Y the vast majority of people on here r wrong and ought to hang their heads in sham

[Interesting Ian]

Originally posted by Wudang
No sorry, Ian but you're skipping between "a random arbitrarily large value" and "the set of arbitrarily large values". What holds true for the former as an element of a set does not hold true for the set as a whole - they're of different orders - think of Russells theory of types, yes? Just because you can get to a randomly selected number in finite time, or by counting, does not mean that you can exhaust the set in the same manner.

I think I understand what you mean. And it's irrelevant. See above post.

BTW I know no maths at all. Never heard of any of this stuff before Monday. Don't know what Russells theory of types are. But guessing my above post addresses it.

[Kerberos]

Originally posted by Interesting Ian

So this is like our problem; no matter how far I can go up the probability will be above 0, equates that no matter how big a number I name there will be a bigger one.

But nevertheless they are all finite!

Now the fact that they are all finite means that I must be able to get to them by counting

(WOW, think I might know what Scribble means by "countably infinite" now! LOL)

If I can get to them by counting then I merely need an unlimited/unbounded search, rather than an infinite one! :D (because like the numbers there is infinite number of finite strings)
What exactly is the difference between unlimited and infinite in your mind? In any case for every one of our countable numbers, there is a probability which is greater than zero that your sequence won't turn up. As I explained infinitively close to zero does not in fact equal zero.

[Interesting Ian]

Originally posted by epepke


You underestimate Ian. There are quite a few people here that could be replaced by a Perl script, but Ian isn't one of them. I'm a big fan of Ian, and I think we should be glad he posts here. He brings so many things that we would otherwise have to go out and look for examples for.

There's only one person I've ever encountered who comes close to Ian's class: Scott Erb. He still exists; you can find him on a google search on USENET. But ten or more years ago, he came up with what I consider to be one of the most important discoveries of postmodern thought. I still call it Erb's Maneuver.

It works like this: Let's say you're in an argument. You want to prove that someone said a certain thing. So, you take the word that the person used and translate into another language. (It doesn't matter which one.) Then you find the etymology in that language. Then you translate the etymology back into the original language and use that to argue. Brilliant!

But Ian comes up with stuff that is nearly this good every day.

You disappoint me epepke. It is the precise converse of what you state. Merely saying "not so, it's you" doesn't alter reality and is moreover very childish.

And you're wrong about this problem. Sorry.

[Interesting Ian]

Originally posted by davefoc
My apologies, I just read through quite a bit of this thread and I still don't quite understand what the various positions are.

Is Ian saying that in an infinite string of random digits the probability that any arbitrary string will occur within the infinite string is 1?

If this isn't what Ian is saying how is what Ian saying different from that.

It need only be finite.

[Interesting Ian]

Originally posted by scribble


You're absolutely right. I was being overly fanciful in my description. I was just trying to point out that he's fond of using words he doesn't understand, and then he thinks that's some kind of defense when we tell him what he said is nonsense.

LOL You really haven't got a clue have you. I remember your total and unbelievable lack of comprehension in all the philosophy debates. You're even worse than the average skeptic, and trust me, that's saying something!! :eek:

[Interesting Ian]

Originally posted by Cecil
No, I stand by my earlier comments.

It was a joke. :p Lighten up. :D

(or maybe I should use :mad: instead? :D)

Oh right, ok :)

I thought you might be pissed off with me for being so arrogant all the time.

I'm only arrogant on here though. It's a reaction to skeptics continually saying all believers are stupid, and I'm particularly stupid, and doing so when they clearly haven't understood my arguments. Hell, it even happens with maths! :confused: :eek:

I'm not arrogant in real life though :) Well ok, a bit, but I've learnt to hide it LOL

[Wudang]

Originally posted by Interesting Ian


I think I understand what you mean. And it's irrelevant. See above post.

BTW I know no maths at all. Never heard of any of this stuff before Monday. Don't know what Russells theory of types are. But guessing my above post addresses it.

I can't see any connection between your referenced post and mine. I say nothing about recurrent numbers. I am saying that just because you can pick one and count to it and repeat for another does not mean you can count to them all, because the concept of "all" doesn't apply.
As for Russell here (http://plato.stanford.edu/entries/russell-paradox/http://) or here (http://forums.philosophyforums.com/showthread.php?threadid=576) is more about logic than maths as such.
You're "guessing"?
If I can get to them by counting then I merely need an unlimited/unbounded search, rather than an infinite one! (because like the numbers there is infinite number of finite strings)
Here's the error - for any random selected number you are correct that you can eventually count to it. But you can never say that you've counted them all because the concept of all doesn't apply. A number one type of thing, the set of all such numbers is a different type of thing. You have to understand that there is a logical difference between saying "for any x" and "for all x".

[Interesting Ian]

Originally posted by Kerberos

First of all it should be made clear that infinitely close to 0, or unlimitedly close to 0 is exactly 0. I do not believe anyone does, or could deny this.
--------------------------------------------------------------------------------


You're wrong, it can, is and should be disputed since it's not true. There is in fact an entire branch of mathematics devoted to calculating what happens when you get infinitively close to a number, but don't reach it (called calculating border values in Danish). In most cases there is no difference but in some there are. For example 0^0=1, but lim(0^x) for x-->0 that is 0^x for x approaching zero is 0. I haven't read the entire thread so this might already have been pointed out..

Well I've never read a maths text book in my life.

But!!

1/infinity cannot equal a finite value!! :eek:

[Interesting Ian]

Originally posted by Wudang


Here's the error - for any random selected number you are correct that you can eventually count to it. But you can never say that you've counted them all because the concept of all doesn't apply. A number one type of thing, the set of all such numbers is a different type of thing. You have to understand that there is a logical difference between saying "for any x" and "for all x". [/B]

I'm not saying you can count to them all. Just you can count to any one of them.

[Interesting Ian]

Originally posted by jj


You just admitted that it MUST be possible, if the generator is random. Not very likely, perhaps, but POSSIBLE. The generator could always randomly choose to leave out any number you name. Yes, the probability is dreadfully small for any finite number of attempts, but it IS finitely small.

As I said before, the LIMIT is zero as you APPROACH infinity.

That is not the same as p==0.

Oh dear, you still don't understand :(

[Wudang]

Originally posted by Interesting Ian


I'm not saying you can count to them all. Just you can count to any one of them.

Cool, that's one of the things that wasn't clear. So what you say applies to any given number but does not apply to the infinite set of numbers. See? Infinity applies to the set of numbers, not to any given element within that set or even to every element.

[Interesting Ian]

Originally posted by jj


(missed that first time around)

I am applying no "intuition". I am simply showing that such a string (no, for simplicity's sake I did not show the most likely) CAN exist. There is no "intuition" necessary. For any number of attempts as you APPROACH infinity, the probabilty remains above zero, reaching zero IN THE LIMIT. Small, certainly. age-of-the-universe-and-beyond small, indeed, but still finitely above zero for any large but finite string.

Who's talking about any large but finite string?

[Interesting Ian]

Originally posted by Wudang


Cool, that's one of the things that wasn't clear. So what you say applies to any given number but does not apply to the infinite set of numbers. See? Infinity applies to the set of numbers, not to any given element within that set or even to every element.

See what? How does this refute my original argument. Let's stay focussed shall we?

[Interesting Ian]

Originally posted by epepke


Thanks.

I liked wollery's explanation, too, with the random number generator that generates a random number in the range [0..1].



Oh dear :( See post above somewhere for refutation of this.



Actually, it's even worse than that. A generator that generates a rational number in that range would also have a probability of 0 of generating any particular number. And there are infinitely more real numbers than rational numbers.

I was just trying to explain the kind of fun mathematicians can have with this stuff.

I expect Ian could use a little lie-down at this point. As opposed to a simple lie.

What am I lying about epepke? You've got it wrong.

[Interesting Ian]

Originally posted by Cecil
Re the intuition thing, I wasn't talking about you specifically, but that's beside the point. :D

I said above that what is true for one PARTICULAR item in a set (oracle in this case) is not necessarily true for a random item from the (infinite) set.

I understand that by extending the string digit by digit, the probability of finding your given substring never "becomes" 0 at any point. Clearly, no matter how many times you extend it there is always a non-zero probability that you have found it so far.

However, by the very nature of your random digit generator, it must necessarily contain any given substring AT SOME POINT (though unspecified until you find it).

And at any point should be stressed.

Yes, I'm in entire agreement with this post.

[Wudang]

We're not even close to focussed as you keep playing fast and loose with definitions. To jj you say "Who's talking about any large but finite string?" to me you say ""I'm not saying you can count to them all. Just you can count to any one of them." i.e a large but finite string. Elsewhere you keep saying "unlimited but not infinite" and then treating "unlimited" as being infinite.
I'm saying that your argument that you can count to any limitted number has no bearing on the original question.

Reminds me of the story of Syd Barrett's last composition for Pink Floyd called "Have you got it yet?". Every time they tried the tune it was different.

[Wudang]

However, by the very nature of your random digit generator, it must necessarily contain any given substring AT SOME POINT (though unspecified until you find it). Whoa! Necessarily only if we allow it to be arbitrarily long - in effect , infinite.

[Interesting Ian]

Originally posted by scribble


But it might never come up. The probability of it coming up approaches 1 so ***** quickly that it would be beyond amazing if it never did... but the math says you cannot promise that it will ever come up.

If you think you can promise that it'll come up, tell me when. Yeah, it's a loaded question, but what else can I say?

In some unspecified finite time.

[Interesting Ian]

Originally posted by Wudang
We're not even close to focussed as you keep playing fast and loose with definitions. To jj you say "Who's talking about any large but finite string?"



Yes, no-one is.




to me you say ""I'm not saying you can count to them all. Just you can count to any one of them." i.e a large but finite string.

[quote]

Yes your search cannot go on forever.

[quote]
Elsewhere you keep saying "unlimited but not infinite" and then treating "unlimited" as being infinite.



No, unlimited as in unbounded. A different thing from infinity. Yes?



I'm saying that your argument that you can count to any limitted number has no bearing on the original question.



I've already explained why it does, [b]several times. Maybe you can understand 69Doge, Vorticity's, Cecils arguments better. Their position is identical to mine but they can express themseves better in mathematical terminology.

[Interesting Ian]

Originally posted by Cecil
You claim that an unbounded string of tails is possible because at each flip, the coin has a chance to land tails. If it lands tails at every flip, by chance, then presto! All tails.

The problem is, this cannot happen. After any specified finite number of flips the coin might have landed tails each time, but I can always say to you: "okay, flip it one more time", without end. The coin cannot land tails EVERY time, an infinite number of times. In fact, any sequence you care to name in ADVANCE will never turn up, since no matter how many matches you get at the start, there are still an infinite number of matches left you have to make. The fact that SOME sequence has to turn up is irrelevant. After it is picked the probability of it being chosen is meaningless. You could not have predicted it.

This probably makes no sense; Vorticity and 69dodge are both far more eloquent than I.

No Cecil, it makes absolute sense. I couldn't agree with you more.

[apoger]

>We're not even close to focussed as you keep playing fast and loose with definitions.

Ian has draw a number of people into a discussion and then dragged things out by "playing fast and loose with definitions"?

What shocking and unprecedented behavior! ;)

[slimshady2357]

Originally posted by Cecil
You claim that an unbounded string of tails is possible because at each flip, the coin has a chance to land tails. If it lands tails at every flip, by chance, then presto! All tails.

The problem is, this cannot happen. After any specified finite number of flips the coin might have landed tails each time, but I can always say to you: "okay, flip it one more time", without end. The coin cannot land tails EVERY time, an infinite number of times.

This is the problem for me. You MUST be willing to search for an infinite amount of time for the probability to become zero. To quote you again:

The coin cannot land tails EVERY time, an infinite number of times.

As 69dodge pointed out in the other thread, the probability that you WILL have to search forever is 0. However, if you are not WILLING to search an infinite amount of finite strings from the start (i.e. from the start you are saying there will be some of the finite strings you WILL NOT SEARCH, then mathematically, you cannot say the probability of finding the string is 1.

Adam

[Paul C. Anagnostopoulos]

I'll repeat what I said in the other thread that Ian started on exactly the same subject:

Does a term like infinite search even make sense? Mathematicians are careful to define sets by specifying the actual members of the set, not by specifying generators to create the sets (except where the latter is unambiguous and nonparadoxical). This is how paradoxes like the town barber are eliminated.

It seems to me that some people are mixing set definitions and generators willy-nilly. If you have the set of all possible strings of digits, Ian's target will be in there somewhere. If you try to generate the set randomly, you may never find the target.

Don't we need to establish whether we are defining the set or generating it?

~~ Paul

[Interesting Ian]

Originally posted by slimshady2357


This is the problem for me. You MUST be willing to search for an infinite amount of time for the probability to become zero. To quote you again:



As 69dodge pointed out in the other thread, the probability that you WILL have to search forever is 0. However, if you are not WILLING to search an infinite amount of finite strings from the start (i.e. from the start you are saying there will be some of the finite strings you WILL NOT SEARCH, then mathematically, you cannot say the probability of finding the string is 1.

Adam

But these finite strings he will not search are not specified upfront. It is unknown what these unsearched finite strings are until the right string is found.

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos
[B]I'll repeat what I said in the other thread that Ian started on exactly the same subject:



Yes, sorry about that. I thought I'd start another thread with an appropriately enticing title to get yet even more people on my side.



Does a term like infinite search even make sense?



We're talking about an unlimited search, not an infinite one.

[Eos of the Eons]

What do you think of my new avatar "smarypants"

[DickK]

Originally posted by Eos of the Eons
What do you think of my new avatar "smarypants" Nice, but I hope you haven't just misspelt smearypants :p "Is it a bird, is it a plane, no it's Smearyp...", on the other hand it does sort of look like a Batpooch.

[Skeptic]

But what is it really? Let's say we have a string of digits that doesn't just approach but really is infinitely long. We can pretend this; we're not constrained to the real world. We could imagine some sort of magical machine (there's an actual name: oracle) which produces one digit in a half second, the next digit in a quarter second, and so on and so forth and let it run forever.

Hmmmm........ what digit is the machine producing after exactly one second?

(Just kidding. Yes, I know what an orcale machine is...)

[Eos of the Eons]

Originally posted by DickK
Nice, but I hope you haven't just misspelt smearypants :p "Is it a bird, is it a plane, no it's Smearyp...", on the other hand it does sort of look like a Batpooch.


LOL!

oops, you know, Smartiepants.

If you know smarties, they might get smeary...

Oh boy, this is very naughty.

Okay, you know I meant Smartypants, but I like your take better :D


Batpooch? http://www.members.shaw.ca/eostory/HMMM14.GIF

No no, he's the sockinator

[Paul C. Anagnostopoulos]

Ian said:
We're talking about an unlimited search, not an infinite one.
Ian's method of responding to a post: Read the words until you find one that doesn't match your personal definition, point that out, and quit.

Gee, I wonder why we never get anywhere?

So Ian: Are we starting with a predefined set of sequences, or are we generating them? Can you answer the question?

~~ Paul

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos

Ian's method of responding to a post: Read the words until you find one that doesn't match your personal definition, point that out, and quit.

Gee, I wonder why we never get anywhere?

So Ian: Are we starting with a predefined set of sequences, or are we generating them? Can you answer the question?

~~ Paul

We are generating them.

[(S)]

Originally posted by 69dodge
Is that correct?

The rationals are countable, and my understanding is that probability is supposed to be countably additive, i.e., the probability of a countable union is the sum of the individual probabilities. If each rational number in [0, 1] is assigned a probability of 0, then so must the entire range have probability 0. But that can't be; the entire space should have probability 1.

Looking at it another way, can you describe in any detail how a random number generator could guarantee that its output be rational while still being uniformly distributed, which I guess was your intention?

The problem is, you're working with infinite numbers of things. Normal rules have to be adjusted. The exact flaw is saying, "A particular random rational number can occur once, and there are an infinite number of rational numbers, so therefore the probability of that number occuring is zero," THEN going and using that /zero/ in an infinite sum. An inifinitesimal, all by its lonesome, might as well be zero. But when you take an infinite number of them together, they might not be. They can add up to zero, or they can add up to one, or they can add up to an infinity. This is how an infinitesimal number is different from zero.
This underscores the biggest mistake people make when working with infinite numbers: You can't make the intuitive simplifications and expect the answer to come out right.

[EDIT]
It gets more fun. Suppose you want to determine the probability of choosing an odd number as a random pick from the set of even & odds. Well, there a 1/alelph-0 chance of picking any particular odd number, you might say. And there are aleph-0 odd numbers [there is a 1-to-1 correspondence between the set of all natural numbers and the set of odd numbers], so that means there's a probability of 1 that you might pick an odd number -- oops. Or maybe you could look at it as being a 0 chance of any particular even number being generated. There is, therefore, a 0 chance of an even number being generated at all, and a 1 probability that you might pick an odd number -- oops.
ERGO, CONCORDANTLY, VIS-A-VIS, it is certain that you will pick an odd number from the set of all natural numbers.

[T'ai Chi]

Originally posted by jj

Troll. Make a mistake, try to cover up for it. Sorry, you're WhoChi Hammegkson as far as I'm concerned.

Some homework for you. You can PM me and I'll grade your work:

a) Z_1 consists of what number(s): {insert number(s) here}
b) Evaluate the following: 1 mod 1 =
c) Evaluate the following: 2 mod 1 =
d) Evaluate the following: 3 mod 1 =

I'm wrong? Really? I'm interested in your answers to a) and b) in that case. :)

[jj]

Originally posted by Cecil
Maybe this will clarify:

If you start flipping a fair coin, you will EVENTUALLY flip a head. Maybe not in 1000 years, maybe not in a googolplex years. Eventually.

Restating that, an important fact in things like network synchroniziation, etc, in fact, you can NEVER guarantee a positive outcome in finite time under any situation.

You can make the probability astonishingly low. That's not zero.


The only way for this to be false is for the coin to land tails every single time you flip it, forever. When the coin does that, it's not a fair coin anymore.

Wrong, it's just 'incredibly unlikely" that it will never give you "heads" in any finite number of trials.

It IS possible that in any finite time you will get no "heads" or whatever.

It is UNLIKELY, to say the least, but it is POSSIBLE.
If you flip a head, then you will have flipped a head in some finite amount of time.

And it is likely that you will do so, however, you can not GUARANTEE that outcome from a fair coin, ever.

If you don't flip a head, then and only then will you have to flip an infinite number of times. But the only way for you to flip an infinite number of tails is for the coin to land tails with probability 1 each time.
That's circular. Now you're setting probability after the fact from observation.

In some cases, that's all one can do, but in such cases there is always a measurement error... Again, this error LIMITS at zero, not is EQUAL to zero.

[Skeptic]

Well, it's a number with these properties. There are numbers that are not greater than it but not less than 0. However, it is less than any such number but not less than 0. In the set of real numbers, we'd have to conclude that it was, identically, 0. But it isn't in that set. The set it's in has to be a set in which the operators >, <, <=, and >= don't work the way they do with real numbers.

Yes, indeed, but I doubt very much Interesting Ian was deliberately trying to get our attention to the fact that non-zero infinitesimals which do not obey the Archemidian axiom can be defined in non-standard analysis, or that the property of addition and multiplication on such sets can be defined in a way that is closed and makes them a commutative ring...

If you use the English language in this non-standard way to mean elements of THOSE sets, then of course you can have such creatures as "inifnitesimally small but not zero" numbers, the number "infinity" which is a number greater than all other numbers (look at the set with denumerable cardinality aleph-null and ordinality omega+1, for instance, call the last number "infinity", and extend addition in the obvious manner), probability functions that assigns non-zero infinitesimals to elements of the possible events Omega, and so on and so forth.

But "Ineresting Ian" was obviously speaking about the usual set of real numbers when he made his statements about probability. Dealing with the real numbers, infinity is not a number, and the only number that is infinitesimally close to 0 is 0 itself. He was just wrong here, since his intuitions about infinitesimals and infinity are muddled, as is his idea about what probability functions are (leading to his continued confusion of "probability 0" with "logical impossiblity").

This in itself is hardly an indictment of Ian--the vast majority of humanity are confused about these issues if they think about them at all. But what IS annoying about him is that he keeps insisting that his vague, confused intuitions of what "must be the case", or on what an inaccurate English term "really means" trumps what all mathematicians in the world believe.

Ian is a typical crank: he REALLY THINKS his vague intuitions are some sort of "deep insight", and that those who disagree are all stupid (or, presumably, in a conspiracy to hide "the truth"). So far, he hasn't accused the world's mathematicians of "hiding the truth about probability to protect their dogmans and their jobs", but that's no doubt coming.

[Interesting Ian]

Originally posted by Skeptic

Yes, indeed, but I doubt very much Interesting Ian was deliberately trying to get our attention to the fact that non-zero infinitesimals which do not obey the Archemidian axiom can be defined in non-standard analysis, or that the property of addition and multiplication on such sets can be defined in a way that is closed and makes them a commutative ring...

If you use the English language in this non-standard way to mean elements of THOSE sets, then of course you can have such creatures as "inifnitesimally small but not zero" numbers, the number "infinity" which is a number greater than all other numbers (look at the set with denumerable cardinality aleph-null and ordinality omega+1, for instance, call the last number "infinity", and extend addition in the obvious manner), probability functions that assigns non-zero infinitesimals to elements of the possible events Omega, and so on and so forth.

But "Ineresting Ian" was obviously speaking about the usual set of real numbers when he made his statements about probability. Dealing with the real numbers, infinity is not a number, and the only number that is infinitesimally close to 0 is 0 itself.



Well well well, a direct contradiction to what you said in the other thread!! :eek:

naughty naughty {shakes head sadly}



He was just wrong here, since his intuitions about infinitesimals and infinity are muddled, as is his idea about what probability functions are (leading to his continued confusion of "probability 0" with "logical impossiblity").



Regarding the logical impossibility business you're a liar. Moreover, if I'm so confused, how come I'm right and you're wrong? Just look at my arguments below, and understand, and you should realise why you're wrong.



This in itself is hardly an indictment of Ian--the vast majority of humanity are confused about these issues if they think about them at all. But what IS annoying about him is that he keeps insisting that his vague, confused intuitions of what "must be the case", or on what an inaccurate English term "really means" trumps what all mathematicians in the world believe.



I doubt good mathematicians would take your side. But let's have some references.



Ian is a typical crank: he REALLY THINKS his vague intuitions are some sort of "deep insight", and that those who disagree are all stupid (or, presumably, in a conspiracy to hide "the truth").



I really do think I am correct. I'm not just winding you up. Besides, would you also believe that 69Dodge, Vorticity and Cecil are also all winding you up?

However, I don't think it's stupid not to see it straight away. But people are taking their time! :eek:


So far, he hasn't accused the world's mathematicians of "hiding the truth about probability to protect their dogmans and their jobs", but that's no doubt coming. [/B]

Nope skeptic. I'm afraid I'm definitely right about this. Address my arguments below. Forget the unlimitedly close to zero and address this:

What we're considering here is an infinite number of finite strings

So for example

1) 6

2) 64

3) 649

4) 6492

etc etc

Now this is analogically akin to the idea we have an infinite number of positive integers.

Yet every single one of these infinite number of integers is finite!

So just as anyone of these positive integers can be reached by an unlimited search (we do not need an infinite search!).

So can any one of these infinite number of strings be reached by an unlimited search (we do not need an infinite search!)

This then necessarily means my original contention was correct and that an appropriate string with the requisite sub-string can be found in an unlimited search (not infinite!)

And the found string is necessarily finite.

You see? :)

And regarding your accusation that I equate probability = 0 with logical impossibility.

You're a bare faced liar

I said impossible, not logically impossible. 2 utterly completely different things. By impossible I mean it couldn't possibly happen.

To say what I did earlier about this logically impossible stuff:

Such a string, or series of coin flips cannot transpire. It has p = 0

It's like we cannot name a number between 0 and 1 with infinite number of decimal points after it, and then for it to actually happen.

People are getting confused because some number has to turn up, so they use this "backwards" reasoning to suppose that before number was generated there was a possibility (by which I mean p > 0) that it couild have happened. It couldn't.

[scribble]

Vorticity and others, I think I've done my best job so far at explaining why Ian is wrong. Again I apologise for the difficult time I have translating the math into english. I invite you to view the Other Thread.

[Paul C. Anagnostopoulos]

Ian said:
We are generating them.
Okay now, just trying to get myself reoriented here. Are we generating one sequence or many?

~~ Paul

[scribble]

Originally posted by Skeptic

But "Ineresting Ian" was obviously speaking about the usual set of real numbers when he made his statements about probability. Dealing with the real numbers, infinity is not a number, and the only number that is infinitesimally close to 0 is 0 itself. He was just wrong here, since his intuitions about infinitesimals and infinity are muddled, as is his idea about what probability functions are (leading to his continued confusion of "probability 0" with "logical impossiblity").
[/B]

Skeptic, as usual, I wish your post had been written by myself.

[Interesting Ian]

Originally posted by scribble
Vorticity and others, I think I've done my best job so far at explaining why Ian is wrong. Again I apologise for the difficult time I have translating the math into english. I invite you to view the Other Thread.

We have no disagree with each other at all. There is nothing that Vorticity and 69Dodge have said that I disagree with. They have also said they agree with me. As for Cecil, I almost agree with everything he says apart from possibly one very small minor thing.

[Interesting Ian]

Originally posted by scribble
Originally posted by Skeptic

But "Ineresting Ian" was obviously speaking about the usual set of real numbers when he made his statements about probability. Dealing with the real numbers, infinity is not a number, and the only number that is infinitesimally close to 0 is 0 itself. He was just wrong here, since his intuitions about infinitesimals and infinity are muddled, as is his idea about what probability functions are (leading to his continued confusion of "probability 0" with "logical impossiblity").

Scribble
Skeptic, as usual, I wish your post had been written by myself. [/B]

And he directly contradicts what he said in the other thread :rolleyes:

[scribble]

Originally posted by Interesting Ian


We have no disagree with each other at all. There is nothing that Vorticity and 69Dodge have said that I disagree with. They have also said they agree with me. As for Cecil, I almost agree with everything he says apart from possibly one very small minor thing.

Vorticity is also wrong. I'm not famililiar with 69dodge's most recent claims but if they are they same as Vort's, which they were, they are incorrect.

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos

Okay now, just trying to get myself reoriented here. Are we generating one sequence or many?

~~ Paul

An infinite number of finite strings. That makes it the easiest to understand.

[Paul C. Anagnostopoulos]

Ian said:
An infinite number of finite strings. That makes it the easiest to understand.
You cannot generate an infinite number of anything, you can only have them. And I thought you were eschewing the word infinite, claiming that unbounded is different?

If you have the set of all finite-length digit strings, an infinite number of them will contain a 5. No searching is required. But you cannot generate that set.

~~ Paul

[scribble]

You cannot generate an infinite number of anything, you can only have them.


Paul, I'd pay you ten bucks to tell people I said this. That's just a beautiful sentence.

[Interesting Ian]

Originally posted by Skeptic
This in itself is hardly an indictment of Ian--the vast majority of humanity are confused about these issues if they think about them at all. But what IS annoying about him is that he keeps insisting that his vague, confused intuitions of what "must be the case",


But the intuition is obviously the other way. People who do not think about the probem will think like you. I certainly did at first until I did some thinking.

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos
[B]
You cannot generate an infinite number of anything, you can only have them. And I thought you were eschewing the word infinite, claiming that unbounded is different?

The search is unbounded. This of course means that an infinite number of strings will not be generated since you will find the required sub-string with probability = 1 in a finite number of strings.

[scribble]

you will find the required sub-string with probability = 1 in a finite number of strings.


Ian, you're ignorant of math. You'd do the thread a service at this point to shut up and leave.

You will find the required substring with a probability of 1 only in an infinite number of strings. There is no way to translate your gobbeldygook into sense; you've strung together words that have no meaning to you.

The only meaning I can read in them, no matter how I try is, "An infinite length search is of finite length."

You shouldn't have to wonder why I won't debate you. I've told you what's required to change that.

I've made my case in other places. All you get from me are at this point are the above assertions.

[LuxFerum]

Originally posted by Interesting Ian
So as the numbers increase we get closer and closer and closer to p = 0. But your argument is that for any number, it will always be a smidgen above p = 0 :) We need an infinite number.

No!!

Why not?

Well just consider all positive integers. There is an infinite number of them, but they are all finite.
No, this is where you are wrong.
There is an infinite number of positive integers.
The first is 1, then 2,3,...... always adding one, if you add 1 for an infinite numbers of times( you already agree that there is an infinite numbers of positive integers) you will get infinity.

So they are not all finite. Therefore you need an infinite search.







Originally posted by Interesting Ian
(WOW, think I might know what Scribble means by "countably infinite" now! LOL)
If you want to know that see this
http://mathworld.wolfram.com/CountablyInfinite.html
http://mathworld.wolfram.com/UncountablyInfinite.html
An this too:
http://mathworld.wolfram.com/InfiniteSet.html


Originally posted by Interesting Ian
I hate to be arrogant guys knowing nothing about maths. But I'm afraid I'm simply not wrong about this.

Sorry about being arrogant but I have to be truthful

It would be very easy for me to lie and say I'm wrong. And everyone would clap me on the back and say they admire me for admitting defeat.

But I can't lie. I'm afraid I'm not wrong :(

Sorry.
Just please read those links and say if you agree with what they say or not.

[Skeptoid]

Ian,

Suppose the world's greatest living mathematician were to read this thread and was able to somehow translate your assertion into a mathematically meaningful statement and then proceeded to rigorously prove "Ian's Theorem" to everyone's satisfaction. So what? Where are you going with all this?

[69dodge]

69dodge: The rationals are countable, and my understanding is that probability is supposed to be countably additive, i.e., the probability of a countable union is the sum of the individual probabilities.

Cecil: Ranges can have non-zero probability while every specific number has zero probabily. This has to do with the idea that every range contains an infinite number of numbers within it. You can draw an analogy to the number line; points have 0 length yet ranges can have non-zero length.

(S): The problem is, you're working with infinite numbers of things. Normal rules have to be adjusted. The exact flaw is saying, "A particular random rational number can occur once, and there are an infinite number of rational numbers, so therefore the probability of that number occuring is zero," THEN going and using that /zero/ in an infinite sum. An inifinitesimal, all by its lonesome, might as well be zero. But when you take an infinite number of them together, they might not be. They can add up to zero, or they can add up to one, or they can add up to an infinity.There are different sizes of infinity. Countably infinite is different from uncountably infinite.

I was referring specifically to a union of countably many items. A range of the real number line, on the other hand, contains uncountably many points. A union of countably many sets of measure zero has measure zero. A union of uncountably many sets of measure zero can have measure zero, or measure one, or infinite measure, or any other measure in between.

The Math Forum has a nice article (http://mathforum.org/library/drmath/view/63983.html) about some of this stuff.

[LuxFerum]

Originally posted by 69dodge
There are different sizes of infinity. Countably infinite is different from uncountably infinite.
But they are both infinity.

[Loki]

Scribble,

"The only winning move..."

(and to contradict that comment) :

------------------------------------------------

Ian,


From the proof you are so obviously proud of (you repeat it often enough)

So can any one of these infinite number of strings be reached by an unlimited search (we do not need an infinite search!)
So you agree that this search you speak of is "not infinite", athough it's a search within an 'infinite pool' of possible targets. So "unlimited " in this context has, at the very least, the property "not infinite"? This seems clear and unambiguous. So, would anyone care to offer a mathematical formula that will produce a probability of '1', and which involves retrieving a non-infinite number of (finite length) strings from an infinite pool of such strings? Anyone other than Ian, that is....

[scribble]

Originally posted by Loki
[B]Scribble,

"The only winning move..."


Damn skippy. I must be crazy.


So, would anyone care to offer a mathematical formula that will produce a probability of '1', and which involves retrieving a non-infinite number of (finite length) strings from an infinite pool of such strings? Anyone other than Ian, that is....

I may be crazy, but I'm not insane.

[epepke]

Originally posted by Skeptic
Yes, indeed, but I doubt very much Interesting Ian was deliberately trying to get our attention to the fact that non-zero infinitesimals which do not obey the Archemidian axiom can be defined in non-standard analysis, or that the property of addition and multiplication on such sets can be defined in a way that is closed and makes them a commutative ring...

Probably, but somebody might be interested.

But "Ineresting Ian" was obviously speaking about the usual set of real numbers when he made his statements about probability. Dealing with the real numbers, infinity is not a number, and the only number that is infinitesimally close to 0 is 0 itself. He was just wrong here, since his intuitions about infinitesimals and infinity are muddled, as is his idea about what probability functions are (leading to his continued confusion of "probability 0" with "logical impossiblity").

Clearly. I was just pointing out a different angle, on the off chance that someone else might be interested in what mathematicians do when they get problems like this.

But what IS annoying about him is that he keeps insisting that his vague, confused intuitions of what "must be the case", or on what an inaccurate English term "really means" trumps what all mathematicians in the world believe.

Yes, but that's part of his charm and why I like him so much.

I once read a science fiction story. I think it may have been by Frederik Pohl. In it, someone looks for the secret to immortality. He succeeds completely. However, one of the effects of this is that it completely prevents the formation of long-term memories. The idea being that forming long-term memories is a sort of aging of the brain, and this was stopped, too. This person becomes a massive boon to psychologists, sociologists, and anthropologists, who can use him for experiments. They can try out all sorts of experiments on him and modify them slightly the next day, certain that there is no contamination on the part of the subject. (Of course, to make the story work, the guy had enough short-term memory to be able to participate in a conversation lasting a few hours, which stretches the premise somewhat.)

Ian is a lot like that. He's like what you would get if you took a genius-level individual, dipped him in liquid nitrogen at age 8 or 9, and then hooked up the brain to some fantastic device gotten through the Stargate or something that could read all its contents and express them using language abilities similar to an adult.

Because I haven't seen Ian say anything that I wasn't absolutely convinced of at some point during my pre-pubescent existence. From reading books by and about people like Feynman, Einstein, and even Nietzsche, as well as what I've seen people say on geek fora, I conclude that this is pretty common amongst people who are reasonably intelligent.

Example. For anyone who was interested in electronics as a kid. Admit it. At one point in your life you wondered what would happen if you took a motor and a generator and hooked up both the shafts and the wires. Maybe you were 6 at the time, but you thought about it.

With Ian, it's like that terribly exciting moment in time, but frozen forever, constantly repeated to us as a public service, lest we forget.

[scribble]

Originally posted by epepke
Because I haven't seen Ian say anything that I wasn't absolutely convinced of at some point during my pre-pubescent existence.


I've had this same feeling myself. You put it much better than I could have.

[Paul C. Anagnostopoulos]

Scribble said:
Paul, I'd pay you ten bucks to tell people I said this. That's just a beautiful sentence.
Let me know when the check's in the mail.

Ian said:
The search is unbounded. This of course means that an infinite number of strings will not be generated since you will find the required sub-string with probability = 1 in a finite number of strings.
So far you have said (i) we are generating the strings; (ii) we are generating an infinite number of finite strings; (iii) the search is unbounded; (iv) an infinite number of strings will not be generated.

Do you care you restate the problem so that it doesn't contain blatant contradictions?

~~ Paul

[Interesting Ian]

Originally posted by epepke


Ian is a lot like that. He's like what you would get if you took a genius-level individual, dipped him in liquid nitrogen at age 8 or 9, and then hooked up the brain to some fantastic device gotten through the Stargate or something that could read all its contents and express them using language abilities similar to an adult.

Because I haven't seen Ian say anything that I wasn't absolutely convinced of at some point during my pre-pubescent existence.



You need to understand my arguments epep. Not simply have an emotional empathy with their conclusions. :)

Give me any evidence or indication that you do. Hey! It needn't be philosophy. Let's choose your own geek subject. Enlighten me as to the errors in my reasoning in this maths conundrum..

But first tell me whether you are denying even an infinite search will generate the requisite sub-string.

[Wrath of the Swarm]

Originally posted by Paul C. Anagnostopoulos
If you have the set of all finite-length digit strings, an infinite number of them will contain a 5. No searching is required. And an infinite number of them will not contain a '5'. So?

The interesting questions arise when we attempt to compare the set of strings that contain the digit '5' and the set of those that don't. I suspect they're the same size, but I don't know enough transinfinite math to have an intelligent opinion on the matter.

[Interesting Ian]

Originally posted by Wrath of the Swarm
And an infinite number of them will not contain a '5'. So?

The interesting questions arise when we attempt to compare the set of strings that contain the digit '5' and the set of those that don't. I suspect they're the same size, but I don't know enough transinfinite math to have an intelligent opinion on the matter.

And this is off topic in anycase i.e it's irrelevant to my question. Oh I know, I realise that people don't understand that ;)

[Kerberos]

Originally posted by Interesting Ian

Well I've never read a maths text book in my life.
You might want to try it someday.
Originally posted by Interesting Ian

But!!
sigth!
Originally posted by Interesting Ian

1/infinity cannot equal a finite value!! :eek:
That's correct; 1/infinite doesn't equal anything at all. Infinite isn't a number, and thus you can't divide by infinite. What you can do is to divide x and see what happens when x approaches infinite (or zero) in this case 1/x --> 0 for x-->infinite. You can't divide by either zero or infinite.

[Paul C. Anagnostopoulos]

Wrath said:
The interesting questions arise when we attempt to compare the set of strings that contain the digit '5' and the set of those that don't. I suspect they're the same size, but I don't know enough transinfinite math to have an intelligent opinion on the matter.
I'm not sure why that comparison is interesting. All I'm saying is that, if you have the set of all finite-length strings, there is no need to wonder whether you have one with a 5 in it. If you want to pick out one such string, that's another matter.

~~ Paul

[Paul C. Anagnostopoulos]

Ian, just in case you missed it up above:

So far you have said (i) we are generating the strings; (ii) we are generating an infinite number of finite strings; (iii) the search is unbounded; (iv) an infinite number of strings will not be generated.

Do you care you restate the problem so that it doesn't contain blatant contradictions?

~~ Paul

[Paul C. Anagnostopoulos]

I said:
I'm not sure why that comparison is interesting. All I'm saying is that, if you have the set of all finite-length strings, there is no need to wonder whether you have one with a 5 in it. If you want to pick out one such string, that's another matter.
Wrath et al, at this point I no longer have any idea what problem we're discussing, which is why I'm trying to get Ian to state it clearly. So this whole set of strings thing might be a red herring.

~~ Paul

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos
Ian, just in case you missed it up above:

So far you have said (i) we are generating the strings; (ii) we are generating an infinite number of finite strings; (iii) the search is unbounded; (iv) an infinite number of strings will not be generated.

Do you care you restate the problem so that it doesn't contain blatant contradictions?

~~ Paul

I'm not sure if I shouldn't just leave this board. Pyrrho has threatened to ban me in the other thread just because I was expressing feelings.

Anyway, we don't ever actually generate an infinite number of strings. The strings generated will be finite, but an unspecified finite number. And yes, of course the search is unbounded. I've made all this explicit time after time.

[Paul C. Anagnostopoulos]

Okay, so you are generating finite-length strings, one at a time, in an effort to find one with a 5 in it.

Do you agree that this is the same thing as simply generating random digits until a 5 is generated?

~~ Paul

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos
Okay, so you are generating finite-length strings, one at a time, in an effort to find one with a 5 in it.

Do you agree that this is the same thing as simply generating random digits until a 5 is generated?

~~ Paul

Replied in other thread. I wonder if we should ask moderators to merge the 2 threads? Hmmmm

[NoZed Avenger]

Is it me, or is this a multi-page, mega-wordy example of the gambler's fallacy at work?

N/A

[Paul C. Anagnostopoulos]

I don't think so, Nozed. I think everyone agrees that the probability of each and every coin flip is .5.

I think the issue boils down to Ian's personal concept of unbounded.

~~ Paul

[NoZed Avenger]

Originally posted by Paul C. Anagnostopoulos
I don't think so, Nozed. I think everyone agrees that the probability of each and every coin flip is .5.

I think the issue boils down to Ian's personal concept of unbounded.


I'm not sure how to separate that out:

Everyone says that they agree that the chance of getting a heads is .5 for every flip, but the position taken after that implicitly is that a heads "just has to come up sometime."

If unbounded is less than infinite, then that is the position being taken by Ian: A heads just has to emerge some time before infinity . . . or a "5" just has to appear at some point prior to infinity.

[Wrath of the Swarm]

It most certainly does not. The chance of a '5' not coming up draws closer and closer to zero the more the coin is flipped -- but it never becomes zero. It's always possible to have a straight run of fives all the way through.

Can't we just conclude (yet again) that the Interesting Idiot has once again suckered us into paying attention to him? C'mon, let's move on to a more interesting topic and thread.

[wollery]

Originally posted by Interesting Ian
And regarding your accusation that I equate probability = 0 with logical impossibility.

You're a bare faced liar

I said impossible, not logically impossible. 2 utterly completely different things. By impossible I mean it couldn't possibly happen.

To say what I did earlier about this logically impossible stuff:

Such a string, or series of coin flips cannot transpire. It has p = 0

It's like we cannot name a number between 0 and 1 with infinite number of decimal points after it, and then for it to actually happen.

People are getting confused because some number has to turn up, so they use this "backwards" reasoning to suppose that before number was generated there was a possibility (by which I mean p > 0) that it couild have happened. It couldn't. Okay, I think that the above statement has finally resolved the whole reason for there being a misunderstanding here. The really sad thing about this is that, if I'm correct about the misunderstanding then we have all been right all the time, just talking at cross purposes.

Please read all of this post and make note of my careful use of language.

Ian, from the above I think that what you are saying is that if we were to try this experiment in real time in the real world, ie create a random string of digits and just keep going until we get a 5, then we would get a 5 at some point.

Some people are probably going to think me crazy for saying this, but....

I agree with Ian that if we were to actually conduct this experiment in the real world this would be the case. The reason I think this, is that the probability of it not being the case is so small that in any real experiment it almost certainly will happen.

Ian, please note that I am not saying that this means that it has probability 1, or that it is absolutely certain. What I am saying is that it is logically possible for the experiment to be conducted without ever finding a 5, but so improbable that in any such real experiment it would almost certainly never occur.

Your problem through all of this is that you have been confusing mathematical probability with working out whether or not something will actually happen. Mathematical probability, as I (and others) have stated, relies on taking into account all logical possibilities. But you have been talking about what would most likely happen in a real experiment. These are not the same thing!

If this is the case then I apologise if I was rude to you, we have simply misunderstood one another. Your lack of mathematical education meant that we were not able to properly explain our ideas to each other in a common language!

To summarise ('cause I think this post needs it) you were realistically correct, but mathematically wrong.

[Interesting Ian]

Originally posted by wollery
Okay, I think that the above statement has finally resolved the whole reason for there being a misunderstanding here. The really sad thing about this is that, if I'm correct about the misunderstanding then we have all been right all the time, just talking at cross purposes.

Please read all of this post and make note of my careful use of language.

Ian, from the above I think that what you are saying is that if we were to try this experiment in real time in the real world, ie create a random string of digits and just keep going until we get a 5, then we would get a 5 at some point.

Some people are probably going to think me crazy for saying this, but....

I agree with Ian that if we were to actually conduct this experiment in the real world this would be the case. The reason I think this, is that the probability of it not being the case is so small that in any real experiment it almost certainly will happen.

Ian, please note that I am not saying that this means that it has probability 1, or that it is absolutely certain. What I am saying is that it is logically possible for the experiment to be conducted without ever finding a 5, but so improbable that in any such real experiment it would almost certainly never occur.

Your problem through all of this is that you have been confusing mathematical probability with working out whether or not something will actually happen. Mathematical probability, as I (and others) have stated, relies on taking into account all logical possibilities. But you have been talking about what would most likely happen in a real experiment. These are not the same thing!

If this is the case then I apologise if I was rude to you, we have simply misunderstood one another. Your lack of mathematical education meant that we were not able to properly explain our ideas to each other in a common language!

To summarise ('cause I think this post needs it) you were realistically correct, but mathematically wrong.

OK, if I'm wrong tell me what the mathematical probability of not ever getting the digit "5" is in an infinite string of random digits.

[Interesting Ian]

Originally posted by Wrath of the Swarm
It most certainly does not. The chance of a '5' not coming up draws closer and closer to zero the more the coin is flipped -- but it never becomes zero. It's always possible to have a straight run of fives all the way through.



In other words p > 0 that one can have an infinite random string all consisting of 5's?




Can't we just conclude (yet again) that the Interesting Idiot has once again suckered us into paying attention to him? C'mon, let's move on to a more interesting topic and thread.

I believe there's a new rudeness rule, although possibly it might only apply to me.

[wollery]

Originally posted by Interesting Ian
OK, if I'm wrong tell me what the mathematical probability of not ever getting the digit "5" is in an infinite string of random digits. If we say that the number of possible infinite strings with "no 5" is A and the total number of possible infinite strings is B, then the mathematical probability is calculated as P=A/B. (Note, Ian, that in order to calculate the mathematical probability we have to include all possibilities)

However, since there are an infinite number of possible infinite numbers with no five, and an infinite (but larger) number of total possible infinite numbers, the mathematical probability is P=infinity/infinity which is incalculable, but > 0 and < 1. That's the problem with dealing with infinities, as you have been told several times.

[wollery]

Originally posted by Interesting Ian
OK, if I'm wrong tell me what the mathematical probability of not ever getting the digit "5" is in an infinite string of random digits. Am I take it from this quote that you are maintaining that you are mathematically correct and that you do understand how to calculate probabilities? In that case answer the question I have posed to you in both of these threads. It's simple enough, and I don't care which thread you do it in.

I'm willing to take the time and effort to show you where you are going wrong, the real question is, are you willing to learn something you didn't know and face the fact that you may be wrong, or are you so arrogant that you believe that you know more than anyone else?

[Interesting Ian]

Originally posted by wollery
Am I take it from this quote that you are maintaining that you are mathematically correct and that you do understand how to calculate probabilities? In that case answer the question I have posed to you in both of these threads. It's simple enough, and I don't care which thread you do it in.

I'm willing to take the time and effort to show you where you are going wrong, the real question is, are you willing to learn something you didn't know and face the fact that you may be wrong, or are you so arrogant that you believe that you know more than anyone else?

I'm sorry Wollery. I can't. I've been threatened with suspension if there is anymore rude or insulting language from me. It's talking about intellectual topics and people denigrating my intelligence which is causing me to lose my temper. Therefore I won't be posting to this forum, or R & P, or Paranormal forums for the immediate future.

Yes I still believe I am right. Thanks to Vorticity and 69dodge I've even gained further insight into why I am right. I've learnt a hell of a lot of maths over the past 5 days.

So even should it transpire I am wrong, in a sense it's been worthwhile.

[T'ai Chi]

Originally posted by jj

There is no 1 in mod 1. Oops, you did it again.

I guess you don't understand what "=" means.

I said: "1 = 0 in mod 1".

Because if you divide 1 by 1 there is 0 remainder. Therefore, 1 = 0 in mod 1.

Duh.

[wollery]

Originally posted by Interesting Ian
I'm sorry Wollery. I can't. I've been threatened with suspension if there is anymore rude or insulting language from me. It's talking about intellectual topics and people denigrating my intelligence which is causing me to lose my temper. Therefore I won't be posting to this forum, or R & P, or Paranormal forums for the immediate future.Sorry to hear that, I really was hoping that we could resolve this.

Yes I still believe I am right. Thanks to Vorticity and 69dodge I've even gained further insight into why I am right. I've learnt a hell of a lot of maths over the past 5 days.As posted above I also believe that you are right, in a sense, but not in the sense that you believe you are right (if that makes sense:)). Glad to hear that you have learnt something.

So even should it transpire I am wrong, in a sense it's been worthwhile. That's absolutely the right attitude to take.

[jj]

Originally posted by T'ai Chi


I guess you don't understand what "=" means.

I said: "1 = 0 in mod 1".

Because if you divide 1 by 1 there is 0 remainder. Therefore, 1 = 0 in mod 1.

Duh.

Except, there is no "1" symbol base one and you know that. Your statement is suborned. How typical for WhoChi Hammegkson.

[sorgoth]

Originally posted by gnome

My own contradiction follows:

Premise 1: At n=1, there exists a random sequence that doesn't contain contains a "5" until at least the second position.

Premise 2: For every n greater than 1, there exists a sequence that doesn't contain a "5" until position n+1 or further.

By mathematical induction, there is no n at which the number 5 must have appeared in that position or sooner.

So since no position "n" offered need ever bear fruit, so to speak, it can be said that it is possible for the sequence not to contain "5".

Thank you! I finally understand this kind of thing!

[T'ai Chi]

Originally posted by jj

Except, there is no "1" symbol base one and you know that.


Where TF did I say that?? Please, post my exact quote, if you can.


Your statement is suborned. How typical for WhoChi Hammegkson.

1 does equal 0 mod 1 though, as I said the very first time.

[epepke]

Originally posted by Interesting Ian
Give me any evidence or indication that you do. Hey! It needn't be philosophy. Let's choose your own geek subject. Enlighten me as to the errors in my reasoning in this maths conundrum..

I already have, in simple language that (no personal offense intended, Cleopatra) even a lawyer can understand. You're applying an operation over a set of numbers which is closed over the real numbers but isn't closed over the numbers to which you are applying it. Then you're asking which real number is the result. It isn't any of them, because the operation isn't closed over the real numbers.

But first tell me whether you are denying even an infinite search will generate the requisite sub-string.

The problem as you have posed it, requiring a true or false answer is undecidable Ian. You have to get into set theory first to know what that means.

I'll give you a simpler example, which you might think about, if you have the inclination. _Take the set of English words, including hyphenated words that any native English speaker can understand.

Define a property over these words, a unary property, of self-description. A word or hyphenated word is self-descriptive if it describes itself.

"Short," for example, is a self-descriptive word, assuming that five letters fits our criteria of short. "Polysyllabic" is self-descriptive. "Monosyllabic" and "long" are not self-descriptive.

By the logic you like so much, a word is either self-descriptive or it isn't. That is, it is closed over the set {true, false}. And law of the excluded middle and all that. So all such words should be either in the set of self-descriptive words or its complement.

However, what about the word "self-descriptive"? It's undecidable. If the word is self-descriptive, then it's self-descriptive. However, if it isn't self-descriptive, then the idea that it is self-descriptive is wrong, so that confirms it isn't self-descriptive.

How about "non-self-descriptive"? Anyone can figure out what that is supposed to mean. But if it's self-descriptive, then it isn't, and if it isn't, then it is. So it can't be either self-descriptive or not.

[scribble]

Epepke, your last post was as brilliant as any other I've seen you make, although I wonder whether Ian will further confuse the issue.

Good analogy, at any rate - and is anyone but me listening?

[jj]

Originally posted by scribble
Good analogy, at any rate - and is anyone but me listening?

Well, yeah, I am, but watching Epepke vs. Interesting Ian is sorta like watching a Titan and a midget wrestle.

[scribble]

I just came across this review and link to five free calculus textbooks online.

I know at least one person who has participated in this thread who could sure benefit from reading one of them...

http://books.slashdot.org/article.pl?sid=04/03/04/028253&mode=nested

[epepke]

Originally posted by scribble
Epepke, your last post was as brilliant as any other I've seen you make, although I wonder whether Ian will further confuse the issue.

Thank you. I'm a big fan of Ian and his constancy, and I fully expect him to find some way of completely missing the point, although it may take him some time to find it.

[epepke]

Originally posted by jj
Well, yeah, I am, but watching Epepke vs. Interesting Ian is sorta like watching a Titan and a midget wrestle.

Thank you, but let's not insult dwarfs. Many of the little people are quite good at wrestling.

[Interesting Ian]

Originally posted by epepke


Thank you. I'm a big fan of Ian and his constancy, and I fully expect him to find some way of completely missing the point, although it may take him some time to find it.

epepke, I am definitely right. I have given my argument. Take a look at the other thread where I make my point more explicit and other people make the same point as me but proving it mathematically.

I have not replied to you since there is nothing more to say if you think that p > 0 for an infinite string.

If you point out that it is logically possible that required sub-string will obtain in an infinite string, I simply retort so what? It's logically possible that from tomorrow that you will lose your head, but your cognitive faculties will remain intact. So what? That's not interesting. We're talking about the real world which obeys mathematics.

I didn't even bother reading the stuff about hyphenated words or whatever you're dribbling on about. It's a complete non-sequitur and has nothing whatsoever to do with my original question.

I won't bother reading any responses by you. Read my posts. Both in this thread but especially in the other (http://www.randi.org/vbulletin/showthread.php?s=&threadid=36384)

[The Central Scrutinizer]

Originally posted by Interesting Ian


epepke, I am definitely right. I have given my argument. Take a look at the other thread where I make my point more explicit and other people make the same point as me but proving it mathematically.

I have not replied to you since there is nothing more to say if you think that p > 0 for an infinite string.

If you point out that it is logically possible that required sub-string will obtain in an infinite string, I simply retort so what? It's logically possible that from tomorrow that you will lose your head, but your cognitive faculties will remain intact. So what? That's not interesting. We're talking about the real world which obeys mathematics.

I didn't even bother reading the stuff about hyphenated words or whatever you're dribbling on about. It's a complete non-sequitur and has nothing whatsoever to do with my original question.

I won't bother reading any responses by you. Read my posts. Both in this thread but especially in the other (http://www.randi.org/vbulletin/showthread.php?s=&threadid=36384)

This sounds like you are admitting you are wrong. Epepke kicked your ass.

[slimshady2357]

Originally posted by Interesting Ian
I didn't even bother reading the stuff about hyphenated words or whatever you're dribbling on about. It's a complete non-sequitur and has nothing whatsoever to do with my original question.


You cannot possibly know if something is relevant or not if you have not even read it.

Adam

[Interesting Ian]

Originally posted by The Central Scrutinizer


This sounds like you are admitting you are wrong. Epepke kicked your ass.

Hmmm . . your reading comprehension ain't too good is it.

[Interesting Ian]

Originally posted by slimshady2357


You cannot possibly know if something is relevant or not if you have not even read it.

Adam

Don't be absurd. If I started describing in great detail about what I've had to eat today, would you have to read and absorb everything I said to know I'm not talking about the issue in question? A very quick skim read is sufficient to determine I'm not.

[epepke]

Originally posted by Interesting Ian
epepke, I am definitely right. I have given my argument. Take a look at the other thread where I make my point more explicit and other people make the same point as me but proving it mathematically.

Thank you for responding.

I have not replied to you since there is nothing more to say if you think that p > 0 for an infinite string.

See, folks? I knew Ian would come up with a way to miss the point. You never disappoint me, Ian.

I didn't even bother reading the stuff about hyphenated words or whatever you're dribbling on about. It's a complete non-sequitur and has nothing whatsoever to do with my original question.

Nice call. From the Boomer Bible, Willie, chapter 38:

1. Do not be tempted to increase your store of knowledge,
2. Because new knowledge can contradict old knowledge,
3. And contradictions lead to thought.

[slimshady2357]

Originally posted by Interesting Ian


Don't be absurd. If I started describing in great detail about what I've had to eat today, would you have to read and absorb everything I said to know I'm not talking about the issue in question? A very quick skim read is sufficient to determine I'm not.

I'm not being absurd, you are.

You stated quite clearly that you did not read the material, you did not say you skim-read it.

And besides, your comparison is not even close. His post is quite relevant to the topic of discussion.

I have no intention of demonstrating this, as you have shown no intention of trying to understand how it could be so. When you don't even read someone's post and yet claim it is not relevant, I hope you can see why some people would choose not to engage you in discussion anymore.

Adam

[Interesting Ian]

Originally posted by epepke



See, folks? I knew Ian would come up with a way to miss the point. You never disappoint me, Ian.


In essence I am maintaining that P = 0 for an infinite string not to contain appropriate substring. Or for example P = 0 that an infinite number of coin flips will always result in tails.

Nececessarily you must maintain that either P > 0 or that at least it might be.

Necessarily therefore, it you think this this is irrelevant, then you have failed to understand my original question.

[Interesting Ian]

Originally posted by slimshady2357
[B]

I'm not being absurd, you are.



No, you are.



And besides, your comparison is not even close. His post is quite relevant to the topic of discussion.



I'm afraid it isn't.



I have no intention of demonstrating this, as you have shown no intention of trying to understand how it could be so. When you don't even read someone's post and yet claim it is not relevant, I hope you can see why some people would choose not to engage you in discussion anymore.



Fine, put me on ignore then.

[Interesting Ian]

Originally posted by epepke

Nice call. From the Boomer Bible, Willie, chapter 38:

1. Do not be tempted to increase your store of knowledge,
2. Because new knowledge can contradict old knowledge,
3. And contradictions lead to thought. [/B]

That's interesting. I don't think, and yet you've never beaten me in an argument, even in your own pet geek subject of maths :rolleyes:

[epepke]

Originally posted by Interesting Ian
That's interesting. I don't think, and yet you've never beaten me in an argument, even in your own pet geek subject of maths :rolleyes:

I appear to have done so except in your eyes, which are connected to a brain that knows next to nothing about mathematics.

This in itself wouldn't be so bad if you displayed some willingness to learn about mathematics, or didn't make claims that you couldn't support. There is no dishonor in ignorance.

[alfaniner]

Originally posted by epepke


... There is no dishonor in ignorance.

No, only in willful, stubborn ignorance, such as Ian constantly shows...

[wollery]

Originally posted by alfaniner


No, only in willful, stubborn ignorance, such as Ian constantly shows... I'm particularly impressed by the way he dismisses any argument that may damage his position as "irrelevant to the subject".:p

[epepke]

Originally posted by alfaniner
No, only in willful, stubborn ignorance, such as Ian constantly shows...

There's a simpler name for that: stupidity.

I like to give people the benefit of the doubt. Either they learn something or you get to watch them work far harder not to learn something than the effort it would take to learn it.

[gnome]

Originally posted by Interesting Ian


In essence I am maintaining that P = 0 for an infinite string not to contain appropriate substring. Or for example P = 0 that an infinite number of coin flips will always result in tails.

Nececessarily you must maintain that either P > 0 or that at least it might be.

Necessarily therefore, it you think this this is irrelevant, then you have failed to understand my original question.

If you insist on looking at it that way, I contend the probability is positive but not quantifiable.

[Quixote]

geek subject of maths

I think it's high time someone addressed the real issue here; why do the British insist on adding an "s" to "math"?

[scribble]

Boing!

Ian thinks if he REALLY BELIEVES he is right, then everything else is irrelevant.

Good luck getting through to that.

[scribble]

Originally posted by epepke
This in itself wouldn't be so bad if you displayed some willingness to learn about mathematics, or didn't make claims that you couldn't support.


Exactly!


There is no dishonor in ignorance.

Contradiction: What Ian displays is dishonor in ignorance. It's possible to show "honorable" ignorance but he hasn't got that down yet.

So it's more true to say: "There doesn't need to be dishonor in ignorance." 'Cause as Ian shows, sometimes there is.

[epepke]

Originally posted by gnome


If you insist on looking at it that way, I contend the probability is positive but not quantifiable.

I'm kind of glad that Ian is no longer reading my posts, because that means I can present the proofs to people who I'm pretty sure don't have squirrels running around inside their heads. They're easy. I'll use plain ASCII, as I don't know what browsers everyone has (oo is infinity).

There are three things to prove:

A) p is such that there exists no real number r such that r <= p and r > 0
B) p is not equal to 0
C) There exists no real number that satisfies both 1 and 2

Proofs:

Take p(k) as the probability of not finding a 5 in a string of k digits. This is obviously .9^k I'll just call .9 "b" so I don't have to type so much. Besides, all that really matters is that b is in the range (0..1). (The parantheses mean that 0 and 1 are not included in the set, just everything in between.)

So p(k) = b^k. (equation 1)
Also, p(k)^(1/k) = b (equation 2, follows directly from 1)

Now, it's clear that the limit as k -> oo of p(k) is 0.

Assume that we can really calculate x^k as k really reaches infinity. To avoid having to posit that infinity is a real number, we'll just just an Oracle machine (a valid concept from computational theory) to do an infinite number of multiplications. Asssume this operation is closed over the set of real numbers. Run it on p(k) and call the result p.

A) p is such that there exists no real number r such that r <= p and r > 0

Assume that there is such a number r. Then there must be some finite number j such that r^(1/j) > b. Since r^(1/j) is greater than b, b^j < r. Since k >= j at some point, r cannot be less than or equal to p.

B) p is not equal to 0

Assume p = 0.

From equation 2, p(k)^(1/k) = b. Since we have assumed that p is a real number, and the operations are known to be closed over the real numbers, then we can take p^(1/k) and compare it with b. We might think we need another Oracle machine to do root extraction for when k gets to infinity. However, if p is 0, then any root of p also has to be 0. But we know that b is not 0. So p can't be 0.

So there's something wrong. You can take the view of the realists and say that the Oracle machine is a bad idea, and p never gets to 0 but only approaches it.

Or you can take the view of the hyperrealists and say that the Oracle machine is fine, and there's a value, but it isn't a real number.

It doesn't matter which route you take. It's [b]undecidable[b].

I hope this doesn't have any major errors; if it does, I'll put an addendum later. But now I'm getting bleary-eyed, and the roast is about done.

[Interesting Ian]

Originally posted by epepke
[B]

I appear to have done so except in your eyes, which are connected to a brain that knows next to nothing about mathematics.



But I really don't need to know much about maths. I can recognise a logical absurdity when I see it. I just find your insults amusing. You've developed this irrational dislike of me. Well, fair enough. I've done my best to make peace. But you won't have any of it. So be it. I know I'm right about this question. You don't want to accept it; ok so be it.



This in itself wouldn't be so bad if you displayed some willingness to learn about mathematics, or didn't make claims that you couldn't support. There is no dishonor in ignorance.

We all have finite lives ;) There are many books to read, and much to do in life before we die ;)

[gnome]

Originally posted by Interesting Ian
But I really don't need to know much about maths. I can recognise a logical absurdity when I see it. I just find your insults amusing. You've developed this irrational dislike of me. Well, fair enough. I've done my best to make peace. But you won't have any of it. So be it. I know I'm right about this question. You don't want to accept it; ok so be it.

You don't see the arrogance in claiming you don't need to know math to contradict mathematicians on a mathematical topic?

[Paul C. Anagnostopoulos]

Quixote said:
I think it's high time someone addressed the real issue here; why do the British insist on adding an "s" to "math"?
Because it is short for "mathematics"?

Ian said:
We all have finite lives There are many books to read, and much to do in life before we die.
But if you don't have time to read a maths book, why do you have time to have this conversation, which involves, coincidently enough, maths?

~~ Paul

[Hellbound]

Thought I'd toss in a layman's two cents on this issue.

Coin toss. Probability of heads is .5, or one out of two.

Adding the probabilities of all possible outcomes one gets 1. i.e., the chance of tails (.5) plus the chance of heads (.5) equals 1.

What this means is that for every toss, some outcome must result. Tails and heads are equally likely.

Now, over two tosses, there are now four possibilities. Each option (HH, HT, TH, TT) has a .25 probability. Or, there are four options of an outcome, and we know that the probability of some outcome =1. Therefore, 1/4 = .25.

Now, say we have 100 tosses. That's equal to 1267650600228229401496703205376, or 2^100 possbile outcomes (forgive me for not listing them). Each individual outcome has a probability of 1/(2^100), or 7.8886090522101180541172856528279e-31. Now, if one takes the probability of a single outcome and multiplies it by the number of outcomes, one gets 1 again.

So, simple formula. Assuming that your generator is absolutely random, then the number of possible outcomes for any given sequence of length n is:

10^n.

The chance of any specific outcome being generated is:

1/(10^n)

I've used 10 instead of 2 here because there are ten digits, as opposed to two options for the coin toss.

Now, for ANY rational number n, 10^n*(1/(10^n)) equals one. If ANY of these probabilities were zero, then they would NOT add up to one. ONLY if n was an infinity could this not be true. So, for any non-infinite string of randomly generated digits, there exists a sequence in which any given digit is not seen. I would argue that EVEN IN INFINITE SERIES of random digits there are infinite strings that would not contain the specified digit. The reasoning is simple.

A random string of all 1's would be infinitly improbable, but not impossible. The probability is zero ONLY because there are an INFINITE number of possible combinations when n=infinity. EVERY possible combination, including those that contain a specified number or string of numbers, has a probability of zero. Now, there are admittedly MANY more possibilities that have a 5 than those that do not, but even then there DO exist an inifinite number of infinite strings do not contain a 5.

Basically, there are an infinite number of infinite random strings that either contain or do not contain any given string or digit. It is BECAUSE there are an inifinite number of possible infinite strings that the probability of any specific given infinite string (i.e., infinite 1's) is zero.

However, the chance of an infinite sequence of some sort being the result is still 1. So one of these zero probability infinite strings has to be produced. It is just as likely to be all 1's as a repeating row of 1234567890 or any other combination.

Infinite possible results means each result has a probability of zero. These ONLY are zero when there are an infinite number. Even then, this does NOT mean they won't occur. So, II, in ANY given case, I would still say that, sadly, you are wrong.

Take a math class. You'd be amazed.

[Loki]

Ian,

We all have finite lives/
Well, each life is finite, but there are (perhaps) an infinite set of possible lives?

Let's see - we give birth to an incarnation of Ian, and check to see if he's an idiot. If he is, we kill him and reincarnate another one. We keep doing this until eventually we get an Ian who's polite, intelligent, and modest. What is the probability that you'll always be a prat? I believe it approaches "1", but I'm not sure I can prove that mathematically.

There are many books to read, and much to do in life before we die
Surely you mean "There are many books to read, and much to do in this life before we die again". I guess you can always study math(s) in your next life? Perhaps you studied it in your last life, and this "conviction" you have about being correct on a subject you apparently know nothing about is just a bit of "information leakage" from your previous incarnation as a maths lecturer??

[Interesting Ian]

Originally posted by gnome


You don't see the arrogance in claiming you don't need to know math to contradict mathematicians on a mathematical topic? [/B]

One doesn't need to know anything about maths to recognise a logical absurdity when one sees one.

[teddosan]

One doesn't need to know anything about maths to recognise a logical absurdity when one sees one.

I have been kind of standing on the sidelines watching this argument for some time now. I usually am content with just watching, but the above quote has been repeated a few times now by II, and no one has addressed it.

All I want to say is that many things which seem "logically absurd" are nonetheless correct... You DO have to know math to recognize a true mathematical logical absurdity. Especially in probability, many things that can be proven true go completely against intuition and "logic."

It's like saying "I don't need to be a physicist to know that relativity is imposssible. It doesn't make sense."

[Quixote]

Originally posted by Interesting Ian


One doesn't need to know anything about maths to recognise a logical absurdity when one sees one.

Although a great deal of mathematics has been discussed, I don't see the dispute as being about mathematics per se. Everyone seems to agree that the probability of generating a 5-less (or head-less) sequence is 0. (Or they will once I point out the flaw in epepke's proof.) The only disputes I see are philosophical and semantic.

[Interesting Ian]

Originally posted by teddosan


It's like saying "I don't need to be a physicist to know that relativity is imposssible. It doesn't make sense." [/B]


Ummmm . .I'm afraid not. People who say such are simply wrong. The world can be what it likes, notwithstanding the whining of the materialists.

You clearly fail to understand the concept of a "logical absurdity". Sure, we can have an infinite number of tails by flipping a coin an infinite number of times. But there would be a logical inconsistency in saying that there is half a chance heads will turn up on any flip.

[Interesting Ian]

Originally posted by Quixote
[B]

Although a great deal of mathematics has been discussed, I don't see the dispute as being about mathematics per se. Everyone seems to agree that the probability of generating a 5-less (or head-less) sequence is 0.

No, they are saying I am wrong in this! :eek: This is what the argument is about and nothing else.

[gnome]

Originally posted by Quixote


Although a great deal of mathematics has been discussed, I don't see the dispute as being about mathematics per se. Everyone seems to agree that the probability of generating a 5-less (or head-less) sequence is 0. (Or they will once I point out the flaw in epepke's proof.) The only disputes I see are philosophical and semantic.

I certainly don't agree.. what about my proof? I thought of a shorter way of saying it...

If you can't propose a finite number "n" where the "5" (or other small sequence) must appear before the n'th position in the series, if no matter how high "n" is, you still can't guarantee the "5" appeared yet, then that is true for the entire series.

[LuxFerum]

Originally posted by Interesting Ian
No, they are saying I am wrong in this! :eek: This is what the argument is about and nothing else.
What if you get this infinite string:

[... 1 2 4 5 5 5 1 5 6 7 8 9 7 6 7 8 ....]

but you start to search in the "6" and go moving to the right, but there is no more 5 in the right side.
Would you need a infinite search or not?

[Quixote]

Originally posted by epepke


I'm kind of glad that Ian is no longer reading my posts, because that means I can present the proofs to people who I'm pretty sure don't have squirrels running around inside their heads. They're easy. I'll use plain ASCII, as I don't know what browsers everyone has (oo is infinity).

There are three things to prove:

A) p is such that there exists no real number r such that r <= p and r > 0
B) p is not equal to 0
C) There exists no real number that satisfies both 1 and 2

Proofs:

Take p(k) as the probability of not finding a 5 in a string of k digits. This is obviously .9^k I'll just call .9 "b" so I don't have to type so much. Besides, all that really matters is that b is in the range (0..1). (The parantheses mean that 0 and 1 are not included in the set, just everything in between.)

So p(k) = b^k. (equation 1)
Also, p(k)^(1/k) = b (equation 2, follows directly from 1)

Now, it's clear that the limit as k -> oo of p(k) is 0.

Assume that we can really calculate x^k as k really reaches infinity. To avoid having to posit that infinity is a real number, we'll just just an Oracle machine (a valid concept from computational theory) to do an infinite number of multiplications. Asssume this operation is closed over the set of real numbers. Run it on p(k) and call the result p.

A) p is such that there exists no real number r such that r <= p and r > 0

Assume that there is such a number r. Then there must be some finite number j such that r^(1/j) > b. Since r^(1/j) is greater than b, b^j < r. Since k >= j at some point, r cannot be less than or equal to p.

B) p is not equal to 0

Assume p = 0.

From equation 2, p(k)^(1/k) = b. Since we have assumed that p is a real number, and the operations are known to be closed over the real numbers, then we can take p^(1/k) and compare it with b. We might think we need another Oracle machine to do root extraction for when k gets to infinity. However, if p is 0, then any root of p also has to be 0. But we know that b is not 0. So p can't be 0.


So there's something wrong. You can take the view of the realists and say that the Oracle machine is a bad idea, and p never gets to 0 but only approaches it.

Or you can take the view of the hyperrealists and say that the Oracle machine is fine, and there's a value, but it isn't a real number.

It doesn't matter which route you take. It's [b]undecidable[b].

I hope this doesn't have any major errors; if it does, I'll put an addendum later. But now I'm getting bleary-eyed, and the roast is about done.

I do not believe that every root of 0 is 0. That intuitively obvious assertion doesn't necessarily hold "at" infinity. Consider. Above, you had the Oracle machine perform repeated exponentiations on b.
Exponentiation, for any base other than 0 or 1, is a 1-to-1 function, that is, if x is not 0 or 1, x^y1 = x^y2 only if y1 = y2.
So the Oracle acting on exponentiation produces a 1-to-1 function. Let's call it g(x).
1-to-1 functions have inverses, so g has an inverse, h(y).
By the definition of inverse function, h(g(x)) = x.
So, since under the current set of assumptions, g(b) = 0, then h(0) = h(g(b)) = b.

[Quixote]

Originally posted by gnome


I certainly don't agree.. what about my proof? I thought of a shorter way of saying it...

If you can't propose a finite number "n" where the "5" (or other small sequence) must appear before the n'th position in the series, if no matter how high "n" is, you still can't guarantee the "5" appeared yet, then that is true for the entire series.

That same reasoning could be used to prove a false statement, so I must conclude that it is flawed.

Consider the intervals of the form (0,1/n), where n is an integer. Let Insct stand for set intersection, so I don't have to worry about special fonts. (0, 1/2) Insct (0, 1/3) is not empty. (0, 1/2) Insct (0, 1/3) Insct (0, 1/4) is not empty. You can't name an integer n such that (0, 1/2) Insct (0, 1/3) Insct ... Insct (0, 1/n-1) Insct (0, 1/n) is empty. And that's true for the entire sequence.

Can we conclude that the intersection of all the intervals in the sequence is not empty? No, because for any number p, there exists an integer m, such that 1/m < p, so p is not in (0, 1/m) and therefore not in the intersection of all the intervals.

[epepke]

Originally posted by Quixote
I do not believe that every root of 0 is 0. That intuitively obvious assertion doesn't necessarily hold "at" infinity.

Well, now you're putting words in quotes.

I was very careful to use the Oracle machine as the only "weird device." I did, however, shorten the proofs and skip over some details. But you could take [b]any[b] root extraction method you have, so long as it is provably convergent, and plug it into an Oracle machine.

[epepke]

Originally posted by Huntsman
Now, for ANY rational number n, 10^n*(1/(10^n)) equals one. If ANY of these probabilities were zero, then they would NOT add up to one. ONLY if n was an infinity could this not be true. So, for any non-infinite string of randomly generated digits, there exists a sequence in which any given digit is not seen. I would argue that EVEN IN INFINITE SERIES of random digits there are infinite strings that would not contain the specified digit. The reasoning is simple.

This is good reasoning, and it makes perfect sense.

The trouble is that you can also generate proofs according to which the probability is undefined.

Ideally, to prove that there isn't something funny going on, you'd have to prove that there exists no proof that does not contradict the proof you've given. Or else stick to operations that are algebraically complete, that is, closed and guaranteed to generate a solution over the set of numbers and operations you've chosen with a finite number of operations.

However, to prove something funny is going on, all you have to do is come up with two proofs that contradict each other.

[epepke]

Originally posted by Interesting Ian


No, they are saying I am wrong in this! :eek: This is what the argument is about and nothing else.

The assertion that you are right is undecidable.

The assertion that you are wrong is undecidable.

That's the answer.

I can't think of a way to make it simpler than that.

[Quixote]

Originally posted by epepke


Well, now you're putting words in quotes.

I was very careful to use the Oracle machine as the only "weird device." I did, however, shorten the proofs and skip over some details. But you could take [b]any[b] root extraction method you have, so long as it is provably convergent, and plug it into an Oracle machine.

I put "at" in quotes, because, unless you are talking about a set that contains infinity (and you made it clear that you were not), I have no clear idea what "at infinity" means.

I didn't need the Oracle machine. I simply used the inverse of your exponentiation function. That function mapped b to 0, so its inverse, by definition, maps 0 to b.

[woodguard]

I think the easiest way to fix the problem is the universe has no memory.
It cannot remember past events. If you used one set of number or 10e99, the past events have no value only the current one.

The past events are only recorded in you mind not in the universe.

[epepke]

Originally posted by Quixote [quote]I didn't need the Oracle machine. I simply used the inverse of your exponentiation function. That function mapped b to 0, so its inverse, by definition, maps 0 to b.

I don't have an exponentiation function. I just have an Oracle machine that repeatedly multiplies. I used exponentiation notation as a shorthand for repeated multiplication (we are, after all, dealing with an integer number of multiplications here). It was just easier than writing b*b*b*b...

I do not assume an exponentiation operator over infinity. I was quite careful about this. I don't even assert that an exponentiation operator with something raised to an infinite power exists, because infinity is not a real number, and there's no guarantee therefore that exponentiation is closed over it. That's because I'm limiting my assumptions to Ian's. If he had made more assumptions, it would be a lot easier to disprove.

Hence he Oracle machine, which can multiply numbers an infinite number of times. No need for infinite exponentiation at all. Every step is simple multiplication.

So there does not exist an infinite exponentiation operator you can take the inverse of.

On the other hand, thank you for thinking about this stuff. It's really quite fun, and the journey is far more important than the result.

Also, of course, there's another way to go about it, which you have pointed to. Just for fun, let's assume that your argument is completely correct. Let's assume that the Oracle machine is an operator with an inverse. And, just as you have said, the inverse operation when applied to p gives you a b = 0.9. Take all of this as given.

Remember, though, that the proof you are referring to was only to prove that p was not 0.

So, now let's define P as the probability of not finding a 5 or a 3. In this case, corresponding to a b, there would be a B of 0.8.

Now run B through the Oracle machine. Assume we get 0 for P, as well as assuming that we get 0 for p. The Oracle machine hasn't changed, only its input.

So now we have P = 0 and p = 0, and 0 = 0, so P = p. Run p back through your inverse operation, and as you have said by definition you get 0.9 for b. Now run P back through the operation, and you should get 0.8 for B. But P = p, and as you have said, by definition, you're going to get 0.9. Now, it seems prety clear that 0.8 is not the same as 0.9, doesn't it? Which means that it is impossible for p = P. Which means that, even if one is 0, the other can't be 0. So you've proved that at minimum, either p or P can't be 0.

You can do this as many times as you like. All of the p, P, p', P', p1, p2, p3, whatever you like, corresponding to b, B, b', b1, b2, b3 all have to be different if the b's are different. Which means that at most one of them could be 0. Which means all the others aren't.

So even if you decide that p is 0 for b, and the inverse operation on p gives you b, all other values, which work exactly the same but have different b's, cannot be 0.

Here's where we start getting into undecidability. You can pick one as 0 and assert your inverse operation as valid, but in the case of digits, that means that nine others don't work. It's easy to see how to extend the problem to a range with a very large or even infinite number of elements, none of which can fit your axioms except one. Which one? It's undecidable.

Or if you want to limit things to the real numbers, you could say that the result of the inverse Oracle machine is undefined, except that it is in the range [0..1) (Here the square bracket means that it can be 0.) Which is the case.

[Quixote]

You're right that .9 and .8 are both mapped to 0 under the implied function. So it has no inverse defined at 0. The fact that no root function, the inverses of exponentiation functions, exists that will map 0 back to .9 is not surprising.

I don't even assert that an exponentiation operator with something raised to an infinite power exists, because infinity is not a real number, and there's no guarantee therefore that exponentiation is closed over it.

But infinite multiplication is? :confused:

Hence he Oracle machine, which can multiply numbers an infinite number of times. No need for infinite exponentiation at all. Every step is simple multiplication.

All the Oracle concept does is disguise the use of limits. Saying that the Oracle operating on .9 yields the number p is simply another way of saying that the limit of .9*.9*.9*... is p as the number of multiplicands increases without bound (or, as some prefer to say it, as that number approaches infinity). And the limit of .9*.9*.9*... as the number of multiplicands increases without bound is 0. That's why there's no way you're going to find a contradiction by assuming p=0.

[epepke]

Originally posted by Quixote But infinite multiplication is? :confused: [/b]

Absolutely not! In fact, the whole point of my proof was to point out that it isn't closed. The way you do these things is by assuming what you want to disprove is true, following it to its conclusion using ideas that are known to be true, and if you get a contradiction, you know that what you want to disprove is false. But you can't assume two or five or fourteen things, because when you find a contradiction, then you don't know which of the assumptions is wrong, just that at least one of them is wrong.

You threw a monkey wrench into that by positing an inverse Oracle machine, which makes it more difficult to prove, but still possible. This is a rare case. Usually it's pretty hard to do.

All the Oracle concept does is disguise the use of limits.

As I pointed out much earlier in the thread, the use of limits to do valuable work using real numbers and still keep all the operations closed is just peachy. But limits aren't part of the algebra. (Yes, preaching to the choir, assuming they know what an algebra is when you put an article in front of the word.)

When people use limits, they're careful to say that it's a limit. This works fine, lasts a long time. So long as you don't get confused about what you are saying.

But this isn't what Ian was saying. Ian was not saying that the limit of probability was zero; he was saying there exists an actual number really at infinity which is identically zero. Which is pretty much what you expect from a philosopher who doesn't know much about math but who is terribly impressed with himself anyway.

I made an earlier post that seemed to make sense to people who did know something about math. But that's just preaching to the choir. Ian just proudly didn't read it and announced that I had not "beaten" him. Apparently, "beating" people is something that is important to him, though it bores me. Nevertheless, as I was thinking about it, I set out to produce some actual proofs. Even if it doesn't convince anybody, it's a valuable intellectual exercize.

Introducing the Oracle machine was the only way I could think of to translate Ian's vague assertions into something that has actually been used in mathematics and has a solid body of theorems behind it. Because you have to do that with mathematics; you can't just wave your hands and get excited like you can with philosophy. (Well, you can do that too, I guess, but it doesn't do much.) Still, I kept it as minimal as possible: repeated multiplications.

At least I think I showed that if there be such a number as Ian posits, it cannot satisfy the Archimedian axioms and therefore cannot be a real number, which means that if treated as a real number, whether it is equal to 0 or greater than 0 is at best undecidable within the normal algebra we're used to over the real numbers. Although I didn't use the name "Archimedean axioms" because, again, that would be preaching to the choir, and they know all this crap already.

[Quixote]

But this isn't what Ian was saying. Ian was not saying that the limit of probability was zero; he was saying there exists an actual number really at infinity which is identically zero. Which is pretty much what you expect from a philosopher who doesn't know much about math but who is terribly impressed with himself anyway.

It seems to me that Ian has amended (or clarified, depending on one's point of view) his position and that he now is (or always was) referring to mathematical probability, a calculation that, in this case, involves either evaluating .9ⁿ as n increases without bound or the "summing" of the infinite series .1 + .09 + .081 + ... .9^n-1*.1 ... . Either way, we're dealing with limits.

If limits are not brought into the discussion, then you're right. p cannot be 0 because .9 times any positive number yields a positive number, never 0.

[gnome]

Originally posted by Quixote


That same reasoning could be used to prove a false statement, so I must conclude that it is flawed.

Consider the intervals of the form (0,1/n), where n is an integer. Let Insct stand for set intersection, so I don't have to worry about special fonts. (0, 1/2) Insct (0, 1/3) is not empty. (0, 1/2) Insct (0, 1/3) Insct (0, 1/4) is not empty. You can't name an integer n such that (0, 1/2) Insct (0, 1/3) Insct ... Insct (0, 1/n-1) Insct (0, 1/n) is empty. And that's true for the entire sequence.

Can we conclude that the intersection of all the intervals in the sequence is not empty? No, because for any number p, there exists an integer m, such that 1/m < p, so p is not in (0, 1/m) and therefore not in the intersection of all the intervals.

An interesting reversal. I must consider at greater depth when it's not the wee hours of the morning.

[peptoabysmal]

Congratulations, Ian.

You've invented the world's worst random number generator.

[Dancing David]

Originally posted by Interesting Ian


No, they are saying I am wrong in this! :eek: This is what the argument is about and nothing else.

Ian, they are really talking about the question you asked! It is amazing the conclusion is not drawn yet, but an answer will be suggested as most likely. Isn't this great, they are really trying to answer the question. You have got them all worked up.

I don't know dude, A Whole New Branch of Mathematics is Being Born and it will be name for you!

Do you want them to be called Ian sets or Interesting sets?

Sit back and enjoy the show!

[Interesting Ian]

Originally posted by Dancing David


Ian, they are really talking about the question you asked! It is amazing the conclusion is not drawn yet, but an answer will be suggested as most likely. Isn't this great, they are really trying to answer the question. You have got them all worked up.

I don't know dude, A Whole New Branch of Mathematics is Being Born and it will be name for you!

Do you want them to be called Ian sets or Interesting sets?

Sit back and enjoy the show!

Interesting Ian sets

[gnome]

Originally posted by Quixote


That same reasoning could be used to prove a false statement, so I must conclude that it is flawed.

Consider the intervals of the form (0,1/n), where n is an integer. Let Insct stand for set intersection, so I don't have to worry about special fonts. (0, 1/2) Insct (0, 1/3) is not empty. (0, 1/2) Insct (0, 1/3) Insct (0, 1/4) is not empty. You can't name an integer n such that (0, 1/2) Insct (0, 1/3) Insct ... Insct (0, 1/n-1) Insct (0, 1/n) is empty. And that's true for the entire sequence.

Can we conclude that the intersection of all the intervals in the sequence is not empty? No, because for any number p, there exists an integer m, such that 1/m < p, so p is not in (0, 1/m) and therefore not in the intersection of all the intervals.

Ok... my analysis of this:

I think the erroneous conclusion here is of a different nature...

In the original example, if we wish to propose that an infinite, random sequence MUST contain a certain number or sub-sequence sooner or later, then we are proposing that the number must eventually appear in a finite position. However, I show that there is no such finite position.

In your example, a stronger conclusion is required to reach the proposition in question.... my proof goes no further than what would be, in your example, the statement that there is no n for which (0, 1/2) Insct (0, 1/3) Insct ... Insct (0, 1/n-1) Insct (0, 1/n) is empty.

This would be completely valid proof against a proposition that there did, indeed, exist such an n.

[Quixote]

then that is true for the entire series.

I interpreted that clause to mean you were making a statement about the entire (infinite) set of finite sequences, not simply any finite subset of the the entire set. I was cautioning against the following (apparently valid) invalid argument.

1) Each finite subset of the infinite set A has property P.
2) Therefore, the set A has property P.

But, if you were talking about finite sets only, my concern was misplaced. On the other hand, if you were talking about finite sets only, weren't you rehashing old ground?

[gnome]

Originally posted by Quixote


I interpreted that clause to mean you were making a statement about the entire (infinite) set of finite sequences, not simply any finite subset of the the entire set. I was cautioning against the following (apparently valid) invalid argument.

1) Each finite subset of the infinite set A has property P.
2) Therefore, the set A has property P.

But, if you were talking about finite sets only, my concern was misplaced. On the other hand, if you were talking about finite sets only, weren't you rehashing old ground?

Not so... I am talking about an infinite set... but to claim that something "Must" appear in an infinite set, is to claim that it exists a finite number of positions into the sequence... because there is no such thing as a number "existing" at an infinite number of positions into the sequence.

[Quixote]

Not so... I am talking about an infinite set... but to claim that something "Must" appear in an infinite set, is to claim that it exists a finite number of positions into the sequence... because there is no such thing as a number "existing" at an infinite number of positions into the sequence.

But knowing that something must appear in a sequence is not the same as knowing where it will appear in the sequence. For example, start with the set of all infinite sequences of the digits 0 - 9 that contain a single 5. For any number n, we can find infinitely many sequences in that set that do not have a 5 in or before the nth position.

Now select one at random (they will not all be equally likely, but a reasonable probability function exists for the set). We can accurately predict that the sequence has a 5 in it; they all do. But we don't know what position the 5 is in.

[gnome]

Originally posted by Quixote


But knowing that something must appear in a sequence is not the same as knowing where it will appear in the sequence. For example, start with the set of all infinite sequences of the digits 0 - 9 that contain a single 5. For any number n, we can find infinitely many sequences in that set that do not have a 5 in or before the nth position.

Now select one at random (they will not all be equally likely, but a reasonable probability function exists for the set). We can accurately predict that the sequence has a 5 in it; they all do. But we don't know what position the 5 is in.

We don't have to know where it will appear for the argument to be valid... it is sufficient to know that the position in the sequence must be a finite number for the proposition (call it P) to be true... and then to prove by induction that there is no finite number that it must appear before in the sequence... we have then proven ~P

[Quixote]

We don't have to know where it will appear for the argument to be valid... it is sufficient to know that the position in the sequence must be a finite number for the proposition (call it P) to be true...and then to prove by induction that there is no finite number that it must appear before in the sequence ... we have then proven ~P (Emphasis added.)

1) I'm not sure we're talking about the same proposition. You posted your proof in response to my statement that "everyone seems to agree that the probability of generating a 5-less (or head-less) sequence is 0." I assume that your post was intended to refute that statement. However, you seem to be addressing a different statement.

2) Look again at my example above, the one about the set of all infinite sequences which contain a single 5. It is obvious that there is no integer n, such that, given a randomly selected sequence S, from the set, a 5 must appear in S before the nth position. You don't even need induction for that. It is, however, equally obvious that the selected sequence has a 5 in it, because they all do.

[gnome]

Originally posted by Quixote
(Emphasis added.)

1) I'm not sure we're talking about the same proposition. You posted your proof in response to my statement that "everyone seems to agree that the probability of generating a 5-less (or head-less) sequence is 0." I assume that your post was intended to refute that statement. However, you seem to be addressing a different statement.

That is precisely what I am refuting. I am saying it is possible to generate a sequence without a 5 anywhere in it.

2) Look again at my example above, the one about the set of all infinite sequences which contain a single 5. It is obvious that there is no integer n, such that, given a randomly selected sequence S, from the set, a 5 must appear in S before the nth position. You don't even need induction for that. It is, however, equally obvious that the selected sequence has a 5 in it, because they all do.

What is "obvious" is the most common source of errors in mathematical proofs. Generally if you intend to prove that something is true for all "n" you use induction to do it. There have been attempts to state this induction explicitly, but for purposes of informal discussion we do seem agreed that "there is no n"

(THERE IS NO SPOON!!!)

It is not obvious to me that every infinite random sequence must have a "5" in it. Let's take it step by step. Do you agree with my premise that if a "5" must appear, it must appear in a finite position?

[69dodge]

Originally posted by gnome
Generally if you intend to prove that something is true for all "n" you use induction to do it.for all integers n, yes. But all integers are finite. So, that doesn't help you if you want to prove that the something is true for infinity also.Do you agree with my premise that if a "5" must appear, it must appear in a finite position?That premise is stated somewhat ambiguously. I can think of two things it might mean, only one of which I agree with. Specifically,<blockquote>it must appear in a finite position</blockquote>might mean<blockquote>with probability 1, there exists an integer n such that it appears at position n,</blockquote>or it might mean<blockquote>there exists an integer n such that, with probability 1, it appears no later than position n.</blockquote>I agree with the first interpretation but not with the second. You seem to think the two are equivalent.

[Quixote]

That is precisely what I am refuting. I am saying it is possible to generate a sequence without a 5 anywhere in it.

I agree. That doesn't change the fact that the probability, computed with mathematical rigor, that a sequence with a 5 in it can be generated, is 0. My position is that a probabilty of 0, in this instance, does not imply impossibility. That doesn't bother me, because this instance, Ian's thought experiment, requires either infinite time or some mechanism for accomplishing infinitely many discrete operations in finite time. That the theory is not a perfect fit to reality (or rather unreality) is not surprising.

It is not obvious to me that every infinite random sequence must have a "5" in it.

But my example doesn't involve every infinite random sequence. It involves only the infinite random sequences which contain exactly one 5. It is obvious that each of those sequences contains a 5. It is also obvious that for any integer n, there exist infinitely many sequences in that set which do not have a 5 in any position from 1 to n.

[Quixote]

Proving that, for each integer n there exists an object X with property P is not the same as proving that there exists an object X such that for each integer n, X has property P.

For example, for each integer n there is an integer m such that m>n. It is not true that there is an integer m such that, for each integer n, m>n.

[epepke]

Originally posted by Quixote
I agree. That doesn't change the fact that the probability, computed with mathematical rigor, that a sequence with a 5 in it can be generated, is 0.

The limit of a probability is 0, but limit is not an operation defined within the algebra of real numbers.

The algebra of real numbers consists of the real numbers and a finite number of multiplications, divisions, additions, subtractions, and root extractions.

[Quixote]

Originally posted by epepke


The limit of a probability is 0, but limit is not an operation defined within the algebra of real numbers.

The algebra of real numbers consists of the real numbers and a finite number of multiplications, divisions, additions, subtractions, and root extractions.

Kolmogorov's probability axioms state that probabilities are "countably additive", that is that the probability of the union of a countable set of mutually exclusive events is equal to the "sum" of the probabilities of those events. The "sum" of a countable set of numbers exists only if the partial sums converge and is defined as the limit of the partial sums.

The probility, P, that a five will appear in some sequence is the countable sum of the probabilities that the first 5 will be in the 1st position, P(N=1), plus the probabilty that the first 5 will appear in the 2nd position P(N=2), etc.

P(N=1) = 1/10
P(N=2) = 1/10 * 9/10
P(N=3) = 1/10 * 9/10 * 9/10

P(N=n) = 9 ^n-1/10 ⁿ for each n.

The partial sums are 1/10 + 9/10² + ... + 9 ^n-1/10 ⁿ

The sum of the infinite series, i.e., the limit of the sequence of partial sums as n increases without bound, is 1.

I do not contend that P=1 means that it must occur under the idealized circumstances of the thought experiment, but there is no doubt that P = 1.

[Loki]

Quixote,

Perhaps you've expressed something in a way that can actually help decide what Ian meant?? It seems to me that there isn't really much of a mathematical difference going on here, just different attempts to translate Ian's query into the "appropriate" mathematical question.

You just said :

The sum of the infinite series, i.e., the limit of the sequence of partial sums as n increases without bound, is 1.
I agree with this (doesn't everyone?) The bolded parts, to me, can be related fairly clearly back to Ian's original statement, in which he wants 'an unlimited but not infinite search'. It seems to me that your "without limit" is what Ian means by "unlimited", and your "limit of the sequence" is what Ian means by "infinite". So if I put your terms into Ian's original expression I get :

If the given sub-string does not occur in a given finite string, then one simply keeps making this finite string longer and longer until it does! If the string (length) is without limit (albeit not the limit of the sequence), any given sub-string must occur.
In which case, it seems to me that Ian is either wrong, or talking gibberish (depending upon how you want to read "albeit not the limit of the sequence").

Would you agree?

[gnome]

Originally posted by 69dodge
for all integers n, yes. But all integers are finite. So, that doesn't help you if you want to prove that the something is true for infinity also.

Nothing is true for "infinity" itself, because that is not a number... but we are checking if something is true for an infinite set...

That premise is stated somewhat ambiguously. I can think of two things it might mean, only one of which I agree with. Specifically,<blockquote>it must appear in a finite position</blockquote>might mean<blockquote>with probability 1, there exists an integer n such that it appears at position n,</blockquote>or it might mean<blockquote>there exists an integer n such that, with probability 1, it appears no later than position n.</blockquote>I agree with the first interpretation but not with the second. You seem to think the two are equivalent.

I will withdraw any argument that I assert the second premise.... I am satisfied if it is accepted that a search through the infinite sequence for the "5" or other small sequence is not guaranteed to terminate.

[epepke]

Originally posted by Quixote
Kolmogorov's probability axioms state that probabilities are "countably additive", that is that the probability of the union of a countable set of mutually exclusive events is equal to the "sum" of the probabilities of those events. The "sum" of a countable set of numbers exists only if the partial sums converge and is defined as the limit of the partial sums.

The probility, P, that a five will appear in some sequence is the countable sum of the probabilities that the first 5 will be in the 1st position, P(N=1), plus the probabilty that the first 5 will appear in the 2nd position P(N=2), etc.

P(N=1) = 1/10
P(N=2) = 1/10 * 9/10
P(N=3) = 1/10 * 9/10 * 9/10

P(N=n) = 9 ^n-1/10 ⁿ for each n.

The partial sums are 1/10 + 9/10² + ... + 9 ^n-1/10 ⁿ

The sum of the infinite series, i.e., the limit of the sequence of partial sums as n increases without bound, is 1.

I do not contend that P=1 means that it must occur under the idealized circumstances of the thought experiment, but there is no doubt that P = 1.

This is better, but already you have a problem. Kolmogorov's axioms don't tell you how to compute the sum of a countably infinite set. For certain sets you define a sum as a limit, there's that word "limit" again. Since I've already agreed that the limit of the earlier probability was 0, I'm not sure what you're arguing about.

As was pointed out much earlier in the thread, a random number generator that generated a random number in the range (0..1) would have a probability of generating some random number of 1 and a probability of generating any particular random number of 0. Now, Kolmogorov's third axiom, which you are referring to, requires that the events be countable and disjoint. Countable obviously doesn't work for the real numbers. We could use rational numbers in the range (0..1), but it's not so easy to show that they are disjoint. So let's take the set of numbers of the quotient of m/n such that m and n are both prime, m >= 3, and n > m. Those are guaranteed to be disjoint, as they cannot be reduced. Since Euclid proved that the number of prime numbers is infinite, we can count it with aleph-null. Doesn't really matter if we stick to Euclid primes.

Given a random number generator that generates numbers of this set, the probability of its generating any particular number is 0. However, the probability of its generating some number is 1. So by Kolmogorov's third axiom, 0 + 0 + 0 + 0... = 1, which doesn't work.

Yet it is possible to define a bijection of that set and the set of infinite strings of digits for which you could apply Kolmogorov's third axiom.

What to conclude? I conclude that limits are a useful trick according to which one can get useful work done over the real (or in this case rational) numbers, but not being part of the algebra, it quite lacks in telling you what the actual number is. Since it's undecidable within the algebra of real numbers, I can't tell you that it isn't 1 (or 0), but nor can I say that it is 1 (or 0).

[Quixote]

Since I've already agreed that the limit of the earlier probability was 0, I'm not sure what you're arguing about.

I don't know exactly what "the limit of the probability" means. A probability is a real number between 0 and 1. There is a single real number between 0 and 1, inclusive, that is the probability of generating a sequence with a 5 in it. That number happens to be the limit of a sequence of related probabilities, but I see no reason to refer to it as something other than the probability of generating a sequence with a 5 in it.

Now, Kolmogorov's third axiom, which you are referring to, requires that the events be countable and disjoint. Countable obviously doesn't work for the real numbers.

But it does work for intervals of real numbers. The real numbers in [0,1] with a 5 as the first digit after the decimal point comprise the interval [.5, .6). Those with a 5 in the second position comprise the 9 intervals [.05, .06), [.15, .16), [.25, .26), [.35, .36), [.45, .46), [.65, .66), [.75, .76), [.85, .86), and [.95, .96). Similarly, for each integer n, there is a finite set of intervals containing numbers whose nth digit is a 5.

The probability function based on the ratios of the lengths of those intervals to the interval [0,1] are

P(N=1) = 1/10
P(N=2) = 1/10 * 9/10
P(N=3) = 1/10 * 9/10 * 9/10

P(N=n) = 9 n-1/10 n for each n.

which, not merely coincidentally, are the same probabilities associated with the finite sequences generated by randomly selecting a digit from 0 to 9 under you get a 5.

As noted in my post above, the sum (actually the limit of the partial sums) of P(N=1) + P(N=2) + ... is 1.

[Quixote]

I conclude that limits are a useful trick according to which one can get useful work done over the real (or in this case rational) numbers, but not being part of the algebra, it quite lacks in telling you what the actual number is.

A trick? :jaw: