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.<\/p>\n
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.<\/p>\n
So once again, I offer to prove it but only on condition you answer my questions. It’s up to you.<\/p>\n
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.<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 181 (MarkCC @ 2\/2\/2010, 3:22pm)<\/b><\/p>\n
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.<\/p>\n
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.<\/p>\n
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?<\/p>\n
But fine, I’ll give you your answers.<\/p>\n
(1) Yes, I’ll agree that all real numbers are *representable* using infinite decimal notation.<\/p>\n
(2) I’ll also agree that taken to infinity, your tree contains all real numbers between zero and 1.<\/p>\n
(3) Yes, I’ll agree that all finitely representable numbers can be enumerated from your tree using a breadth-first traversal.<\/p>\n
(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.<\/p>\n<\/blockquote>\n
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<\/em>; that anyone can post anything they want, any time they want. In the history of this blog, I have never<\/em> deleted a non-spam comment; and I’ve banned a total of three people for egregiously inappropriate behavior.<\/p>\n\nComment 182 (John Gabriel @ 2\/2\/2010, 3:45pm)<\/b><\/p>\n
Very good. Now let’s address your issue with question number 4 so that we can move on.<\/p>\n
“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*.”<\/p>\n
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<\/p>\n<\/blockquote>\n
In a trend that will be repeated frequently, he follows it immediately with other comments:<\/p>\n
\nComment 183 (John Gabriel @ 2\/2\/2010, 3:49pm)<\/b><\/p>\n
That last sentence should have read:<\/p>\n
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.<\/p>\n
If I cease to respond today, it is because I am in a different time zone. So I’ll continue to respond tomorrow.<\/p>\n<\/blockquote>\n
\nComment 184 (John Gabriel @ 2\/2\/2010, 3:51pm)<\/b><\/p>\n
I will not respond to anyone else but you.<\/p>\n<\/blockquote>\n
From here on, it rapidly gets boringly repetitive. John is absolutely convinced that the fact that he’s got a representation<\/em> of real numbers means that he’s got an enumeration<\/em> of real numbers. He continually insists that he can do a “left-right traversal” of the infinitely long<\/em> paths of his tree. I try to point this out:<\/p>\n\nComment 185 (MarkCC @ 2\/2\/2010, 4:01pm)<\/b><\/p>\n
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.<\/p>\n
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.<\/p>\n
But in the case of the infinite tree, traversing the “leafs” makes no sense. It’s not something that you can meaningfully do.<\/p>\n<\/blockquote>\n
He responds in typical fashion:<\/p>\n
\nComment 187 (John Gabriel @ 2\/3\/2010, 2:25am)<\/b><\/p>\n
\tMChu: Looks like we have to first agree on definitions. You claim:<\/p>\n
“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.”<\/p>\n
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.<\/p>\n
Where do you see anything in that definition that implies finite time?It’s not there.<\/p>\n
You claim the number of steps in a top-down traversal can be found in finite time.<\/p>\n
*Finite time* has nothing to do with enumeration. 1\/3 can be found in a finitenumber 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.<\/p>\n
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.<\/p>\n
You also claim:<\/p>\n
“But in the case of the infinite tree, traversing the “leafs” makes no sense. It’s not something that you can meaningfully do.”<\/p>\n
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!”<\/p>\n
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.<\/p>\n
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.<\/p>\n
Another way of stating Countability is by saying, “A set is countable if one can write out its members”.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
So, in order to proceed, I must have your acknowledgement that it is possible to perform left-to-right finite traversals.<\/p>\n
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.<\/p>\n<\/blockquote>\n
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 somehowmake my mathematical arguments less valid?<\/p>\n
And of course, John never replies with one comment. There are always multiple follow-ons.<\/p>\n
\nComment 188 (John Gabriel @ 2\/3\/2010, 2:28am)<\/b><\/p>\n
Sorry, I need your acknowledgement to:<\/p>\n
So, in order to proceed, I must have your acknowledgement that it is possible to perform left-to-right infinite traversals.<\/p>\n<\/blockquote>\n
\nComment 189 (John Gabriel @ 2\/3\/2010, 2:41am)<\/b><\/p>\n
That last comment was too long. Let me summarize it.<\/p>\n
One can perform an infinite left to right traversal as follows:<\/p>\n
Find the first digit of the decimal representation and stop.<\/p>\n
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.<\/p>\n
As simple as that. So now we have taken care of your finite time issue. YES? May I continue?<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 190 (MarkCC @ 2\/3\/2010, 8:11am)<\/b><\/p>\n
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.<\/p>\n
More importantly, it doesn’t matter<\/em>. 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.<\/p>\nOn 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<\/em> on a level in left to right order.<\/em><\/p>\n 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”.<\/p>\n
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.<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 191 (John Gabriel @ 2\/3\/2010, 10:01am)<\/p>\n
<\/b><\/p>\n
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*.<\/p>\n
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.<\/p>\n
Now read carefully: Don’t you dare address me ever again as you have in your previous comment!<\/p>\n
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?<\/p>\n
Now let’s proceed again:<\/p>\n
You say:<\/p>\n
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”.<\/p>\n
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.<\/p>\n
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.<\/p>\n
Ok, stupidicus? Now, may I continue? Abuse me one more time verbally and piss on you!!!!<\/p>\n<\/blockquote>\n
And, as always, one comment isn’t enough for John.<\/p>\n
\nComment 192 (John Gabriel @ 2\/3\/2010, 10:05am)<\/b><\/p>\n
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).<\/p>\n
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!<\/p>\n
YES or NO to question number 4? Spare me any other shit talk. I don’t want to discuss anything else.<\/p>\n<\/blockquote>\n
\nComment 193 (John Gabriel @ 2\/3\/1010, 10:20am)<\/b><\/p>\n
Just a point of clarification:<\/p>\n
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.<\/p>\n
I am not completing the infinite traversals – there is no sense in even thinking about such an absurdity.<\/p>\n<\/blockquote>\n
Back to me. At this point, I was starting to lose my temper, and stopped bothering to try to be polite.<\/p>\n
\nComment 194 (MarkCC @ 2\/3\/2010, 10:26am)<\/b><\/p>\n
@191:<\/p>\n
See, this is exactly what I predicted. You just regurgitate the same nonsense, and throw insults the moment I disagree with you.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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?<\/p>\n
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<\/em> 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?<\/p>\n 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?<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 195 (John Gabriel @ 2\/3\/2010, 10:29am)<\/b><\/p>\n
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.<\/p>\n
So if there isn’t anything else, I think you are ready to agree with point 4.<\/p>\n<\/blockquote>\n
\nComment 196 (John Gabriel @ 2\/3\/2010, 10:32am)<\/b><\/p>\n
A infinite left-right traversal is starting off at any one of the top nodes and visiting only one node from each level.<\/p>\n<\/blockquote>\n
\nComment 197 (John Gabriel @ 2\/3\/2010, 10:34am)<\/b><\/p>\n
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.<\/p>\n<\/blockquote>\n
\nComment 198 (John Gabriel @ 2\/3\/2010, 10:37am)<\/b><\/p>\n
Left-right:<\/p>\n<\/p>\n
0.1 – 4 – 5 – ……<\/p>\n<\/p>\n
Top-down:<\/p>\n<\/p>\n
0.1<\/p>\n
0.2<\/p>\n
0.3<\/p>\n
…..<\/p>\n<\/blockquote>\n
\nComment 199 (John Gabriel @ 2\/3\/2010, 10:42am)<\/b><\/p>\n
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.<\/p>\n
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.<\/p>\n<\/blockquote>\n
\nComment 200 (John Gabriel @ 10:59am)<\/b><\/p>\n
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.<\/p>\n
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.<\/p>\n<\/blockquote>\n
Finally, I manage to get a response in.<\/p>\n
\nComment 201 (MarkCC @ 2\/3\/2010, 11:24am)<\/b><\/p>\n
@200:<\/p>\n
In one message, you say: <\/p>\n
\n4. 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?\n<\/p><\/blockquote>\n
In another, you say:<\/p>\n
\nn 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.\n<\/p><\/blockquote>\n
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. <\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 202 (John Gabriel @ 2\/3\/2010, 12:41pm)<\/b><\/p>\n
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.<\/p>\n
2. Yes, if you are enumerating (writing down) the digits of a specific number, then you are traversing a single path.<\/p>\n
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. <\/p>\n
4. Enumerating multiple numbers: This requires parallel traversals that all take place at the same time.<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 203 (John Gabriel @ 2\/3\/2010, 12:50pm)<\/b><\/p>\n
“If you’re enumerating multiple numbers, then your traversal is not ever visiting the nodes corresponding to the numbers with infinite representations.”<\/p>\n
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.<\/p>\n
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.<\/p>\n<\/blockquote>\n
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<\/em>.<\/p>\n\nComment 204 (MarkCC @ 2\/3\/2010, 1:09pm)<\/b><\/p>\n
No, it *won’t* visit the nodes. It will never reach them.<\/p>\n
It comes back to exactly where we started: I said that your problem is that you can represent nodes, but not actually enumerate them.<\/p>\n
What you’re doing now is playing games with terminology. You’re trying to redefine the meanings of “enumeration” and “traversal”.<\/p>\n
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 neveractually finish reaching the nodes that they supposedly enumerate.<\/p>\n
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”.<\/p>\n
You’ve got a representation<\/em> 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.<\/p>\n<\/blockquote>\n And John’s next comment confirms my enlightenment.<\/p>\n
\nComment 205 (John Gabriel @ 2\/3\/2010, 1:54pm)<\/b><\/p>\n
The nodes *are* visited, but you are not able to see it. So this is why we are having this discussion.<\/p>\n
Enumerate means to count off one by one.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
MChu: And it involves “enumerating” nodes that will never actually get enumerated,<\/p>\n
No. You are not enumerating *nodes*, you are enumerating decimal numbers. Big difference. The nodes are the building blocks of each decimal number.<\/p>\n
MChu: …by using an infinite number of processes which will never actually finish reaching the nodes that they supposedly enumerate.”\n<\/p>\n
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.<\/p>\n<\/blockquote>\n
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<\/em> generate 1\/3, the tantrums start again.<\/p>\n And of course, there’s always the sequence of multiple followups.<\/p>\n
\nComment 206 (John Gabriel @ 2\/3\/2010, 1:58pm)<\/b><\/p>\n
MChu: You still can’t enumerate the real numbers. You can represent them, but you can’t enumerate your representations.<\/p>\n
If I can represent them, I can enumerate them. Representation Enumeration<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 207 (John Gabriel @ 2\/3\/2010, 2:02pm)<\/b><\/p>\n
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.<\/p>\n
Hate quoting from Wikipedia but this definition is correct.<\/p>\n<\/blockquote>\n
\nComment 208 (MarkCC @ 2\/3\/2010, 2:09pm)<\/b><\/p>\n
@205:<\/p>\n
\nTell 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.\n<\/p><\/blockquote>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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<\/sub> the tree in a way where you visitJG<\/sub> all of the nodes, and thus you can enumerateJG<\/sub> all of the nodes by sorta pointing in their direction.<\/p>\n But back in the real world, visitJG<\/sub> != visit, traverseJG<\/sub> != traverse, and enumerateJG<\/sub>!=enumerate.<\/p>\nthe real numbers without actually enumerating them”, then you can<\/p>\n<\/blockquote>\n
(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.)<\/em><\/p>\n\nComment 209 (MarkCC @ 2\/3\/2010, 2:19pm)<\/b><\/p>\n
Yes John, an enumeration is a complete listing of all of its elements. Which is really just begging the question: what is a listing?<\/p>\n
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.<\/p>\n
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”!)<\/p>\n
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.<\/p>\n
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?<\/p>\n<\/blockquote>\n
\nComment 210 (John Gabriel @ 2\/3\/2010, 2:34pm)<\/b><\/p>\n
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.”<\/p>\n
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.<\/p>\n
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.<\/p>\n
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.”<\/p>\n
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!<\/p>\n
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.<\/p>\n
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. <\/p>\n<\/blockquote>\n
\nComment 211 (John Gabriel @ 2\/3\/2010, 2:46pm)<\/b><\/p>\n
By the way, I have never redefined anything. Please don’t go down that road. You are smarter than this, I think.<\/p>\n
Just stay with the argument. If you do not understand I will try to explain. Just ask.<\/p>\n<\/blockquote>\n
\nComment 212 (John Gabriel @ 2\/3\/2010, 2:51pm)<\/b><\/p>\n
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”!)<\/p>\n
Once we agree on all four points, the above statement is what I shall proceed to prove.<\/p>\n<\/blockquote>\n
\nComment 213 (MarkCC @ 2\/3\/2010, 3:05pm)<\/b><\/p>\n
I’m not disputing that you can represent 1\/3 as an infinitely long decimal string 0.333…..<\/p>\n
What I’m disputing is that you can produce an enumeration of the nodes of your tree which will ever<\/em> include that string. The fact that you’ve got a representation for the real numbers does not<\/em> imply that you’ve got an enumeration<\/em> of the real numbers. Your representation is not<\/em> enumerable.<\/p>\nYou 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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
In your system, you can’t specify the natural number that corresponds to 1\/3 in your “enumeration” – because 1\/3 will never be<\/em> enumerated by your “traversal”.<\/p>\n The fact that you can represent the members of an infinite set does not imply that you can enumerate them.<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 214 (John Gabriel @ 2\/3\/2010, 3:23pm)<\/b><\/p>\n
MChu: “You can whine and complain all you like – but it’s not going to change anything.”<\/p>\n
Taking cheap shots at me does not help. In fact it makes you look stupid.<\/p>\n
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.”<\/p>\n
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!<\/p>\n
MChu:”Your representation is not enumerable.”<\/p>\n
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:<\/p>\n
MChu: “His enumeration is based on trees. You can create an infinite tree of the decimal representation of the numbers between zero and one.”<\/p>\n
My, oh my, but you have such a short memory, don’t you?<\/p>\n
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.<\/p>\n
Now, you have to answer YES to question 4 and understand why the answer is correct otherwise I am wasting my time with you.<\/p>\n
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.<\/p>\n<\/blockquote>\n
Once again, we encounter the fundamental problem. John simply does not understand that the fact that a set is representable<\/em> doesn’t mean that it’s enumerable<\/em>. 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<\/em> all of the real numbers using infinitely long representations as a proof that I accept that you can enumerate<\/em> those infinitely long representations.<\/p>\n\nComment 215 (John Gabriel @ 2\/3\/2010, 3:32pm)<\/b><\/p>\n
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.<\/p>\n
1. Do you agree that every real number in the interval (0,1) can be represented in decimal?<\/p>\n
\tYES or NO<\/p>\n
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.<\/p>\n
\tYES or NO.<\/p>\n
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.<\/p>\n
\tYES or NO.<\/p>\n
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?<\/p>\n
YES or NO.<\/p>\n
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.<\/p>\n
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.<\/p>\n
I now am beginning to see I was mistaken.<\/p>\n<\/blockquote>\n
\nComment 216 (MarkCC @ 2\/3\/2010, 3:42pm)<\/b><\/p>\n\nOh 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!\n<\/p><\/blockquote>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
You keep responding to tiny pieces of my criticism that you can misrepresent, while ignoring the substantial parts.<\/p>\n
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.<\/p>\n
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.<\/p>\n
So come on, put up or shut up. Where does 1\/3, or 1\/9, or 1\/\u03c0, 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.<\/p>\n<\/blockquote>\n
\nComment 217 (John Gabriel @ 2\/3\/2010, 4:37pm)<\/b><\/p>\n
I am not surprised at your outburst and false accusations.<\/p>\n
I have informed you that in order for me to proceed, you have to consent with the 4 statements I put to you.<\/p>\n
Look at you: Come on John! Tell me, what number corresponds to 1\/3, or pi or e!<\/p>\n
Let me ask you: Come on Mark, tell me, what number corresponds to the smallest rational number greater than pi?<\/p>\n
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?<\/p>\n
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.<\/p>\n
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.<\/p>\n
Your arguments are lame.<\/p>\n<\/blockquote>\n
\nComment 218 (John Gabriel @ 2\/3\/2010, 4:42pm)<\/b><\/p>\n
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.<\/p>\n
This is what enumeration is all about. Just another gaping hole in your understanding!<\/p>\n<\/blockquote>\n
I’ll give John one small point here. My understanding is that he’s proposing his tree as a way of generating a specific<\/em> 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<\/em> 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<\/em> number with an infinite representation. He can’t, because it won’t.<\/p>\n\nComment 219 (MarkCC @ 4:50pm)<\/b><\/p>\n
\tJohn:<\/p>\n
You can “inform” me of anything you want. It doesn’t change reality.<\/p>\n
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.<\/p>\n
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.<\/em><\/p>\n In return, you come back with something that by definition<\/em>, doesn’t exist, and ask me to show it to you. There is no rational number closest<\/em> to \u03c0. It doesn’t exist. And you know that.<\/p>\n But you’re asking for the impossible, because you can’t do what I asked.<\/em> If you had an enumeration, you could specify where things occur in it. By definition<\/em>, if you had an enumeration, you could specify where things occur in it. But you don’t.<\/p>\n 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”.<\/p>\n
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.<\/em><\/p>\nJohn, you are wrong.<\/em> Your representation is not enumerable. A “left to right” traversal – or in fact any<\/em> traversal of your tree will never reach a single number with an infinite representation.<\/p>\n<\/blockquote>\n\nComment 220 (John Gabriel @ 2\/3\/2010, 5:03pm)<\/b><\/p>\n
Nonsense. I do not have to provide a particular natural number that corresponds to any of the real numbers in my tree.<\/p>\n
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.<\/p>\n
No where in the definition of countability is there a requirement to produce a particular natural number for any of the rational numbers.<\/p>\n
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.<\/p>\n
Alas, you do not.<\/p>\n<\/blockquote>\n
\nComment 221 (John Gabriel @ 2\/3\/2010, 5:03pm)<\/b><\/p>\n
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.<\/p>\n
Evidently not according to your paragraph 10!<\/p>\n<\/blockquote>\n
Again with the same old problem. He just can’t wrap his head around the fact that representation does not<\/em> imply enumeration.<\/p>\nComment 222 (John Gabriel @ 2\/3\/2010, 5:09pm)<\/b><\/p>\n
MChu: There is no rational number closest to \u03c0. It doesn’t exist. And you know that.<\/p>\n
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.<\/p>\n<\/blockquote>\n
\nComment 223 (John Gabriel @ 2\/3\/2010, 5:11pm)<\/b><\/p>\n
MChu: There is no rational number closest to \u03c0. It doesn’t exist. And you know that<\/p>\n
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!!<\/p>\n<\/blockquote>\n
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.<\/p>\n
\nComment 224 (Scully @ 2\/3\/2010, 10:03pm)<\/b><\/p>\n
Quite possibly the dumbest thing I have ever read:<\/p>\n
MarkCC: There is no rational number closest to \u03c0. It doesn’t exist. And you know that.<\/p>\n
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.<\/p>\n
okay, really Gabriel?<\/p>\n
We shall show, John Gabriel is wrong, and there is no rational closest to pi.<\/p>\n
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<\/p>\n
Really? Try not to make it that easy…<\/p>\n<\/blockquote>\n
\nComment 225 (Scully @ 2\/3\/2010, 10:05pm)<\/b><\/p>\n
this pi less than s less than r, contradction<\/p>\n<\/blockquote>\n
And of course, John’s fit and leavetaking – which includes an absolutely spectacular<\/em> 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.<\/p>\n\nComment 226 (John Gabriel @ 2\/4\/2010, 3:51am)<\/b><\/p>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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*.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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!<\/p>\n<\/blockquote>\n
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. <\/p>\n
You see, John is special<\/em>. 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<\/em> his own rules, wherever he goes, and everyone else is obligated<\/em> to follow them. He can insult you<\/em>, 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<\/em>.<\/p>\n John: you’re a crank. In fact, you’re a pathetic<\/em> crank.<\/p>\n","protected":false},"excerpt":{"rendered":"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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[11],"tags":[],"class_list":["post-842","post","type-post","status-publish","format-standard","hentry","category-cantor-crankery"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p4lzZS-dA","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/842","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/comments?post=842"}],"version-history":[{"count":0,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/842\/revisions"}],"wp:attachment":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/media?parent=842"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/categories?post=842"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/tags?post=842"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}