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 visitingevery nodeon a levelin 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

entirelyirrelevant. 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

representationwhich 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 “&hArr”

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 traverse_{JG}

the tree in a way where you visit_{JG}all of the nodes, and thus you can

enumerate_{JG}all of the nodes by sorta pointing in their direction.But back in the real world, visit

_{JG}!= visit,

traverse_{JG}!= traverse, and enumerate_{JG}!=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 willeverinclude that string. The fact that you’ve

got a representation for the real numbers doesnotimply that you’ve

got anenumerationof the real numbers. Your representation is

notenumerable.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 1/3will never beenumerated 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 becauseyou 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

closestto π. It doesn't exist. And you know that.But you're asking for the impossible, because

you can't do what IIf you had an enumeration, you could specify where things occur in

asked.

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 thatyou are

wrong.John,

you are wrong.Your representation is not enumerable. A

"left to right" traversal - or in factanytraversal 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 thatIn 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 piReally? 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.