LoFi version for PDAs  Help Search Members Calendar 
Welcome Guest ( Log In  Register )  Resend Validation Email 
Pages: (381) « First ... 274 275 [276] 277 278 ... Last » ( Go to first unread post ) 
Add reply · Start new topic · Start new poll 
meBigGuy 
Posted on Dec 3 2007, 06:38 AM


Advanced Member Group: Members Posts: 1454 Joined: 24August 07 Positive Feedback: 75.56% Feedback Score: 32 
bzzzzzt WRONG AGAIN The notation to the right of the equals sign in [1'"] = .000....1 is meaningless (bogus, invalid, a violation). Infinitesimals are not numbers, they are a CONCEPT. You can put nothing after ... (the alternate meaningless notation is 0.0r1) If you allow this, you can prove all sorts of bogus things like 1 is not always 1. Just like division by infinity, it is not allowed. 0.9r is not a power series, is not a limit, is not a sequence, is not a converging anything. It is simply a number. You can represent it by such things, or set such things equal to it, but you must be careful. Your manipulations of series and equations have nothing in common with the actual value of the number 0.9r. Lets go back to basics. Please respond to the following questions. (really, please!) Let's pin down where the fundamental disagreement starts. Can you respond simply to the following 3 questions/statements. Don't try to prove anything just yet. Just help me understand where our differences lie. 1. Does 0.3r always and exactly equal 1/3 (since 0.3r is defined as the decimal notation for the number that is given by the operation one divided by three). If you say no, we will proceed from this to #3 below. 2. If 0.3r = 1/3 exactly and always, then are the following 2 equations true: 0.3r * 3 = 0.9r and 1/3 * 3 = 1 If so, then you must agree that 0.9r = 1, unless you have some argument with those operations. If you disagree, say No to #2 and proceed to #3. 3. If you say no to 1 or 2, do you think 10*0.3r 0.3r= 3? because if you do, that only works if 1/3 = 0.3r. If you do not, then you must think 0.3r only equals 0.3r sometimes, depending on what you did previously, or that 0.3r is not really the notation for a real number (which, unfortunately, it is DEFINED to be). SO, please comment on where we agree or disagree. I find it much easier to work with the equation 0.3r = 1/3 when trying to resolve all the series/converging/limit stuff. It doesn't require the leap that two different number notations that are so seemingly different can be equal (as 0.9r = 1 does). (although there is no real difference) This post has been edited by meBigGuy on Dec 3 2007, 06:46 AM  Proud recipient of negative feedback from:
ZarabattyStDullas AKA TrOUTAlphahahahahaBenTheBoyRabbitchNOMbskulfivedohNUTSPrincess.BlueballsCecil.P.NoScience PJParent001TheEnd(it) AKA Robin ARSEons (2 OF WHICH KNOW ANY SCIENCE) 

Send PM · Send email ·

Raphie Frank 
Posted on Dec 3 2007, 07:39 AM


Advanced Member Group: Power Member Posts: 2204 Joined: 25March 07 Positive Feedback: 55.56% Feedback Score: 7 
Dear MeBigGuy, Without reference to the 3 questions, which I will try to answer later, you ask where we agree or disagree and I believe that where we disagree is that the concept of infinity, and indeed it IS only a concept, cannot be treated AS IF it were a number, as imaginary in nature as it may be. This is also related to the concept of modular arithmetic and/or statements you have made to the effect that they are offtopic. With all due respect I disagree that one can not treat numbers AS IF the end circled back around to the beginning, in ourobourus snake chasing it's tail manner, which is related to the entire infinity issue, and also to the quotation I include from a post I made last June. In any case, the notion of numbers "wrapping around" back to the beginning (but at higher or lower "orders of exponentiation"), akin to curved space/time, for instance, is where a Lucas number such as 199 [=L_11 = (phi^11  phi^11)] becomes quite interesting to me. Treat the 9 on the end AS IF it existed in a "curved space" and 199 == 1+99 == 100, which from an ethnomathematical perspective is congruent with the Greek "Rho" == 100 == which represents "radius" in a system of spherical polar coordinates. SEE: http://en.wikipedia.org/wiki/Rho_(letter) It is a philosophical conceit to be sure, but the question is not if it is right or wrong, but if it is useful or not. I have found it to be quite useful, indeed, in discerning numerical relationships and correspondences. Best, Raphie ============================================ Dear Precursor562, You write  In order for [1'] to equal [1"] (which is required for .9r to equal 1) then [1'"] would have to equal 0 but where [1'"] actually equals 1/10^(n) as n>infinity, the actual value of [1'"] can never equal 0. So as a result [1'] can never equal [1"].  [1'] can indeed never equal [1"], but this does not mean it cannot be CONGRUENT with it in exponential manner. (/edit I might add that there is an element of "Yin Yang" here. 1', 1"" and 1"" are all required for "completion.") The following is a false statement... 1 = 10 = 100 But the the following is a TRUE statement... 1 == 10 == 100 because one can 1::1 map out a bijective relationship between 1, 10 and 100 and the integers in a very simple manner as follows: for p_x = index position; x = positive integers; n = 1 p_1 = 1 > 1 > 10^0 > 10^(1)+n > 10^(i^2)+n > 10^(i^2)+1 > 10^((i^2)+p_1) p_2 = 2 > 10 > 10^1 > 10^(1)+n+n > 10^(i^2)+n+n > 10^(i^2)+2 > 10^((i^2)+p_2) p_3 = 3 > 100 > 10^2 > 10^(1)+n+n+n > 10^(i^2)+n+n+n > 10^(i^2)+3 > 10^((i^2)+p_3) The best analogy I can offer at the moment (and I have to run out...), with respect to "exponentiation" as "orders of progression" would be to make reference to the keys I hold in my hand as I prepare to leave my loft. My house keys are ganged together on a "loop" of metal, the key ring, that overlaps with itself. If I look at it on edge, it looks something like this: +++++++++++++ +++++++++++++ More later... Best, Raphie P.S. I am working on something in response to the questions RPenner raised regarding infinity and will post a link to it here once I post it online. This post has been edited by Raphie Frank on Dec 3 2007, 07:56 AM  Reality is always bending itself for us. sometimes it bends itself to amuse us, sometimes to teach us, sometimes to confuse us. It bends itself overtly and covertly. the bending takes many different forms  sometimes visual, sometimes spiritual, sometimes we feel vertigo that has nothing to do with any physical circumstances...  Egg Theorem


meBigGuy 
Posted on Dec 3 2007, 09:24 AM


Advanced Member Group: Members Posts: 1454 Joined: 24August 07 Positive Feedback: 75.56% Feedback Score: 32 
No No No I will not fall prey to this off topic nonsense. bite tongue  bite tongue  bite tongue must resist whew  made it  that was close.  Proud recipient of negative feedback from:
ZarabattyStDullas AKA TrOUTAlphahahahahaBenTheBoyRabbitchNOMbskulfivedohNUTSPrincess.BlueballsCecil.P.NoScience PJParent001TheEnd(it) AKA Robin ARSEons (2 OF WHICH KNOW ANY SCIENCE) 

Send PM · Send email ·

Gorgeous 
Posted on Dec 3 2007, 11:08 PM

An actual person Group: Members Posts: 2933 Joined: 11January 07 Positive Feedback: 73.33% Feedback Score: 51 
It's probably about time to ask the experts...
http://news.bbc.co.uk/2/hi/science/nature/7124156.stm g.  "Science is the belief in the ignorance of experts."
(Richard Feynman, The Pleasure of Finding Things Out (1999) In order to fool others, we must firstly be able to fool ourselves: Who ya gonna fool? © If I sit atop a hill looking down into the valley below, I see waves, I feel waves, I smell and hear waves. The crops in the fields below sway in waves just like the water of the ocean does, and sound waves come and go. From this simple and empirical premise alone, the WaveStructure of Matter is just leaps and bounds ahead in terms of plausible description for that which we observe. It thus comes as no surprise whatsoever, for those whose minds are fixed on Reality, to learn that the REAL 'equations' will also match up...If your 'math' or your 'physics' does not plausibly explain that which we observe empirically, it has not yet reached the same level of understanding that WSM presents: http://www.spaceandmotion.com/ 
Send PM · Send email ·

Precursor562 
Posted on Dec 3 2007, 11:22 PM


Advanced Member Group: Members Posts: 1333 Joined: 20June 06 Positive Feedback: 42.5% Feedback Score: 46 
That's what I get for randomly picking numbers without considering the other random numbers.
OK fine. [1'] = 1 [1''] = .9r [1'''] = X where X = 1/10^n as n>infinity And I mean the exact value of X and not it's limit.
I'm gonna throw something by you. 1/4 = .2 with a remainder of 2. Now you could keep going but lets leave it like that. I ask you. Does that remainder hold any value? Now take 1/3. You get .3 with a remainder of 1. Does 1/3 = .3? So does that remainder have to have value? .3r is the direct result of a process that does not end. Go ahead and work out 1/3 using long division and try to work out the remainder so you no longer have one. You will quickly find that no matter how long you try (even if you did it for an infinite length of time, and even if such time passed instantaneously), never will that remainder disappear. It only becomes another infinitesimal whose value is 1/3rd that of the infinitesimal found between .9r and 1. I believe that the equation that you will find is... S(n) = .9 + .09 + .009 +... S(n) = .9r as n>infinity Please provide a link if incorrect but I don't recall ever seeing... Lim S(n) = .9 + .09 + .009 +.... Lim S(n) = .9r as n>infinity I do recall seeing.... Lim S(n) = 1 I also still haven't seen a straight up answer for... How can you get a finite number (with a finite value) as a sum for a summation that is infinite in length, has no end, does not end, has an infinite number of terms and so no last term? Where each term is a finite number with finite value (so no saying 0 + 0 + 0 +....).  Time is the wisest counsellor


Send PM · Send email ·

NeoDevin 
Posted on Dec 4 2007, 12:41 AM


Advanced Member Group: Members Posts: 456 Joined: 9October 07 Positive Feedback: 90% Feedback Score: 22 
By taking a limit. See p.221 of the notes I linked to before. This is first year university calculus, I don't know why it's so hard for you. 

Send PM · Send email ·

meBigGuy 
Posted on Dec 4 2007, 02:43 AM


Advanced Member Group: Members Posts: 1454 Joined: 24August 07 Positive Feedback: 75.56% Feedback Score: 32 
All infinitesimals = 0 in the real number system. If you substitute 0 everywhere you erroneously make one, it all works out.
In my mind it is common knowledge that 0.9r = 1. That they do not use that notation? I can't say why. There is no real number between 0.9r and 1. Only "infinitesimals" which = 0. The real number system says the 0.0r1 notaion is invalid, bogus, ambiguous, illegal.
There is no such thing as an infinitesimal real number. PERIOD. The use of infinitesimal real numbers will cause ambiguous results. Infinitesimal real numbers have no use in the set of real numbers. You lose nothing by removing them. You gain consistency by removing them. That is why they are removed. Consider my example where 0.3r = 0.3r SOMETIMES. How is one going to resolve that? There is no use for infinitesimal real numbers other than 0. The infinitesimal is a CONCEPT like infinity. You need to take a pill for the infinitesimal real number disease. http://en.wikipedia.org/wiki/Infinitesimal Look carefully at my 0.3r = 0.3r SOMETIMES example. How does one resolve that? This post has been edited by meBigGuy on Dec 4 2007, 02:43 AM  Proud recipient of negative feedback from:
ZarabattyStDullas AKA TrOUTAlphahahahahaBenTheBoyRabbitchNOMbskulfivedohNUTSPrincess.BlueballsCecil.P.NoScience PJParent001TheEnd(it) AKA Robin ARSEons (2 OF WHICH KNOW ANY SCIENCE) 

Send PM · Send email ·

Precursor562 
Posted on Dec 4 2007, 03:37 AM


Advanced Member Group: Members Posts: 1333 Joined: 20June 06 Positive Feedback: 42.5% Feedback Score: 46 
Once again. Basic stuff. Actual does not always equal limit. Example. f(x) = 2x + 3 x>4 Lim f(x) = 11 What does f(x) actually equal? f(x)>11 See the difference? I'll spell it out. The Limit of f(x) equals 11 The value of f(x) approaches 11 In order for the value of f(x) to equal 11 (reach its limit), X must equal 4 (X must reach its limit). f(x) = 1/10^(x) x>infinity Here X has no limit. Lim f(x) = 0 but the value of f(x) can never be 0. The limit of f(x) equals 0 The value of f(x) approaches 0 The value of f(x) can never become 0 So limit Lim S(n) = .9 + .09 + .009 + .... Lim S(n) = Lim .9(n) = Lim .9r = 1 as n>infinity Actual Value S(n) = .9 + .09 + .009 +.... S(n) = .9(n) = .9r as n>infinity  Time is the wisest counsellor


Send PM · Send email ·

NeoDevin 
Posted on Dec 4 2007, 06:16 AM


Advanced Member Group: Members Posts: 456 Joined: 9October 07 Positive Feedback: 90% Feedback Score: 22 
Here, I will do a proof for you, perhaps you will find something in it to argue with. (I'll number the steps so you can point to exactly which step you think is wrong) Conditions 1. Choose any convergent sequence S(n) (heck, make it any strictly increasing positive bounded/convergent sequence). (theorem 2.3, page 40 of the notes I linked to is the proof that any monotonic bounded sequence is convergent) 2. Choose some x such that x > S(n) for all n. Theorem 3. I will propose that x must be greater than or equal to lim_{n>infinity} S(n). Proof 4. Assume not, that is, assume lim_{n>infinity} S(n) = L > x 5. Then define d = L  x > 0 (means that L = x + d) 6. Then, we choose the epsilon (e) in the definition of the limit L at the top of page 32 of those notes to be d/4 7. Remember, in order for it to be convergent, it must be true for ever e > 0, d/4 is certainly greater than 0, so we must be able to find an N such that n > N implies S(n)  L < e = d/4 8. We have restricted S(n) to be positive and strictly increasing, so we can say that L > S(n) for all n, and therefore we can say that S(n)  L = L  S(n) 9. But this means that for every n > N we have L  S(n) = x + d  S(n) = (x  S(n)) + d < e = d/4 10. But we have chosen x such that x > S(n) for all n, therefore x  S(n) > 0, and therefore: L  S(n) = (x  S(n)) + d > d 11. Therefore we have that: L  S(n) < d/4, but also L  S(n) > d... A contradiction. 12. Our original assumption must be wrong, L cannot be greater than x, and therefore the limit L must be less than or equal to x, if x and S(n) satisfy the conditions listed above. Ok, that ends the proof, if you have any problems with it, please let me know which step the occur in, and I will try to fix my error. Application 13. Let S(n) = 0.9 + 0.09 + 0.009 + ... 9*10^(n) = 1  10^(n) > 0 (positive) [Edit] 13.1. S(n+1)  S(n) = 9*10^(n1) > 0, S(n+1) > S(n) (strictly increasing) 13.2. S(n) < 1 for all n (bounded) [/Edit] 14. Let x = 0.9r 15. Then clearly we have that x  S(n) = 9*10^(n1) + 9*10(n2) + ... (ad infinitum) Is greater than zero. 16. This means that x is greater than S(n) for all n. 17. According to your post, lim_{n>infinity} S(n) = 1 (quoted at top) 18. Therefore L = 1 (as defined in the proof above) 19. Therefore, according the the little theorem I just proved above, we have that x = 0.9r must be greater than or equal to L = 1 Unless you are about to start arguing that 0.9r is greater than 1, I think that does it. If you have a problem with any of the steps I've outlined above, feel free to point out exactly which step. This post has been edited by NeoDevin on Dec 4 2007, 06:20 AM 

Send PM · Send email ·

prometheus 
Posted on Dec 4 2007, 02:40 PM

Annoyed by you. Group: Power Member Posts: 1140 Joined: 1November 07 Positive Feedback: 78.26% Feedback Score: 103 
Here's another proof using a geometric series:
1) let 0.9r = s = 0.9 + 0.09 + 0.009 + ... 2) s = 9/10 + 9/100 + 9/1000 + ... 3) s/10 = 9/100 + 9/1000 + ... 4) s  s/10 = 9/10 5) s(1  1/10) = 9/10 6) s * 9/10 = 9/10 7) s = 1  Hac in hora sine mora corde pulsum tangite.  O Fortuna from Carmina Burana
For precept must be upon precept, precept upon precept; line upon line, line upon line; here a little, and there a little:  Isaiah 10:28 
Send PM · Send email ·

StevenA 
Posted on Dec 4 2007, 05:24 PM


Forum countermafia Group: Members Posts: 2630 Joined: 20February 06 Positive Feedback: 51.85% Feedback Score: 70 
Your mistake is at step 4. s=9/10 ss/10=9/109/100=91/100!=9/10 s=9/10+9/100 ss/10=(9/10+9/100)(9/100+9/1000)=901/1000!=9/10 s=9/10+9/100+9/1000 ss/10=(9/10+9/100+9/1000)(9/100+9/1000+9/10000)=9001/10000!=9/10 ... s=9/10+9/100+9/1000+... ss/10=(9/10+9/100+9/1000+...)(9/100+9/1000+9/10000+...)=9...1/1...!=9/10 The geometric series was created by using geometric approximations and wasn't derived from logical derivations of mathematical identities. In other words, the "proof" for this summation relies upon "eyeballing" a geometric construction and assuming it entirely fills some volume, but there's no proof that this is true and mathematically we find it's only an approximation with a continual infinitesimal error. This post has been edited by StevenA on Dec 4 2007, 05:28 PM 

Send PM · Send email ·

NoCleverName 
Posted on Dec 4 2007, 05:31 PM


Advanced Member Group: Power Member Posts: 2641 Joined: 21November 06 Positive Feedback: 86.67% Feedback Score: 90 
Laughingly wrong. 

Send PM · Send email ·

NeoDevin 
Posted on Dec 4 2007, 05:32 PM


Advanced Member Group: Members Posts: 456 Joined: 9October 07 Positive Feedback: 90% Feedback Score: 22 
There's no mistake in step 4 of his proof, the geometric series is a well defined mathematical identity. Just because you don't know how to sum an infinite series, doesn't mean it's not defined mathematically. Care to try and find any errors in my proof? Or any real errors in prometheus's? 

Send PM · Send email ·

Precursor562 
Posted on Dec 4 2007, 09:25 PM


Advanced Member Group: Members Posts: 1333 Joined: 20June 06 Positive Feedback: 42.5% Feedback Score: 46 
OK... S(n) = .9 + .09 + .009 + .0009 +....
X would have to equal 1 or greater.
Which simply proposes that X is either the limit or is greater than the limit.
That goes against your proposition. Which you show with the next steps and state in step 12. All it does is define the limit.
That's the same as... let X = .9r 10X = 9.9r 10X  X = 9.9r  .9r 9X = 9 X = 1 Which is a complete farce. It treats .9r as equaling 1 before you start. That is because 1*10 and 1+9 will give you the same number. The same can not be said with .9r. If it were the case then that would mean .9r * 2 should equal 2. 1) .9r = 9/10 + 9/100 + 9/1000 + .... 2) .9r * 2 = 18/10 + 18/100 + 18/1000 + .... Take the partial sums of 1 then 2 .9, .99, .999, .9999, etc. 1.8, 1.98, 1.998, 1.9998, etc. See the difference? Or how about.... .99 * 10 = 9.9 .99 + 9 = 9.99 Does 9.9 = 9.99? .999 * 10 = 9.99 .999 + 9 = 9.999 Does 9.99 = 9.999? 1.0 * 10 = 10 1.0 + 9 = 10.0 Does 10 = 10.0? 1.00 * 10 = 10.0 1.00 + 9 = 10.00 Does 10.0 = 10.00? So can you see the difference? It's comes down to the difference of infinities and infinity being a concept and not a quantity of digits found after the decimal.  Time is the wisest counsellor


Send PM · Send email ·

NeoDevin 
Posted on Dec 4 2007, 09:48 PM


Advanced Member Group: Members Posts: 456 Joined: 9October 07 Positive Feedback: 90% Feedback Score: 22 
Proof by contradiction in mathematics goes like this: 1. Assume the opposite of what you are trying to prove (in this case, I was trying to prove that if x > S(n) for all n, then lim S(n) <=x, so I assumed that x < lim S(n)) 2. Show that it leads to a contradiction. (steps 5 through 11) 3. Conclude that your assumption must be wrong, and therefore what you're trying to prove must be correct. In symbolic logic We try to prove a statement P Assume !P (not P) Show that !P => Q & !Q for some other statement Q. Therefore !P must be wrong, and P must be right. Aside from that the first part of the post was the general proof, and then the second part was where it applied to the problem.
You would argue that S(n) is greater than 0.9r for some n? If so, the find the n! I showed in step 15 that 0.9r is greater than S(n) for any n. You yourself claim that any number satisfying x > S(n) for all n would have to be equal to 1 or greater? So which is it? Is 0.9r equal to 1 or greater? As far as prometheus's proof, it's a perfectly valid proof, and you would understand that if you knew anything about infinite series. 

Send PM · Send email ·

Pages: (381) « First ... 274 275 [276] 277 278 ... Last » 
Add reply · Start new topic · Start new poll 