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neon snake
Veteran

Joined: 18 Mar 2004
Posts: 70
Location: Chelmsford, UK

 Quote: None of these have anything to do with the facts that: 1. In a infinite fully (unflawed, unlimited) random stream of numbers, you will always find every single permutaion of finite length of numbers, an infinite number of times.

Seeing as this is how I started this discussion, I'll get behind this one again (The whole summing infinite numbers thing, whilst I remember enough of it to know that they can equal finite sums, is too far out of memory for me to offer convincing proofs). Plus, they are two very different fields of mathematics.

thebruce -

Make the following assumptions for me.

1. The string of digits is infinitely long.
2. Each of the digits 0-9 has a chance (any chance, doesn't have to be equal) of appearing as the next digit.

Because of these assumptions, the number will not be an infinitely long string of 5s.

Even if the chance of a 1 being the next number is only 1%, in an infinitely long string, it will appear at some point. Imagine 'testing' for the next number 40,000 times. You can see that a 1 should appear 400 times (Obviously, that won't be exact, but you can see the point).

You should be able to see that even microscopic chances are made an awful lot more likely when you are performing the trial an infinite amount of times! Basically, when the number of trials makes the jump from 'many, many times' to 'infinite', the chance of success jumps to 100% (actual 100%, not approaching or essentially 100%).

As I said earlier, I think this is a difference in education - I spent a lot of time learning probability.

 Quote: 2. In a infinite, either non-random, or not-fully random stream of numbers, each permutation of finite length will either exist (100%), or not exist (0%). This is commonly reffered to as the 0 1 law.

Indeed.

 Quote: 3. We (in this thread), have come to the conclusion that e is not fully random, and therefore, each finite string has a chance of either 0 or 1 to appear in e.

Good call, Avenger. Having linked the 0 1 law to e, this now seems obvious.

Cheers.[/b]
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'To know, yet to think that one does not know is best. Not to know, yet to think that one knows will lead to difficulty.'
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Posted: Mon Jan 03, 2005 3:35 pm
thebruce
Dances With Wikis

Joined: 16 Aug 2004
Posts: 6865
Location: Kitchener, Ontario

Re: Some Clarifications

 ariock wrote: The Infinite series that Chris posted earlier depends on using angles measured in RADIANS, not degrees. If you put 30 into it, you will get garbage. If you put a number like pi/2 into it, the series will converge to 1 at infinity.

(please note for future reference, 'converge to 1 at infinity')

 Quote: Second. thebruce, you have posted a couple of links to sites that supposedly refute the Identity of 0.99~ = 1. This is not actually the case. Neither site uses a valid mathematical proof to disprove the assertion. They merely use some philosophical hand-waving that, unfortunately for them, is trumped by mathematical proofs.

Mathematical proofs that don't offer precise values. Which many here claim don't matter in math, because proofs can result in multiple possible answers. If that's so, then there's no point in arguing because the statement is looking for precise values, which either converging, or approaching, this value cannot give. The most precise that equation can get is the statement that the value approaches a number. 0.9~ -> 1

 Quote: Seriously, the second link you posted actually uses the line, ""=" in this case does NOT mean the same as." This is, in all honesty, a joke. Now = doesn't mean equal?

The same thing I was saying, and which has been condeded to - A -> B does not mean A = B. He wasn't saying that = has multiple meanings, but that equation should have been written as approaching, or converging, not equals.

 Quote: Another interesting point from your first link: Harold says, "So, 0.3 repeating times 3 should equal 1, correct? But, it doesn't. It gives you 0.9 repeating. But we all know three thirds is a whole, so what happened?" What did happen? Unfortunately, Harold doesn't tell us.

Because he's not trying to state what happened, simply why the statement doesn't work.

 Quote: 1/3=0.33~ 3*0.33~=0.99~ 3*1/3=1 Therefore 0.99~=1

Bear with me here - what we're saying is that that statement is essentially like saying
orange = fruit
orange = colour
therefore fruit = colour
jeer at that as you may, but the point to that is simply that you can't calculate a decimal value based on an infinite decimal. You can only work with what would become an infinite decimal value - that is an equation which results in the infinite value - ie you can't calculate with 0.3~, but you can work with 1/3 instead. You can't multiply 0.3~ * 3, because you would never return a final value. You can multiply 1/3 * 3 because you can reduce the equation to 1. I don't know of any math that writes an equation, officially, with a decimal value including the notation for repeating or infinity. It's always, from my knowledge, replaced by a symbol for transcendental numbers (number which can't be written as decimal or fraction), or converted to a fraction for irrational numbers (which can't be written as decimal), assuming my terminology is correct.

 Quote: Note that the symbol ~ indicates that the digit preceding it is repeated infinitely. This is NOT an approximation. It is not like saying 0.99(with another 50 billion 9's following it) when to say that invites the criticisms you have made. 0.99~ does not say that. It says explicitly and by definition that there MUST be an INFINITE number of trailing 9s. Since there is no such thing as "Infinity plus 1," our proof is logical and valid.

And thus any calculations involving a repeating decimal will be converted to a symbol or a fraction if it's to be further calculated, so as to remove the necessity to bother with ~ in decimal notation.
The correct progression is 1/3 = 0.3~ and 2*1/3 = 2/3 = 0.6~ and 3*1/3 = 3/3 = 1. You can't calculate 0.3~*n. The calculation would never end, it becomes an endless loop. So 0.3~*3=0.6~ is not a calculation that can be made, only a derived assumption.

 Quote: Another quote from Harold: "Thirds exist, but not for the number 10." ...But to do the same thing with 1 cookie between 2 friends is eqally impossible. And yet you have no problem in saying that 2*0.5=1. If you use one as proof of the problem of thirds, then there must also be a problem with halves.

I believe we covered this earlier in the thread... if you check back to some pre 10 pages, we went through the problem with division to create infinite and finite amounts...

 Quote: Now HERE is where the dichotomy occurs. I say that if we assume that this continues infinitely that we will have an infinite string of 3s and that there will therefore be no remainder. Do you believe in an infinitely small remainder? What would that mean? -infinity? or 0? or an infinitesimal?

Derived from an assertion made above, not by me, mathematically, you can take the decimal notation 0.3~ and define it as the sum of 3/(10^n) where n = 1 to infinity. The limit of that equation is 1/3. The more we calculate, the closer and closer we are to 1/3. But no matter how long we calculate, we'll never reach 1/3. By process, the answer to the equation converges at 1/3. In other words, we can never calculate 1/3 because for as long as we calculate it, we'll never end up with the precise value of 1/3. We will always require an additional value in order to reach 1/3. And yes, that additional value becomes so small, that at infinity, it becomes infinitesimal.

 Quote: Let's assume an infinitesimal. We will call it (tesm). So: 1/3=0.33~+(tesm) 3*{0.33~+(tesm)}=0.99~+3*(tesm) so 0.99~+3*(tesm)=1??

No. Because by definition, 1/3=0.3~. We simply cannot further calculate any equation if we use the decimal notation 0.3~.
But, 1/3 = 0.3 + sum(3/(10^n) where n=2 to infinity)
to better understand the 'tesm', work with the infinite series sum:
0.5+0.25+0.125+0.0625+etc. sum((1/2)^n as n=1 to infinity)
to equate to one, at any point, the final term must be added twice. Otherwise, there will always be a remainder.
So...
sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1
If you replace X with infinity, you end up having to add (1/2)^infinity in order for the equation to equate to one. as X approaches infinity, (1/2)^X becomes infinitesimal, but it never becomes 0.
This is the exact process I'm speaking of.
Assuming X = infinity, the following are both true:
sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1
and
sum((1/2)^n as n=1 to X) -> 1
The first, with any value, including infinity, is always true.
Assuming X = a finite value:
sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1
and
sum((1/2)^n as n=1 to X) = Y
Are both true. The first will always be true, the latter is the difference between an infinite sum and a finite sum.

If you're discussing 0.9~, the same result:
if at any point, the remainder is 1-(9/(10^X)) then,
sum(9/(10^n) as n=1 to X) + (1-9/(10^X)) = 1
if X = infinity, then the remainder becomes infinitesimal, but never becomes 0.

 Quote: And here is the thing. It really, honestly is zero. At infinity, there is no remainder. Because there can't be. A remainder implies stopping. Since there is no stopping, there can never be a remainder.

This is 'the crux of the matter'. Somehow, a definite, required remainder is just removed, because of a definition. The remainder is infinitesimally small. It's not 0, nor is it a tangible number. For all intents and purposes, many maths will consider this to be 0. But when it comes down to it, the correct representation is a number which approaches, but never reaches, 0.

 Quote: By the way, Calculus is based on the concept of infinite series of very small numbers adding up to discrete values. And even though Mathematics has symbols for "less than," "greater than," and "approximately equal to," you will not find them used in the calculus.

Once again, when coming to practical values for use in life, the results cannot be infinite. They must be rounded to a sufficiently accurate precision. Call me an engineer... *shrug* I can't work with an infinite value when I'm programming, nor can NASA when developing space routes. Calculus may consider infinite as a concept, but the result can never be a finite precise value if infinitesimal is considered 0.

Granted, it might take a universal scale before the A to B claim is shown to be inaccurate, but at some point, assuming that the limit of a number is the number, the result will turn up imprecise.

 neon snake wrote: Make the following assumptions for me. 1. The string of digits is infinitely long. 2. Each of the digits 0-9 has a chance (any chance, doesn't have to be equal) of appearing as the next digit. Because of these assumptions, the number will not be an infinitely long string of 5s.

Sorry, can't agree with that assertion. Just because each digit has an equal chance of showing, that is a 10% chance of showing, does not mean that every digit WILL appear. Otherwise the chance of each digit appearing after some number of digits will be 100%. If you're saying that the chance of the next digit being any of 0-9 also changes by digit, then there will essentially always be a chance that the next digit will not be any specific digit 0-9. The whole concept of random means that there is equal chance of any digit appearing at any point, which includes the chance that a digit will not appear. 9 of 10 digits will not appear, digit by digit. So in a random number, there is always a 90% chance that a specific digit will not appear, or a 10% chance that it will.

Digit by digit, the chance that 5 will not have appeared after 1 digit is 9/10
After 2 digits, 81/100 (19 combinations will produce at least 1 5 between 00 and 99).
I believe the equation for calculating the chance of any specific digit appearing at least once for any length of random number was stated earlier in the thread. The result was an infinitesimal chance that a number would not have been shown at an infinite length.

Flip it, for the chance of '5' being shown for every digit. After 1 digit, it's 1/10. After 2 digits, 1/100 (between 00 and 99, 55 only appears once), etc. It never becomes 0. An infinitely small chance, infinitesimal, but not 0 - for precise calculation. Or, essentially 0%.

 Quote: You can see that a 1 should appear 400 times (Obviously, that won't be exact, but you can see the point).

No, because you're saying that the certainty of a number appearing in a 40,000 digit number is 100%. That's not a random number, because you claim to know that it will appear. ie, you've added a definition to the recurrance of digits, so in the end it's no longer a totally random number.

 Quote: You should be able to see that even microscopic chances are made an awful lot more likely when you are performing the trial an infinite amount of times! Basically, when the number of trials makes the jump from 'many, many times' to 'infinite', the chance of success jumps to 100% (actual 100%, not approaching or essentially 100%).

We're just going in circles here... You can never perform a trial an infinite amount of times. No trial will ever return a 100% chance of success, if the chance is calculated on an infinite equation that never reaches a specific value. No matter how many times you attempt the trial, you will always have a fraction of a chance of failure. For practical purposes, you cannot result in a precise finite value from an infinite equation.

Quote:
 Quote: 3. We (in this thread), have come to the conclusion that e is not fully random, and therefore, each finite string has a chance of either 0 or 1 to appear in e.

Good call, Avenger. Having linked the 0 1 law to e, this now seems obvious

Is this not one thing I've said since the beginning?
The chance of any sequence of numbers appearing in e (that is, any length and any digits composed in the sequence) is 0% or 100%. It's not 100%. It's not 0%. It's both, but it's not. In other words, it's an uncalculatable chance. That's like coming to the conclusion of 'all possible values of x'. Practically, this does not return a specific result. In other words, to state the value of the chance of an occurance as a specific value when there are multiple possibilities, is not practically feasible. In some maths, such as calculus, or others, an answer such as +/- 1 is perfectly valid, because it remains an equation which can be used for other purposes. But it does not return a specific value.

By definition we are working with the requirement of knowing a specific value. 'The chance of any sequence appearing in an infinite truly random number is...'. Can only have one answer. A finite absolute chance. In this case, that's not feasible. But, given a certain mathematical basis the answer can be 0% or 100%. And that's perfectly feasible, but not practical.
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Posted: Mon Jan 03, 2005 4:46 pm
Chris K
Boot

Joined: 15 Dec 2004
Posts: 37

You seem to have invented this whole 'A => B, A = B' argument thing.

There is no such dispute in mathematics.

Everything everyone other than you has said in this thread is mathematically proven. It is not in dispute that infinite series can have finite sums.

Find one link to support the claim that there is any dispute over this matter.

How much maths have you actually studied? What exams etc. have you taken?

Posted: Mon Jan 03, 2005 5:00 pm
Guest777
Guest

x = 0.9~
y = 1 - x
.:. y = 0.1~
also y + x = 1

10x = 9.9~
9x = 9.0
x = 1

y + x = 1
0.1~ + 1 = 1?

Posted: Mon Jan 03, 2005 5:28 pm
Sasuntsi Davit
Unfettered

Joined: 04 Dec 2004
Posts: 352
Location: London, UK : Yerevan, Armenia

 Quote: Flip it, for the chance of '5' being shown for every digit. After 1 digit, it's 1/10. After 2 digits, 1/100 (between 00 and 99, 55 only appears once), etc. It never becomes 0. An infinitely small chance, infinitesimal, but not 0 - for precise calculation. Or, essentially 0%.

no actually, the chance of 5 appearing as the next digit does not change since its appearance is independant of any event, it remains 1/10. It is just the way we humans perceive chance that makes it seem unlikely for 5 to show up as the next digit because it has just appeared. It is like the lottery...the chance of 1,2,3,4,5,6 appearing as lotto numbers has the same probability of 7, 13, 45, 31, 24, 1 appearing, yet how many people do you see buying lotto tickets with the former set of numbers...?

edit---

Posted: Mon Jan 03, 2005 6:39 pm
thebruce
Dances With Wikis

Joined: 16 Aug 2004
Posts: 6865
Location: Kitchener, Ontario

yeah, I wasn't referring to the chance for each digit, but the chance of a specific digit appear before any specific position. ie 81/100 is the chance that 5 will appear within the first 2 digits.
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Posted: Mon Jan 03, 2005 7:11 pm
WolverineFan
Decorated

Joined: 30 Dec 2004
Posts: 192
Location: Michigan/USA

 Guest777 wrote: x = 0.9~ y = 1 - x .:. y = 0.1~ also y + x = 1 10x = 9.9~ 9x = 9.0 x = 1 y + x = 1 0.1~ + 1 = 1?

Ok, so I know practically nothing about math (some of thebruces' arguments actually made sense to me ) but there's a huge gaping flaw here:
x = .999 (not repeating)
y = 1 - x
y = .001 NOT .111

so the corrected line above would be something like:
y = 1 - x
y = .000~1

but if you accept the concept of infinity, you'd never get to the 1 so it's actually 0 which I suppose is further evidence to support the other side's argument that .9~ = 1.

I know I should stay out of this... But that was just begging for a correction

While I'm posting anyway, I might point out that thebruce is using infinity to justify his claims even as he essentially disputes the concept of infinity. Pretty cool.

Also, if
 thebruce wrote: Because by definition, 1/3=0.3~

then
0.3~ = 1/3 (= is transitive)
Also
1/3 * 3 = 0.3~ * 3

You agree that
1/3 * 3 = 1
so
0.3~ * 3 = 1

Or does 1/3 not actually equal 0.3~? It just approaches it.

I think you'd be better off staying away from infinity entirely and just saying you don't believe in the concept. If you accept the concept of infinity your argument falls apart. If not, you're golden

Posted: Mon Jan 03, 2005 8:38 pm
thebruce
Dances With Wikis

Joined: 16 Aug 2004
Posts: 6865
Location: Kitchener, Ontario

 WolverineFan wrote: Ok, so I know practically nothing about math (some of thebruces' arguments actually made sense to me ) but there's a huge gaping flaw here: x = .999 (not repeating) y = 1 - x y = .001 NOT .111 so the corrected line above would be something like: y = 1 - x y = .000~1 but if you accept the concept of infinity, you'd never get to the 1 so it's actually 0 which I suppose is further evidence to support the other side's argument that .9~ = 1. I know I should stay out of this... But that was just begging for a correction

hehe I decided not to comment on that post because I thought others would see the flaw. So I'm glad someone replied. Anyway, as above, as 0.9~ approaches 1, the remainder approaches 0. No, you can't express an infinitesimal number in decimal form, but that doesn't mean it equals 0. As 0.9~ -> 1, the difference -> 0. Beyond that, you have accept that A->B so A=B in order to state that 0.9~=1 or that an infinitesimal number = 0.

 Quote: While I'm posting anyway, I might point out that thebruce is using infinity to justify his claims even as he essentially disputes the concept of infinity. Pretty cool.

Well, I'm not disputing infinity. My point basically comes down to a matter of infinity or not infinity. 0 or 1 basically. It either exists or doesn't. An infinite equation that cannot be reduced further will result in an infinite answer, as it pertains to absolute, precise values, not derived math, in which the A to B claim may be totally valid.

Quote:
Also, if
 thebruce wrote: Because by definition, 1/3=0.3~

then
0.3~ = 1/3 (= is transitive)
Also
1/3 * 3 = 0.3~ * 3

You agree that
1/3 * 3 = 1
so
0.3~ * 3 = 1

I'm saying you can't run calculations on infinite decimal values. In math, those numbers are represented as fractions or symbols such as e and pi so as to be able to work with them in equations retaining their precision.

 Quote: Or does 1/3 not actually equal 0.3~? It just approaches it.

By definition, 1/3 = 0.3~, because we know that the remainder never changes, so the 3 continues for inifinity. the statement 1/3=0.3~ is perfectly valid. What we can't do is perform a calculation on 0.3~, whereas we can with 1/3. Of course, that claim will be argued by saying that because we know if *2 that each of the digits will become 6, then 0.3~*2 = 0.6~. But that's not a calculated result, it's a result that only works by explaining it. Whereas I can calculate 0.3 * 2 because I come to the end of the calculation, 0.3~*2 I will never come to the end of the calculation.
So again, I'm saying the same thing in different words.

1/3 = 0.3~
*2..... *2 is N/A
2/3 = 0.6~
*3..... *3 is N/A
3/3 = 1

 Quote: I think you'd be better off staying away from infinity entirely and just saying you don't believe in the concept. If you accept the concept of infinity your argument falls apart. If not, you're golden

I believe in the concept of infinity. Simply saying I have a word with an inifnite of h's means I believe in the concept of inifinity. Heck, I know there are an infinite number of integers, because you just keep adding one to get a new number. there is no maximum integer where adding 1 is not possible.
But that doesn't mean the argument falls apart. My argument simply states that the infinite cannot result in the finite in a balanced equation, for precise, single results.

Really I'm just going in circles now... let's please end this debate... none of us can express what we know any differently than we already have. I'm not trying win the debate, I'm saying it's a stalemate. Let's shake hands and continue with our lives. If I am wrong, and you know it, then what do you care? Let other people decide for themselves. I'm only human, and if I feel my claims are right, it's hard for me to back down if the last word paints me as a fool. But I can back down if I'm shown to be wrong. I can also live with the end to a debate where neither side "wins", because I can see that both sides here are true depending on the perspective one takes in reviewing the evidence.

I respect all the minds that were involved in this discussion... it was quite intelligent, and I've learned a lot about the maths and proofs out there as they apply to their own areas of math. So I'm extending my hand to respectfully agree to disagree, and let the matter drop, ended in this thread.
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Posted: Mon Jan 03, 2005 9:15 pm
ariock
Has a Posse

Joined: 11 Aug 2004
Posts: 750
Location: SF East Bay

Re: Some Clarifications

thebruce wrote:

 Quote: 1/3=0.33~ 3*0.33~=0.99~ 3*1/3=1 Therefore 0.99~=1

Bear with me here - what we're saying is that that statement is essentially like saying

orange = fruit
orange = colour
therefore fruit = colour

I decided to cut to the chase here. It appears that you have no idea what "=" means. This is where you are having trouble. In MATH we have the IDENTITY. If A=B then B=A. You posit that orange = fruit. Does fruit = orange? Does Colour = orange? Here is the part that will blow your mind: In math, if orange = fruit and orange = colour, then you can be sure that fruit = colour.

What you are doing is investing your beliefs and bias in a realm where it has no business. You take into your equation a bias that there is more than one colour....that there is more than one kind of fruit. In your mind, your fruit = colour proof is ridiculous. In math, you have defined fruit, color, and orange, and how they interrelate, and therefore, your proof is VALID. Fruit = colour. I am merely viewing them as variables. If Orange=452, then colour=452 and fruit=452.

It is precisely your beliefs outside of mathematics that are causing you not to understand math. And honestly, if you don't understand it, you shouldn't be going on and on for pages about it.

 thebruce wrote: The most precise that equation can get is the statement that the value approaches a number. 0.9~ -> 1. The same thing I was saying, and which has been condeded to - A -> B does not mean A = B.

I think this has been taken out of context. A->B doesn't mean A=B in all cases.... For example in the equation y=1/x, as x approaches 0 from the right, y approaches Infinity. But as x approaches 0 from the left, it approaches Negative Infinity. So at 0, it doesn't make sense to say that y=infinity OR -infinity.

thebruce wrote:
 Quote: But we all know three thirds is a whole, so what happened?" What did happen? Unfortunately, Harold doesn't tell us.

Because he's not trying to state what happened, simply why the statement doesn't work.

No. He is saying that it doesn't work. He, like you, is saying it doesn't add up. And no matter how many times we throw proofs at you, you dismiss them with philosophy and anything BUT math instead of math.

 thebruce wrote: you can't calculate a decimal value based on an infinite decimal. You can only work with what would become an infinite decimal value - that is an equation which results in the infinite value - ie you can't calculate with 0.3~, but you can work with 1/3 instead. You can't multiply 0.3~ * 3, because you would never return a final value.

I can't? 0.33~*3=0.99~=1.
0.33~/3=0.11~.
0.33~*0.33~=0.11~ (notice a similarity in those last two?)
0.66~*10-0.66~=6
There. I just did.

 thebruce wrote: The correct progression is 1/3 = 0.3~ ....SNIP....you can take the decimal notation 0.3~ and define it as the sum of 3/(10^n) where n = 1 to infinity. The limit of that equation is 1/3. The more we calculate, the closer and closer we are to 1/3. But no matter how long we calculate, we'll never reach 1/3....SNIP....No. Because by definition, 1/3=0.3~

So which is it? is 0.33~ equal to it or not? You appear to have it both ways, depending on which way your fancy takes you. In math, when you define something, it stays defined.

 thebruce wrote: Assuming X = infinity, the following are both true: sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1 and sum((1/2)^n as n=1 to X) -> 1 The first, with any value, including infinity, is always true.

I put it to you that if x=infinity,

sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1 = sum((1/2)^n as n=1 to X)

because 1/2^infinity = 1/infinity = 0

So lets take a line from 0 to 1 and split it up an infinite number of times. Oh wait, that has already been done. It is called a number line. and each of the infinite points on that number line between 0 and 1 has a width of zero. hm....an infinite number of zero width points...and they have a length of one.

 thebruce wrote: . We will always require an additional value in order to reach 1/3. And yes, that additional value becomes so small, that at infinity, it becomes infinitesimal.

If there is an additional quantity required for 0.3333~ to precisely equal 1/3 then to say 1/3=0.333~ is incorrect. We are talking MATH here, and not inches. If you want to talk about math, then you need to speak MATH.

If they are NOT equal then some quantity must make up the difference. Thus we have your "infinitesimal."

I notice that you ignored my other calculations where I included your "infinitesimal." If you are going to define a number, you must then evaluate its properties. I am going to restate them here:

0.99~+3*(tesm)=1??
0.99~+7*(tesm)=1???

Therefore 3*(tesm)=7*(tesm) Therefore 3=7???

Your infintesimal math doesn't add up. You cannot define an undefinable number. Let me reiterate the most obvious way that that math works. It works if (tesm)=0. The only other possibility is if you allow various sizes of infinitesimals. And doesn't that go against the definition of an infinitesimal to have an infinitesimal that is larger than another infinitesimal?

ALL OF THIS because you don't think 0.999~=1. For your philosophical disagreement, we must define an unknowable, variable length infinitely small number. But it isn't zero! .....even though it possesses all of the properties of zero. And by the way, this significant caveat of decimal notation isn't mentioned in any mathematics textbooks or anywhere else. Good luck with that.

 CALCULUS Fourth Edition by Larson/Hostetler/Edwards (c) 1990, Page 1 wrote: Rational numbers can be represented either by terminating decimals such as 2/5=0.4, or by repeating decimals such as 1/3=0.333...=0.3~. Real numbers that are not rational are called irrational. They cannot be represented as terminating or repeating decimals. To represent an irrational number, we usually resort to a decimal approximation. For example, SQRT(2) (wavy=) 1.4142135623

The wavy = is the symbol for approximately that I mentioned in my previous post. Notice there is no wavy = for 0.3~.

Posted: Tue Jan 04, 2005 1:56 am
Chris K
Boot

Joined: 15 Dec 2004
Posts: 37

thebruce, can you please stop saying 'as 0.9~ approaches 1...'!

You cannot have that!

This is a very immature mistake. You basically think of 0.9~ as a number that keeps on going, ie. you imagine more and more 9s being put on the end. You think, the more accurate you take it, the closer it gets to 1.

NO! NO! NO!

It is a number, it has a value, therefore it cannot approach
----------------------
Can you please provide evidence that you can't work in decimal with recurring numbers?

Are you saying you can't do stuff like this:

0.4~ * 2 = 0.8~

???

What's wrong with that? I didn't see 'maths break down' when I used recurring decimals.

Posted: Tue Jan 04, 2005 2:38 am
da5id
Veteran

Joined: 20 Oct 2004
Posts: 80

last night i had a vision in my mind: in a not all too distant future, some ppc folllower will say to another such person, "do you remember the E-Numbers-thread and how they fought about mathematics for pages and pages? THOSE were the days!" and both will get this distant look, trying to remember what they did or where they were when they read those posts (and pretended to understand them).

Posted: Tue Jan 04, 2005 2:43 am
chrome_halo
Boot

Joined: 30 Dec 2004
Posts: 55
Location: Bedfordshire UK

1nf1n1tyyyyyyyy................

I would just like to say that I 've found this thread highly entertaining, at least it's given us something to think about while we wait for the action to kick off. I no longer count electric sheep to get to sleep, instead I keep cutting a line in half and pondering infinity. No matter what the facts may be it makes good reading. Keep debating but please remain friendly, we're all in this together.
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Posted: Tue Jan 04, 2005 7:17 am
Sasuntsi Davit
Unfettered

Joined: 04 Dec 2004
Posts: 352
Location: London, UK : Yerevan, Armenia

thebruce wrote:

yeah, I wasn't referring to the chance for each digit, but the chance of a specific digit appear before any specific position. ie 81/100 is the chance that 5 will appear within the first 2 digits.

isn't the chance of 5 appearing at least once within the first two digits 3/100 rather than 81/100?

Posted: Tue Jan 04, 2005 7:38 am
WolverineFan
Decorated

Joined: 30 Dec 2004
Posts: 192
Location: Michigan/USA

 thebruce wrote: ...I'm extending my hand to respectfully agree to disagree, and let the matter drop, ended in this thread.

I too extend my hand. It's impossible to come to an agreement when we don't have a common language.

I agree with your point that machines don't handle infinity well. On the other hand, try adding .111 to itself in Java a few times someday and you'll see they don't handle  finite [/edit] rational decimals all that well either sometimes

I agree to disagree on the rest.

Posted: Tue Jan 04, 2005 9:40 am
thebruce
Dances With Wikis

Joined: 16 Aug 2004
Posts: 6865
Location: Kitchener, Ontario

Re: Some Clarifications

Ariock, it's good to have you on board

First off, I'm going to respond to the last post first...
 WolverineFan wrote: I too extend my hand. It's impossible to come to an agreement when we don't have a common language. I agree to disagree on the rest.

Excellent.

 Quote: I agree with your point that machines don't handle infinity well. On the other hand, try adding .111 to itself in Java a few times someday and you'll see they don't handle  finite [/edit] rational decimals all that well either sometimes

haha well pentium had that problem as well, no? Where were you when you first heard the joke:
Q: How many Pentium designers does it take to screw in a light bulb?
A: 1.99995827903, but that's close enough.
(or something like that)

Anyway, I hereby, following this post, resign myself to agreeing to disagree. I respectfully ask that we all agree to do the same. If I'm wrong, or just plain nuts, which some seem to think, I'm wrong, and so are the masses of people not in this community who debate the same issue. Or, it's an issue where the fundementals of the debate itself are flawed, and an agreement can never be reached because the framework for each side is different. I don't resort to personal attacks, so if you agree and withdraw, please let the matter drop and let's treat each other with respect (I'll treat individuals with as much respect as those individuals treat me with) when it comes to the rest of the ARG which I'm sure will post some nice challenges to all the intellectual minds we've conversed with here. Most likely, I'll be stumped on 95% of the arg Hey, some of the ilovebees puzzles were pretty complex, and it was impressive to see how fast some of them were solved... when the PMs are surprised, you know there's a good community at work! And from my 1/2 year of experience here, Unfiction is an amazing community - respectful, intelligent, fun and crazy... quite a mixing pot to appear in one solid community. I consider many here friends... especially online hehe

Either way, I hereby withdraw from the debate, because it's going nowhere. To know more, simply google the topic on the net, and everything we've discussed will be covered somewhere in someone else's debate...

 ariock wrote: I decided to cut to the chase here. It appears that you have no idea what "=" means. This is where you are having trouble. In MATH we have the IDENTITY. If A=B then B=A. You posit that orange = fruit. Does fruit = orange? Does Colour = orange? Here is the part that will blow your mind: In math, if orange = fruit and orange = colour, then you can be sure that fruit = colour.

What I was drawing a parallel to, given the those names probably didn't give the best example, was that you can't run calculations using infinite decimals. They must be represented in a workable form first in order to use them. ie, you can't run 0.3~ * 3 because the calculation would never end. What's ~ * 3? So saying 1/3=0.3~ is fine. You can multiple 1/3 by 3, but you can't actively calculate 0.3~ * 3. That's my point. So you can't state that 0.3~ * 3 = 0.9~. Logic by variable may essentially show 1/3=0.3~, 3/3=0.9~, but mathematically the decimal half can't be calculated.

If that's what it comes down to - philosophy - then that's what we're arguing here, in which case it's pointless to continue.

 Quote: What you are doing is investing your beliefs and bias in a realm where it has no business.

What beliefs? In absolute math? That there is the infinity and there is finite? 0.3~ is an irrational number - a number that cannot be written as a decimal. So how can you say you can calculate any equation using a decimal number which cannot exist? 0.3~ is a representation of 1/3, because you cannot write 1/3 as a decimal. In order to multiply 0.3~ by n, you assume to be able to calculate every digit. But we know we can calculate every digit in 1*3/3, and return a result. This seems to be where our 'beliefs' diverge, and why for absolute precision, I cannot accept (along with many people in the world) the 'proof' that 0.9~=1 by multiplying in infinite decimal value.

 Quote: It is precisely your beliefs outside of mathematics that are causing you not to understand math.

My beliefs outside of math have absolutely nothing to do with how I treat math. Math is absolute. Math can be derived, but it is no longer absolute. ie, practical math vs math theory, or whatever terminology you want to use. Practical math requires an answer that is defined by the problem at hand. Most times it needs a single, precise value. Other times it may only need a sufficiently accurate value, and it may also ask for a set of any possible value. When I'm not working with a practical problem where the answer has a clearly defined definition (ie the chance of something happening is...) then I will gladly accept the non-absolute, the derived, math theory, etc.
Regardless of my terminology, with absolution, 0.9~ cannot = 1. In theory, it can. In other words, I completely agree with what you, or anyone here is saying, except that it is not a practical, absolute answer or solution.

 Quote: I think this has been taken out of context. A->B doesn't mean A=B in all cases.... For example in the equation y=1/x, as x approaches 0 from the right, y approaches Infinity. But as x approaches 0 from the left, it approaches Negative Infinity. So at 0, it doesn't make sense to say that y=infinity OR -infinity.

So you just said that = can have multiple meanings. How do I know one case from another if A->B means A=B only sometimes? I'm saying, where A->B, let it remain at A->B, and state that the limit of the equation=C. y=1/x where x->+0 means y->infinity. the balance is - AS x approaches 0 from the right, SO does y approach positive infinity. AS x approaches 0 from the left, SO does y approach negative infinity.

 Quote: No. He is saying that it doesn't work. He, like you, is saying it doesn't add up. And no matter how many times we throw proofs at you, you dismiss them with philosophy and anything BUT math instead of math.

Because the issue is one of the difference between precision and derivation. For precision, 0.9~=1 is untrue. In theory, 0.9~=1 can be true.

 Quote: I can't? 0.33~*3=0.99~=1. 0.33~/3=0.11~. 0.33~*0.33~=0.11~ (notice a similarity in those last two?) 0.66~*10-0.66~=6 There. I just did.

By your predefined notion that my claim is wrong, which is precisely why I say it doesn't work. So, we can't agree on that.
You can't multiply 0.3~ by n, because you can't calculate every digit. You can multiply 1/3 by n because you can multiply 1*n and 3*1 to arrive at a final fraction which reduces to a finite value. You can't multiply 3/10*n + 3/100*n + 3/1000*n + 3/10000*n + etc to infinity. You can only extrapolate what you would consider to be the result.
In theory, you can arrive at 0.9~=1, but for practical, precise purposes, this cannot be possible in the real world.

 Quote: So which is it? is 0.33~ equal to it or not? You appear to have it both ways, depending on which way your fancy takes you. In math, when you define something, it stays defined.

Yes, we define 1/3 as 0.3~, because we know from demonstration that the remainder when dividing will never change, so we can 1/3=0.3~. But we can't do anything with 0.3~. We can do something with 1/3. We can never write the true value of 1/3 in decimal, but we can represent the true uncalculatable value as 0.3~, just as 1/3 represents the true uncalculatable value. 1/3 is a practical representation we can work with to calculate further, 0.3~ is not.

 Quote: I put it to you that if x=infinity, sum((1/2)^n as n=1 to X) + ((1/2)^X) = 1 = sum((1/2)^n as n=1 to X) because 1/2^infinity = 1/infinity = 0

Again your proof that I'm wrong simply presumes I'm already wrong. My statement says that 1/infinity -> 0 for an absolute value. You're saying 1/infinity=0, therefore you're wrong. ?? That is where we clearly disagree, if we don't agree in the difference between a precise value, and a derived value.

 Quote: So lets take a line from 0 to 1 and split it up an infinite number of times. Oh wait, that has already been done.

Yes, it has, a number of pages ago.
 Quote: It is called a number line. and each of the infinite points on that number line between 0 and 1 has a width of zero near 0. hm....an infinite number of zero infinitesimal width points...and they have a length of one.

Sound like gibberish? sure, because in dealing with infinity, you're not dealing with practical values. You're saying that slitting a 1 unit line into infinite segments essentially means the line actually does not exist. This is not practical math, this is theory, it's philosophy, whatever you want to call it. How can 0=1? Where's problem with the original question? Because somehow a specific amount is being lost. The only way the answer balances to the question is to treat infinity as a representation, a variable. With a unit of a line, splitting up into infinite parts, each part must be infinitely small. It's not practical math.

 thebruce wrote: If there is an additional quantity required for 0.3333~ to precisely equal 1/3 then to say 1/3=0.333~ is incorrect. We are talking MATH here, and not inches. If you want to talk about math, then you need to speak MATH.

I'm not saying that there's a quantity for 1/3 to = 0.3~. I'm saying that as you increase the precision of 0.3333333etc, the remainder becomes smaller and smaller... as you calculate 0.3~ by digit, the remainder in order to truly reach 1/3 approaches 0, but will never reach it. That's not the same as 1/3 = 0.3~ + X. Because 0.3~ IS 1/3. Essentially, the remainder when calculating 1/3 will also alwys be an infinite number. 1/3 = 0.3 + (0.03~), = 0.3+0.03+(0.003~) However you express that in an equation won't change the fact that 1/3 = 0.3~, or the sum of an infinite sequence where the remainder approaches 0.
1/3 = 0.3 + 0.03 + (0.003~)
1 = 0.5 + 0.25 + (0.25)
Every number can be expressed as the sum of an infinite series IF the remainder remains defined as an infinitely small amount. You can derive from that, which is not practical, that the remainder essentially becomes 0, so the infinite sum essentially = the value.

 Quote: If they are NOT equal then some quantity must make up the difference. Thus we have your "infinitesimal."

Exactly. For absolution, you can't simply just hand-wave this unknowable variable to non-existence.

 Quote: I notice that you ignored my other calculations where I included your "infinitesimal." If you are going to define a number, you must then evaluate its properties. I am going to restate them here: 0.99~+3*(tesm)=1?? 0.99~+7*(tesm)=1??? Therefore 3*(tesm)=7*(tesm) Therefore 3=7???

No, because you can't calculate with infinite decimals. But I wouldn't be surprised if in theory someone could come up with a logical proof that equate 3 to 7... *shrug*

 Quote: Your infintesimal math doesn't add up. You cannot define an undefinable number. Let me reiterate the most obvious way that that math works. It works if (tesm)=0. The only other possibility is if you allow various sizes of infinitesimals. And doesn't that go against the definition of an infinitesimal to have an infinitesimal that is larger than another infinitesimal?

Doesn't it go against the definition of infinity to have an infinity that is larger than an infinity? orders of infinity? is 2x where x approaches infinity not smaller than 3x as x approaches infinity? I believe near the beginning of the thread there was a discussion about working with orders of infinity in math.

 Quote: ALL OF THIS because you don't think 0.999~=1. For your philosophical disagreement, we must define an unknowable, variable length infinitely small number. But it isn't zero! .....even though it possesses all of the properties of zero.

A definition, not an absolute. 0 is 0. 0 is nothing. Infinitely small may be considered 0, but it is not nothing.

 Quote: And by the way, this significant caveat of decimal notation isn't mentioned in any mathematics textbooks or anywhere else. Good luck with that.

Yet this whole debate is still taking place elsewhere in the world. There are professors who would teach why 0.9~=1 is true, and others who would not. If it was so absolute, then there would be no question and the argument would have been dispelled long ago.

 Quote: The wavy = is the symbol for approximately that I mentioned in my previous post. Notice there is no wavy = for 0.3~.

Of course not, because 1/3 = 0.3~. I don't dispute that. I do dispute that you can calculate absolute values in decimal notation using 0.3~

What mathematical process do you use to calculate 0.3~ * 3?
 Code: 0.3333~     * 3 ------- 0.9999X

what's X? Infinity? So 3 * ~ is ~? ~ isn't a decimal value, how can you multiply a symbol by a number? 0.3~ isn't algebraic notation, there's no such thing as 3~ like there is 3a. ~ isn't a variable. If you work in decimal, you're working with digits. If you treat ~ as a variable, it's value isn't static, it's value depends on how many of it's containing digits you write beforehand. So really, 0.3~ = 0.9(3~), so you write that as an equation and you get the 3 * the sum of an infinite series, which approaches a value if you leave out the remainder, or = a value if you add the remaineder.

Let's change the 0.3~ to try to reduce it a bit to multiply. Choose a form:
sum + remainder = 0.3+(0.03~)
or maybe just splitting the digits, so maybe you can at least prove that 3*3=9
0.3 followed by (3~)
multiply either of those, you get
= 0.9 + 3*(0.03~) or
= 0.9 followed by (3*3)~
how would you calculate that?

Here's the thing... was that entire paragraph gibberish and invalid math and enormously bad form? Of course... because you can't actively calculate decimal values that are infinitely long. But you can algebraicly work with finite values and variables in order to reduce the complexity to come to a result, or results, or a subsequent infinite representation. That's where the absolute precise calculations end, and you cross into theory, the non-practical.

What I'm describing is where you retain ALL input in an equation, without including any math proof that removes any part of a value, no matter how large or how small. In this way, an infinitely large equation will return an infinitely large result, and an infinitely small equation will return an infinitely small value. After that, you take the step of equating infinitesimal with 0, in theory.

 Quote: This is a very immature mistake. You basically think of 0.9~ as a number that keeps on going, ie. you imagine more and more 9s being put on the end. You think, the more accurate you take it, the closer it gets to 1.

So 0.9~ is not a number that keeps going and going? So it has an end? Is it then not an infinite number?

 Quote: "do you remember the E-Numbers-thread and how they fought about mathematics for pages and pages? THOSE were the days!" and both will get this distant look, trying to remember what they did or where they were when they read those posts (and pretended to understand them).

in another forum with friends of mine, someone started a thread about 'where were you when...' we had a lot of fun with that

 Quote: No matter what the facts may be it makes good reading. Keep debating but please remain friendly, we're all in this together.

Very much echoed here! This is intellectual stimulation

Quote:
 Quote: yeah, I wasn't referring to the chance for each digit, but the chance of a specific digit appear before any specific position. ie 81/100 is the chance that 5 will appear within the first 2 digits.

isn't the chance of 5 appearing at least once within the first two digits 3/100 rather than 81/100?

Actually no, because there are 19 numbers in 100 possible values.
[0-9]5 = 10, 5[0-9] = 10, - 55 which appears in both, =19.
your chance would be the chance of 5 appearing at least once in 2 digits with one other number. given 5 and 1, you can have 51, 15, 11, 55, or 3/4 chance.

Actually, allow me a correction... 81/100 is the chance that 5 will not appear at least once in the 2 digits. it's a 19/100 chance that it will.

The chance that 5 will appear anywhere in the 2 digits is a different chance. That's one that doesn't matter on the position of the digits. so there aren't 100 possibilities, there are 55 digit combinations, where 5 appears 10 times - a 5 and any other digit, so a 10/55 chance.

Well I can't remember which is the actual chance that 5 would appear at least once within 2 previous digits, I can't remember if that problem depends on the order of the digits or not. It's either 19/100 or 10/55... either way, as the number of digits increase, the chance decreases, but the chance never reaches 0 for all digits being 5 - there will always be a 1/X chance, and X approaches infinity, 1/X approaches 0.
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Posted: Tue Jan 04, 2005 11:19 am
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