gabriel @ 2/2/2010, 2:04pm)
You can send your answers to my email address and I will respond here on your web page.
john underscore gabriel at yahoo dot com
And then shortly after that:
Comment 178 (John Gabriel @ 2/2/2010, 2:24pm)
1. Do you agree that every real number in the interval (0,1) can be represented in decimal?
YES or NO
2. Do you agree that my tree contains every real number in the interval (0,1)? Don't concern yourself about finite/infinite numbers at this time. We don't care about enumerating the numbers at this stage, only that if we know a number, we can find it in the tree.
YES or NO.
3. Do you agree that if we traverse the tree in each level from top to bottom that we can be sure to enumerate all the finitely (not the repeating decimals or irrational numbers) represented numbers in decimal? I know there are duplicates but we shall not worry about this right now.
YES or NO.
4. Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?
YES or NO.
Once you answer these questions, I shall proceed. I am expecting a YES answer on all of them. Nothing secretive here. Besides, I think you will agree that these assumptions are reasonable
You can already see the flavor of what's coming. John is rather a control freak, who is fond of dictating terms to people, and who demands a kind of respect that he's unwilling to grant to anyone else. My initial response was, essentially, to tell him to drop dead:
Comment 179 (MarkCC @ 2/2/2010, 2:30pm)
I'm not interested in playing games. I've read your "proof". I've told you what's I think is wrong with it. You've refused to address that. And I've seen how you respond to criticism - both from me, and from other people who disagree with you.(A) This is a blog. There are lots of readers. If you want to have this discussion, it's not just with me: it's with all of the people who read the blog.
(B) Why on earth would I believe that if I start playing your game that you're going to say anything different from your Knol article?
(C) If it's just going to be the same as your Knol piece, why should I believe that if I continue to disagree with it, that you're going to respond any differently?
You've demonstrated that you're a juvenile who throws tantrums whenever anyone dares to disagree with your obvious brilliance. Even this "offer" of yours is just more of that: yeah, you'll explain why I'm wrong - provided I'm willing to do exactly what you want, when you want, how you want. The moment I do or say anything you don't like, you'll just start up your tantrums again.
One thing that I'm proud of on this blog is that I've got a history of admitting my errors. Just go back and look at the history of the blog. I've made my share of mistakes. And I've always done my best to admit them, and correct them. And I've done it without trying to hide it: I've always made the correction, and inserted extra text to explain that the original post contained an error.
If you're really interested in defending your proof, go ahead and do it here, in the open, in the comments. If you want, I'll even set up a new top-level post specifically for your arguments. But I won't play games with sending you private mails between each step, and I won't tolerate you throwing insults at readers who point out problems with your argument.
This seemed like a reasonable response to me. Course, I'm not batshit insane.
Comment 180 (John Gabriel @ 2/2/2010, 2:52pm)
See what I mean? I started to prove this to you but look how you responded. I am not playing games with you. I am not prepared to continue unless you answer my questions. The reason for this is obvious: if you do not answer satisfactorily then I cannot continue the proof because you can always waver later on.
The reason I asked you to email me is because this page takes forever to load. But it's okay. Just place your responses here and I shall continue to respond whenever I can.
So once again, I offer to prove it but only on condition you answer my questions. It's up to you.
Before you accuse me, take a good long hard look in the mirror. You are guilty of everything you have accused me of. You can say whatever you dislike about me but I despise anyone who calls my intelligence into question. So, want to move ahead? Answer my questions. They're easy. YES or NO. We shall address every issue you have as we go along.
As you can see, John is rather sensitive about anyone who questions his intelligence. Doing so is clearly absolutely out of bounds in his universe. But he can't go for two minutes without attacking the intelligence of anyone who doesn't buy into his vastly inflated self-image.
Comment 181 (MarkCC @ 2/2/2010, 3:22pm)
John, *this is a blog*, not private email. I don't track stats that closely - but each comment thread is followed by *at least* several dozen users. You're *not* just talking to me; you're talking to *all of the people reading the thread*. I'm not the only one who can respond, disagree, and criticize: anyone who reads this blog can.
Playing this little game of "you have to do it my way, and answer my questions, or I'll take my crayons and go home" is bullshit. It's just an excuse.
Are you going to throw more of your hissy-fits when some commenter who *didn't* send you per-comment answers points out a problem?
But fine, I'll give you your answers.
(1) Yes, I'll agree that all real numbers are *representable* using infinite decimal notation.
(2) I'll also agree that taken to infinity, your tree contains all real numbers between zero and 1.
(3) Yes, I'll agree that all finitely representable numbers can be enumerated from your tree using a breadth-first traversal.
(4) NO, I do *not* agree that you can do a "left to right" traversal of the infinitely-long representations. This is the problem with your whole damned argument: you're mixing together notions from finite representations with infinite representations. You cannot do an ordered traversal of the leaves of an infinite tree. It's *meaningless*. What's the left-neighbor of 1/3 in your tree? You cannot specify it - it doesn't really exist: there simply is no real number which is "closest" to 1/3 without being 1/3. But a left-to-right traversal supposes that there *is*. And that's the problem. You're trying to get a result using a property of a finite tree, when that property doesn't exist on a tree extended to infinity.
This is something that's going to come up again and again. John really doesn't understand what a blog is. He never seems to grasp that the comments on my blog are an open forum; that anyone can post anything they want, any time they want. In the history of this blog, I have never deleted a non-spam comment; and I've banned a total of three people for egregiously inappropriate behavior.
Comment 182 (John Gabriel @ 2/2/2010, 3:45pm)
Very good. Now let's address your issue with question number 4 so that we can move on.
"This is the problem with your whole damned argument: you're mixing together notions from finite representations with infinite representations. You cannot do an ordered traversal of the leaves of an infinite tree. It's *meaningless*."
Well, if you can't do this, then I do not see how you agreed with 2? Let me clarify it for you: at this time we do not care about ordered traversals, only that the traversals are possible. Cantor's original argument says nothing about order of numbers, only that these have to be placed into a one to one correspondence
In a trend that will be repeated frequently, he follows it immediately with other comments:
Comment 183 (John Gabriel @ 2/2/2010, 3:49pm)
That last sentence should have read:
Cantor's original argument says nothing about order of numbers, only that these have to be placed into a one to one correspondence with the natural numbers.
If I cease to respond today, it is because I am in a different time zone. So I'll continue to respond tomorrow.
Comment 184 (John Gabriel @ 2/2/2010, 3:51pm)
I will not respond to anyone else but you.
From here on, it rapidly gets boringly repetitive. John is absolutely convinced that the fact that he's got a representation of real numbers means that he's got an enumeration of real numbers. He continually insists that he can do a "left-right traversal" of the infinitely long paths of his tree. I try to point this out:
Comment 185 (MarkCC @ 2/2/2010, 4:01pm)
Re 184: Fine, but I will still welcome comments from anyone who has anything to say, and I will *not* tolerate any abusive behavior towards them.
Re 182: You can enumerate things in a breadth-first fashion easily. (0, 0.1, 0.2, 0.3, 0.4, 0.5, ..., 0.9, 0.11, 0.12, 0.13, ...). There are an infinite number of values - but the number of steps that it takes to get to any one of them in the traversal is finite. That's *exactly* what it means to be countable: it's an infinite set, so you'll never stop enumerating them; but you can pick out any particular value, and it will be enumerated after a finite amount of time.
But in the case of the infinite tree, traversing the "leafs" makes no sense. It's not something that you can meaningfully do.
He responds in typical fashion:
Comment 187 (John Gabriel @ 2/3/2010, 2:25am)
MChu: Looks like we have to first agree on definitions. You claim:
"That's *exactly* what it means to be countable: it's an infinite set, so you'll never stop enumerating them; but you can pick out any particular value, and it will be enumerated after a finite amount of time."
This is not true. An infinite set is *countable* if and only if its members can be placed into a one-to-one correspondence with the natural numbers.
Where do you see anything in that definition that implies finite time? It's not there.
You claim the number of steps in a top-down traversal can be found in finite time.
*Finite time* has nothing to do with enumeration. 1/3 can be found in a finite number of steps as I demonstrated. Naturally, one is not going to follow the full path for the traversal because this is physically impossible. But it does not matter because you agree that 0.3333..... is the decimal representation of 1/3 in base 10.
One does not care too much *where* in the tree/list the numbers are as much as one cares that these numbers are *in* the tree.
You also claim:
"But in the case of the infinite tree, traversing the "leafs" makes no sense. It's not something that you can meaningfully do."
If this is the case, then you can throw Cantor's diagonal argument out the window and we are done. Cantor's diagonal argument works on the premise that any list provided will not contain every real number. Now if we know that 1/3 is in our list (hey, it's just 0.333...), then what you are claiming is the same as Cantor telling us "But you can't meaningfully write down 1/3!"
Nonsense! If 1/3 can't be in our list, then there is no use taking up Cantor in his challenge, is there? I can't think of a more ridiculous excuse than this.
So now we know that the number 0.333... can be in out list. It cannot be written down or enumerated in a finite time - well, not in base 10 at any rate.
Another way of stating Countability is by saying, "A set is countable if one can write out its members".
Well, to this with the rational numbers, it can be done in finite time because they are well defined - N x N plus 0 defines the rational numbers. Since we cannot apply this product to all real numbers (as you know most real numbers cannot be represented as ratios), we need to find another way to represent real numbers. This other way is my tree.
In fact, if we cannot find a way to represent all real numbers, then it is not true that every real number can be represented using decimals.
So, do you now agree that one can perform a left-to right infinite traversal? Hey, let me put it to you this way: I can perform an infinite left-to-right traversal of 1/3 in finite time. How? I find the path that starts with a digit 3 and then I stop! Why? I know that every possible *permutation* is in my tree. I don't need to continue any further. I could apply this same process to pi or e or sqrt(2) or any other irrational number I like, provided I know the first few significant digits.
So, in order to proceed, I must have your acknowledgement that it is possible to perform left-to-right finite traversals.
BTW: How do I know you are Mark Chu and you know I am John Gabriel? This was another reason I wanted to confirm via email. I had another scheme but I would have to tell you in an email how we could know who the real poster is. Otherwise I'll proceed here as I have been.
This marks the beginning of something that he's going to harp on. He's got rather an obsession with identity. He keeps trying to demand that I somehow prove that I'm really Mark Chu-Carroll. I don't get this at all: what difference does it make? If I were, like many of my blogging friends, posting under a pseudonym, would that somehow make my mathematical arguments less valid?
And of course, John never replies with one comment. There are always multiple follow-ons.
Comment 188 (John Gabriel @ 2/3/2010, 2:28am)
Sorry, I need your acknowledgement to:
So, in order to proceed, I must have your acknowledgement that it is possible to perform left-to-right infinite traversals.
Comment 189 (John Gabriel @ 2/3/2010, 2:41am)
That last comment was too long. Let me summarize it.
One can perform an infinite left to right traversal as follows:
Find the first digit of the decimal representation and stop.
That's it. I can rest assured that number is in my tree because the tree contains every possible *permutation* of the digits 0-9 in the decimal system.
As simple as that. So now we have taken care of your finite time issue. YES? May I continue?
I was, frankly, stunned when I saw this last one. He thinks that you can meaningfully traverse all of the nodes of a tree by visiting one of their parents, saying "Yup, there's children down there", and then declaring that entire subtree traversed.
Comment 190 (MarkCC @ 2/3/2010, 8:11am)
How do you know I'm Mark Chu? You don't, because I'm *not* Mark Chu. I'm Mark Chu-Carroll. And what the hell difference does it make whether I send you email? I could *still* be sending from a newly created fake email address.
More importantly, it doesn't matter. You don't seem to understand that this is a blog. This is not a private conversation between two people. This is a public forum where anyone who wants to participate can participate.
On to substance: we clearly disagree about what "traversal" means. Traversing a tree means visiting its nodes in some sequence. A left-to-right traversal means visiting every node on a level in left to right order.
Traversal of a node doesn't mean saying "It's down there somewhere". It means visiting the node. You can't touch a parent of a node in a tree, say "I know the child is down there somewhere, so I've traversed it".
So no, you haven't taken care of anything. You're just redefining "traversal" in a non-sensical way in order to make your "proof" work.
And now things start to get amusing. John starts throwing one of his temper tantrums, as I predicted back at the beginning. I'm still frankly rather mystified by the particular form of the tantrum: I'm don't see anything in that comment that addresses him in an insulting fashion. I think that my comment above is actually very reasonable and polite. But John disagrees.
Comment 191 (John Gabriel @ 2/3/2010, 10:01am)
Nonsense. You have a PhD? From where did you get your degree? Must be somewhere in the US. Allow me to educate you: A tree traversal does not say anything about visiting nodes in any order or sequence except that the nodes are visited exactly *once*.
That different ways of visiting the nodes exactly once are called traversals and can be accomplished in systematic ways. This however is irrelevant because I visit each node in my tree exactly once and this process is called a traversal. Rather than call it a pre-order or post-order or b-tree traversal, I call it top-down or left-right. See, this is one of the reasons I wanted to verify you are who you say you are. You evidently know very little about computer science.
Now read carefully: Don't you dare address me ever again as you have in your previous comment!
You DO NOT tell me what to do and you DO NOT get to say how things are done where I am concerned. Do I make myself clear?
Now let's proceed again:
You say:
Traversal of a node doesn't mean saying "It's down there somewhere". It means visiting the node. You can't touch a parent of a node in a tree, say "I know the child is down there somewhere, so I've traversed it".
As I just explained to you. I have a systematic way of visiting each node once. As for saying "I can't touch the parent node in a tree and say the child is down there somewhere" - OH, YES I CAN. This is part of the beauty in computer science algorithms - they are predictable. Tree structures are designed the way we want them to be. I wrote complicated balanced tree algorithms in Assembly x86 way back in 1980. Trust me, if I could not say the "child" is down there somewhere, I would have had major issues. Do you know anything about algorithms and optimisation? Probably not much.
So, don't spew out any more nonsense at me. I am infinitely more intelligent than you can ever hope to be. Yes, I know it's arrogant. Do you know what? F..k the lot of you at Google. You are the biggest morons I have ever come across in my entire life. The sad thing is you don't realize it because your ego is way out of proportion with your intelligence.
Ok, stupidicus? Now, may I continue? Abuse me one more time verbally and piss on you!!!!
And, as always, one comment isn't enough for John.
Comment 192 (John Gabriel @ 2/3/2010, 10:05am)
In order to complete my proof, it is imperative that you agree with point number 4 (it does not require the axiom of choice - I don't believe in the axiom of choice).
After this, I have one more question and then I shall complete the proof. Once the proof is completed, you shall apologize for calling me a crank and update your webpages to reflect the fact that you have been wrong yet again!
YES or NO to question number 4? Spare me any other shit talk. I don't want to discuss anything else.
Comment 193 (John Gabriel @ 2/3/1010, 10:20am)
Just a point of clarification:
In a infinite left-right traversal, I am referring to a traversal that results in only one real number. By stepping down the tree, the next real number and so forth.
I am not completing the infinite traversals - there is no sense in even thinking about such an absurdity.
Back to me. At this point, I was starting to lose my temper, and stopped bothering to try to be polite.
Comment 194 (MarkCC @ 2/3/2010, 10:26am)
@191:
See, this is exactly what I predicted. You just regurgitate the same nonsense, and throw insults the moment I disagree with you.
Let me explain something to you again. This is a blog. This is not a private conversation. This is a public forum. You don't get to make up your own rules about how the forum works. If you don't like it, tough.
And it's really quite amusing to see you throwing tantrums because you think I didn't address you correctly. I'm honestly not even sure about exactly what set off that tantrum: but you want to be addressed in some particular way, but you can't even be bothered to get my name right. I'm supposed to be properly respectful and deferential towards you, to the point of rewriting the rules of how I handle comments on my blog; but you can't have the simple, trivial decency of getting my name right.
I still fail to understand your obsession with identity. How would my sending you private email prove who I am? It takes 30 seconds to set up a new gmail address. I could set up a dozen variants on Mark Chu-Carroll on gmail. I could set up a gmail address claiming that I was John Gabriel, and send you email from that. What would it prove? Email can't prove identity. It can't prove that I have a PhD. It can't prove that I don't have a PhD. It can't prove that this is my real name, and it can't prove that it's not. And as I keep trying to explain, this isn't a private conversation. This is taking place on a public forum, with somewhere between a couple of dozen and a couple of hundred readers, any of whom are welcome to participate. So even if there were some way for you to verify my credentials, what difference would it make?
More importantly, this is an argument about math. If you're wrong about a piece of math, the identity of the person who points out that error is entirely irrelevant. It's math: a critique of a proof doesn't become correct because the person who wrote it has a particular degree. Greg Chaitin, who I admire greatly, wrote his first major mathematical critique of Kolmogorov when he was in high school. Is that original critique less correct than the exact same criticism written down after he got his PhD?
Tell me, John. How exactly is touching the parent node of an infinite subtree equivalent to visiting every child of that subtree exactly once? What does "left to right" mean if not, well, left to right?
At this point, I think we crossed our comments - I think that he was working on the following couple of comments at the same time that I was posting #194 above.
Comment 195 (John Gabriel @ 2/3/2010, 10:29am)
Yet another thought: If you are thinking that I would not be able to complete the left-right traversals, well, you would not be able to complete pairing the natural numbers either.
So if there isn't anything else, I think you are ready to agree with point 4.
Comment 196 (John Gabriel @ 2/3/2010, 10:32am)
A infinite left-right traversal is starting off at any one of the top nodes and visiting only one node from each level.
Comment 197 (John Gabriel @ 2/3/2010, 10:34am)
A top-down traversal on the other hand is finite and remains within one level with the first digit being the parent node in that level.
Comment 198 (John Gabriel @ 2/3/2010, 10:37am)
Left-right:
0.1 - 4 - 5 - ......
Top-down:
0.1
0.2
0.3
.....
Comment 199 (John Gabriel @ 2/3/2010, 10:42am)
You do realize that to accomplish all the left-right traversals means starting off at a top node. But this is okay because we can leave a marker at the last node we forked off somewhere.
Although nodes are revisited this way, it does not matter because each traversal is for a different number. What I am saying is that for one traversal, each node is visited *exactly once* - this is in order. Think of a formula parsing tree that contains many different formulas. We could do one traversal to calculate one function and start another traversal to calculate another function.
Comment 200 (John Gabriel @ 10:59am)
One need not be concerned about not being able to move from one node to the next. One can think of infinitely many parallel traversals all taking place at the same time.
This may not have been a mental block for you but I just thought I'd address it to make sure we covered every possibility.
Finally, I manage to get a response in.
Comment 201 (MarkCC @ 2/3/2010, 11:24am)
@200:
In one message, you say:
4. Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?In another, you say:
n a infinite left-right traversal, I am referring to a traversal that results in only one real number. By stepping down the tree, the next real number and so forth. I am not completing the infinite traversals - there is no sense in even thinking about such an absurdity.How can you claim to "enumerate" the irrational numbers, without being able to complete traversals? Or are you claiming that you can enumerate the digits of any particular irrational number? If the latter, than what on earth does "left to right" traversal have to do with it? If you're enumerating the digits of a specific number, then you're traversing a single path - no "left-to-right". If you're enumerating multiple numbers, then your traversal is not ever visiting the nodes corresponding to the numbers with infinite representations.
Now we go in circles. How do you enumerate a number? Why, you use a left-right traversal. What does left-right traversal mean? A process by which you enumerate the numbers by visiting their parents.
Comment 202 (John Gabriel @ 2/3/2010, 12:41pm)
1. One enumerates an irrational number by a given left-right traversal. The completion is imaginary for one does not write down all the digits of pi say, this would be absurd because it is impossible. However, the imagined completion is the number pi.
2. Yes, if you are enumerating (writing down) the digits of a specific number, then you are traversing a single path.
3. Left-to-right is just what I call the traversal where one visits exactly one node in each top-down level and it results in one real number either rational or irrational. If one does not like the name left-right, you can change it to whatever you like. It's immaterial.
4. Enumerating multiple numbers: This requires parallel traversals that all take place at the same time.
So now, "left-right traversals" have become not left-right traversals, but infinite numbers of parallel traversals. But they're still left-right. I was never able to get him to define just what the heck he meant by that. My suspicion is that he doesn't actually know himself.
Comment 203 (John Gabriel @ 2/3/2010, 12:50pm)
"If you're enumerating multiple numbers, then your traversal is not ever visiting the nodes corresponding to the numbers with infinite representations."
Yes, it will visit the nodes - we just won't wait around for it to finish. See what I mean? Once the number's digits are on the path, we bid it adieu. Take the simple example of 1/3. We start at the digit 3 and we know there is an infinite string containing only the digit 3.
This is the imaginary completion. It is absurd to think you can find the last node because there *is no last node*. However, you cannot disqualify the argument for this reason. Cantor's challenge to provide an imaginary list includes numbers that are infinitely represented.
This one was actually somewhat enlightening. John's "enumeration" involves an infinite number of things that will only be in the enumeration after an infinite amount of time. In other words, he counts this as being included in his enumeration, despite the fact that his enumeration procedure will never produce those numbers.
Comment 204 (MarkCC @ 2/3/2010, 1:09pm)
No, it *won't* visit the nodes. It will never reach them.
It comes back to exactly where we started: I said that your problem is that you can represent nodes, but not actually enumerate them.
What you're doing now is playing games with terminology. You're trying to redefine the meanings of "enumeration" and "traversal".
You're claiming that you can do a "left-right traversal" which isn't actually a traversal. It involves "visiting" nodes without ever actually visiting them. And it involves "enumerating" nodes that will never actually get enumerated, by using an infinite number of processes which will never actually finish reaching the nodes that they supposedly enumerate.
And none of that has anything to do with Cantor. Cantor doesn't generate an enumeration. Cantor says "Given a supposed enumeration of the real numbers, I can show you a real number which isn't in that enumeration".
You've got a representation which includes all of the real numbers. But you can't enumerate the values in it. As I said at the very beginning, representation is not enumeration. You still can't enumerate the real numbers. You can represent them, but you can't enumerate your representations.
And John's next comment confirms my enlightenment.
Comment 205 (John Gabriel @ 2/3/2010, 1:54pm)
The nodes *are* visited, but you are not able to see it. So this is why we are having this discussion.
Enumerate means to count off one by one.
I *do not* want to write out or count the digits of any traversal. The digits only have meaning as a *collective*, that means taken to infinity, they represent the decimal number. Again, take the example of 1/3. Can you enumerate the digits of the decimal representation 0.333... ? No. Suppose now that each digit 3 is contained in a node. One left-right traversal is an *enumeration* of 1/3. It's that simple.
MChu: You're claiming that you can do a "left-right traversal" which isn't actually a traversal. It involves "visiting" nodes without ever actually visiting them.
Not so. It is exactly a traversal. As I explained, one does not have to wait around for ever to be convinced that every node is visited exactly once for each real number represented by a given left-right traversal.
MChu: And it involves "enumerating" nodes that will never actually get enumerated,
No. You are not enumerating *nodes*, you are enumerating decimal numbers. Big difference. The nodes are the building blocks of each decimal number.
MChu: ...by using an infinite number of processes which will never actually finish reaching the nodes that they supposedly enumerate."
Tell me, what node is one supposed to reach in the case of 1/3? Do you realize how ridiculous this last statement of yours is? There is no last node.
You can see that I'm hitting close to home. He starts getting insulting any time you get close to his errors. He can't enumerate 1/3 - so when I start pushing on the fact that his enumeration will never generate 1/3, the tantrums start again.
And of course, there's always the sequence of multiple followups.
Comment 206 (John Gabriel @ 2/3/2010, 1:58pm)
MChu: You still can't enumerate the real numbers. You can represent them, but you can't enumerate your representations.
If I can represent them, I can enumerate them. Representation Enumeration
Here's the crux of his argument - which is exactly what I said in the original post that spawned this stupid discussion. He believes that being able to represent numbers implies that you can enumerate them. I'm not quite sure what "Representation Enumeration" means; my guess is that he left out a "⇔" symbol.
Comment 207 (John Gabriel @ 2/3/2010, 2:02pm)
In mathematics and theoretical computer science, the broadest and most abstract definition of an enumeration of a set is an exact listing of all of its elements (perhaps with repetition). Wikipedia.
Hate quoting from Wikipedia but this definition is correct.
Comment 208 (MarkCC @ 2/3/2010, 2:09pm)
@205:Tell me, what node is one supposed to reach in the case of 1/3? Do you realize how ridiculous this last statement of yours is? There is no last node.Tell me, what node is one supposed to reach in the case of 1/3? Do you realize how ridiculous this last statement of yours is? There is no last node. That's exactly the point. Your mechanism will never produce 1/3 in its enumeration. It will produce an infinite succession of closer and closer approximations to 1/3 - but it will never produce the real number 1/3.
Your enumeration will only ever really generate finite length numbers. Anything which requires an infinitely long representation cannot be generated by an enumeration from your tree. None of the numbers with infinitely long representations are reachable by traversal.
Like I keep saying: You're redefining "enumeration", "traversal", and "visit". You can't enumerate the real numbers with your tree, because you can't traverse the tree to visit any number with an infinite representation.
If you're allowed to change the meaning of visit to "not really visit, but be able to sorta point in the direction of", and traverse as "sorta point in the direction of all of the parts of the tree", then sure, you can traverseJG the tree in a way where you visitJG all of the nodes, and thus you can enumerateJG all of the nodes by sorta pointing in their direction.
But back in the real world, visitJG != visit, traverseJG != traverse, and enumerateJG!=enumerate.
the real numbers without actually enumerating them", then you can
(That last line was an editing error which should have been deleted. But in the interests of keeping the conversation exactly as it occurred, I'm leaving it as is.)
Comment 209 (MarkCC @ 2/3/2010, 2:19pm)
Yes John, an enumeration is a complete listing of all of its elements. Which is really just begging the question: what is a listing?
You're essentially claiming that a "listing" is in some sense equivalent to a predicate. That is, if you can define a predicate describing a set of numbers, then you can enumerate those numbers.
Meanwhile, the rest of the world of computer science and math gives enumeration a rather different meaning: an enumeration of a set is a 1:1 mapping from a subset of the natural numbers to members of the set. In the case of an infinite set, it's generally a mapping from the complete set of natural numbers to the members of the infinite set. (Hell, it's even implicit in the word "enumeration"!)
If you have an enumeration of a set, one of the properties of it is that you can describe what natural number will be mapped to what element of the set. If you really have an enumeration, then finding the natural number associated with a particular member of your set is trivially computable.
But numbers with infinitely long representations aren't enumerable in your representation. If you search for them, you'll never find them. They'll never show up in the list. There's a theoretical representation of every real number - but it's not a representation that will ever appear in an enumeration. In your system, there is no natural number N such that 1/3rd is the Nth element of the list. So in what sense is 1/3 actually a member of your list?
Comment 210 (John Gabriel @ 2/3/2010, 2:34pm)
MChu:"That's exactly the point. Your mechanism will never produce 1/3 in its enumeration. It will produce an infinite succession of closer and closer approximations to 1/3 - but it will never produce the real number 1/3."
In other words, 0.333... is not equal to 1/3 in your opinion? No. You agreed in Point 1 that all real numbers can be represented in decimal. So we are not going back to that one. Each time I explain to you, you keep going back and forth.
You've already agreed to this. And now you don't agree with it any more? So, if you don't agree with point number 1, you may as well discard Cantor's diagonal argument because it is about an imaginary list that can contain infinitely represented numbers.
MChu:"But numbers with infinitely long representations aren't enumerable in your representation. If you search for them, you'll never find them. They'll never show up in the list. There's a theoretical representation of every real number - but it's not a representation that will ever appear in an enumeration."
The fact that I have imagined these representations means they are very *real*. And of course they appear in a list but I haven't got there yet because you still can't see that point number 4 is true. In fact, you are now disputing that you agreed with point number 1!
See why I must have you answer these questions first? If you don't see that these four statements are true, we cannot proceed with the next question and finally the proof.
As I see it now, you are disagreeing for the sake of disagreeing. I think you know you are wrong. Well, for one thing you are contradicting yourself.
Comment 211 (John Gabriel @ 2/3/2010, 2:46pm)
By the way, I have never redefined anything. Please don't go down that road. You are smarter than this, I think.
Just stay with the argument. If you do not understand I will try to explain. Just ask.
Comment 212 (John Gabriel @ 2/3/2010, 2:51pm)
MChu: Meanwhile, the rest of the world of computer science and math gives enumeration a rather different meaning: an enumeration of a set is a 1:1 mapping from a subset of the natural numbers to members of the set. In the case of an infinite set, it's generally a mapping from the complete set of natural numbers to the members of the infinite set. (Hell, it's even implicit in the word "enumeration"!)
Once we agree on all four points, the above statement is what I shall proceed to prove.
Comment 213 (MarkCC @ 2/3/2010, 3:05pm)
I'm not disputing that you can represent 1/3 as an infinitely long decimal string 0.333.....
What I'm disputing is that you can produce an enumeration of the nodes of your tree which will ever include that string. The fact that you've got a representation for the real numbers does not imply that you've got an enumeration of the real numbers. Your representation is not enumerable.
You can whine and complain all you like - but it's not going to change anything. If you claim that 1/3 appears in an enumeration of the nodes of your tree, then either you're lying, or you don't actually understand what an enumeration is, or you have some other meaning of enumeration that's different from the rest of us.
If you have an enumeration, then specifying what natural number maps to 1/3 must be possible. By the definition of an enumeration, there must be exactly one natural number that maps to 1/3 in your enumeration.
In one of the canonical ways of enumerating the rationals, 1/3 corresponds to 5: {0, 1, 2, 1/2, 3, 1/3, 2/3, 4, 1/4, 3/4, ...}. Given any rational number, I can find its position in that enumeration. It might take me a while to do it - if you give me a number like 623784/98237421, it'll take me a long time to find it - but I can.
In your system, you can't specify the natural number that corresponds to 1/3 in your "enumeration" - because 1/3 will never be enumerated by your "traversal".
The fact that you can represent the members of an infinite set does not imply that you can enumerate them.
Once again, whenever you get close to pinning John down on one of his errors, he starts throwing tantrums and flinging insults. Typical crank behavior: get too close to their errors, and they start getting angry.
Comment 214 (John Gabriel @ 2/3/2010, 3:23pm)
MChu: "You can whine and complain all you like - but it's not going to change anything."
Taking cheap shots at me does not help. In fact it makes you look stupid.
MChu: "What I'm disputing is that you can produce an enumeration of the nodes of your tree which will ever include that string. The fact that you've got a representation for the real numbers does not imply that you've got an enumeration of the real numbers."
Oh yes, it does. Suppose I represent the first six natural numbers of my set by a, r, b, w, e, t. Now, I create a bijection: f(1)=a, f(2)=r, f(6)=t. I have enumerated my set. In fact, the it is *required* that one can represent the members of a set. What you have written is the dumbest thing I ever heard! How on earth can you enumerate anything when you don't even know what it is?! Get real!
MChu:"Your representation is not enumerable."
Well, you are not able to comprehend even the most basic concepts. I still have to show that [0,1) is enumerable but I cannot do this unless you *understand* that the representations in my tree diagram represent every real number. In fact, you have already admitted this in your introduction. See paragraph 10:
MChu: "His enumeration is based on trees. You can create an infinite tree of the decimal representation of the numbers between zero and one."
My, oh my, but you have such a short memory, don't you?
Please be frank with me: do you really have a PhD in computer science? I am even doubtful that you are who you claim to be. You are contradicting yourself over and over again.
Now, you have to answer YES to question 4 and understand why the answer is correct otherwise I am wasting my time with you.
So you see why I did not bother engaging you even over a year ago? I imagines these would be the dumb, thoughtless responses I would receive. My problem is I often give others the benefit of the doubt when it comes to believing they are of reasonable intelligence.
Once again, we encounter the fundamental problem. John simply does not understand that the fact that a set is representable doesn't mean that it's enumerable. If something is enumerable, then it's representable; but the converse isn't necessarily true. The next comment of his continues to hammer on that point: he tries to take my statement that he can represent all of the real numbers using infinitely long representations as a proof that I accept that you can enumerate those infinitely long representations.
Comment 215 (John Gabriel @ 2/3/2010, 3:32pm)
In fact you have not realized it, but by stating what you have in Paragraph 10, you have essentially agreed to all 4 points already. The reason I brought up these points is because I wanted to make sure you understood what you were putting your head on the block for. Evidently, you did not know what you were agreeing with.
1. Do you agree that every real number in the interval (0,1) can be represented in decimal?
YES or NO
2. Do you agree that my tree contains every real number in the interval (0,1)? Don't concern yourself about finite/infinite numbers at this time. We don't care about *enumerating* the numbers at this stage, only that if we know a number, we can find it in the tree.
YES or NO.
3. Do you agree that if we traverse the tree in each level from top to bottom that we can be sure to enumerate all the finitely (not the repeating decimals or irrational numbers) represented numbers in decimal? I know there are duplicates but we shall not worry about this right now.
YES or NO.
4. Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?
YES or NO.
Paragraph 10 confirms all these questions with a YES. Now, I need to know that you know your admission in par. 10 is equivalent to all these 4 points.
Truthfully, the only reason I bothered engaging you is because I took the time to read your introduction and realized you were almost *there* in terms of understanding.
I now am beginning to see I was mistaken.
Comment 216 (MarkCC @ 2/3/2010, 3:42pm)
Oh yes, it does. Suppose I represent the first six natural numbers of my set by a, r, b, w, e, t. Now, I create a bijection: f(1)=a, f(2)=r, f(6)=t. I have enumerated my set. In fact, the it is *required* that one can represent the members of a set. What you have written is the dumbest thing I ever heard! How on earth can you enumerate anything when you don't even know what it is?! Get real!I did not say that you cannot represent an enumerable set. I said that the fact that a sent is representable does not imply that it is enumerable.
As I originally predicted, you're not addressing the substantial criticisms of your construction. You're picking out bits and pieces, mis-representing them, and then using them as a basis for flinging insults.
You have a representation for all of the real numbers. I have never claimed or implied otherwise. From the start, from the original post, I have continually said that you have a representation of the real numbers, but your representation is not enumerable.
You keep responding to tiny pieces of my criticism that you can misrepresent, while ignoring the substantial parts.
So let me repeat: By the definition of an enumeration of a set, given any element of the set S, it's possible - it's easy to discover what natural number corresponds to that element in the enumeration. If you have an enumeration, then specifying what natural number maps to 1/3 must be possible. By the definition of an enumeration, there must be exactly one natural number that maps to 1/3 in your enumeration.
So, John, what number is it? At what natural-number position will 1/3 appear in your enumeration? In fact, go ahead and pick any number you want that has an infinite representation as a decimal, and tell me where it will appear in your enumeration. If you've really got an enumeration, then it must be possible to say where some number with an infinite representation appears in the enumeration.
So come on, put up or shut up. Where does 1/3, or 1/9, or 1/π, or 1/27th, or 1/11th appear in your enumeration? What natural number maps to any number with an infinite representation? If you can't answer that, you don't have an enumeration.
Comment 217 (John Gabriel @ 2/3/2010, 4:37pm)
I am not surprised at your outburst and false accusations.
I have informed you that in order for me to proceed, you have to consent with the 4 statements I put to you.
Look at you: Come on John! Tell me, what number corresponds to 1/3, or pi or e!
Let me ask you: Come on Mark, tell me, what number corresponds to the smallest rational number greater than pi?
Would you be able to answer? NO. Yet you have the audacity to tell me that you are able to enumerate all rational numbers. So tell me then, where is this rational number in your list and tell me which natural number it is mapped to?
Unlike like you, I know that this smallest rational number greater than pi will be in your list and that some natural number will *eventually* be assigned to it.
Do I tell you that the rational numbers are not denumerable because you cannot provide this information to me? No, because I understand that it can be in your list.
Your arguments are lame.
Comment 218 (John Gabriel @ 2/3/2010, 4:42pm)
Let me correct you: I do not have to provide a natural number that corresponds to any of the reals in my tree. All I have to do is show you that it is possible to assign a natural number to every real in my tree and then I am done.
This is what enumeration is all about. Just another gaping hole in your understanding!
I'll give John one small point here. My understanding is that he's proposing his tree as a way of generating a specific enumeration of the reals. If that's the case, then it's simple to provide the position of any particular real number. But if he's not arguing that he's got a specific enumeration, then he doesn't need to be able to assign a position to a particular real number. However, he keeps harping on the fact that you can traverse his tree using a "left-right traversal", which he implies is a deterministic process - which would imply that there is a specific enumeration associated with that "left-right traversal". If that's the case, then it should be possible to say just when that traversal will produce some number with an infinite representation. He can't, because it won't.
Comment 219 (MarkCC @ 4:50pm)
John:
You can "inform" me of anything you want. It doesn't change reality.
You want me to agree with you that your tree is traversable in a particular way - but it isn't. And as usual, you ignore the point, throw insults, and whine.
I'm giving you the freedom to pick any number with an infinite representation in your tree, and tell me where it appears in your enumeration. I'm doing that, because you and I both know that you can't do it, and that the reason you can't do it is because you don't have an enumeration.
In return, you come back with something that by definition, doesn't exist, and ask me to show it to you. There is no rational number closest to π. It doesn't exist. And you know that.
But you're asking for the impossible, because you can't do what I asked. If you had an enumeration, you could specify where things occur in it. By definition, if you had an enumeration, you could specify where things occur in it. But you don't.
You keep wanting me to admit to agreeing with you that you can enumerate these things. But you can't. They're not enumerable. This is the crux of your argument, which is why you demand that I accept it. But it's not true. You cannot enumerate any numbers with infinite representations by traversing your tree - unless, of course, you redefine the words "traverse" and "enumerate".
And all that you can do is constantly play games like this - throw insults, shout, whine, and ignore the point of anyone who disagrees with you. You're a crank all right - and in typical crankish fashion, you demand respect that you haven't earned, and throw tantrums anytime anyone points out that you are wrong.
John, you are wrong. Your representation is not enumerable. A "left to right" traversal - or in fact any traversal of your tree will never reach a single number with an infinite representation.
Comment 220 (John Gabriel @ 2/3/2010, 5:03pm)
Nonsense. I do not have to provide a particular natural number that corresponds to any of the real numbers in my tree.
All I have to do is show that it is possible to assign a natural number to every real in my tree and then I am done. Then the real numbers in (0,1] are denumerable.
No where in the definition of countability is there a requirement to produce a particular natural number for any of the rational numbers.
I could denumerate my rational numbers by placing them into a one-to-one correspondence with the even natural numbers or the prime numbers. There is *no* requirement to produce a particular value. If you knew any mathematics, well then you would know this fact.
Alas, you do not.
Comment 221 (John Gabriel @ 2/3/2010, 5:03pm)
MChu:John, you are wrong. Your representation is not enumerable. A "left to right" traversal - or in fact any traversal of your tree will never reach a single number with an infinite representation.
Evidently not according to your paragraph 10!
Again with the same old problem. He just can't wrap his head around the fact that representation does not imply enumeration.
Comment 222 (John Gabriel @ 2/3/2010, 5:09pm)
MChu: There is no rational number closest to π. It doesn't exist. And you know that.
Oh yes there is. But don't worry, you can't find it. You can't do a lot of things in mathematics, but that does not prevent you from producing useful results and theorems.
Comment 223 (John Gabriel @ 2/3/2010, 5:11pm)
MChu: There is no rational number closest to π. It doesn't exist. And you know that
In fact at some or other position in your enumeration of rational numbers, it must appear. For if it does not, then your rational numbers are also not denumerable!!
At this point, another commenter jumps in. And as I predicted, he's going to use that to throw a tantrum because his rules are being broken, and therefore he's going to take his very special toys and go home.
Comment 224 (Scully @ 2/3/2010, 10:03pm)
Quite possibly the dumbest thing I have ever read:
MarkCC: There is no rational number closest to π. It doesn't exist. And you know that.
JG: Oh yes there is. But don't worry, you can't find it. You can't do a lot of things in mathematics, but that does not prevent you from producing useful results and theorems.
okay, really Gabriel?
We shall show, John Gabriel is wrong, and there is no rational closest to pi.
Proof: We shall prove by contradiction and assume r is in Q, such that r is the first rational after pi. Consider the interval (pi...r), then by Archimedes principal there exists a rational s in the interval, thus pi
Really? Try not to make it that easy...
Comment 225 (Scully @ 2/3/2010, 10:05pm)
this pi less than s less than r, contradction
And of course, John's fit and leavetaking - which includes an absolutely spectacular demonstration of arrogance. John is the smartest person in the whole world; he's only ever met one person who's as smart as he is. That's why we can't see the obvious brilliance of his magnificent proof: because we're just not at smart as him.
Comment 226 (John Gabriel @ 2/4/2010, 3:51am)
MChu: Let me put it to you another way: Suppose that you could enumerate the real numbers. Just "suppose". Now, if I were to ask you what natural number corresponds to pi, would you be able to tell me? NO.
Now, let's see if we can approach this in another way because your mental faculties are obviously limited. From my tree it is evident there are two distinct *infinities* (hate to use this word but anyway): the one infinity describes the top-down traversals and the other, the left-right traversals.
Try to imagine two bottomless wells: The one well is a reservoir for the top-down numbers. In other words, each time we visit one of these numbers, it goes into the reservoir. The other well is also a reservoir that stores all the numbers from the left-right traversals in a first-in-first-out fashion. The left-right reservoir is populated with infinitely many left-right traversals taking place at the same time.
So, at any given time, one can get a number from either of the reservoirs. Now, from the top-down reservoir one can produce a one-to-one correspondence with the even natural numbers. From the other reservoir, one can produce a one-to-one correspondence with the odd natural numbers. Result: A list of all the real numbers in [0,1). However, job is still not done. We go through the list removing the duplicates and what is left is exactly all the numbers in [0,1). And we have produced a bijection. Therefore the real number interval according to Cantor is *countable*.
There is an undergrad student here who is one of the biggest fools I have ever had the misfortune of meeting - Scully. Since he has started to comment here, this is my last comment to you.
MChu: You do not call anyone a crank for any reason whatsoever. You can call me a Cantor disputer or even better refuter. However, you have no right to call me a crank because you are unable to understand my arguments. Just so you are aware, there are mathematics professors who agree with me. My knol had a very high rating before it was voted down by fools the likes of Scully. This insignificant worm does not posses even 1/100 of my intelligence. You probably have less.
In fact the word crank comes from a German word meaning sick. Let me assure you, I have only met one person in my almost 1/2 century of existence who was intellectually on par with me.
As I mentioned earlier and have proved from this dialogue, you are not only incapable of engaging me, you are rude, arrogant and insulting. MChu, be prepared for insults when you call anyone a crank. You are the real crank!
It all ends just as I predicted at the start. John can't address criticisms of his proof. Every time he encounters something that he can't actually address, he throws another little tantrum and tosses around a few insults. Then he bitterly complains about how insulting I am towards him. And finally he storms off, because some other commenter, who I have no control over, decided to post a comment that broke John's special rules.
You see, John is special. He's the smartest person he's ever met. He doesn't have to follow the same rules as everyone else. In fact, he gets to make his own rules, wherever he goes, and everyone else is obligated to follow them. He can insult you, because he's much more special and intelligent than you are - but you're not allowed to do something as crude and horrible as point out that he's wrong.
John: you're a crank. In fact, you're a pathetic crank.
Physical Science
Technology
Comments
I'm sorry, Mark. I almost lost the will to live half-way through reading this. You have my sympathies for getting involved.
Posted by: Bob O'H | February 4, 2010 1:35 PM
Sadly, Dr. Chu-Caroll that is what many scientists / mathematicians / engineers encounter. The web is littered with websites of cranks who believe that they have trisected an angle, or who claim that they have discovered a new ground state of the hydrogen atom, and don't get me started on the crazy nutbar who decided that applying dental x-rays to a sample of hafnium 178 in an isomeric state caused the state to decay and a burst of gamma rays 60 times more powerful than the input x-rays to be release.
I believe that you dealt with this crank as well as anyone could, you tried to explain to him where he was wrong, and I believe that, even in places where you said you were getting angry, you were quite patient and polite.
Posted by: James T Matta | February 4, 2010 1:40 PM
Wouldn't this guy's argument imply that the set of real numbers is recursively enumerable?
Posted by: Tyler DiPietro | February 4, 2010 1:42 PM
John's case is a very sad one, where arrogance and poor communication skills has allowed him to seal himself off in a miniature world of crankism.
There is a silver lining though - even those of us not well-versed in mathematics can still recognize cranks from their classic behavioral traits.
Posted by: SteveWW | February 4, 2010 1:42 PM
WOW! I can't believe I read the whole thing. BUT... This did a better job of explaining the deconstruction of his proof than the original article, so thanks for the effort.
Posted by: iHunger | February 4, 2010 1:47 PM
I'm still not sure how you have so much time for beating up on cranks, but none for replying to legitimate queries and corrections.
Posted by: Ivan | February 4, 2010 1:50 PM
@3:
Yes. That's basically what he's arguing.
Posted by: Mark C. Chu-Carroll | February 4, 2010 2:00 PM
Mr. Gabriel is a child. I couldn't stomach reading the majority of that discourse for the exact reasons you point out. As far as I'm concerned, the credibility of Mr. Gabriel vanishes the instant he plays the "I'm smarter than you so you have to do what I say". Even if he was right, I *still* wouldn't care.
It's a shame that banning him would make him a martyr.
Many points, incidentally, to Scully's succinct contradiction, though I always find it amusing when blowhards take points that prove them completely wrong and state that it proves them right. "You can't do a lot of things in mathematics, but that does not prevent you from producing useful results and theorems." Sure, but you don't get to use that as carte blanche to patently ignore logic.
Posted by: Zero | February 4, 2010 2:02 PM
@6:
In case you haven't noticed, I try to write in an informal style to make things as comprehensible as I can to non-mathematicians. I don't think that I've ever written a single substantial post without having someone nitpick about some use of language that they thought was imprecise, incorrect, or objectionable. And I'm sick to death of obnoxious prats who feel like I'm obligated to address every comment they make, and correct every bit of terminology that I use in informal descriptions until it meets their approval.
Posted by: Mark C. Chu-Carroll | February 4, 2010 2:07 PM
*facepalm* This guy is really krank. Mark, you got my deepest respect for staying this calm. I couldn't have done that ;)
Posted by: Cthulhu | February 4, 2010 2:24 PM
"Now read carefully: Don't you dare address me ever again as you have in your previous comment!"
I always boggle at those who show up on somebody else's blog and start laying down rules. I simply can't fathom the arrogance and sense entitlement required for that sort of behavior.
"You DO NOT tell me what to do and you DO NOT get to say how things are done where I am concerned. Do I make myself clear?"
Apparently logic, mathematics, and reality DO NOT get to tell Mr. Gabriel how to do things either.
Thanks for taking the time to debunk this crank, though. My understanding of set theory is not as sharp as I'd like, but I believe reading this dialogue has honed it a bit.
Posted by: Pocket Nerd
| February 4, 2010 2:30 PM
I only have one criticism...why did you ever concede two? It's blatantly false. There are no infinite numbers in N or R, so the finite/infinite thing is irrelevant, but still, the point you kept making when arguing is that number two is false, and then number 4 doesn't matter. As you said 1/3 is not in his tree.
Posted by: Charles Siegel | February 4, 2010 2:46 PM
@12: It *is* in his tree, but it will never be enumerated :) that's the difference John is missing.
Posted by: Sam K. | February 4, 2010 3:00 PM
Ahh, so his tree includes limit nodes, not just the nodes after finitely many steps? I must have missed that (I've seen the same "proof" before, but without limit nodes)
Posted by: Charles Siegel | February 4, 2010 3:05 PM
@12:
Whether two is true depends on just how you define "tree". In set theory, you can define trees (and other structures) with infinitely long paths or components. The best example that I know of how to do that is in John Conway's book on the surreal numbers - his construction of the surreals using the infinite construction technique. In the surreal numbers, you can't define the fraction 1/3 using less than ω steps. But 1/3 is a surreal number.
John Gabriel defines his tree in a way where you can interpret his construction in the infinite set-theoretic sense. (Of course, he probably doesn't deserve even that path; in this comment thread, he disavowed the axiom of choice - and I don't think that the infinite constructions work without the axiom of choice.)
Posted by: Mark C. Chu-Carroll | February 4, 2010 3:08 PM
Mark, Yeah, I know that you can define trees that way (and I'm not entirely sure how much choice comes into it, I haven't sat down and read Conway's book carefully yet) but I had mistaken (or perhaps merely interpreted) his tree as not being of that sort. If we allow nodes at omega, then sure, of course the tree contains all the reals, many times for some of them. But the tree itself is no longer denumerable, and that's a point I've focused on with people who've come up with this "proof" and were acting in good faith (there are some people who are merely confused, rather than cranks, and fortunately, they're usually happy to have their confusion cleared up)
Posted by: Charles Siegel | February 4, 2010 3:15 PM
Oh man, you have infinitely (heh) more patience than I do. After somebody saying, "Don't you dare address me ever again as you have in your previous comment" on my blog, I'm pretty sure I'd devolve into "I'll address you however I want, dickweasel!" and proceed to call him nothing but "dickweasel".
Also, that guy's a dickweasel.
Posted by: jtradke | February 4, 2010 3:21 PM
if you ask me, the guy is 100% troll. And even if he isn't, he's clearly trying to anger you by deliberately getting your name wrong among other more blatant insults. Not that I don't understand the value of demonstrating why he's wrong but you've done that many times already and I honestly think he's only trying to get your goat at this point, even if he does believe he's right. This is beyond beating a dead horse. This is incinerating a dead horse and then nuking the ashes. Thanks for the posts anyway, I enjoy your blog.
Posted by: pjb | February 4, 2010 3:27 PM
I couldn't get through more than about a quarter of the crap. How can you do it Mark?
Posted by: Thony C. | February 4, 2010 3:32 PM
I guess I'm also an idiot in John's world since I believe in reality.
Posted by: foole | February 4, 2010 3:32 PM
@9
Welp, you wanna be a mathematician, you gotta act like one, that's all I'm sayin.
But I'm sure it's more fun, and draws more traffic, to pick on the mentally disordered.
Posted by: Ivan | February 4, 2010 3:32 PM
@15
Oh, and you're welcome.
Posted by: Ivan | February 4, 2010 3:38 PM
I can't help but feel that your failure to relate to John echoes his own failure to develop a robust mental representation of infinites. Well done on making him look daft - it's good that reality isn't overturned during these skirmishes - but if we are to get anywhere as a race I'd like to bring the cranks with us into the higher dimensions, otherwise our production of proof after proof may be nothing more than a list of real numbers.
Obviously I've done rather a lot of acid.
Posted by: fraac | February 4, 2010 3:42 PM
@19:
Well, it doesn't hurt that I've got pneumonia, and don't really feel well enough to do actual work. As a result, I'm bored, and doing this didn't exactly strain my abilities.
Posted by: Mark C. Chu-Carroll | February 4, 2010 3:42 PM
From what I can tell, the mapping wouldn't be a bijection even if you did assume his premises. It isn't one-to-one since numbers appear more than once in the list (e.g., 1/2 has two representations, 0.5 and 0.4999...).
Posted by: Tyler DiPietro | February 4, 2010 3:49 PM
@21:
I never said I was a mathematician. Never. In fact, I have repeatedly pointed out that I an not a mathematician.
Even if I were, I wouldn't be obligated to satisfy the arbitrary demands for terminological changes from random twits. Like I said before, I'm sick to death of obnoxious idiots who believe that they've got the right to demand that I do what they want. There's never been a single post in nearly four years on this blog where some nitpicking ass didn't find something to complain about. You don't need to read the blog if you don't like it. But coming back into multiple comment threads to gripe because I didn't make the wording change that you demanded? That pretty much makes you an obnoxious jackass.
Finally, I don't post based on what will bring in traffic. Hell, I can't even begin to predict which posts are going to get traffic and which aren't. But it's the new trendy insult: you don't like the fact that I don't write my blog the way you want me to - therefore I must just be traffic whoring. It can't possibly be because I think your demands are silly. It can't possibly be because I write for fun, and I pick what I feel like writing about. No, it's all about gaming the traffic.
Please. Take your complaints, and go find some other blogger to pester.
Posted by: Mark C. Chu-Carroll | February 4, 2010 3:51 PM
Posted by: Bayesian Bouffant, FCD | February 4, 2010 3:57 PM
Bayesian Bouffant, FCD:
I don't know the answer to that one, but I'm going to suggest referring to the name "John Gabriel" as a "crank handle."
Thank you, I'll be here all week.
Posted by: Sharkey | February 4, 2010 4:14 PM
On his Knol page he warns anyone believing in any type of religion to read his Knols at their own risk.
That said -- he argues like a Creationist: Given [I am right], if you do not agree, it is because you do not understand.
I can kind of follow WHY he thinks he's correct, but the issue of enumeration does seem to be his downfall. On the upside -- I feel like I understand Cantor's theorem better now, so thanks! :)
Posted by: Aaron | February 4, 2010 4:15 PM
btw -- you still haven't proven you're not Bob Barker. Perhaps you are really an AI program coded by John Gabriel, and you pass the Turing test with flying colors so bright that he can lose an argument to his own genius creation.
Posted by: Aaron | February 4, 2010 4:17 PM
@9, 26
I'll admit, I may not always make my points tactfully, but DAYum...
Posted by: Ivan | February 4, 2010 4:21 PM
I *suspect* that "left-to right traversal" doesn't mean a traversal of the tree but a traversal of the digits.
ie for pi, the traversal is
3
.
1
4
1
5
9
etc.
Not that that helps any.
What you've run into is a common problem, and there has been at least one good psychological paper on it.
Basically, those who are very skilled in something tend to underrate their aptitude in something (because they know who the really good people are), and those who aren't very skilled overrate their aptitude because the same lack of aptitude means they also lack the aptitude to rate where they are in the continuum.
It's not that he is being arrogant when he says he has only met one person who is smarter, it's that he is unable to evaluate which people are smarter.
Posted by: Eric | February 4, 2010 4:21 PM
Gabriel:
> Nonsense. You have a PhD? From where did you get your degree? Must be somewhere in the US. Allow me to educate you[...]
Too bad you couldn't go to a country with reputable academic institutions, Mark, and that you had to settle for this 3rd rate system. I can't believe you turned down that gracious tutoring offer.
Posted by: B-Con
| February 4, 2010 4:43 PM
Wow, you were amazingly patient with this guy. I took a quick trip to his knol site. Most of his arguments seem to boil down to "The actual math disagreed with what I want to believe, so I'll just made up my own stuff." One example is that he wants to believe that the decimal system of numeration provides unique representations of numbers, so any demonstration that it does not must be false. A visit to his site is an instructive example of crankery at it's best.
Posted by: VinceH | February 4, 2010 4:53 PM
It is not the correctness of the argument that seems to define many of these cranks, it is their irrational behavior and their emotional outbursts.
Mathematics is certainly the topic on which we should be able to discuss things objectively, and gradually all reach a consensus on what is right (and what is acceptable). There is some room for personal preference, you could prefer NF Set Theory to ZFC Set Theory for example, but the underlying nature of our abstractions is mostly unambiguous and non-debatable.
What I find scary, is not the hordes of cranks trying to break down the doors, but that some of them seem to already be on the inside. That is, in many of these discussions, the irrational and emotional outbursts come from BOTH sides, those that are cranks on the outside and those that are defending the established view.
One apparent weakness of ALL people is that we have very strong tendencies to believe that we know way more than we do. That our knowledge is deeper than it really is. Discussion is an excellent way to truly confront that weakness, but only if one is willing to stay objective. Willing to be wrong.
Some people genuinely want to learn new or different things, most are just trying to prove that what they already know is the truth (regardless of what it actually is). Knowledge (right or wrong) is sticky. And that can be a big problem.
Sometimes the established knowledge is incorrect, it can and does happen. Rarely, but it is possible. Still we'll never know the state of our beliefs if we can't hold civilized discussions.
If everyone that holds a non-standard opinion was quickly dismissed as a crank our knowledge and understanding would stagnate. Big new leaps of inspiration come from surprisingly unexpected directions.
Sometimes a crank is a crank, sometimes he (or she) is another Cantor. We always need to separate the ideas from the people, otherwise our reactions are driven off the wrong stimulus. Dismissing an idea only because it contradicts an authority or because the person presenting is annoying is just more crankery. It just holds us back.
Posted by: Paul W. Homer | February 4, 2010 4:56 PM
"One example is that he wants to believe that the decimal system of numeration provides unique representations of numbers"
Wow, does this mean that he is going to take on limits, the Calculus and Newton next?
Actually, I would pay to see Gabriel and Newton debate live on stage.
Posted by: KeithB | February 4, 2010 5:01 PM
And the moral of the story is... do not feed the trolls
Posted by: Curt | February 4, 2010 5:12 PM
Mark,
Without belittling your level of experience with cranks, it seems to me that John Gabriel suffers from a pathological personality disorder.
I would suggest that a public "debate" is not the appropriate way to engage him, and may be fueling his disease. Whenever you encounter such cases (and it seems that you do quite often) I believe an evaluation of whether a logical response is the ethical thing to do is warranted.
Joao
Posted by: Joao | February 4, 2010 5:13 PM
I get it -- a left-to-right traversal is where you write the numbers down all on one line, and a top-down traversal is where you write numbers on new lines!
Posted by: John Fouhy | February 4, 2010 5:21 PM
tl;dr - bloke's clearly a nutter. surprised you had the patience to deal with him as long and as reasonably as you did, you're clearly a better man than I.
Posted by: col | February 4, 2010 5:41 PM
John Gabriel, you might be interested in Alan Turing's paper "On Computable Numbers" if you haven't read it already. It sounds like the idea you are alluding to when you refer to algorithms. It's true that every number you can think of can be represented by a finite program. There's a program to generate the digits of 1/3, pi, e, and "almost all" real numbers. The set of all (finite) programs in a given language is numerable, because you can convert them to binary and then list them in order. However, there do exist non-computable numbers such as Chaitin's constant. It's a fascinating topic. Mark has written some posts on this topic in the past, and I also recommend Chaitin's book "Meta Math!".
I realize you will only respond to Mark C., but if you want vindication, you can respond to this challenge. Write out the first few elements in your infinite enumeration of [0,1] separated by commas. I'll give you one number in particular that is not in your list. Don't worry, I won't cheat by just picking a number not listed in the fist few terms. I know that the finite list represents a pattern that can be extended indefinitely. Cantor's argument guarantees a number that will never be included in your list, no matter how far you extend it. But you have to define one list in particular: You can't cheat by going back and changing it. If you do I will produce a new number not in the modified list.
Posted by: Jair | February 4, 2010 5:46 PM
For anyone interested in reading Turing's paper "On Computable Numbers", I would highly recommend Charles Petzold's book "The Annotated Turing: A Guided Tour Through Alan Turing's Historic Paper on Computability and the Turing Machine". Petzold does a pretty good job of explaining the actual paper and includes interesting tidbits about Turing. The whole paper is presented with annotations (hence the title) to help explain what paper is saying.
Posted by: Jason Jackson | February 4, 2010 6:16 PM
Posted by: Scott Simmons | February 4, 2010 7:14 PM
I'm struck by the way Gabriel conflates mathematical proof with functional algorithm: he keeps refering to the time something takes, and the space it occupies, and it's clear that he's thinking like an engineer, not like a logician.
Posted by: Ahistoricality | February 4, 2010 8:46 PM
@44:
I think that's unfair to engineers. :-)
Seriously - I'm an engineer, not a mathematician or a logician. And in my experience, one of the hallmarks of a good engineer is recognizing the difference between theory and practice. A good engineer knows the theory, and uses the theory - but they don't confuse it with practice.
Theory describes what you can do, and how things work in the abstract, without the constraints of reality. Practice is all about understanding the constraints, and finding a way to work within them. You can't do the practical side without knowing the theory: it tells you what's possible, and how it works. But you also can't do the practical without understanding how the constraints of reality affect what the theory tells you you can do.
John pushes things from the practice side into the theory without understanding that he's mixing up incompatible concepts into a giant mishmash of rubbish. And as an engineer, I consider that to be an indication in reality, he understands neither the theory nor the practice.
Posted by: Mark C. Chu-Carroll | February 4, 2010 9:05 PM
http://knol.google.com/k/kurt-deligne/funny-shit-that-john-gabriel-says/2oygz0grkt42u/1#edit
Posted by: Kurt Deligne | February 4, 2010 9:24 PM
This suddenly makes maths seem a lot more interesting. Thanks!
Posted by: simba | February 4, 2010 9:44 PM
What's wrong with conflating mathematical proofs with functional algorithms? That's the insight of the Curry-Howard correspondence: that (intuitionistic, usually) propositions and proofs correspond to types and terms in a functional language/lambda calculus. And consistency is equivalent to all functions in the corresponding theory terminating in a finite amount of time.
There are people who work on constructive mathematics that stick to this view, and they've demonstrated that a lot of the results in classical mathematics can work in constructive mathematics. And as a bonus, whenever they prove ∀x. P(x), they give you the ability to compute a function from values x to proofs that P holds for x. And when they prove that ∃x. P(x), they've computed some x, together with a proof that P holds for that x (and as a combination, when they prove that ∀x. ∃y. P(x,y), they give you a computational procedure that given an x, computes a y with evidence that P(x,y) holds).
So, there's nothing wrong with thinking of proofs algorithmically. You just have to not be a crank about it. Cantor's proof is even algorithmic; you can formalize it in some programming languages, and have it checked by a computer.
Posted by: Dan Doel | February 4, 2010 9:45 PM
I never even took calculus, so this is all well above my head. From reading websites and books I can follow along a bit and I more or less understood what you two were arguing about and why he was wrong. That being said, I still found two different pieces of his "logic" to be breathtakingly stupid. Someone noted above that he argues like a creationist; specifically, he seems to be making the ontological argument when he says something to the effect of "if i can imagine it in my set, then it must be real" (can't find the exact quote and there's no way I'm re-reading all of that to find it), as though his imagination is an arbiter of reality. He also seems inexplicably hung up on the 4 premises he lays out originally and you agreeing with them. It's as though, if you agree with him, then reality itself will change because of that. It's absolutely hilarious that he keeps insisting that it's absolutely imperative that you accept premise 4. Premise 4 is, as you point out, exactly what's wrong with his theory, but if your feeble mind would just accept it, you'd see he's right!
Posted by: Ryan | February 4, 2010 10:00 PM
I wholeheartedly agree with John Gabriel's comments. Furthermore, it is simple to enumerate any number. If I choose to use a silly limited tree with only 0-9 in it, then I'm limited by my own folly. If, however, I am intelligent and have a PhD from a real country, I make up my own tree and include branches labeled "1/3", "π", "John Gabriel", and anything else irrational I can think of.
Before you disagree, please swear you are you and get the post notarized (assuming notaries exist in your 3rd rate US legal system).
Posted by: Michael Maguire | February 4, 2010 10:48 PM
I think that people like John Gabriel are fascinating in a way. It's not too difficult to understand how his arguments are wrong and why his point 4 cannot be true. I think that it's safe to assume that JG is capable of understanding why he is wrong, which makes it really strange that he should try to defend his view so vigorously. I cannot help but to agree with Joao (#38) that some sort of a mental disorder is likely. I wonder if any psychologists have ever made a study of people like these. It would make for an interesting read.
Posted by: Flaky | February 5, 2010 12:53 AM
At 190 was because you were coming close to saying "I'm going to allow anyone to comment whether you like it or not." I think he got mad at that because he specifically said he didn't want anyone else to comment. Brushing that off was a sign of disrespect pure and simple.
Insults math-related are also 'disrespectful' but he could write those off as being a product of stupidity, and therefore not entirely your fault.
Posted by: TruthIsSubjective | February 5, 2010 3:20 AM
In regards to his obsession with identity, it may be that he wanted to believe that you were an imposter rather than that you were really this hostile to his arguments.
Posted by: TruthIsSubjective | February 5, 2010 3:25 AM
@52 That's absurd. Demanding that no one else respond to his comment on someone's public blog is about as disrespectful as you can get.
Posted by: BF | February 5, 2010 4:50 AM
If you cannot represent the numbers, then how can you enumerate them? Surely you have to know what you are counting?
Posted by: Anonymous | February 5, 2010 5:03 AM
I first noticed Mark Chu Carroll’s (henceforth MChu) blog regarding me being a crank over a year ago. I read some of his blogs and quickly realised that he had nothing of any real worth to say.
At first I thought to respond but decided after careful consideration my efforts would be wasted. MChu had clearly made up his mind. If he were interested in a debate, he would not have started out by calling me a crank.
In MChu’s world, if you have a PhD (does not matter in what subject), you are allowed to call anyone who does not have a PhD names of your choice. For some unknown reason, it does not occur to MChu that calling someone a crank is a significant insult. In his position, entitlement allows him to be insulting and obnoxious. Oh, he works for Google, so everyone thinks he must be bright.
People like MChu project themselves onto others. The attributes they hate most in others are those only they want to posses: authoritarian, arrogant, dismissing and condescending.
A few days ago I decided to respond even though I subconsciously knew it would not really change much if anything at all.
As expected, MChu was disdainful, arrogant and insulting. I quickly reprimanded him and let him know that I would not accept his insults (being called a crank is an insult) nor would I respect him for having a PhD. I do not believe in respect because of entitlement. MChu quickly realized he was not dealing with a fool or an amateur. His reputation was on the line. This is why he chose to respond.
Well, rather than admit he is wrong, he chose to follow a path of argument for the sake of argument. At first he wrote that he would agree to all my conditions if I accepted the Axiom of infinity. I ignored him because this axiom is not required to prove my argument. In fact, it is not required in mathematics at all.
So, he agreed to the first 3 points but decided to make an issue of point number 4. Only thing is, without his knowing, he had not only agreed to point number 4, but he had agreed to all four points.
Realizing he was on the verge of being exposed for the fool that he is, he decided to harp on the fact that representation is not the same as enumeration. Well, of course it is not. However, one cannot enumerate what one cannot represent. He fails to see the absurdity of his argument!
How can you count anything whose form or representation you do not know? How absurd! Does MChu realize this? Obviously not.
As an example of how dumb this argument is, one can talk about enumerating particles of dark matter. If you don’t know what it is you are enumerating, how do you enumerate it?!
I do not believe the real numbers are either countable or uncountable. I have only shown that according to the father of fools - Georg Cantor, that the real numbers are countable using his assumptions and his definition.
I do not even believe the real numbers are well-defined but this is not at issue here.
http://scienceblogs.com/goodmath/2010/02/_so_remember_back_in.php
The real fool and the real crank is MChu!
Posted by: John Gabriel | February 5, 2010 5:20 AM
John is funny. If representation and enumeration are the same thing then his tree is unnecessary; we can represent all the reals by their infinite decimal representations, so hey, they must be enumerable! Pffft.
What natural number N corresponds to the real 1/3 in John's enumeration?
Posted by: Stephen Wells
| February 5, 2010 5:41 AM
Seeing math errors exposed is sometimes fun, sometimes interesting, so thanks. But this discussion became a waste of time very soon, when all "arguments" began to repeat over and over (not to speak of the insults).
@ John Gabriel: I think we can all agree that you are wasting your time too. You are free to believe this is because the rest of us are stupid, I on the other hand believe your pride prevents you from seeing the error you make (and, in fact, math is not the easiest of all things, so even professional mathematicians with phds and all do make mistakes: you would not need to be ashamed). But you do understand you are not convincing anybody, so why not simply turn to something else?
Posted by: q | February 5, 2010 5:45 AM
Johnny, Johnny, do you understand that just because something is formally representable, that doesn't mean it's enumerable? No? Wow, that's dim.
Posted by: Stephen Wells
| February 5, 2010 5:51 AM
If nothing else, this kind of thing really helps get to the crux of why your intuition is wrong on some of these things (because it's intuitive thinking that normally lies at the heart of the mistakes people are making when they think they have proven / disproved centuries of accepted theories).
And I guess that's the crux of this. These kind of theories only hold together while you can keep all the ideas nebulous and abstract, and then, because stuff is fuzzy, you can glide over the inconsistencies.
Tying things down with concrete examples will show you where and how you are wrong. But some people are so convinced that they are right that they see these clear counter-examples as proof that they have been misunderstood. And a misunderstanding is hardly a contradiction is it? And so they will never learn...
Posted by: Joe Kilner | February 5, 2010 6:43 AM
@56...
John... if you're going to mangle Mark's name, can you at least do us the favor and do it in an entertaining way? I suggest "MChu-bacca"
(sorry... are you sure it's not troll-feeding time??)
But seriously Mark... hurry and wake up... I wanna see this crank bitch-slapped around again... This post of yours is a wonderful response to the typical antagonistic asshole*, and it's a great example of why your blog is one of my favorites.
*my humble apologies for being redundantly redundant... lol
Posted by: doctorgoo | February 5, 2010 8:14 AM
The hilarious thing is that his breadth-first enumeration method doesn't even include 1/3. It produces values arbitrarily close, but it will NEVER reach 1/3. Forget its nearest neighbor and all the other obfuscation.
I don't even need Cantor.
Posted by: Matt Skalecki | February 5, 2010 8:36 AM
Mark's argument is that representation does not imply enumeration. Your response is that enumeration implies representation, which of course doesn't contradict Mark's point at all. Might I suggest that, in future, you formulate your counter-arguments such that they address the argument actually made? Failing to do so gives the impression that you have failed to understand the argument, and are just flailing around like a gibbering imbecile. I'm sure you wouldn't want anyone to think that.
Posted by: MartinM | February 5, 2010 8:56 AM
To all the fools that responded:
My argument does not suggest representation implies enumeration. If Mark had agreed to the 4 conditions (which are true), I would have proceeded to demonstrate how the real numbers are countable according to Cantor's original argument.
Posted by: John Gabriel | February 5, 2010 9:31 AM
Dear Mr. Gabriel,
We give up. You win. I'm sure your Abel Prize is in the mail, so please enumerate the moments until it arrives.
Posted by: Tomato Addict | February 5, 2010 9:34 AM
John Gabriel wrote:
I think most people he has met have just indulged him in his delusions. From what I've seen here, it's probably much easier than dealing with him when he's been proven wrong.
John Gabriel wrote:
It appears he has no sense of irony either.
Posted by: Deen | February 5, 2010 9:36 AM
@56:
John, do us all a favor and stop wasting both your and our time.
I'm not the one who invoked my PhD repeatedly. In fact, I'm the one who repeatedly pointed out that it doesn't matter. In math, credentials don't matter. What matters is the correctness of the arguments. If a second grader were able to show an argument that something is wrong with Cantor's diagonalization, it would be no less true than if a PhD from the best university in the world made the same argument.
Where did I ever say anything about the axiom of infinity? The discussion is all there in the post above. Please, show us, where I said anything about it?
I wasted a ridiculous amount of time responding to you patiently and politely. In response, you've been rude, insulting, condescending, and generally obnoxious. Don't bother responding with an argument - the entire discussion is here, in its entirety, to be seen by anyone who cares to look.
You're basically an almost perfect prototype of a typical crank. You've got pretty much every single one of the classic attributes. The incredible arrogance: you're the smartest person you've ever met; you've only ever met one person who could even match your
amazing brilliance. The temper tantrums when presented with an argument that you can't respond to. The endless excuses about the foolishness of your opponents, even though you can't respond to their arguments. It would be sad if you weren't also such a total asshole.
By the way, I looked at some more of your knols suggested by one of my readers. On top of your other wonderful attributes, you're also apparently a pathetic little jew-hater. As it happens, I'm a Jew. So hey, there's another reason for you to hate me.
Posted by: Mark C. Chu-Carroll | February 5, 2010 9:51 AM
@48:
There's nothing wrong with that. My point isn't that you shouldn't draw the connection between theory and practice. It's that a good engineer recognizes that one of the biggest parts of doing things in practice is dealing with the constraints of the real world. But those constraints don't apply to the theory. So there are limits on what you can do in the real world that don't correspond to limits of what you can do in theory; and there are things that make sense in theory, but which don't work well in practice, because they don't fit the constraints of reality.
For example, back before I wrote this blog, I used to participate on usenet. I got into a long argument with a guy who insisted that the halting problem is solvable. The argument was that every real computer has a finite amount of state - and that, therefore, they're all finite state machines, and the halting problem is solvable for FSMs. In theory, he's absolutely correct. Our computers are limited to finite state. In practice, while their state is limited, it's large enough that it might as well be infinite. For example, ignoring anything except built-in memory, the computer that I'm typing this on has 234359738368 possible states. In theory, that's finite, and you can write a halting solver. In practice? Forget it.
A good engineer knows and uses the theory. But they don't forget the distinction between the unlimited world of theory, and the highly constrained world of reality.
Posted by: Mark C. Chu-Carroll | February 5, 2010 10:04 AM
Yes, you are just another stupid, arrogant jew. Well-said. By your own admission of course.
This is the reason I did not respond a year ago because I know you are a moron.
It's hilarious how you carry on about the same crap all the time:
"I wasted a ridiculous amount of time responding to you patiently and politely. In response, you've been rude, insulting, condescending, and generally obnoxious."
"The temper tantrums when presented with an argument that you can't respond to. The endless excuses about the foolishness of your opponents, even though you can't respond to their arguments. It would be sad if you weren't also such a total asshole."
You arrogant, conceited little jewish swine! You call me names and it's okay. F..k you! Who the hell are you little yehuda, to call me a crank?!
Filthy swine!!!! Oh, let me see your response:
As I expected John, you are just another crank. Blah, blah, blah.
Whenever you are refuted, you simply revert back to your stale old argument. You know, I remember reading a similar story in Mein Kampf about how often Hitler would refute the stupid jews and two weeks later they were back at square one again. Seems very little has changed eh?
Listen asshole, I am ashamed that I have Jewish blood. I don't consider myself a filthy dirty arrogant fool like you. I consider myself Greek.
Yes, go ahead now and play the victim card. Ha, ha, ha. CRANK!!!!!!!!
Posted by: John Gabriel | February 5, 2010 10:07 AM
When a Jew loses an argument, he plays the anti-semitic card or the Holocaust card.
Posted by: Anonymous | February 5, 2010 10:10 AM
WOW!
I'm overwhelmed by the intellectual level of Mr Gabriel's arguments.
Posted by: Thony C. | February 5, 2010 10:14 AM
Frankly I don't give a crap what this fool thinks. What has upset me is the fact that he calls me a crank. EVEN IF I WERE WRONG (WHICH I AM NOT), WHAT RIGHT DOES THIS STUPID, JEWISH C..T HAVE TO CALL ME A CRANK?!!!! WHAT RIGHT DOES ANYONE HAVE?
If you want to publish a view that opposes someone else's, there is a right way to do it. He could have called the title of his crappy blog "John Gabriel - Cantor Doubter", John Gabriel's Anti-Cantor argument.
The swine has no right to call me a CRANK!!!!!
NO RIGHT. I am righteously indignant. The rest of you pathetic assholes should be pointing this out to moron MCHu. Not blowing his dick!!!
Posted by: Anonymous | February 5, 2010 10:20 AM
@69: ...
Posted by: Deen | February 5, 2010 10:20 AM
Cranksplosion!
Posted by: Deen | February 5, 2010 10:37 AM
#69
Wow...just, wow.....
Posted by: Feynmaniac | February 5, 2010 10:47 AM
Who am I to call you a crank?
John, as I've explained to you numerous times... We're talking about math here. Credentials don't matter.
When it comes to math, you are a crank! I don't need any special credentials to make that statement. All I need is a reasonable argument to support my claim.
I'll also just note that I never invoked anything about antisemitism in our actually discussion. I wouldn't, because it's irrelevant. You could be the most antisemitic son-of-a-bitch in the history of the world, and it wouldn't affect the math. You could also be the most saintly, wonderful person in the history of the world, and it wouldn't affect the math. When it comes to the math, you're a total, raging, idiotic crank.
The fact that you are also a raging loonie of an antisemite is just icing on the cake. And I couldn't resist pointing it out now, just to watch you go totally batshit over the fact that you've been arguing with a Jew.
Posted by: Mark C. Chu-Carroll
| February 5, 2010 10:48 AM
Stunning mathematical arguments in #69.
I for one am convinced.
Arguing with and bested by an "arrogant, conceited little jewish swine".
I bet that stings.
Posted by: Rev. BigDumbChimp
| February 5, 2010 10:58 AM
Mark, I think that you errored slightly in answering an unqualified yes to John's second question (that his tree contains a represents every real number). It DOES, but only if you consider (potentially infinite) paths to be the representation, because no single node on the tree represents a number with a nonterminating decimal representation (whether it be rational (1/3) or irrational (pi)).
This is the crux of the issue, because the breadth first enumeration DOES make a one-to-one mapping between the nodes and the naturals.
Posted by: Matt Skalecki | February 5, 2010 10:59 AM
Anyone who is having trouble understanding Cantor's argument should check out Asimov's essay on the subject in "Asimov on Numbers". Asimov is really good about explaining the concepts at a high school level or so (I first read the book when I was in HS), and making it fascinating.
Posted by: Andrew | February 5, 2010 11:25 AM
Dr. Chu Carrol, please! Your patience so far has been exemplary and truly impressive. Working in research for a very short time myself, I have to agree with #38 and #51; I have already received emails from no less than three obsessives trying to claim things from Einstein being wrong to the Earth, Moon and Sun all weighing a single gram. These people produce astounding amounts of very basic nonsense to try to prove something which they genuinely believe to be true and then try to attract the attention of academia. You've given him your time and reasonable politeness; he will never, ever change. I think that continuing to engage with these sorts of obsessed people might well be to the detriment of their health.
Following the recent outburst of anti-Semitism, you have surely reached the point of blocking this poster. If I may suggest it, you don't /need/ to respond or explain after that point; you are plainly dealing with someone who is simply fraught with aggression.
Posted by: MJ | February 5, 2010 11:31 AM
The problem in Gabriel's argument is middle-school-level logic: he's assuming that if a statement is true, its converse is also true. This doesn't work. Knowing that p implies q tells me nothing about the truth value of "q implies p."
Posted by: Vicki | February 5, 2010 11:33 AM
Good GOD! Ok, it's one thing to be wrong. We have all done that. It's another thing to be arrogant. I think we've all done that too, but most of us recognize it as a flaw, and work to diminish it.
This embrasure of arrogance is astonishing, and so entertaining that I must wonder: can it be real? Or is Mr. Gabriel in fact a performance artist of some sort?
And yes, I say Mr. Gabriel. I do not believe for a moment that you successfully completed a juried doctorate at a respectable university. No professors would endorse such a thing!
Posted by: AnyEdge | February 5, 2010 11:36 AM
As I predicted. You lost the argument and what better way to save face - play the Jewish card.
No, you have no right to call me a crank, not for math, not for religion, not for anything.
Anyone who reads this will know you are a bigoted son of a bitch who doesn't know shit about computer science, never mind mathematics.
Posted by: John Gabriel | February 5, 2010 11:52 AM
Delusion, thy name is John Gabriel.
Posted by: Rev. BigDumbChimp
| February 5, 2010 11:58 AM
As a non math guy, I struggled through the article. I was rewarded for my efforts in the comments.
Posted by: Rob H | February 5, 2010 12:02 PM
!!!!
Wow!
Haha, this just got real interesting.
Posted by: mike | February 5, 2010 12:14 PM
"I was rewarded for my efforts in the comments."
Exactly.
Mark,
This has made my day.
Posted by: mike | February 5, 2010 12:18 PM
What is it with cranks and projection?
Posted by: MartinM | February 5, 2010 12:26 PM
Aw come on, John. You can do better than that! Where's that vastly superior intellect of yours? Surely you can do better that a baseless accusation of bigotry.
You're not just a crank, you're a pathetic crank. Watching you degenerate into rage and spittle is both entertaining, and incredibly enlightening. Each comment you post just further demonstrates just what an incredible, amazing, pathetic crank you are. Seriously - I don't think I've seen anyone - not even John Davidson - degenerate into such rampantly pathetic spittle-flinging insanity as you're doing here. Please, keep it up. You're doing a great job of teaching all of us pathetic losers just what a true crank looks like.
Posted by: Mark C. Chu-Carroll | February 5, 2010 12:27 PM
Synchronicity, thy name is John Gabriel:
http://www.penny-arcade.com/comic/2004/03/19/
Posted by: Sharkey | February 5, 2010 12:37 PM
I don't regularly read this blog, and the math discussed in this post is somewhat above me, but someone I follow on twitter linked here and after reading most of John's comments (you're right - they get very tedious, very quickly) I have something to say.
Mark - your mini bio on the left and the way you express yourself makes it somewhat clear that you must be an intelligent guy. How you haven't managed to figure out that John Gabriel is a troll eludes me. As you might have heard before, the solution is just not to feed him.
Posted by: Omri | February 5, 2010 12:38 PM
Trust me, John's arguments make no more sense even if you know the math.
Posted by: Davis | February 5, 2010 12:41 PM
MChu:
I have a much higher readership than you do and I am going to expose you for the fraud and crank that you are:
http://knol.google.com/k/john-gabriel/a-crank-mark-chu-carroll-calls-me-a/nz742dpkhqbi/21?hd=ns#
Posted by: John Gabriel | February 5, 2010 12:53 PM
Minor correction Davison...
But please don't repeat his name more than twice. If you do he appears.
But that's a good comparison from the small about of crank-ness that John "Those damn Jews don't know math" Gabriel seems to be displaying here.
Posted by: Rev. BigDumbChimp
| February 5, 2010 1:01 PM
Be sure to check out this knol about Mr. Gabriel too, titled "Funny shit that John Gabriel says":
http://knol.google.com/k/kurt-deligne/funny-shit-that-john-gabriel-says/2oygz0grkt42u/1#
Wow... just... this guy is a complete crankapotamus!
(Hey! If Deen @74 can say "Cranksplosion!" ... then I should be allowed to say "crankapotamus!")
Posted by: doctorgoo | February 5, 2010 1:14 PM
Go read his blog: and note that he disables all comments so that he cannot be contradicted.
http://mathphile.blogspot.com
It's HILARIOUS.
Posted by: HA HA HA | February 5, 2010 1:20 PM
@91: Judging by the amount of text generated by this guy, I don't think he is a troll. It really looks more like a mental disorder.
@93: You now have one more reader! Your pages make for a very amusing read, indeed. Another entry in the 'cranks' folder.
Posted by: boris | February 5, 2010 1:21 PM
Yikes. I just went and rad his Knols and the Davison comparison is more than close, it's frighteningly right on target.
He even quotes himself at the end of his rants.
Posted by: Rev. BigDumbChimp
| February 5, 2010 1:23 PM
rad = read
typos for the win
Posted by: Rev. BigDumbChimp
| February 5, 2010 1:25 PM
Hmm, so what's Gabriel's crank ranking vis-à-vis Davison? Inquiring minds want to know the structure of the cranktinuum! (And does the Cranktinuum Hypothesis hold?)
Posted by: Ivan | February 5, 2010 1:37 PM
@Ivan #100,
I'm not sure about the Davison scale, but it's looking to be something like 0.7 timecubes. Other people are free to dispute that, of course.
Posted by: Shawn Smith | February 5, 2010 1:40 PM
Wow, I wish this thread of Gabriel's allowed comments:
http://knol.google.com/k/john-gabriel/are-you-a-crank-when-they-can-t/nz742dpkhqbi/22#
Yes, John, by definition, you are a crank. The photo in Webster's dictionary, next to the "crank" entry should be a photo of you.
John then goes on to compare himself to Copernicus and Newton.... what, no Galileo-like persecution complex for John??
This really is funny stuff here!
John, some serious advice: Step back for a moment. Put aside your rage and your bigotry. Look at what Mark and others are saying about your math, and take their criticisms seriously.
A non-crank is capable of admitting when they are wrong. If you want to prove that you are not a crank when it comes to this topic, then I encourage you to rethink your math.
Also, your anti-semitism makes it near-impossible to put up with you. Such attitudes have no place in any professional environment. If you expect people to take YOU seriously (as opposed to your math), then you should really get an attitude adjustment.
Your name is attached to these bigotted statements. Any employer of yours, current or future, might Google them some day. Is this really how you want people to think of you? It is plainly obvious from the tone, writing style and grammar that you are writing as "anonymous" in comments 70 and 72.
Shame the fuck on you, John Gabriel.
Posted by: doctorgoo | February 5, 2010 1:45 PM
@92:
So is that how the superior intellect responds to an open, uncensored, unlimited discussion of his crackpottery? By running away with his tail between his legs, back to a safe forum where no one is allowed to comment or criticize?
And so it end, exactly the way I said it would: when you couldn't respond to criticism, you threw your tantrums, tossed around your insults, and ran away.
Posted by: Mark C. Chu-Carroll | February 5, 2010 1:48 PM
Rev, you foolish fool! You've named him twice! Counting Mr. Chu-Caroll, that's 3 times!
I'm getting the hell out of here.
Posted by: Paul | February 5, 2010 1:55 PM
Running away from who? A retard like you?
I refuted every one of your arguments you stupid bastard! And what did you do? You decided it was time to play the Jewish get-out-of-embarrassment free card - anti-semitism.
Fuck you asshole! I did not bring up your Jewishness - you did! I piss on the cunt that spewed you forth you vile, disgusting slob!!!
Posted by: John Gabriel | February 5, 2010 1:57 PM
That knol is a work of art. I know nothing convinces me of the validity of a mathematical proof more than pink explanatory text.
Posted by: Paul | February 5, 2010 1:59 PM
I'm just waiting for him to say he loves it so!
Posted by: Rev. BigDumbChimp
| February 5, 2010 2:00 PM
[inappropriate comment pretending to be from John Gabriel deleted.]
Posted by: John Gabriel | February 5, 2010 2:01 PM
Running away? How could I suggest such a thing? After all, a superior intellect like you would never do something like that, right? After all, taking a discussion from an open, uncensored, public discussion forum to one where no one is allowed to comment or point our your errors - why, that would be cowardly. And a proud genius like you would never act like such a coward, would you?
Heh. Just keep it up, John. The more you shoot your mouth off, the more you prove what a completely loonie crackpot you are. You're a sad, sad case John.
Posted by: Mark C. Chu-Carroll | February 5, 2010 2:03 PM
Mark, please! I know you're angry but /ban him/. It's the kindest thing to do - this anger isn't constructive!
Posted by: MJ | February 5, 2010 2:05 PM
Well, the real question is, can you enumerate that?
Listen Johnny boy, you have made your thoughts clear on Jews before, he just pointed it out and you fell hook line and sinker for it.
Not a very bright move there little guy.
Posted by: Rev. BigDumbChimp
| February 5, 2010 2:06 PM
Put a whole lot of dimwits together and what do you get? Mark Chu Carrolls anti-mathematical blogs!
Once MChu can no longer discuss the maps, he announces he is Jewish. Ah, I am so sorry. Am I supposed to feel sorry for you? Fool.
Posted by: John Gabriel | February 5, 2010 2:06 PM
MChu: Oh you self-righteous, pompous ass!
The minute you decided to continue insulting me, you asked to be abused and now I am showing you what it is like to be abused. Please now, stop pretending to be so gracious. You are a deceitful fool.
Posted by: John Gabriel | February 5, 2010 2:11 PM
[inappropriate comment pretending to be from John Gabriel deleted.]
Posted by: John Gabriel | February 5, 2010 2:13 PM
I can see the whole picture now. You thought I was going to bother with the other idiots on your site? Pfah, they are your croonies. Do you think anyone with a modicum of intelligence can't see this? My, you are dumber than I had realized.
Posted by: John Gabriel | February 5, 2010 2:14 PM
As comment 32 points out, John Gabriel suffers from the Dunning-Kruger effect.
Also, Mr Gabriel, if you fit these criteria, you are a crank, completely independently of whether that word comes from German krank ("ill"). As already hinted at in comment 67, "crank" isn't an insult like "poopyhead" or "asshole", it's a technical term with something close to a definition. It's completely and utterly ridiculous to claim certain people have or have not a right to call certain other people cranks – everyone has the right to point out a mere fact.
While I am at it, you should learn some biology. Especially human genetics.
Heh. Newton, too, had a huge ego! That would be very entertaining indeed :-)
Except if you count at infinite speed, which Mr Gabriel clearly believes he can. Then it will reach 1/3 after ∞/∞ seconds… whatever that is, LOL. Mr Gabriel appears to believe it's 0 or 1 or something.
"After infinite time" and "never" is the exact same thing, Mr Gabriel. That's also why 0.999… and 1 is the exact same thing.
Posted by: David Marjanović
| February 5, 2010 2:16 PM
From John's Knol:
Seriously: 1 sentence between "I never claimed X" to "X".
Wow.
-Richard
Posted by: Richard | February 5, 2010 2:19 PM
And you sure as hell did.
John someone with your supposed intelligence shouldn't let their buttons be pushed so easily.
So very very easily.
Step away from the computer little guy. It'll do your blood pressure some good.
Posted by: Rev. BigDumbChimp
| February 5, 2010 2:22 PM
@114:
Masquerading as John is not acceptable. Even for someone who is as much of a total schmuck as John, that's not OK. I'm going to remove the comment.
Posted by: Mark C. Chu-Carroll | February 5, 2010 2:29 PM
I literally laughed out loud, and the door is open. Good that everyone else is already gone… I hope…
ROTFL!
Sounds about accurate.
Posted by: David Marjanović
| February 5, 2010 2:31 PM
@119, then delete 108 as well. That was me too.
Posted by: That's Reasonable | February 5, 2010 2:34 PM
@110:
Honestly, I'm not angry. I did get angry during the "debate", when he kept throwing insults every time he couldn't actually respond to the argument. But now... I'm somewhere between amused and bemused: It's like watching a train-wreck in slow motion - watching him descend lower and lower into spittle-flecked insanity.
I'd almost like to do another post about him - I hadn't read his Knol pieces on the problems with the real numbers, or the problems with calculus, or the problems with set theory. It's amazingly foolish crackpottery. But given his raging insanity, I don't think I want to provoke him quite that much. (But really - he insists that there's no such thing as an empty set, because every set contains itself! I mean, imagine how hard it is to resist going off on that!)
Posted by: Mark C. Chu-Carroll | February 5, 2010 2:37 PM
@166: 'Except if you count at infinite speed'. No, the tree simply does not contain a node with the value 1/3, so it cannot be reached at any speed. Paths of the tree represent decimal sequences, which may be infinite and therefore correspond to real numbers in the range [0,1]. But of course according to JG, 1 isn't there at all, because 0.999... isn't 1.
Posted by: Flaky | February 5, 2010 2:37 PM
While I understand that internet dialog is sometimes abrasive, and I certainly understand the frustration of dealing with someone so intractably misinformed, I think you would have been better off taking a higher road here.
As you say, it was foolish to get drawn in. Failing that, it was also I think a mistake to get personal. Attack the ideas, not the person. Invoking the term 'crank' is a personal attack. Again, I understand the frustration, but I think you might have been feeding the troll.
Posted by: Mikem
| February 5, 2010 3:00 PM
Mikem @124...
How is that a personal attack? It is the correct term to use to describe Mr. Gabriel.
Let me ask you something... if I said you were ignorant on Topic X (whatever that might be)... is that an insult? I'll readily admit that I'm ignorant on many topics in which I have no experience nor formal education.
In the same way, I cannot think of a more concise way to describe Mr. Gabriel except for the word "crank".
The fact that he doesn't like being described as such doesn't mean that the definition of "crank" fits him perfectly.
Posted by: doctorgoo | February 5, 2010 3:04 PM
:-D :-D :-D
No, see comments 102 and 116 and references therein.
Where I come from, we feed trolls till they cranksplode, so I don't see the difference :-)
Posted by: David Marjanović
| February 5, 2010 3:15 PM
@doctorgoo:
"Let me ask you something... if I said you were ignorant on Topic X (whatever that might be)... is that an insult?"
If someone thought, felt, or believed that they were an knowledgeable or an expert on Topic X, and particularly if they said this out loud, and especially if they bragged about it, then YES, calling them "ignorant" is an insult. A pretty big one.
Mostly, the connotations of the word 'ignorant' are negative. It's what educated people accuse un-educated of being. It's what city people accuse country people of being. Generally it is not used neutrally or as a complement, even if it is true.
And of course there is always the SNL quote "Jane, you ignorant slut ..." from Dan Aykrod. Definitely not meant to be a complement :-)
Posted by: Paul W. Homer | February 5, 2010 3:35 PM
All Ideas are not equal. Some ideas are just plain wrong. People who cling to these just wrong ideas aren't always just trying to forward an idea, many times they are being cranks.
Posted by: Rev. BigDumbChimp | February 5, 2010 3:40 PM
MarkCC:
Better or worse than this guy?
Posted by: Blake Stacey | February 5, 2010 3:57 PM
Mark, you have to keep this guy around. He's a fucking trip. I'll tell you more in the next jewsletter, and we can make a plan at the next jew meeting---that is, if we can turn away from all those delicious babies.
Posted by: Jewy McJewstein | February 5, 2010 4:05 PM
Let me put this in a format Mr. Gabriel can understand.
John, you must say yes to the following statements:
1. The elements of the power set (the set of all subsets) of a finite set of order greater than 2 (since you don't think the empty set is a "real" set) cannot all be labeled using only the members of the set. For example, the non-empty subsets of {1,2,3} are {1}, {2},{3},{1,2},{1,3},{2,3},{1,2,3} Since the original set has only three elements we cannot name all of the seven subsets using just 1, 2, and 3.
2. The same result can be extended to any non-finite set, such as the set of natural numbers. *
3. The set of real numbers contains the set of all sums of distinct ternary unit fractions (fractions with numerator 1 and denominator an integer power of three) (1/3+1/9) is one such sum but (1/3+1/3) is not.
4. No two different sums of ternary unit fractions are equal and every sum of distinct ternary unit fractions is a real number.
5. These sums can be put into 1-1 correspondance with the subsets of the set of powers of three.
6. Since the set of powers of three is countable, the set of subsets of these is of greater cardinality and therefore cannot be countable.
7. As these numbers constitute only a subset of the set of real numbers, the real numbers cannot be countable.
John, since I am a genious and you are not you must say yes to all of these statements.
*The argument goes as follows: Suppose you can label all of the subsets using only the elements of the set. Let Y={All the set names where the set name is a member of the set.} N={All the set names where the set name is not a member of the set.} Y and N are both subsets of our infinite set. All set names are in either Y or N and the two sets do not intersect. Assuming all subsets have been labeled, either N's name is in N or Y . This gives a contradiction as N's name in N means that N's name is not in N. And N's name in Y means that N's name is in N and therefore N and Y intersect, again a contradiction. The conclusion is that the original assumption, that we can label the subsets using the elements of the set, must be false.
Posted by: Vhurtig | February 5, 2010 4:12 PM
[In blue (Indicating John's narrative of the argument)] If I can represent them, I can enumerate them. Seriously: 1 sentence between "I never claimed X" to "X".
Wow. -Richard
You quote me out of context but that is to be expected. That statement was made to MChu so that he might allow me to complete my argument.
And yes, in order to enumerate the real numbers, one would first have to represent them. What are you wowing about idiot?
Posted by: John Gabriel | February 5, 2010 4:28 PM
"MarkCC:
I'd almost like to do another post about him - I hadn't read his Knol pieces on the problems with the real numbers, or the problems with calculus, or the problems with set theory. It's amazingly foolish crackpottery."
Carry on little Jew boy. I have already consulted an attorney to see what I can do about you calling me a crank - you fucking dirty asshole!!! If I can help it I am going to make you eat shit - you fucking bastard!!! After I am done with you, you will think twice about badmouthing your pussy cat.
Filthy kike!!!!
Posted by: Anonymous | February 5, 2010 4:31 PM
10 bucks says that the attorney will laugh in his face, and will call John a crank too!
LOL
Posted by: doctorgoo | February 5, 2010 4:35 PM
@Vhurtig, in your footnote, it reads like N is the empty set. I think I understand what you intended, and I think this is a valid rephrasing.
"Suppose you can label all of the subsets of a set A using only the elements of the set (such as P(N) being enumerable).
Let Y={All the labels of subsets such that the subset label IS a member of said subset.}
Let N={All the labels subsets where the subset label IS NOT a member of said subset.}
Y and N are both subsets of our infinite set. All subset names are in either Y or N and the two sets do not intersect. Then N and Y are sets containing elements of A and are themselves subsets of A.
Assuming all subsets have been labelled, either N's label is in N or Y . This gives a contradiction: N's name in N means that N's name is not in N. And N's name in Y means that N's name is in N and therefore N and Y intersect, again a contradiction. The conclusion is that the original assumption, that we can label the subsets using the elements of the set, must be false."
Posted by: MJ | February 5, 2010 4:39 PM
I wonder what the hate speech laws in the UK are like. I'm sure MarkCC's attorney would let him know...
Posted by: Sharkey | February 5, 2010 4:41 PM
@132 John Gabriel
"...representation does not imply enumeration. I never once claimed this was the case... If I can represent them, I can enumerate them."
Richard gave the context, and how does context even matter?
Call me an idiot too! I'm one of Mark's 'croonies' (sic).
Posted by: mike | February 5, 2010 4:46 PM
So what would JG say about the proof that the reals are uncountable which uses the Baire category theorem?
Posted by: Anonymous | February 5, 2010 4:47 PM
The fact that you can't tell the difference between those two statements is staggering. Linear Algebra and Differential Equations is the furthest I've bothered studying math, and it's even obvious to me why you're off your rocker.
Posted by: Paul | February 5, 2010 4:53 PM
@Anonymous #133,
If that was Herr Gabriel in #133, any law that would punish Mark Chu-Carroll for calling John Gabriel a crank could just as easily punish John Gabriel for calling Mark Chu-Carroll a "retard", an "asshole", and a "vile, disgusting slob!!!" (#115) It makes me wonder if #133 is a Poe or sockpuppet.
Posted by: Shawn Smith | February 5, 2010 4:55 PM
Hee-hee, now there's a cure for your rockin' pneumonia, and I bet it helps your boogie-woogie blues, too. ;-)
Hey, speaking of music, haven't seen a Friday Random Ten in a while. You can't be too ill to enjoy music, can you?
Posted by: Jud | February 5, 2010 5:04 PM
Is there a Poe equivalent for a Math crank? A Gabriel perhaps?
Posted by: iHunger
| February 5, 2010 5:08 PM
Paul @127:
"If someone thought, felt, or believed that they were an expert"... but was in fact, woefully ignorant on the topic at hand... then yes, indeed the word "ignorant" is correctly applied against them... regardless if his feelings are hurt by the honest truth.
In these cases you provide, one can easily see that "ignorant" is being incorrectly (and unfairly) applied across the board. This makes this an insult instead of a truthful statement of fact.
Just like the word "crank"... it describes John (WRT this topic and perhaps others) perfectly. The fact that this truthful statement upsets him does not mean that it's unfair to call him that.
Posted by: doctorgoo | February 5, 2010 5:12 PM
Scott Simmons:
Actually, I meant to say that almost all real numbers are not computable. Thanks for pointing this out.
Posted by: Jair | February 5, 2010 5:21 PM
MJ @134
I agree, your wording is much clearer than what I used.
Posted by: vhurtig | February 5, 2010 5:28 PM
@doctorgoo:
I was always taught that ALL words have both denotations and connotations. That is, the dictionary definition of the word is only a subset of it's usage and meaning.
There are lots of factual statements that one can make about people that are or will hurt their feelings, regardless of whether they are true or not.
Most (all) people are in denial of some aspect of their personality or self. When I was young I was fond of saying that there are three views of a person: a) how they see themselves, b) how their friends see them, and c) how the world sees them. For most (all) of humanity, these three views are not particularly aligned. And it's in these mis-aligned places that you can really hurt someone.
For instance, I like to live happily under the delusion that I'm right up there in looks with Brad Pitt and friends. While I know it's not true, since I don't have to look at myself all day long it's a useful, confidence building falsehood. One that I truly appreciate my friends playing along with (white lies can be acts of kindness).
In traveling, I learned to never say that something was "cheap", because in many parts of the world, particularly in Asia that can be seen as very insulting. It can mean inexpensive, but it can also mean inferior. And it changes region by region.
If you pick 'loaded' words, particularly when you know they are loaded, you can't hide your intentional choice behind the banner of "the truth". If you meant to hurt someone, and you do, then you you've done what you intended; any excuse is just an excuse.
Posted by: Paul W. Homer | February 5, 2010 5:54 PM
I would like to address some arguments about Mark that have gone relatively unchallenged. I am a long-time reader of this blog. Mark makes these arguments perfectly well but it might mean something different coming from a reader.
1. 'Mark flaunts his degree and job.'
I've been reading mark since before he worked at Google and I wasn't even sure about his highest level of education until reading this particular blog. Throughout the history of this blog Mark mentions his degree and job only when it's relevant, which is rare.
2. 'Mark is a popularity whore.'
Mark's Friday Random Ten is one of the most consistent series in this blog and it always receives the least number of comments.
I'm disappointed that this criticism came from Ivan, who believes that Mark didn't correct errors because he is focused on popularity. No, Mark has always maintained that his blogs have unedited minor erros. Mark makes it clear that he is not a mathematician, and his blogs are not being submitted to math journals. The mistakes, typos, etc. leave an artistic originality, and it doesn't alienate us readers with math educations.
3. 'Mark is a mean name-caller'
Mark is only pointing out what should be obvious to anyone. John Gabriel damaged his own reputation more than Mark could. Judging by his tantrums John Gabriel was personally affected before Mark labelled him. Also, I think it's safe to say that most every reader has less kind names for John Gabriel than what Mark has wrote.
Posted by: mike | February 5, 2010 5:57 PM
I'd like to see a video, or even a transcript, of dialogue between John Gabriel upon being pulled over for speeding by a cop. The cop need not be Jewish, nor John Gabriel as drunk as Mel Gibson was that time that he claimed to own all of Malibu.
I imagine that it would start something like this.
Cop: Do you have any idea how fast you were going?
JG: I do not want you to say anything else. Just YES or NO. I am waiting for your response before I continue my proof that I was not speeding. I will ask you a few more questions and then I will show you my proof. Deal?
Cop: I ask the questions here. Playing this little game of "you have to do it my way, and answer my questions, or I'll take my car and go home" is bullshit. It's just an excuse for breaking the law.
JG: "That's *exactly* what it means to be speeding: it's an driving through an infinite set of points, so you'll never stop enumerating them; but you can pick out any particular value, and it will be enumerated after a finite amount of time.
Cop: I don't care if the road is continuous or discrete, and don't care about Zeno's paradox. Neither will the judge.
JG: You do realize that to accomplish all the left-right traversals of this road means starting off at an entrance. But this is okay because we can leave a marker at the last node we forked off somewhere. Like maybe throwing a beercan out the window.
Cop: I don't care about forks in the road. I'm asking you if you know how fast you were going on this road.
JG: Nonsense. You have a badge? From where did you get your badge? Must be somewhere in the US. Allow me to educate you: A road traversal does not say anything about visiting mile markers in any order or sequence except that the mile markers are visited exactly *once*.
Cop: As I originally predicted, you're not addressing the substantial criticisms of your driving. You're picking out bits and pieces, mis-representing them, and then using them as a basis for flinging insults. Step out of the vehcile, sir. Lean over and put your hands on the hood of the car. Or, as they say, "assume the position."
JG: I'm not assuming anything, I have won this argument. See what I mean? I started to prove this to you but look how you responded. I am not playing games with you. I am not prepared to continue unless you answer my questions. The reason for this is obvious: if you do not answer satisfactorily then I cannot continue the proof because you can always waver later on....
[sound of handcuffs being locked onto wrists]
Posted by: Jonathan Vos Post | February 5, 2010 8:37 PM
Bwahaha. Thanks for posting this summary, Mark. Now your cranky crank "friend" can see just how many people read your blog and just how un-private a blog is as a forum for crank proofs that boil down to proving you can enumerate the reals by using words that implicitly assume you can enumerate the reals. Or to assuming that ad hominem attacks constitute a mathematical proof. And why the fetish about decimal numbers? A binary tree would make you the Star at the next crank math convention.
But I'll play. No, I don't agree that you, personally, can represent every number in decimal. Before we continue, could your please write out 1/3 as a decimal number? All of it. Get back to us when you are done. Maybe you'll realize your implicit assumptions about your tree are flawed before you finish this task.
Posted by: CCPhysicist | February 5, 2010 8:50 PM
@97: ... It really looks more like a mental disorder.
Maybe this one: http://en.wikipedia.org/wiki/Narcissistic_personality_disorder
Posted by: Anonymous | February 5, 2010 9:30 PM
I would like to apologize for JG running away from the argument... sorry!
Posted by: Scully | February 5, 2010 10:09 PM
just as an outsider looking in,how were cranks handled before the blog? say in the 50's or 60's did the math prof just toss these people out of his office and that was that?
Posted by: Dave | February 5, 2010 11:46 PM
I am fascinated by cranks, so I found this pretty interesting. I have to say though, I didn't have the patience to read the whole post.
I've always wondered how crankiness happens. How does someone who is, if not a great genius, at least not a complete moron get like this? And I guess now I know the answer, at least in this case: He has a pretty dramatic personality disorder that prevents him from ever considering the possibility that he might be less than perfect or that the people around him are not idiots.
Posted by: Mike Benfield | February 6, 2010 1:49 AM
Even though I have about as much understanding of the mathematics as my brain-damaged cat would, this was absolutely fascinating from a psychological point of view.
Posted by: Buffybot | February 6, 2010 2:22 AM
mike @147
"I'm disappointed that this criticism came from Ivan, who believes that Mark didn't correct errors because he is focused on popularity."
I agree that Mark is not focused on popularity, but I think he sometimes leaves in major errors because he misinterprets the corrections as "nit-picking."
I guess that's bound to happen some of the time, but I'm still shaking my head over the finger tree posts.
Posted by: Brian Jaress | February 6, 2010 2:43 AM
Well that confirms it, J.G. is fucked up.
To be honest though I am much more disturbed by Vorlath(i.e. the other Cantor crank from the previous posts on this blog). He doesn't seem to be ... ah ... emotionally disturbed the way J.G. is, which makes it impossible to understand WHY he persists in trying to refute the diagonalization argument.
Posted by: CranksAreInteresting | February 6, 2010 3:16 AM
MChu:
Now that all your friends have donated their two cents worth, you can feel a lot better about calling me a crank. You should continue to call everyone who disagrees with you a crank. This is the right thing to do. Oh, and if anyone refutes your argument, just tell them you are Jewish. That will be the winning line.
Bravo Mark!
Posted by: John Gabriel | February 6, 2010 3:29 AM
J.G. it's not everyone who disagrees with MarkCC, but rather those who do not understand mathematical logic, namely the subset of individuals, which includes yourself. No offense.
Posted by: Scully | February 6, 2010 4:33 AM
The fact that a weasel like you Scully who has no brain and whose English is worse than someone from Eastern Europe agrees with MChu, gives me comfort.
Scully, even after you graduate (which won't say much for you - Scranton is a 3rd rate American university and most of us know American universities are not that great anymore) and have lived for 60 years, you shall still not be in my league.
Have you noticed that I have been ignoring you - worm?
Posted by: John Gabriel | February 6, 2010 5:17 AM
One shouldn't kick a man when he's down, but I can't resist this.
John Gabriel Wrote:
..and whose English is worse than someone from Eastern Europe..
Given that Joseph Conrad, who is regarded as one of the greatest English novelists, as well as Jacob Bronowski and Arthur Koestler, who are regarded as the two best English popular science writers, are all of Eastern European origin JG's comment comes across as a compliment.
Posted by: Thony C. | February 6, 2010 6:46 AM
MChu: No, it does not end here. It ends here:
http://knol.google.com/k/john-gabriel/mark-chu-carroll-who-is-a-crank-calls/nz742dpkhqbi/21#
Of course there is a good chance you might also delete this comment.
Posted by: John Gabriel | February 6, 2010 7:37 AM
JG @161,
Reread comment 103:
As for what you said:
He didn't delete the link to this post the first time you posted it... and he didn't delete it when I linked to it as well @102, either. So he's not going to delete this one either. Why should he care if you respond like that??
Posted by: doctorgoo | February 6, 2010 7:56 AM
@155:
A proper correction to the finger tree posts is coming. I really did screw that one up - and I've been waiting to have time to properly write up both how the things really work, and what I got wrong. It's taken much longer that I wanted, because I just don't have as much free time as I'd like to work on it.
Posted by: Mark C. Chu-Carroll | February 6, 2010 8:59 AM
@161:
John, the only comments I've deleted were the ones that unfairly attacked you. You're the one who can't face uncensored forums where people might disagree with your loonie crankery.
Posted by: Mark C. Chu-Carroll | February 6, 2010 9:01 AM
Remember, you're dealing with superior knowledge like this:
http://knol.google.com/k/is-0-999-equal-to-1#
Posted by: Peter Hurford | February 6, 2010 9:08 AM
J.G., I told you all you would do is mock my school and that's what you did. Like I've said before, you mock eastern europe, but look at the Russian, Polish, and Hungarian undergrad and grad math contests. If you could even understand half of the problems, you would be smarter than I thought. Likewise, in America, we have the Putnam. Those who score on the Putnam are problem solving mathematicians, and guess which one of us did that?
(In case you were wondering, it wasn't you.)
Posted by: Scully | February 6, 2010 10:37 AM
@161 "MChu: No, it does not end here. It ends here:
...Comments have been disabled on this knol"
It couldn't possibly end there. You must convince each other through open discussion that one of you is right. John Gabriel, if you are to convince Mark that he is wrong then you must properly address his mathematical criticisms. But you can't receive criticism if you don't allow comments.
John Gabriel, this points to the main difference between Mark's blog and yours. Many readers, including myself, disagree with Mark a lot. Mark *invites* our criticism and then works to resolve disagreements.
By the way, your mathematical criticism of Mark has been resolved. It's clear to almost everyone exactly how you are wrong, why you are wrong, why you think you are right and why you are unwilling to accept that you are wrong. That is why Mark is calling you pathetic. We readers now are intimately familiar with the ugly way you think. Just consider what it takes for a reader to agree with you now: we must have prejudice against jews, we should be convinced by you (without comment) that most mathematicians are wrong, and we should know that we are likely not as smart as you.
If you know more than us and are smarter than us, there is nothing we can teach you. Stop wasting your time reading us and write more knols. Forget about us with our third-rate American educations.
Posted by: mike | February 6, 2010 11:08 AM
Posted by: Scott Simmons | February 6, 2010 11:19 AM
@165: That stuff is just so funny! He starts with the assumption 'All rational numbers can be represented uniquely in the decimal radix system.' Which of course, if true, would trivially mean that 1 is not equal to 0.999... And he backs that up by saying 'If the above assumptions are true, and there is no reason to believe they are not...'
This is quite good stuff too and shows where that business with 0.999... comes from: http://knol.google.com/k/how-we-got-radix-systems#
Posted by: Flaky | February 6, 2010 11:38 AM
"John, the only comments I've deleted were the ones that unfairly attacked you. You're the one who can't face uncensored forums where people might disagree with your loonie crankery."
MChu: Ha, ha, you pathetic crank! If you had done this, you would have deleted most of the comments on this site because anyone who reads them shall see that blundering idiots post non-math related comments and those who do, don't have a clue whereof they speak.
Well, I guess once you play the Jew card, you're off the hook eh? Gosh, I must remember to play that card next time. Only thing is, I don't get into arguments where I am not sure of what I say.
Posted by: John Gabriel | February 6, 2010 12:31 PM
"Here's the crux of his argument - which is exactly what I said in the original post that spawned this stupid discussion. He believes that being able to represent numbers implies that you can enumerate them. I'm not quite sure what "Representation Enumeration" means; my guess is that he left out a "⇔" symbol."
I wrote representation {is required for} enumeration. It was modified by you! You knew you were in a pickle, so you did a Fagan on me. So what's new?
Posted by: John Gabriel | February 6, 2010 12:35 PM
MChu: You are such a snake. You kept back all the comments until you were no longer able to engage me. Then you decided to open the floodgates and release all these morons to confuse the matter and swing things in your favour. Pfah! Knucklehead!
Yes, dumbo. I tried to tell you over and over again that unless you can represent what it is you are counting, you can't even begin to talk about enumeration.
Now you have already admitted the four points are true but you are too thick to realize it. If this were not so pathetic, it would be funny.
Posted by: John Gabriel | February 6, 2010 12:39 PM
John, you're welcome to believe whatever you want. But the fact remains that I have allowed you to say whatever you want in comments on this blog, without deleting or editing anything, no matter how offensive you've been. And yet, you still make accusations that I'll delete your comments, along with the strong implication that I've deleted your comments in the past.
I'd also like to ask just when I "played the Jew card". I told you that I was Jewish, because I know how you hate Jews - but I've never even implied that it had anything to do with the mathematical argument. As I've said, over and over again: who I am is irrelevant to the math. What my credentials are is irrelevant to a mathematical argument. What my religion is, what my skin color is, what my nationality or ethnic background is, what my education level is, what my profession is - they're all irrelevant. In a mathematical argument, there's only one thing that's relevant: the math. You're the one who's obsessed with credentials and identity. You're the one who's thinks that there's any relevance whatsoever whether or not various historical mathematicians were Jewish or not. I keep on saying - in mathematical discussions, the only thing that matters is the math. Whether a mathematical argument is made by a third grader, or a fields-medal winning PhD makes absolutely no difference. The only thing that matters is the content of the mathematical argument - the identity of the arguer is absolutely meaningless, completely irrelevant, and has absolutely no role to play in an evaluation of the correctness of the argument.
Your problem is that you can't win an argument on the math. But being a spectacularly arrogant crank, you can't even begin to face the fact that when confronted with valid mathematical arguments that critique your crankery, you can't address those arguments. And since you can't address them, you throw tantrums, call people names, and play all sorts of games to try to hide from the fact that you're wrong.
I think that a big part of why you're so livid over this blog, why you keep coming back to fling more insults, is simply because you can't handle the fact that you made yourself look like an ass, and you can't do anything about it. Your normal method is to write your little Knol pieces, and put them up without allowing comments. You're using to being in complete control. On your Knols, no one is allowed to say anything critical of you, no matter how ridiculously wrong you are.
You came to this blog with a set of demands about how the debate would work. And as I predicted, you threw a tantrum and stormed off when someone didn't follow your rules.
John, the truth of the matter is, you're an ass. You're a two-bit crank who's incapable of having a rational discussion with anyone who isn't willing to kiss your ass. You've made an argument that you can't possibly defend, because it's wrong. But in typical
crank fashion, you just make all sorts of excuses, fling insults, and throw tantrums.
But at the end of the day? You're still just another pathetic crank. And your juvenile behavior reduces you to a pathetic laughingstock.
If you weren't such an ass, I'd pity you.
Posted by: Mark C. Chu-Carroll | February 6, 2010 12:53 PM
@171:
John, do us all a favor, and fuck off. I've never edited a users comments without clearly flagging the fact that I'd done so. And the only times I've done so were when it was a case of a commenter being abusive towards another commenter.
You're welcome to call me names all you want. But lying to save face is absolutely not acceptable. I have never edited any comment that you've posted. Not once; not ever. I have not changed a single character. The only comments in this thread that I've changed were the two that someone posted masquerading as you - and I clearly marked those as edited.
Not only are you a pathetic crank, you're a pathetic crank who's willing to lie in order to save face. You just keep getting more, and more pathetic. Keep it up, boy, keep it up.
Posted by: Mark C. Chu-Carroll | February 6, 2010 1:01 PM
LOL! I love that he erupts after a Density of Q in R argument. Makes you wonder if he's ever studied analysis or just heard about it. Good work Scully! :)
Posted by: chowno | February 6, 2010 2:32 PM
MChu: I never said I hate Jews you fucking cunt. Nowhere have I ever said anything like this. You lost the argument and so snake that you are, you decided to divert the focus of attention on something as worthless as your nationality.
Asshole: I don't hate anyone because of what ethnic group they belong to. I hate conceited fucking assholes like you!!!
You have lied and misrepresented anything and everything you could.
You better be worried asshole because I promise you if there is anything I can do to get you for this slander and libel, I am going to make you eat shit! You pathetic Jewish cunt!!!
Posted by: John Gabriel | February 6, 2010 3:08 PM
MCHu: Dirty oily haired slob! You the liar accusing me? Snake!!!
You define CRANK!!!!
Posted by: John Gabriel | February 6, 2010 3:10 PM
MChu: What does you being a stupid Jew have to do with any of the argument - nothing. There is only one reason you brought it up: You are thicker than a brick and you know it!
And the nice thing is that any one with half a brain knows what a reptile you are now. You can't so anything about it. I have created a knol on you and it already has surpassed 300 hits in less than 2 days. Trust me, by the time I am done with you (you devious dog!), you will be sorry you called me a crank! You will be even more sorry you played the Jewish card.
FOOL!!!
Posted by: John Gabriel | February 6, 2010 3:17 PM
MChu: I lied?!!!
Representation Enumeration. What happened to the words in between? They were swallowed up by some black hole you think? Yes, the black hole that is your head.
Now you accuse me of lying. Fuck you!
Posted by: John Gabriel | February 6, 2010 3:21 PM
John is certainly fixated on representation. The fact is, you don't need a representaion for all of the real numbers to show that they are not denumerable. All you need is a representation of a non denumerable subset.
An interesting note about the Cantor diagonal number, although one only needs to show one number not enumerated to reach a contradiction, the set of Cantor diagonal numbers (numbers whose nth digits are different from the nth digit on the nth number in the list) is itself uncountable.
Posted by: vhurtig | February 6, 2010 3:23 PM
John,
Do you maybe sorta think he got the idea that you don't like Jews from your own Knols? In your own "Wise Men and Fools" knol you specifically identify a large portion of the "fools" as Jews.
http://knol.google.com/k/john-gabriel/list-of-wise-men-and-fools/nz742dpkhqbi/72#
Posted by: Anonymous | February 6, 2010 3:31 PM
Stop it! Pleaseeee!!
Posted by: sadaf | February 6, 2010 3:32 PM
Not sure why, but my last post, #181, came up as anonymous.
Posted by: vhurtig | February 6, 2010 3:36 PM
@147
Yes, I insinuated that, but I did it by stating a fact: the crank posts draw the majority of traffic around here. On the other hand, it's clear that Mark just likes talking to cranks anyway, and the null hypothesis would be that's it just a happy coincidence that a lot of other people, like me, enjoy watching train wrecks.
To me, at this point, the train wreck consists in the blogger calling both the crank AND certain members of his educated audience "asses".
Posted by: Ivan | February 6, 2010 4:27 PM
@165:
"Is 0.999... = 1 ?"
Wow... might this turn into a repeat of the PolyMath saga from 3 and a half years ago?
http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html
You know... JG just appears to want to be famous for his genius. So he picks a topic that he's familiar with, but less understood by others. He then takes a minimally supportable -- but ultimately incorrect -- position on something and then tries to convince the uneducated that he's absolutely correct and all the experts are wrong.
While I haven't seen him actually invoke Galileo (yet), he practically presents himself as a martyr whenever he gets cornered. If he isn't a crank/denialist/whatever... then I don't know what is!
Posted by: doctorgoo | February 6, 2010 5:41 PM
Have you read the gibberish on his blog? He writes things like the limit of f(x) as n goes to infinity, with a straight face, in a context where it's clearly not just a typo. I would be hard pressed to say that he even knows the correct definition of any mathematical terms, let alone that he's familiar with any topics. But perhaps in the past... which would make his case all the more sad.
(OH NOEZ! I've violated his copyright by quoting from his blog! Without permission!)
Posted by: Ivan | February 6, 2010 6:06 PM
Mark,
My math isn't strong enough to understand all of the arguments going on here, but this guy is without question an obnoxious jerk.
For the sake of your sanity and that of those who enjoy visiting Good Math, Bad Math to learn a little of the world's mathematical coolness, cut this troll loose. Ignore him for long enough and perhaps he can find someone else to bother.
Posted by: MatthewG | February 6, 2010 6:13 PM
I agree with Ivan in 186. Many of his nonsense can be pointed out by referencing definitions. His name-dropping list of mathematicians doesn't even include people like Gauss, Galois, Riemann, ...
Posted by: mike | February 6, 2010 6:30 PM
Many people are mathematical cranks while being uneducated in it - John is but one example. He seems to have the "value-added" characteristic of using insulting language and massive denial to shut down all valid corrections. If he really is a software developer, it must be a joy working with him.
Posted by: dean | February 6, 2010 6:51 PM
@175: Thanks. It is rather fun to prove him wrong.
Posted by: Scully | February 6, 2010 7:19 PM
LOL @186 & 188...
Yes... absolutely you are correct. I should have had it as:
...with vaguely being defined as "read it on wikipedia" once...
;-)
Posted by: doctorgoo | February 6, 2010 7:57 PM
JG, assuming a vein hasn't exploded in your forehead yet, I'm trying to get an idea of how you enumerate the reals with your tree, and it's not exactly clear to me how you are defining your traversals of the tree. Could you give a few examples? Take a few simple numbers with finite decimal expansions, say, 0.12, 0.1234, 0.4321, and 0.5555. What natural numbers would these map to in your enumeration?
Posted by: pjb | February 6, 2010 8:16 PM
Mark -
Wondering if at this point Mr. Gabriel is wound up enough to take it out on someone in Real Life, and whether that would be good and sufficient reason to go on to a new post.
Posted by: Jud | February 6, 2010 9:16 PM
@178:
300 hits? 300 hits? John, this post has gotten 9000 hits in the last two days. You're bragging about 300 hits? That's just sad.
And little man, you are such a pathetic child that you can't even admit to making a
typo. You're too perfect for that. No, it can't be that you made a mistake typing in the midst of your spittle-flecked fury. No, it must be that for some idiotic reason, I edited one of your comments in a bizarre and utterly pointless way. Because that makes more sense than actually admitting that you're less than perfect.
You are really pathetic.
You're obviously going to come back here over and over again until you get the last world. So go ahead. I'm not going to waste time responding to any more of the personal bullshit.
Posted by: Mark C. Chu-Carroll | February 6, 2010 9:50 PM
I'm a lurker who is delurking to say 1) to Mark that all the race-based hate speech directed at you was awful and it made me feel a little sick to read it. Not cool, crank-man.
2) Math cranks are funny to me. I'm a math GA, teaching college algebra, and I'm used to kids who HATE thinking about math. So the idea that there are people sitting around thinking about math for fun but aren't doing the work to learn it leads to some weird cognitive dissonance on my part. why not just enroll and come to college algebra? it's fun!
3) The eastern european thing made me worried for about half a second that the math crank was actually this eastern european professor in my department, because i've always justified his, erm, eccentricities with the idea that he's actually a brilliant mathematician, and if that weren't in fact the case then i was going to be seriously disillusioned. it sort of sounded like him for a bit with the whole 'american university' thing & i started to freak out a little-but i checked the crank's knol & it's not my prof, so it's all good.
Posted by: Emily | February 6, 2010 10:41 PM
@194
MarkCC, your typo in the end of 194 was either a mistake or genius. Either way, I applaud you.
Posted by: mike | February 6, 2010 11:06 PM
MChu: I never said I hate Jews you fucking cunt. Nowhere have I ever said anything like this. You lost the argument and so snake that you are, you decided to divert the focus of attention on something as worthless as your nationality.
humm, interesting.
You arrogant, conceited little jewish swine! You call me names and it's okay. F..k you! Who the hell are you little yehuda, to call me a crank?!
Whenever you are refuted, you simply revert back to your stale old argument. You know, I remember reading a similar story in Mein Kampf about how often Hitler would refute the stupid jews and two weeks later they were back at square one again. Seems very little has changed eh?
WHAT RIGHT DOES THIS STUPID, JEWISH C..T
Carry on little Jew boy.
Filthy kike!!!!
You pathetic Jewish cunt!!!
What does you being a stupid Jew
John, you seem to be suffering from an acute case of dumbfuckery.
Posted by: Rev. BigDumbChimp | February 6, 2010 11:24 PM
Obviously the end of that should have been blockquoted as well up to the last sentence.
Posted by: Rev. BigDumbChimp | February 6, 2010 11:26 PM
I really really hope this man does not have a spouse or children. He definitely has some mental health issues to address. This part especially sounds like a classic control freak spouse/child/elder abuser:
"Now read carefully: Don't you dare address me ever again as you have in your previous comment!
You DO NOT tell me what to do and you DO NOT get to say how things are done where I am concerned. Do I make myself clear?"
Seriously, who does he think he is? Ooooh, don't piss off the mighty John Gabriel. Get a grip. You're a nobody making threats over an internet debate about a century old proof. From you rude demeanor to your inability to grasp simple English it would be a surprise if anyone everyone responds to you with anything other than derision.
Posted by: penn | February 6, 2010 11:46 PM
Flipping through my old copy of Li and Vitanyi's book on Kolmogorov complexity, I came across a good analogy that illustrates John's fallacy quite vividly.
Chaitin's halting probability, defined as the sigma[2-l(p)], taken over the set of all programs p on the reference prefix machine that halt, is an uncomputable real. However, if you only take the sum over the set of all programs that halt in n steps, you have a computable real that has the halting probability as the limit with n taken to infinity. John Gabriel is basically making an argument that the latter statement is equivalent to saying that the halting probability is computable, which is false.
John Gabriel's enumeration is not an effective enumeration of the reals because you can't pick out an arbitrary real and find the step of the recursion at which it is enumerated, which is something you can do when enumerating the naturals. Just like so, the reason the halting probability is not computable with the above described algorithm is because at no point in the recursion is the halting probability actually enumerated.
Posted by: Tyler DiPietro | February 6, 2010 11:58 PM
@171: John Gabriel corrects a previous statement, resulting in:
"If I can represent them, I can enumerate them. Representation {is required for} enumeration."
Unfortunatly, this creates an already *formally invalid* argument.
The problem arises from {is required for}.
"Representation is required for enumeration." actually means "If something is enumerable then it is representable.". (Representation is assumed to be a *nescessary condition* of Enumeration.)
So, his argument can be translated to:
1."If something is enumerable then it is representable."
2."The real numbers are representable."
Therefor: "The real numbers are enumeratable."
If the real numbers were *not* enumeratable but representable both premisses were true and the conclusion false. Hence the argument is (formally) invalid.
MarkCC's inferring of "⇔" was actually quite a nice thing to do, since he assumed an at least formally valid argument and thereby a level of competence on Gabriel's part that aparently did not exist.
Posted by: MarioF | February 6, 2010 11:59 PM
@152
Before blogs i believe there was little two way (not communication, let's say) exchange.
Most cranks contented themselves with postcards or letters (CAPITALS being common even then) to names plucked from newspapers or college catalogs.
Posted by: jahigginbotham | February 7, 2010 2:00 AM
@79
thanks andrew i will check out that book
Posted by: jahigginbotham | February 7, 2010 2:05 AM
@194 - not only is the 300 hits piddlingly insignificant, I wonder just how many of those hits came from Good Math readers. If he's so amazed at having 300 hits in 2 days, I'd say chances are it's not his regular readership that has been responsible for reading his vitriolic and offensive dross.
I must just say thanks to you Mark for providing an entertaining and informative blog - not so much this one (although it is hilarious to watch John slowly disintegrate from Crank to completely deranged). I've been reading for a while though, and always enjoy seeing a new article appear, because it's either going to be in a field of maths I'm ignorant of, or one where I have some knowledge, but, well, the more the merrier.
Posted by: Anonymous | February 7, 2010 2:08 AM
MChu: Gosh, I struggled to find your comment through all the riff-raff of your colleagues at Google. It was hard because it is so similar to their crap.
1. You called me a crank. For a whole year, I ignored you and the fools that frequent your site.
2. I read your write up and noticed you had written:
"As a representation, taken to infinity, it includes every possible real number."
This is the reason I decided to comment. You see, any intelligent person will immediately realize that this statement implies all the four conditions I tried to explain to you. I thought that perhaps if anyone is close enough to understanding my proof, you might have the best chance. Boy, was I mistaken. I also do not like being called a crank. No one has any right to call anyone else a crank. I don't feel good about the exchanges of foul languages between us. This does not mean I don't dislike you any more. I can't stand you!
Your actions have demonstrated to me that you are a nasty, conniving individual. You will not hesitate to lie or deceive so that you look better.
3. Representation, contrary to what one fool on your site wrote, is not a choice, it is a requirement. In fact, representation once accomplished means one is 90% of the way to enumeration. The rest is elementary details. There are several ways of showing the representation is enumerable. However, one has to clear up any points of issues by setting out the definitions. I could have gone straight into my proof but your *arguments* (most of them related to representation) would have kept surfacing even though you *agreed* that all real numbers are represented in my tree.
Pray tell, if one can't do infinite left-right traversals, then how did you get past this in your mind? For example, you would not be able to represent a single irrational number and many rational numbers in my tree if you could not perform left-right traversals. What is astounding to me is how you managed to arrive at these conclusions in your head? Let's simplify: If you can't do left-right traversals in my tree, then how do you account for any irrational number in the tree? Remember you already agreed these are all in the tree!!!
MChu, I can just see your response: John, this is a blog, this is a forum, I am Jewish, you are a pathetic crank, fuck off, son of a bitch, blah, blah, blah ,blah.
Shame on you!
Posted by: John Gabriel | February 7, 2010 3:35 AM
I wanted to add the following:
1. I do not think all the real numbers can be represented in any radix system.
2. I think the reason that Cantor was unable to
show the real numbers are countable is due to the fact that real numbers are not well-defined.
For example, every rational number can be logically defined in terms of pairs of natural numbers (natural numbers are themselves special ratios that were derived from the concept of ratio, but this is another story). One can't do this with real numbers unless one assumes the same can be represented using radix systems.
3. I DO NOT believe the real numbers are *countable* - surprised? Why? Because these are not *well-defined*. However, if one uses Cantor's reasoning (which is completely beserk), one can show logically as I have, that the real numbers are indeed countable.
I can just see you trying to swallow this one with your limited intellectual faculties.
Posted by: John Gabriel | February 7, 2010 3:45 AM
"I thought that perhaps if anyone is close enough to understanding my proof, you might have the best chance."
Don't do it Mark! He's baiting you!! Don't feed the trolllll.....
Posted by: Anonymous | February 7, 2010 4:33 AM
Hits in the past three days-
IT'S OVER NINE THOUSAAAAAAAAAAND!
Posted by: Anonymous | February 7, 2010 4:59 AM
What? NINE THOUSAND?
Posted by: Anonymous | February 7, 2010 5:56 AM
MChu: In the last 3 days just over 6,000 unique visitors. So I was right all along - this is about traffic to your site.
And what better way than to turn the whole discussion into Hollywood...
Why don't you set up a survey on the page:
Whose winning the argument?
MChu:
Gabriel:
Don't worry, you won't lose. Most of the visitors to your site are the passing through riff-raff.
Posted by: John Gabriel | February 7, 2010 6:21 AM
MChu: May I suggest a name change also:
Hollywood Blogs.
Sheesh, I am certain your hits will skyrocket.
Posted by: John Gabriel | February 7, 2010 6:25 AM
MChu:
"One thing that I'm proud of on this blog is that I've got a history of admitting my errors. Just go back and look at the history of the blog. I've made my share of mistakes. And I've always done my best to admit them, and correct them. And I've done it without trying to hide it: I've always made the correction, and inserted extra text to explain that the original post contained an error.
If you're really interested in defending your proof, go ahead and do it here, in the open, in the comments. If you want, I'll even set up a new top-level post specifically for your arguments. But I won't play games with sending you private mails between each step, and I won't tolerate you throwing insults at readers who point out problems with your argument."
MChu won't tolerate insults and play games. Ha, ha! If he had to remove the comments that are insults to me, there would be very few comments left on the page.
As for games, whose winning? Ha, ha. What a pathetic crank you are MChu. Sooo transparent.
Posted by: John Gabriel | February 7, 2010 6:30 AM
@210 - who exactly was it who first brought up measured page views, as a demonstration of their righteous knowledge and validity?
I wonder....
Posted by: Anonymous | February 7, 2010 7:29 AM
Wait, I must respond to this Anon (I suspect he may be MChu in stealth mode):
I brought up page views, but please tell me what does this have to do with demonstration of righteous knowledge and validity? Have you gone quite mad? Are you deluded?
The number of hits is not an indicator of a web page's quality. It is an indicator of how many people visit that page. From the amount of riff-raff comments on this page, I need say nothing further regarding its quality.
By mentioning that my web pages receive many hits I wanted MChu to know that he will eventually be exposes for the fraud and crank that he really is. Only a crank calls someone else a crank first.
Is there some subliminal message here perhaps MChu?
Posted by: John Gabriel | February 7, 2010 7:59 AM
So, JGab, are you aware of Dedekind Cuts? Perhaps you can tell us why they're flawed?
Posted by: Michael Ralston | February 7, 2010 8:02 AM
@214 haha, no, that definetly isn't MChu posting. That's me. Take a look at the completely different writing style. #204 was another one of mine.
But hey, I'd be honoured to be considered as Mark Chu-Carroll, as I actually have a fair bit of respect for him. After reading your work, I have none for you.
But you attack Mark for posting the number of hits he has, taking it as proof that he is some kind of publicity whore, even though you've already posted, with breathless incredulity, a number of hits, in two days, that you find astounding. The hypocricy is strong in this one.
I'd offer Mark to post my IP log, to prove that I posed 204 and 213, and that I am not Mark, but, well, considering my read on you is that of a hateful, intollerant, biggoted asshole, I'd prefer you not knowing any more information about me. Another reason why I wouldn't offer because it doesn't matter - everyone is convinced you're a complete spanner John. Nobody believes your sad little ramblings. It's all just a strange, incoherent, rambling trainwreck to us all now.
Go home. Get help. Because your great crusade here, to educate the primitive savages, has failed. Take your high horse and fuck right off to some new pastures.
Posted by: Anonymous | February 7, 2010 8:12 AM
John Gabriel:
There's been (at least) 3 ways that various people have tried to show your mistake, and I'll try another way. It's informal, but I'm sure it could be made rigorous by someone. Good luck finding anyone to give you the time of day after this thread, however.
There is no such thing as an integer with infinite digits. Therefore, every integer you use to make a correspondence in your enumeration is finite. Therefore, your enumeration will only get closer and closer to any real with an infinite number of digits in its decimal representation. If you could get to 1/3 in your traversal, you would need an integer with an infinite number of digits, which you don't have.
I think your problem is that you are assuming that integers with infinite digits exist. Therefore, you are placing the reals in correspondence (via hand-waving) with another set S of equal cardinality (where S is the set of infinite digit integers), but that set is not the integers as is commonly understood.
John, you're wrong, but it's not a big deal. Dealing with infinities is difficult, which is why Cantor's result is so illuminating.
Posted by: Sharkey | February 7, 2010 8:15 AM
By the way... this isn't the first time that JG has been called a crank. Apparently before he went onto knol, he was putting his stuff on wiki... until they all got deleted apparently:
http://en.wikipedia.org/wiki/Wikipedia:Votes_for_deletion/John_Gabriel%27s_Nth_root_algorithm
Posted by: doctorgoo | February 7, 2010 8:35 AM
some choice quotes from wiki:
"Delete this crankwork"
"Mathematical Vanity? It's too funny..."
"Delete. We've dealt with this guy's stuff before"
"I think I should also warn that, from experience, [John Gabriel's] email correspondence style can be rather aggressive."
...and this was 5 years ago... he hasn't changed much at all...
Posted by: doctorgoo | February 7, 2010 8:44 AM
On that tree, which supposedly proves countability of the real numbers.
Here is a list that, continued to infinity, would contain *all paths* in the tree from 0.000..., where you always branch to the leftmost successor of every node you visit, to 0.999... where you always brach to the rightmost successor.
I use a kind of "reversed counting" starting with the most significant digit to "create" the list.
01: 0.000...
02: 0.100 ...
03: 0.200 ...
04: 0.300 ...
05: 0.400 ...
06: 0.500 ...
07: 0.600 ...
08: 0.700 ...
09: 0.800 ...
10: 0.900 ...
11: 0.010 ...
12: 0.110 ...
13: 0.210 ...
14: 0.310 ...
...
21: 0.020 ...
22: 0.120 ...
23: 0.220 ...
24: 0.320 ...
...
31: 0.030 ...
32: 0.130 ...
33: 0.230 ...
34: 0.330 ...
... (continue infinitly until 0.999...)
This list is just as good a representation of the 10-adic numbers in the intervall [0,1] as the tree is.
As a bonus one does not have to worry about the traversal anymore, since it has already been done. *All that counts is that all paths/numbers are in the list, doesn't it*.
Now, since the list is in a square shape, one can use the 2nd diagonal argument to "construct" a number that is not in the list. Done.
Conclusion:
By organising 10-adic numbers in a tree one cannot show that those numbers are countable, the cardinality of the set remains unchanged as expected. Cantor's berserk reasoning demonstrates that the tree contains uncountably many paths/numbers.
Posted by: MarioF | February 7, 2010 9:16 AM
MChu: Oh common MChu, surely you can get someone who is not an undergrad student to help you?
Have you read the comments of your morons? Tsk, tsk. I keep forgetting you are a moron!!
This happens to me when I give idiots the benefit of the doubt. Actually I am just too soft and kind-hearted, but that's better than being like most of you - soft in the head. Ha, ha.
Posted by: John Gabriel | February 7, 2010 9:33 AM
@221:
Well yes I am a moron, for actually taking your arguments seriously and addressing them. What a waste of time. Get a life John Gabriel.
Posted by: MarioF | February 7, 2010 9:49 AM
"Oh common MChu, surely you can get someone who is not an undergrad student to help you?"
Believe me, John, most undergrad maths students are working on stuff much, much harder than Cantor's proof of the uncountability of the real numbers. An undergrad student is the perfect person to point out your logical flaws, given that you're making mistakes a first year student ought to be embarrassed about...
Posted by: Richard Hughes
| February 7, 2010 10:19 AM
Indeed, MarioF's little demonstration shows precisely why JGab's tree fails - there are infinitely many numbers that do not show up until after an infinite amount of time, which is precisely the problem.
For any enumeration, if I identify a specific number on that enumeration, it is by the definition of enumeration possible to identify the position in the enumeration where that number falls. Depending on the nature of the number and the nature of the enumeration it may take more time than the heat death of the universe, but it should at minimum be possible to provide a procedure to do so, and to show that that procedure will terminate in a finite (although potentially arbitrarily large) time.
If you cannot do that, then there is no natural number to which my identified number can correspond, which means you don't have a Cantor-refuting bijection.
For JGab's tree "enumeration", or the equivalent list provided by MarioF, there are even many rationals which are not included - 1/3 being the most classic. I suspect (but have not even attempted to prove) that the numbers included in such an enumeration are precisely equivalent to the surreal numbers with finite birthday - anyone else care to take a stab at that?
Posted by: Michael Ralston | February 7, 2010 10:51 AM
@208 Anon
zomg, lol.
I thought it was funny that the proof of .99.. = 1 that JG disputes is the one being circulated around 4chan's /b/
Posted by: Anonymous | February 7, 2010 10:58 AM
Michael Ralston:
I was thinking the same thing; it's a similar idea, building numbers from an infinite tree construct. I think the proof was in Conway's book, but it's been a while since I read it.
Posted by: Sharkey | February 7, 2010 11:00 AM
Michael:
I think those two sets are close, but not identical. The set of surreals with finite birthdays is equivalent to the set of numbers with finite representation as binary decimals, whereas the enumeration of John's tree produces the numbers with finite representation as base 10 decimals.
Posted by: Mark C. Chu-Carroll | February 7, 2010 11:03 AM
Here's another site where people take JGab to task for his cluelessness:
http://www.metafilter.com/88104/Calculus-of-Averages
This stuff is funny!
Mark, I know that others are urging you to move on from the mess that is JGab... but honestly, I'm enjoying this too much (and learning stuff, too!) ... how about some more takedowns of his maths?
Watching him respond to you will be entertainment enough!
Posted by: doctorgoo | February 7, 2010 11:40 AM
There are numbers in your tree, Mr Gabriel, that can never be reached and therefore never be enumerated, even though they are in your tree. They're just infinitely far in the distance. This is why your fourth premise is wrong even though your first three are correct.
Then what should we call you? You fulfill, after all, the definition of "crank".
This is wrong, as has been explained above.
You misunderstand. He has removed the unfair attacks on you.
There is such a thing as a fair attack on your egnorance (correct spelling, look it up).
Also, learn to use the <blockquote> tag.
See, your webpages don't receive many hits. 300 is not many.
Besides, have you tried to check where those visitors came from? I bet all or almost all came from here.
You still haven't understood that crank is a term with a definition.
I already explained it to you: it's not simply an insult like "poopyhead" or "asshole".
BTW, does Greece not count as "eastern Europe"? If Romania and Albania count... Are "southern" and "eastern" mutually exclusive to you?
Posted by: David Marjanović
| February 7, 2010 11:48 AM
MChu:You're obviously going to come back here over and over again until you get the last world. So go ahead. I'm not going to waste time responding to any more of the personal bullshit.
MChu: John, you are wrong. Your representation is not enumerable.
Gabriel: This is all MChu can say: "Your representation is not enumerable." Well dimwit, how can you say on the one hand that my tree contains every decimal number and then you make the following statement:
A "left to right" traversal - or in fact any traversal of your tree will never reach a single number with an infinite representation."
So, why don't you put your bullshit aside and answer the question? How can the tree contain every decimal number and yet - no traversal will ever result in a number with an infinite representation?
To all other morons on this site: Spare me your 2 cents worth. Don't want to hear anything you dimwits have to say because you don't have anything to say. Do you get it?
Posted by: John Gabriel | February 7, 2010 12:28 PM
I'll ignore the usual insults, and focus on the actual substance, as minimal as it is.
Very, very easily.
Any number with an infinitely long representation will never be reached by any traversal in a finite amount of time.
You keep confusing representation with enumeration. If you use a set-theoretic view of the infinite tree, then it contains every possible real number. But that doesn't mean that you can enumerate every member of that set.
This isn't a difficult concept. For a particularly elegant explanation of it, I'd suggest John Conway's book on the surreal numbers. The surreals have very similar properties to your tree. They define a representation that can represent every real number (plus some non-real numbers). But they can't represent all of the reals with finite representations.
You can describe the enumeration of numbers in your tree via traversal using Conway's "birthday" notion. each level of your tree can be though of as a "birthday": the first generation of numbers includes every number who's representation can be represented using things found in the first level of your tree. Then you can create the next generation of numbers by adding digits to the numbers created by the previous generation. So the first generation represents the set {0.1, 0.2, 0.3, ...}; the second generates {0.11, 0.12, 0.13, .., 0.21, 0.22, 0.23, ...}. And so on.
The number of enumeration steps it takes to get to a particular number has a lower bound of 10n, where N is the number of the generation. So the initial generation, the 0th generation in Conway's terms, starts to be enumerated in the very first step. The 1th generation can start to be enumerated after 10 steps. The 2th generation can start to be enumerated after 100 steps.
Now, what happens when we get to numbers with infinite representations? In Conway's term, they belong to the ωth generation. They're never enumerated in a finite amount of time. It takes 10ω steps before you can enumerate them. No matter how long you wait, nothing with length ω will ever be enumerated by traversal - because no matter how long you wait, a traversal can never reach the ωth level of your tree.
Like I keep explaining: the existence of a representation for an infinite set does notimply that that set is enumerable.
Posted by: Mark C. Chu-Carroll | February 7, 2010 12:51 PM
Same old shit again...
You have two ways of representing numbers in the tree, and you're confusing them.
If you represent each number with a path in the tree, you get a represenation of all decimal numbers. But you can't enumerate those.
If you represent each number with a node in the tree, you get a representation for only a countable subset of decimal numbers. You can enumerate that, but your conclusion that "all decimal numbers can be enumerated" does not follow.
Posted by: boris | February 7, 2010 1:06 PM
JG: "I do not even believe the real numbers are well-defined but this is not at issue here."
It seems to me this is a serious issue whenever you are considering any theorem about real numbers, as any statement about something which is not well defined is clearly not a mathematical theorem. I suspect your objections to Cantor's theorem and your objections to the well-definedness of real numbers both stem from the same cause, either you think the axioms of set theory are inconsistent or you plainly reject one of them.
However, until you learn to discuss these ideas in a clear and civilized manner, I doubt anyone in the mathematical community will take you seriously.
Posted by: Robert | February 7, 2010 1:31 PM
What John Gabriel means when he says WELL DEFINED:
http://knol.google.com/k/kurt-deligne/what-john-gabriel-means-when-he-says/2oygz0grkt42u/2#
Please read this before talking to him; you'll save yourself a lot of headache.
Also, for fun:
http://knol.google.com/k/kurt-deligne/funny-shit-that-john-gabriel-says/2oygz0grkt42u/1#
Posted by: Kurt Deligne | February 7, 2010 2:00 PM
JG: How can the tree contain every decimal number and yet - no traversal will ever result in a number with an infinite representation?
MChu: Any number with an infinitely long representation will never be reached by any traversal in a finite amount of time.
JG: So what? Finite time is not required to verify that a given traversal results in a decimal number.
What do you mean "reached by any traversal in a finite amount of time"?
?????What is there to reach???????
One does not care about visiting every node in an infinite traversal - it's not possible. And it does not matter, because we are not interested in the nodes, but in the traversal as a whole. It's the whole traversal that represents the number, not the individual nodes. The sum is greater than its parts...
MChu: You keep confusing representation with enumeration.
JG: Not true. I have stated over and over again, representation is required for enumeration. You appear to be the one who is confused.
I wanted to see if you understood the implications of your agreeing that every real number is represented in the tree. This is required before one can demonstrate how the numbers are enumerated thereafter. The four points are equivalent to that one statement. The proof however is still incomplete. Once this is in place (that is, you understand the 4 points imply every number is in the tree and vice-versa), then the next step is an algorithm or method to enumerate the numbers that are represented in the tree.
Surreal numbers do not exist - don't bother discussing this or introducing anything that is unrelated to the topic. Also, do not suggest or recommend any books to me. Just stick to the subject.
Posted by: John Gabriel | February 7, 2010 2:43 PM
To make it very simple: the tree is not an enumeration. It is a representation. Once we agree the tree represents (contains) every real number in the interval [0,1), then we find a method to enumerate the numbers. There are several ways to do this and I have already shown you one way of doing it.
Posted by: John Gabriel | February 7, 2010 2:48 PM
John Gabriel,
Yes, it all makes sense now! You are absolutely right and you are my hero. XOXOXO
- andrew
Posted by: andrewr | February 7, 2010 2:56 PM
See the problem is that you require us to accept all four of your propositions in order to support your proof, but what everyone is trying to tell you is that number 4 is incorrect precisely because the irrational numbers are not enumerable either:
1) A number is either rational or irrational
2) The rationals are enumerable
3) Therefore, if the reals are not enumerable then neither are the irrationals.
If you want us to accept your fourth proposition then you must demonstrate that the irrationals are enumerable. Of course, if they are then, trivially, so are the reals.
Personally I'm not too comfortable with proposition 2 either. Surely it cannot contain any non-computable numbers, though perhaps Mark can correct me here.
Posted by: Archena | February 7, 2010 3:06 PM
MChu: Why do you allow useless, non-mathematical comments? Read this page and see how many of these comments contain nothing but heckling, non-math related bullshit and what not...
Let me ask you this: How is this working for you? Is it part of your strategy to distract and conquer? It's not working for you. Just hold all your readers comments until our argument is completed. Most of them are imbeciles without half a brain.
Posted by: John Gabriel | February 7, 2010 3:09 PM
238: Therefore, if the reals are not enumerable then neither are the irrationals.
Almost correct. But we have not even begun to enumerate anything yet. And honey, you should have written:
"Therefore, if the irrational numbers are not enumerable, then neither are the real numbers."
@238: If you want us to accept your fourth proposition then you must demonstrate that the irrationals are enumerable.
Point 4 is not about enumeration. It is about representation. Can we represent irrational numbers in my tree? Yes.
Once MChu accepts point 4, then we start the process of enumeration.
Posted by: Anonymous | February 7, 2010 3:17 PM
"One does not care about visiting every node in an infinite traversal - it's not possible. And it does not matter, because we are not interested in the nodes, but in the traversal as a whole. It's the whole traversal that represents the number, not the individual nodes. The sum is greater than its parts..."
Now you're just spouting random bullshit. You clearly don't even understand what an enumeration is.
Posted by: Tyler DiPietro | February 7, 2010 3:17 PM
That's because this is a blog, open to any type of discussion. Many of the posts have been either snickerings of your ideas/posts/comments, or refutations to your ideas/posts/comments.
Look, you are arguing mathematical concepts here. So here are the rules:
1. Any mathematical Axiom is true.
2. Any derivation from Axioms is true.
So, it has been pointed out numerous times here, and I have attempted to do so to you in the past. However, if you want to be taken seriously, than you must state which Axiom you do not use, since all proofs of the uncountability of the real numbers is from the mathematical axioms we use.
For example: The parallel axiom when changed into its different forms generates different geometries.
Posted by: Scully | February 7, 2010 3:19 PM
@241: Now you're just spouting random bullshit. You clearly don't even understand what an enumeration is.
JG: I am not enumerating anything at this stage. Points 1 - 4 DO NOT require enumeration! They are not about enumeration. FOR THE LAST TIME: MY TREE IS A REPRESENTATION!!!
Posted by: Anonymous | February 7, 2010 3:20 PM
"FOR THE LAST TIME: MY TREE IS A REPRESENTATION!!!"
And not a bijection between the natural numbers and the reals.
Posted by: Tyler DiPietro | February 7, 2010 3:23 PM
@239:
John, I don't know why you're incapable of wrapping your head around this:
I do not control who comments on this blog. I don't "hold comments". I don't control comments. I don't decide who gets to comment when. The comments on this blog are an open public forum. They have been since I started the blog nearly four years ago. Anyone can comment any time they want - I don't control it.
The only control that goes on is that ScienceBlogs, which provides hosting for me, has a spam filter running - and certain comments can trip the spam filter. So comments which contain too many links will sometimes be blocked by the filter until I check it.
And that's all I do. I'm not going to change the long-standing policy of my blog because you don't like it.
Posted by: Mark C. Chu-Carroll | February 7, 2010 3:23 PM
Nope, its an if and only if statement like most definitions, but nice try.
This is because the set of Rationals is countable, and the Reals are not. Thus, by definition if you removing countably many things from an uncountable set, you are still uncountable.
Posted by: Scully | February 7, 2010 3:24 PM
@240:
"Point 4 is not about enumeration. It is about representation. Can we represent irrational numbers in my tree? Yes."
...but...
"Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?"
Posted by: Archena | February 7, 2010 3:24 PM
That's it. I shall only respond to MChu. You are welcome to read, but do me a favour and just stay out of the argument. You know too little to contribute anything.
Besides I have my hands full with MChu - he is by no means the brightest.
And I am really not interested in anything you have to say.
Posted by: Anonymous | February 7, 2010 3:24 PM
"Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?"
No, that should be:
"Do you agree that if we only perform left to right (infinite) traversals, that we can represent all the irrational numbers and some of the rational numbers?"
If I wrote that, it was a mistake because MChu had upset me.
Posted by: Anonymous | February 7, 2010 3:27 PM
BTW, I decided to take another look at points 1-4 just for grins. So when you say this:
"Points 1 - 4 DO NOT require enumeration! They are not about enumeration."
There's always this, visible right in the OP:
"Do you agree that if we only perform left to right (infinite) traversals, that we can enumerate all the irrational numbers and some of the rational numbers?"
Fail.
Posted by: Tyler DiPietro | February 7, 2010 3:28 PM
Apparently I did make a mistake. Sorry.
Point 4 should read:
Do you agree that if we only perform left to right (infinite) traversals, that we can represent all the irrational numbers and some of the rational numbers?
I can't imagine why MChu didn't pick this up. Well, I guess I am correct in saying he is not that bright.
Posted by: John Gabriel | February 7, 2010 3:30 PM
John:
Before we go any further, please reconcile these two statements that you've made:
and
You want me to agree with your (incorrect) statement that you can enumerate all of the irrational numbers; and you insist that the statement that you can enumerate all of the irrational numbers is not about enumeration, and does not require enumeration.
No, John. You cannot enumerate all of the irrational numbers. In your tree, you cannot even enumerate all of the rational numbers.
Frankly, I'm tired of wasting my time on this. It's just going in circles. If you don't have anything new to say, I'm not going to respond any more. if you've got anything actually with any actual new content, I'll respond to it - but the actual content of this discussion has long since devolved into "I can enumerate them", "No you can't", "Yes I can (plus some extraneous insults)", "No you can't".
I'm sure that you'll declare victory and spend some more time flinging insults. Enjoy yourself.
Posted by: Mark C. Chu-Carroll | February 7, 2010 3:31 PM
"That's it. I shall only respond to MChu. You are welcome to read, but do me a favour and just stay out of the argument. You know too little to contribute anything."
Translation: BAAAAAAAAWWWWWWW!
"And I am really not interested in anything you have to say."
Here's my caring face. :(
Posted by: Tyler DiPietro | February 7, 2010 3:31 PM
Point 3 is also in error. It should read:
"Do you agree that if we traverse the tree in each level from top to bottom that we can be sure to represent all the finitely (not the repeating decimals or irrational numbers) represented numbers in decimal? I know there are duplicates but we shall not worry about this right now."
Posted by: John Gabriel | February 7, 2010 3:33 PM
Here are the four points again:
1. Do you agree that every real number in the interval (0,1) can be represented in decimal?
YES or NO
2. Do you agree that my tree contains every real number in the interval (0,1)? Don't concern yourself about finite/infinite numbers at this time. We don't care about enumerating the numbers at this stage, only that if we know a number, we can find it in the tree.
YES or NO.
3. Do you agree that if we traverse the tree in each level from top to bottom that we can be sure to represent all the finitely (not the repeating decimals or irrational numbers) represented numbers in decimal? I know there are duplicates but we shall not worry about this right now.
YES or NO.
4. Do you agree that if we only perform left to right (infinite) traversals, that we can represent all the irrational numbers and some of the rational numbers?
YES or NO.
Posted by: John Gabriel | February 7, 2010 3:36 PM
You're whole argument is an error, n00b. Just admit your failure and go back to your epic failboat on Knol.
Posted by: Tyler DiPietro | February 7, 2010 3:37 PM
And if you look at point 2, you should have realized the word was a careless mistake in both point 3 and 4.
But of course you harped on how this is a blog and all the other crap.
Anyway, those are the points. Now do you agree?
Posted by: John Gabriel | February 7, 2010 3:38 PM
John:
This is a public forum, MarkCC allows anyone to comment. How dare you come here, to his blog, and tell him what to do?! It's MarkCC's blog, and if you don't like it then pipe down and GET OUR OWN BLOG!
Jeez.
Posted by: maxh | February 7, 2010 3:38 PM
Well, do you or don't you agree with all the four points now?
BTW: You had agreed with point 3 and it was in error.
Posted by: John Gabriel | February 7, 2010 3:42 PM
"BTW: You had agreed with point 3 and it was in error."
No it wasn't, since you restricted it to finitely representable numbers. Those are enumerated by your tree.
You don't have a clue where you're going at this point, you're just flailing.
Posted by: Tyler DiPietro | February 7, 2010 3:46 PM
DiPietro: Dry up and die fool. Yes, it can be enumerated but I did not want to say that.
It was incorrect because I copied and pasted the text - this is what caused my error.
I intended it to read represented, not enumerated. By the way, it was not you who alerted me to this error, it was Archena. So go and stuff yourself.
Posted by: John Gabriel | February 7, 2010 3:50 PM
John Gabriel:
Point 4 is the fundamental part of your argument, and it's wrong. MarkCC explained why it's wrong, and explained again using a comparison to the surreals. If you can't be bothered to find out about surreal numbers in Conway's book, then what's the likelihood you understand the necessary definitions involved in this problem?
Honestly, this situation is like a software meeting I attended years ago. We argued for days and days over a trivial misunderstanding over a software pattern, and the project lead made no attempt to understand the argument against his incorrect design. Walking away from that project was the best decision I ever made, I'm sure MarkCC feels the same way about finishing this conversation.
Relief.
Posted by: Sharkey | February 7, 2010 3:51 PM
If most of the riff-raff on this site just kept their two cents worth to themselves, this discussion would be a lot easier.
Posted by: John Gabriel | February 7, 2010 3:51 PM
You're a pathetic liar, Gabriel. You only changed your arguments when it was obvious that the whole thing had been exposed as fraudulent by our superior mathematical intellects. You decided to move the goalposts.
"So go and stuff yourself."
How about you lick my balls?
Posted by: Tyler DiPietro | February 7, 2010 3:54 PM
This question is not for John, since he is an insane, rather stupid, crank who doesn't understand math. I'm trying to understand something: is the problem that you can enumerate the nodes, but not the paths, and the tree contains the reals iff you include both paths and nodes?
Anyone (other than Gabriel the crank, of course)?
Posted by: Rilke's granddaughter | February 7, 2010 3:56 PM
MChu: I await your response. Agree to all 4 points or not?
Posted by: John Gabriel | February 7, 2010 4:01 PM
"is the problem that you can enumerate the nodes, but not the paths, and the tree contains the reals iff you include both paths and nodes?"
You can enumerate both, but the problem is that neither enumerates the reals. You can't pick out an arbitrary real and find the point at which it is enumerated, they all only exist in the tree in the limit.
"If most of the riff-raff on this site just kept their two cents worth to themselves, this discussion would be a lot easier."
A lot easier for you to avoid epic amounts of pwnage and butthurt, maybe.
Posted by: Tyler DiPietro | February 7, 2010 4:02 PM
@241
NOW? He only started spouting random nonsense JUST NOW?
It continues to amuse and baffle me that persons who actually understand the mathematical theory can be convinced to carry on an extended conversation with this guy.
He's like one of those chatbots that some people are fooled by, except with more mathematical vocabulary, dickishness, and singleness of purpose. Sure, his sentences are syntactically valid English and sometimes mathematics, but that doesn't mean you can reason with him any more than you can reason with ELIZA or A.L.I.C.E.
But then, maybe this is all just some kind of performance art, and I'm the one who's been fooled. Yeah, that must be it.
Posted by: Ivan | February 7, 2010 4:03 PM
@265:
That comes down to a question about just what axioms and definitions you're using.
According to the simplest definition of the tree, then what you said is absolutely correct. The reals with infinite representations are paths which don't correspond to nodes.
But there's another notion of tree, which is based on ideas from set theory dealing with structures over infinite sets. In an infinite set, you can define an infinitely deep tree - and by using the same kinds of ideas that Cantor used to get to the transfinite numbers, you can talk about nodes where the path to the node is infinitely long. Using the latter, infinite description of the tree, then there's a one-to-one relationship between nodes and paths - and so there are nodes in the tree for every real number.
In the latter notion of tree, the disctinction isn't between nodes and paths, but between nodes with finite-length paths, and nodes with infinite length paths.
John's argument only works for the latter, infinitary concept of tree. If you use the former definition - the finitary tree - then anything with an infinite representation is not even in his tree - at least not considering the usual definition of membership in a tree. (That is, the usual definition of membership in a tree is that X is in a tree if and only if X is a node (or the label of a node) in the tree.)
Posted by: Mark C. Chu-Carroll | February 7, 2010 4:07 PM
Why can't an infinite set of paths represent all the reals?
Posted by: Rilke's Granddaughter | February 7, 2010 4:07 PM
@267
No. If we're talking about the tree whose nodes are all the finite decimal representations (with the usual edges which are given by adding single digits to the end of a number), then the nodes are countable and the paths are UNcountable.
That tree is an ordinary graph-theoretic tree. If you promote it up to a set-theoretic tree, then you can add nodes at the end of every infinite path-- these correspond to the infinite decimal representations.
Posted by: Ivan | February 7, 2010 4:11 PM
For what it's worth, I do agree with these assumptions, but only if we actually define the 'representation' part - each real is represented by a traversal in the tree.
I predict that if we ever get to see that masterpiece, it will use a different representation (probably nodes of the tree), without explicitly stating so.
Posted by: boris | February 7, 2010 4:13 PM
@269. OK, that's clearer, but then I'm not clear where the problem is. Why can't I incude a node in an ennumeration that is reached by an infinite path? I'm a neuroendochrine post-doc, not a mathie, so be gentle with me, please.
Posted by: Rilke's Granddaughter | February 7, 2010 4:13 PM
Sorry, I was thinking of the set of paths defined up to a specific level of the tree. So, disregard that.
Posted by: Tyler DiPietro | February 7, 2010 4:16 PM
@273: Are you asking: Why are the nodes-at-the-ends-of-infinite-paths uncountable?
Posted by: Ivan | February 7, 2010 4:19 PM
MChu: That comes down to a question about just what axioms and definitions you're using.
JG: I am not using any axioms and have already stated the definitions clearly. I would like you to respond with a clear YES. Thereafter I have one more question and I will complete the demonstration.
Posted by: Anonymous | February 7, 2010 4:20 PM
JohnG:
You say:
"Once we agree the tree represents (contains) every real number in the interval [0,1), then we find a method to enumerate the numbers."
Therin lies the rub, you assume that having a representation implies that you can enumate the real numbers. You have blathered on about this in many posts. When are you going to show us this enumeration that us mere mortals cannot comprehend? What's your algorithm? Show me what the 1st, 2nd, 3rd, nth numbers are in your enumeration. Tell me where 1/3 is on your list, is it the first number? The 1000th? The 1.2345*10^65432th? Explain to me how to find the location of 0.10100100010000... In order to have an enumeration every real has to paired with a natural number. If you really have such an enumeration algorithm you can show it to us mere mortals. Maybe you only speak the language of god's and we mere mortals will never comprehend? You keep telling us you've done it but you never show us.
Posted by: vhurtig | February 7, 2010 4:22 PM
Ah, he's not using ANY axioms! Look ma, no hands!
Oh the hilarity. Please, do continue.
Posted by: Ivan | February 7, 2010 4:24 PM
@275. Exactly. I understand how the diagonalization argument works (my, but that's lovely), but I don't know how to extend it to this case without just..., well using it. (I.e. assume that the infinite-path nodes are ennumerable, put 'em in a table, and then diagonalize). Is there another way to thing about it?
Gabriel, you truly are a moron, aren't you?
Posted by: Rilke's Granddaughter | February 7, 2010 4:25 PM
"I.e. assume that the infinite-path nodes are ennumerable, put 'em in a table, and then diagonalize"
That would seem to me to be a typical proof by contradiction, so I don't see what the difficulty is with it. I'll defer to Ivan, though.
Posted by: Tyler DiPietro | February 7, 2010 4:34 PM
@279
Oh, so ONE proof of uncountability isn't enough, you need TWO? (I'm totally kidding.)
The thing is, you can't really approach this problem from the perspective of: Well, I tried THIS enumeration, but that didn't work, so I tried THIS OTHER enumeration, and that didn't work either...
It's not really helpful to notice that certain enumeration strategies never get you to the "infinite nodes". After all, it's quite easy to "mis-enumerate" the natural numbers themselves: the enumeration 0,2,4,6,8,... never reaches any odd numbers.
Posted by: Ivan | February 7, 2010 4:45 PM
@280: I'm not assuming anything wrong with the proof by contradiction. I just want to know if there's another way to show that they can't be ennumerated (a way that even I - smarter than John Gabriel by a long shot, but still not much of a mathelete) can understand.
Posted by: Rilke's Granddaughter | February 7, 2010 4:47 PM
@282:
It's not really that difficult.
To enumerate the numbers in the tree, you need to traverse the tree. Traversing means visiting each and every node in the tree exactly once.
There are two basic ways of traversing a tree: depth-first, and breadth-first. To give you a sense of how they work, here's a simple example:
a / \ b c / \ / \ d e f gIn a depth-first traversal, you go all the way down the left, and start at the bottom: d, b, e, a, f, c, g.
In a breadth-first traversal, you start at the top, and read across each level, one by one: a, b, c, d, e, f, g.
If you use a breadth-first traversal of John's tree, you'll enumerate a lot of numbers. But you'll never get to the nodes that are reached by infinite paths - you'll traverse the nodes that are 1 level deep, then 2 levels deep, etc. But you'll never ever get to the infinite nodes.
If you use a depth-first traversal, you'll never be able to get started. You need to follow the left-most path all the way down to the bottom of the tree. But that's an infinitely long path, so it takes an infinite amount of time to get there. To make it clear what it would mean if you could: if you got to the bottom of the tree, the first thing you'd output in your enumeration would be the smallest number greater than 0. But there's no such thing! (After all - imagine you could. call it t, for "tiny". What would happen if you divided it by 2? You'd have a number smaller than t, but greater than 0. And then, you could divide that by two. and so on.)
Any other traversal of the tree is some variant on either depth-first or breadth-first. And in either case, you can't make it work: either you'll never be able to reach anything at the bottom of the tree, or you'll never be able to reach anything at all.
Posted by: Mark C. Chu-Carroll | February 7, 2010 5:10 PM
Ok, now you've gone and made what you said before seem almost sensible in comparison! You claim that you're not using any axioms?
John, do you understand what an axiom is? If you're not using any axioms, then whatever you're doing, it's not math. If you're not using any axioms, then you cannot talk about numbers - because you need some axioms to define what numbers are. You can't talk about what it means to enumerate something - because to be able to define enumeration, you need to be able to define numbers, sets, and relations - and without any axioms, none of those exist. If you're not using any axioms, then you can't be right or wrong, because in math, "right" means "follows logically from the axioms", and "wrong" means "does not follow logically from the axioms". Without any axioms, there is no logic, no reasoning, no definitions, no proof.
(And no, I do not accept your statement number 4. And it's deeply pathetic to see you trying to worm your way out by changing your definitions retroactively - but as it happens, that definition change doesn't help you. See Chaitin's description of the undescribable numbers for a particularly fascinating explanation of exactly why.)
Posted by: Mark C. Chu-Carroll | February 7, 2010 5:17 PM
@283
I'm not sure what that statement could possibly mean. Are you saying there's some kind of classification theorem for traversals? Cause I'm pretty sure I can think up some pretty bizarre ones.
In the end, the tree structure merely obscures the question. Enumerating the nodes has nothing to do with edge-following, and examining the failure of particular enumerations seems pretty useless.
Posted by: Ivan | February 7, 2010 5:24 PM
@284
Now you've gone and done it! See, by explaining to him precisely what's wrong with "LOOK MA, NO AXIOMS!", you're creating a supercrank!
He'll go and read just enough of wikipedia so that he can add a little nonsense about Zermelo-Fraenkel to his chatterbot script.
Posted by: Ivan | February 7, 2010 5:41 PM
I'm a non-mathematician, so a fair amount of what goes on here is quite new to me. I want to say that I appreciate the time and patience put into this by Mark and all the other folks who know what they are talking about. I've learned a lot here.
A thought occurred to me just now and I would appreciate any comments. It seems to me that JGs scheme to enumerate all the reals is even less effective than has been noted. The ability to enumerate any number is because a special status is given to the zero digit. There is nothing wrong with that, but it is part of the scheme. A number is assigned to 1/4 when you get to 0.250... because we treat the infinite repeating zeros as ignorable. Without that special status to zero there would be no numbers enumerated by this scheme. Without this status to zero, the number assigned to 1/4 would be the non-existent "terminating" digit to 0.250...
I suppose if this is correct that JG could give the enumeration for 1/3 by giving this same status assumed for zero to 3. But then he couldn't give the enumeration for 1/4. The other thing of note is that this scheme far from enumerating all the reals, actually only enumerates a small portion of the rationals.
Posted by: MSR | February 7, 2010 5:45 PM
John - your most recent comments have all been along simmilar lines, complaining about the amount of 'fools' throwing in their useless 'two cents worth', and that they're filling this discussion up with rubbish.
So, what I decided to do was go through all your posts in this thread (so all the Posted by: John Gabriel, ignoring your Anon posts, as I have better things to do, and there's no guarantee those posts are from you, they might be from an imposter, for example) - anyway, here's your tally
Biggoted, racist, insulting or rambling content. Also included here were some of the corrections, namely the ones were you called mark an idiot for not having picked up on your mistakes - 33 posts.
Actual mathematics - 6 posts.
So who's that again who's contributing the rubbish to this discussion?
By the way, I'm #204, #214 and #216 - the one you decided was Mark in Anon mode. *waves*
I'll leave you all with #93, from Mr John Gabriel himself, refuting the idea that he didn't bring up the page counts as an argument ad populum - "I have a much higher readership than you and I am going to expose you for the fraud and crank that you are:"
Have a nice day folks
Posted by: Anonymous | February 7, 2010 6:22 PM
@JohnG, don't worry about what others say. I believe in you. All you have to do is show these hecklers the values of two nodes from your tree that are successive (n, n+1) in your traversal and they will shut up. Once you do that we all will follow you to the gates of hell.
Posted by: reveilebeurt | February 7, 2010 6:27 PM
@289
Whiskey tango foxtrot, dude.
Posted by: Ivan | February 7, 2010 6:34 PM
At the risk of dragging this discussion out further, I am curious to see how John intends to jump from representation to enumeration if Mark were to accept the 4 points. If - as we've now been told - the tree is just a representation, then so what? the enumeration is the hard part.
Also, @Mark: regarding point (2). it was my thinking that since some numbers are not computable, they could never appear in any such tree, even when taken to infinity, because that branch of the tree can't be computed at all. Am I wrong? if not, should (2) also be rejected?
Posted by: Archena | February 7, 2010 6:44 PM
@290. It's a smart move. Since IF John is right, he can do it. If he can't do it, then it proves John is a rather boring, probably autistic, possibly Asperger case, and most certainly a crank.
Since John won't be able to do it, I predict he will ignore it. Cranks are all the same..
Cowards.
Posted by: Rilke's granddaughter | February 7, 2010 6:51 PM
@291:
The fact that they aren't computable doesn't mean that they aren't representable.
The fact that a number is non-computable means that that you can't write a program that specifically generates its representation - not that no representation exists.
For example, Chaitin's ω is a number. It's just a probability between zero and one.
It clearly exists as a real number. If we could figure out what the heck it is, we could write it out as a decimal number. The problem is, we don't know what the digits are.
The undescribable numbers, or uncomputable numbers, are just perfectly normal numbers. The catch is that they're all irrational numbers - so they've got an infinitely long representation which never repeats. The fact that they're non-computable just means that there is no program which generates the sequence of digits which is its representation.
Posted by: Mark C. Chu-Carroll | February 7, 2010 7:01 PM
@292: I realize reveilebeurt was being snarky. I was commenting on the general unintelligibility of the mathematical portion of his comment, although I can guess what was meant.
I think I was slightly wrong earlier. This thread is either going to create a supercrank OR cause a massive cranksplosion on a scale yet undreamt. I, of course, am rooting for the latter.
Posted by: Ivan | February 7, 2010 7:10 PM
@293: that makes sense. Thanks for the explanation.
Posted by: Archena | February 7, 2010 7:11 PM
@292
What does autism, or Asperger's, have to do with anything here? Kindly keep your insults to wide swaths of the world's population to yourself, thank you.
Posted by: Jimbo Jones | February 7, 2010 7:12 PM
@291
Computability is not the same as mathematical existence, except possibly under certain forms of constructivism.
Posted by: Ivan | February 7, 2010 7:19 PM
@294
Much like fire, I figure you have to fight bat sh*t crazy with bat sh*t crazy.
@292
You have exposed my ruse and insulted the autistic and those with Asperger's (although I don't believe that was your intent).
Posted by: reveilebeurt | February 7, 2010 7:21 PM
Ivan:
I'm actually hoping that it will be Cranksgiving.
Posted by: Sharkey | February 7, 2010 7:43 PM
@292: "John is a rather boring, probably autistic, possibly Asperger case, and most certainly a crank."
I don't think he's Autism-spectrum. On one of his Knols he expresses a dislike for doctors, which I find telling.
Posted by: Jon H | February 7, 2010 7:57 PM
@298
Ah, splendid idea. Confuse-a-Crank!
Posted by: Ivan | February 7, 2010 8:00 PM
Yes, one can only hope that we will be given much more of such gems as "I am not using any axioms."
Posted by: Ivan | February 7, 2010 8:06 PM
@302:
Really, I think he's jumped the shark. I mean, where else is there to go after the "no axioms" thing? I don't think that he can possibly top that!
Posted by: Mark C. Chu-Carroll | February 7, 2010 8:09 PM
@303 We shouldn't underestimate this one. His powers are growing stronger by the hour.
Posted by: mike | February 7, 2010 8:30 PM
@303
Do not underestimate the power of the Crank Side of the Force. He might need a little time to complete the transmogrification into Supercrank, though.
By the way, I don't know if you noticed-- Vorlath showed up at Jeffrey Shallit's blog. He gives Jeffrey a link to the wikipedia article "Identity matrix". LOL!
Posted by: Ivan | February 7, 2010 8:49 PM
Mark, how do you know if a discussion like this might change from just words to violence? How do you test the sanity of your correspondent?
Posted by: anon | February 7, 2010 10:00 PM
Mark, if you could recommend a decent book (i.e., legitimate, not crank-worthy, or other such b.s.) on computability and/or Chaitin, that would be greatly appreciated.
Also, I was thinking (in the feed-the-troll sense) that if you *really* want to up the ante on the no axiom crap, why not ask good 'ol Johnny Gab-I've-shoved-my-brains-through-my-own-ass-riel if you accept his 4th point and see where his rabbit hole really goes to :-)
Personally, I think the crankscapades from that would top any cranksplosion we've seen so far.
Keep up the good work folks, this has been one of my most entertaining weekends in a while :-)
Posted by: interested in decent reading materials | February 7, 2010 10:51 PM
"Mark, if you could recommend a decent book (i.e., legitimate, not crank-worthy, or other such b.s.) on computability and/or Chaitin, that would be greatly appreciated."
I'm not Mark, but I can make a couple of recommendations in this area.
First off, if you want to read something by Chaitin, the only thing worth reading is Algorithmic Information Theory. Just one man's opinion, but his popular books are garbage. The best text in this area is Intro. to Kolmogorov Complexity and its Applications by Li and Vitanyi. Warning: both are extremely dry, academic texts, but they're my favorites since I favor accuracy.
For a general introduction to the theory of computation, my personal favorite it Mathematical Theory of Computation by Zohar Manna is the only dedicated text I've read on the subject, and since it's available in Dover paperback, it's quite inexpensive.
Posted by: Tyler DiPietro | February 7, 2010 11:13 PM
Reading Gabriel's blog, I am not impressed. He argues using totally illogical reasoning that 0.999... is not equal 1, rejecting even a beautiful algebraic proof, saying that arithmetic only work on real numbers.
Posted by: Anonymous | February 8, 2010 12:26 AM
@282
Here's a way of looking at it that I came up with. I'm not as math-trained as Mark or Ivan, and this may not be correct, but I thought it was interesting:
Take the tree and consider paths which start at the root and never repeat a node. Every node can be associated with one such path by being the end node of that path (including the root, if you include the path that goes nowhere). But there are some paths without an end node, the paths that go on forever.
You can map the reals to the paths, for example by using the same mapping you used to construct the tree. You can also map the naturals to the nodes, for example make the root be 0, it's children be 1-10, the children of 1 be 11-20, the children of 2 be 21-30, etc.
But you won't be able to one-to-one map the nodes to the paths, or the naturals to the paths, or the naturals to the reals.
The mappings that I've given as examples are the obvious ones to try, and they don't work because some paths don't have an end node. If you try to map naturals directly to paths without using nodes, you run into Cantor's argument in disguise: you can talk about the direction taken on the nth step of the path assigned to n and define a path that always takes a different direction.
There's no need for any of it if you have Cantor's proof, but I felt like this helped explain things to my intuition.
Posted by: Brian Jaress | February 8, 2010 1:48 AM
"(And no, I do not accept your statement number 4. And it's deeply pathetic to see you trying to worm your way out by changing your definitions retroactively - but as it happens, that definition change doesn't help you. See Chaitin's description of the undescribable numbers for a particularly fascinating explanation of exactly why.)"
Well, if you do not accept point 4, then you are a confirmed crank. I have not changed ANY definitions retroactively. I provided perfect explanations for what happened. Listen dimwit, I don't see too well. That's right. It is hard for me to sift through all this shit every time I want to read your worthless response.
If you don't agree to point 4, then my tree does NOT contain every real number and you have contradicted yourself over and over and over again. Anyone will see that you are a fool.
I am not trying to trick you. By agreeing to the four points you are not agreeing the numbers are enumerable. I even changed point 3 which was correct but I did not mean it to be stated that way. The challenge comes in explaining how the numbers are enumerated. I have come up with an even better explanation than the one with the bottomless wells. It is more constructive for dummies like you to understand. If you do not agree with ALL the points, this is a waste of time. You are a hopeless fool.
Posted by: Anonymous | February 8, 2010 2:56 AM
MChu: For the last time: Representation does not imply enumeration BUTTTT it is REQUIRED for enumeration. Get over it!
If you cannot agree to all 4 conditions, it is pointless for me to carry on. I am not trying to trick you. By agreeing to all 4 conditions, you are not agreeing the real numbers are enumerable. That still has to be done.
Look you snake, look at point 2 which I HAVE NEVER CHANGED since the beginning. All I changed were two words (TYPOS) in point 3 and 4. To err is human?... Like you have never erred in your life eh????
My intentions have been clear from the start. You just have not been smart enough to see where I am headed. You will do everything you can to cast yourself in a better light even if it means never reaching the truth.
Posted by: John Gabriel | February 8, 2010 3:03 AM
For the LAST TIME, DO NOT recommend anything for me to read. I have read more than you will ever read in your life. I have completed courses in real analysis and studied a lot more than can be covered in any real analysis course.
DO NOT lecture me. ANSWER the questions ONLY.
DO YOU AGREE THAT ALL FOUR POINTS ARE TRUE?
All I want to see is YES or NO. Choose your response carefully because if you don't, people will see you for what you are... a fake!
Posted by: John Gabriel | February 8, 2010 3:07 AM
Dude. Chill. And fix your caps lock.
Posted by: Colin | February 8, 2010 3:11 AM
JG: I recommend that you read Carry on, Jeeves by PG Wodehouse. It's a great read! A classic example of British literature. It's also available on audiobook if you do a lot of journeys driving - I've experimented with listening to Wodehouse on tape in the car and it's a satisfying experience.
Posted by: MJ | February 8, 2010 3:31 AM
JG: Perhaps try reading some Kerouac?
I find his work to be rather pleasant.
Posted by: Anonymous | February 8, 2010 6:03 AM
You know, JGab's four questions, if you accept certain caveats, are true - if you permit an infinite path to "represent" a number, then if you give me any number we can identify (ie, not Chaitin's constant), I can come up with a procedure that will generate a "roadmap" of the path - for every node along the path, it will tell us what child of that node continues the path. If we pick an irrational or rational whose denominator, in simplest form, has a prime factorization containing numbers other than 2 and 5, that path will never end, but I think it's a well-defined infinite path.
So you know what? I will answer yes to all four of his -revised- question, because I want to see how he tries to get around the problem of paths containing an omega'th birthday.
Posted by: Michael Ralston | February 8, 2010 6:41 AM
Chaitin's constant is neither relevant nor required. Firstly, I am not dealing with probabilities; every aspect of my proof is completely deterministic. Secondly, Chaitin's constant applies only to algorithms that are selected at random.
This idea is just another distraction ploy by MChu. So what's new? I am still waiting for him to figure out that he has already agreed to point 4 without knowing it. He is so afraid that if he accepts point 4, then the real numbers are by default enumerable. As I said, agreeing to the 4 points does not mean the numbers in my tree are enumerable. One still requires a logical method to traverse and retrieve every number from the tree - this is called enumeration. If one cannot retrieve the numbers in a logical deterministic fashion, my argument is dead.
The fool still fails to realize that he is very inferior to me in intelligence. His responses show. He has contradicted himself over and over again. Not agreeing to point 4 but claiming that my tree contains all real numbers demonstrates thought processes that are juvenile.
Posted by: John Gabriel | February 8, 2010 7:10 AM
Can your opinion be defamatory?
No—but merely labeling a statement as your "opinion" does not make it so. Courts look at whether a reasonable reader or listener could understand the statement as asserting a statement of verifiable fact. (A verifiable fact is one capable of being proven true or false.) This is determined in light of the context of the statement. A few courts have said that statements made in the context of an Internet bulletin board or chat room are highly likely to be opinions or hyperbole, but they do look at the remark in context to see if it's likely to be seen as a true, even if controversial, opinion ("I really hate George Lucas' new movie") rather than an assertion of fact dressed up as an opinion ("It's my opinion that Trinity is the hacker who broke into the IRS database").
MChu: I publicly announce to you that if you do not change the title of this page, I am going to file a lawsuit with regards to defamation of character. You are maliciously calling me a crank. This will serve as your only warning.
If the titles of your web pages are not changed within 7 days and you do not publicly apologize for calling me a crank, you leave me no choice but to pursue justice in a court of law.
My lawyer will be in touch with you at any rate.
John Gabriel
Posted by: John Gabriel | February 8, 2010 7:56 AM
Can I get insurance to cover defamation claims?
Yes. Many insurance companies are now offering media liability insurance policies designed to cover online libel claims. However, the costs could be steep for small blogs—The minimum annual premium is generally $2,500 for a $1 million limit, with a minimum deductible of $5,000. In addition, the insurer will conduct a review of the publisher, and may insist upon certain standards and qualifications (i.e. procedures to screen inflammatory/offensive content, procedures to "take down" content after complaint). The Online Journalism Review has an extensive guide to libel insurance for online publishers.
http://www.eff.org/issues/bloggers/legal/liability/defamation
Posted by: John Gabriel | February 8, 2010 8:04 AM
MChu: I strongly suggest you heed my advice or get insurance because I think you are going to need it.
Posted by: John Gabriel | February 8, 2010 8:08 AM
Perhaps instead of threatening to sue people because they called you a bad name, you could demonstrate that you are in fact not a crank by simply proving what you claim. Several people have now accepted point 4, so go for it...
Posted by: Archena | February 8, 2010 8:13 AM
MChu: These demands I've made are not dependent on your YES/NO response. You need to comply immediately with my requests to remove any printed statement in your page titles that suggests I am a crank.
As for continuing the discussion, it is entirely up to you. I couldn't care less.
Posted by: John Gabriel | February 8, 2010 8:14 AM
As I learned in journalism, it's not libel if it's true...
Posted by: Scully | February 8, 2010 8:31 AM
@JohnG
Ha, ha, ha, oh I can't stop, ha, ha, ha, ha...
"Courts look at whether a reasonable reader or listener could understand the statement as asserting a statement of verifiable fact. (A verifiable fact is one capable of being proven true or false.) This is determined in light of the context of the statement."
Looking at the content of your replies and the total BS you have written and called mathematics, I think the context is pretty clear. If a court only found you to be a crank you would be lucky. Be careful, you may find yourself 'resting' in an institution somewhere.
Whatever you do, don't stop replying, I haven't laughed this hard in a long, long time.
Posted by: reveilebeurt | February 8, 2010 8:46 AM
Posted by: dean | February 8, 2010 8:53 AM
John Gabriel has finally made the stupidest statement yet in his quest. Trying to formalize his argument he writes,
http://knol.google.com/k/john-gabriel/are-real-numbers-uncountable/
That's right, the last step in his logically valid proof of the countability of the reals? Assign a natural number to each of the real numbers. :) :) :)
Posted by: Kurt Deligne | February 8, 2010 9:09 AM
Wow! I loved waking up this morning and seeing such nonsense from JGab!
He's just a big bully, completely ignorant of his own ineptitude in math, yet trying to force others to agree with him -- no, bow down to him -- even though his entire premise behind his math is flawed.
Really... JGab isn't really even attempting to make his point (muchless realizing that his math is wrong and learning from it)... for him, this is now a full-fledged pissing contest.
JGab, not that it would happen, but if Mark were to cry "Uncle!" and admit that he is inferior to you in intelligence, would you still be interested in suing him for calling you a crank?
Also, many others here, at various math forums, and at Wikipedia have called you out on your crankitude, on this topic and others... So are you going to sue everybody who has called you wrong, and then when refused to agree with you called you a crank?
Perhaps all those "mentally challenged" math professors and academics who are "afraid" of you -- probably because they merely called you out on your BS -- should sue you instead?
I mean, at least they, and Mark, are factually correct when they call you a crank. But when math professors, students and probably every person except for the "one person in [your] almost 1/2 century of existence who was intellectually on par with [you]" are claiming that what you say is complete BUNK... well then, I don't think a judge would ever listen to you. As scully said, it's not libel if it's true.
I mean, have you ever considered the possibility that you are wrong and that you are a crank?
Posted by: doctorgoo | February 8, 2010 9:25 AM
@318:
John, it continually amazes me that you are unable to grasp the nature of this forum.
This is not a private conversation between the two of us. This is an open forum.
(1) The Chaitin comment had nothing to do with you. It was part of my attempt to answer Archena's question in comment #291.
(2) You do not get to dictate the terms of what gets discussed here.
(3) I will not edit the title or content of any post or comment on this blog in response to idiotic threats.
Posted by: Mark C. Chu-Carroll | February 8, 2010 9:32 AM
As people may or may not know from my previous occasional comments here, I am an alcoholic. Mr. Gabriel reminds me very much of some unfortunate persons that I meet in AA from time to time: incapable of looking at themselves, incapable of admitting wrong doing. It is a sad state of affairs, and when coupled with substance abuse, is usually fatal.
It is a hard road from where Mt. Gabriel is to sanity. I wish him the best.
Posted by: AnyEdge | February 8, 2010 9:57 AM
"Chaitin's constant is neither relevant nor required. Firstly, I am not dealing with probabilities; every aspect of my proof is completely deterministic. Secondly, Chaitin's constant applies only to algorithms that are selected at random."
Everyone in this thread is dumber for having read that, you owe us all some form of restitution. Chaitin's constant is a real, one that can't be identified in your tree. In fact, most reals are uncomputable, presenting yet another problem for your attempt at enumerating the reals.
Posted by: Anonymous | February 8, 2010 10:08 AM
The above comment was by me, forgot to fill the fields out. :(
Posted by: Tyler DiPietro | February 8, 2010 10:10 AM
@327
Kurt, you saw comment #276, right? "I am not using any axioms".
But yeah, that's pretty awesome. Step 4: Beg the question!
@315
EL OH EL!
By the way, if anyone is interested in some on-topic reading, I can recommend Fads and Fallacies in the Name of Science by Martin Gardner. In it you'll find some perfect descriptions of JG.
Posted by: Ivan | February 8, 2010 10:22 AM
Scully @324: "As I learned in journalism, it's not libel if it's true..."
Gabriel's in the UK, which has laws notoriously favorable to the plaintiffs in these cases.
Even if the defendant wins, they'll likely be bankrupt.
Posted by: Jon H | February 8, 2010 10:58 AM
"As I learned in journalism, it's not libel if it's true..." That's not true, especially in UK. If JGab hadn't called MarkCC so many names, there would've been a remote chance that he could've successfully sued in UK. (It doesn't matter that MarkCC or his blog are not in UK. A court could decide that this blog being readable in UK constitutes a publication in UK, which isn't far fetched as similar rulings have been made in the past.) Of course, I'm no lawyer.
Posted by: Flaky | February 8, 2010 11:16 AM
I'd say it's a mistake to try to address the details of the crackpot's errors.
I know you mentioned it to him but it might be better to just concentrate on as simple a proof as possible that he's wrong. In this case it might be:
1. Cantor's diagnol shows that a countable list can't be a complete list of the numbers between 0 and 1.
2. We can make a countable list of all of the rationals. It's not complete but that's OK - there are irrationals too.
3. Assume we can make a countable list of all of the reals. It's not complete, and that's not OK - the reals have no gaps by definition. So that's not a consistent assumption.
Posted by: kinem | February 8, 2010 12:19 PM
I'd say it's a mistake to try to address the details of the crackpot's errors.
I know you mentioned it to him but it might be better to just concentrate on as simple a proof as possible that he's wrong. In this case it might be:
1. Cantor's diagnol shows that a countable list can't be a complete list of the numbers between 0 and 1.
2. We can make a countable list of all of the rationals. It's not complete but that's OK - there are irrationals too.
3. Assume we can make a countable list of all of the reals. It's not complete, and that's not OK - the reals have no gaps by definition. So that's not a consistent assumption.
Posted by: kinem | February 8, 2010 12:19 PM
This was beautiful, the kind of stuff I love (I must be a sick puppy). Thank you for having so much patience that it makes other people sick :).
Posted by: Shirakawasuna | February 8, 2010 12:30 PM