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15th November 2014, 06:24 AM  #41 
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It is well known that mathematicians avoid diagrams as rigorous mathematical proofs because of optical illusions that may involved with them.
Nevertheless, proof without words is accepted if it
Quote:
I claim that there are also symbolic illusions that can be discovered if both visual AND symbolic methods are combined in some mathematical work, for example: In this diagram (where visual AND symbolic methods are involved) 0.111..._{2} = 1 if only N observation of the realline is used. By using the same diagram (where visual AND symbolic methods are involved) 0.111..._{2} < 1 if also R observation of the realline is used. In this case 0.111..._{2} < 1 by exactly 0.000...1_{2}, where the value 0.000...1_{2} is rigorously analyzed as follows: The value 0.000...1_{2} acts differently than value 0, as follows: The ".000..." is used as a N size place value keeper that is inaccessible to "...1_{2}" that is at R size.  The following is an example of symbolic illusion (that is discovered by using also R observation of the realline) simply because n = 0 to ∞ is a handswaving that misses the fact that N<R (as rigorously given by the diagram above (where visual AND symbolic methods are involved)):
Quote:

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15th November 2014, 10:13 AM  #43 
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f(x) = 1 + 1/xTherefore: f > 1 > g for all x in (0, ∞)In your notation, there must be a value L such that: f  L = 1 = g + L as x > ∞But notice that: f = 1 + L as x > ∞The difference separating between f and g is exactly 0 as x > ∞. Therefore: f = 1 = g as x > ∞QED  Let's carry this a little further. Note that we can derive f' and g' in terms of f and g as follows: f'(x) = f(x)  L = 1 + 1/x  LObserve that: f'(1) > f'(2) > f'(3) > . . . > f'(x) > 1 for all x in (0, ∞)Likewise: g'(1) < g'(2) < g'(3) < . . . < g'(x) < 1 for all x in (0, ∞)Because f' > 1 and 1 > g', the following inequality must hold: f' > 1 > g' for all x in (0, ∞)In your notation, there must be a value M such that: f'  M = 1 = g' + M as x > ∞Everything looks good here ^_^ From this, we can infer: f  L = f'  MAnd with a little algebra, we can write: f  f' = L + MBut remember, f' was defined as f  L, and g' was defined as g + L. Let's rewrite the expressions above in terms of f and g: f  (f  L) = L + MSimplifying: f  (f  L) = L + MSo, M = 0 and (L  L) = M implies L = 0. Because L = M = 0, then f(x) = f'(x) and g(x) = g'(x) as x > ∞QED.  You can generalize this further for f''(x) = f'(x)  M, f'''(x) = f''(x)  N, . . . , and so on to show that L = M = N = ... = 0. We assume that 0.000...1 must be some C in an expression f^{a}(x) = f^{a  1}(x)  C for any a > 1, so C = L = M = N = .... = 0. Therefore, there exists is no C = 0.000...1 such that C > 0. QED. 
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15th November 2014, 12:08 PM  #44 
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Awesome!
Edit: Dessi, please be persistent... I am nowhere near as rapierlike in mathematical argumentation, I am just muleheaded persistent when it comes to Doronetics. I am so going to enjoy this thread (together with people like jsfisher we are in this thread like over 7 or 8 years now...) 
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15th November 2014, 12:12 PM  #45 
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Dear Dessi,
All your poofs are based on x > ∞. We know that there are infinitely many levels of infinity, for example: N < P(N) < P(P(N)) < ... etc. ad infinitum. So when you are using ∞, what level of infinity you are using? Again, 0.999..._{10} = 1 if we observe the realline from level N. This is not the case if we observe the realline from level R. In this case 0.999..._{10} is the result of parallelsummation of countable N finite values, where this result is inaccessible to 1 exactly by the value 0.000...1_{10}, such that ".000..." is used as a countable N level place value keeper that is inaccessible to "...1_{10}" that is at uncountable R level. As long as you are not rigorously define ∞, x > ∞ has nothing to do with the fact that there are infinitely many levels of infinity, and my argument is based on this fact. So if you really wish to write any meaningful reply about my argument, you have to do that by explicitly not ignore the fact that there are infinitely many levels of infinity. Can you demonstrate how your proofs hold also if you do not ignore the fact that there are infinitely many levels of infinity? 
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15th November 2014, 12:18 PM  #46 
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15th November 2014, 12:20 PM  #47 
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15th November 2014, 01:52 PM  #48 
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I promise you, I am far from a math noob, but I'm afraid the statement above is unreadable. I don't understand what you are describing.
Its not clear to me how the expression 0.000...1 > 0 can be constructed at all. Could you provide a function which computes C = 0.000...1, then a provide a proof that C > 0.
Quote:
f(x) = 1 + 1/x as x > ∞Let's annotate this graph a little: Note that x > ∞ is the same as saying c > 0, so: f(x) = 1 + 1/x as c > 0 = g(x) 1  1/x as c > 0So, for every M > 0, there exists a number c such that 2c = f(M)  g(M). A c = 0 must exist because: If M > 0, then c = 0. Therefore, existence of a limit L at c > 0 implies the existence of L at x > ∞. The rigorousness of f(x) as x > ∞ is proven. QED
Quote:
Let N be a superset of real numbers R Let f be the injective function from R to N Let P be a proof function on the set of real numbers R Let P' be the proof P(f(R)) Therfore, for every P(R) there exists a P'(N) Which is a very pretentious way of saying my proof in the set of real numbers implies existence of a proof in any superset of real numbers. In other words, the expression C = 0.000...1 where C > 0 cannot be constructed any superset of real numbers, for any N > R, where s refers to the cardinality of the set s. QED. Apologies if I've misunderstood what you are trying to communicate. 
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15th November 2014, 02:58 PM  #49 
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Again, the following quote is true because N<R is true:
Quote:
Dear Dessi, For example: C = 0.000...1_{10} is constructable only if you observe the realline from Runcountable level. By doing this the place value example 0.999..._{10} (which is at most at N countable level of the realline) is a value of its own along the real that is < 1 exactly by C. 
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15th November 2014, 03:20 PM  #51 
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It is constructable only if you observe the realline from R level, and as a result, for example, 0.999..._{10} (which is a value of its own at N level, if it is observed from R level) < value 1, exactly by 0.000...1_{10}
Please do not ignore anything of post http://www.internationalskeptics.com...5&postcount=41 if you really wish to know my theorem. 
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15th November 2014, 03:29 PM  #52 
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The highlighted statement clarifies nothing. I have no idea what you are saying. You are obviously talking about some mathematical abstraction, but I'm afraid I can't guess that abstraction is or how its derived.
Could you tell me what you think the real line is, what R means, what the expression "observe the real line from R level" refers to? After that, can you show how these things allow you to write a function which computes 0.000...1 > 0? 
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15th November 2014, 03:35 PM  #53 
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You never heard of that expression before because it is (probably) novel thing about the realline, that was not used (yet) by traditional mathematicians, even if N<R is an established mathematical fact.
Did you read all of http://www.internationalskeptics.com...7&postcount=40 before you replied? 
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15th November 2014, 03:47 PM  #54 
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15th November 2014, 03:52 PM  #55 
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Me: Could you tell me what you think the real line is
You: The real line is the set of real numbers Me: I understand this. Me: what does R mean? You: R is the cardinality of that set of real numbers Me: I understand this. Me: what does the expression "observe the real line from R level" refer to? You: Me: Me: how does one construct the expression 0.000...1 > 0 at all? You: It is constructable only if you observe the realline from R level Me: Whatever manner one "observes the real line from R level" is most certainly not welldefined mathematical knowledge. 
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15th November 2014, 03:52 PM  #56 
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Please read very carefully the first part of http://www.internationalskeptics.com...0&postcount=49.

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15th November 2014, 03:56 PM  #57 
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Already given in http://www.internationalskeptics.com...postcount=4282.
Only if N<R is also not welldefined mathematical knowledge. 
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15th November 2014, 03:59 PM  #58 
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15th November 2014, 04:07 PM  #60 
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15th November 2014, 04:13 PM  #62 
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N is the cardinality of natural numbers and R is the cardinality of the real numbers, and I use the fact that N<R, or in other words N and R are two different levels of infinity that can be used in order to observe the realline, where N is a countable observation and R is an uncountable observation.

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15th November 2014, 04:14 PM  #63 
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15th November 2014, 04:16 PM  #64 
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15th November 2014, 04:30 PM  #65 
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Ok, N is the cardinality of natural numbers, R is the cardinality of reals. That is exactly the information I wanted.
Yes, N < R, but the language "two levels of infinity" is very imprecise. It is more precise to say that N is enumerable, and R is nonenumerable, meaning: Natural numbers are the set {1, 2, 3, 4, 5, . . . N_{k2}, N_{k1}, N_{k} . . . }, this set is enumerable. In other words, it is possible to name every number in this set, given an infinite amount of time. Rational numbers are the set of all numbers naturals numbers a and b of the form a/b: { { N_{1}/N_{1}, N_{1}/N_{2}, N_{1}/N_{3}, . . . } ∪ { N_{2}/N_{1}, N_{2}/N_{2}, N_{2}/N_{3}, . . . } ∪ { N_{3}/N_{1}, N_{3}/N_{2}, N_{3}/N_{3}, . . . } ∪ . . . } This set is also enumerable. It is also clearly the cartesian product of N, so the cardinality of rationals > cardinality of naturals. Real numbers are not enumerable, which is proved by Cantor's diagonal argument. To put this another way, N ⊆ R or ∀x{x∈N → x∈R}. Formality out of the way, I want to comment on:
Quote:

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15th November 2014, 04:36 PM  #66 
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Counter example: the cardinality of {1, 2, 3} is less than naturals.
Counter example: the cardinality of the countable set of even naturals < the cardinality of naturals. Counter example: the cardinality of the countable set of primes < the cardinality of naturals. Counter example: the cardinality of the countable set of rational numbers > the cardinality of naturals, see theorem Q. 
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15th November 2014, 05:20 PM  #67 
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doronshadmi, thank you for responding to my posts, they provide much needed background to make sense of your diagram.
I'm still struggling with this part:
Quote:
I understand that "proof without words" does not need accompanying text, but could you humor me here and provide an explanation anyway? Can you explain in more detail how one derives 0.111... from the blue and green lines? 
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15th November 2014, 08:07 PM  #68 
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15th November 2014, 09:44 PM  #70 
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"Infinity" is one of those things were intuition betrays you. For example, the even integers and the odd integers are both half the size of the integers, right? Or so it would seem.
For cardinality, the test for equality is can you put the members of the two sets into onetoone correspondence. For the evens vs. integers, there is a simple correspondence between any integer, n, and any even, e: e = 2n. The rational numbers are a bit less obvious, but the link you cited and its Theorem Q provides a correspondence method. The primes are even less so, but it is straightforward to show there is no largest prime, and from that we can conclude the set of primes is infinitely large. The primes being a subset of the integers assures us the primes are merely countably infinite. 
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16th November 2014, 12:55 AM  #71 
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16th November 2014, 04:02 AM  #72 
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17th November 2014, 02:18 AM  #73 
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Ok, let's address the ability to get a given value by using parallelsummation of values.
parallelsummation is a given value, which is the result of at lest two added values, such that the addition is done in parallel (no "process" of oneafterone steps (no serial steps) are involved). It means that only one step is used among more than one added values. The investigated object is [0,1]. Let's investigate some finite example of two added values: 0+1 is some example of how to get value 1 by parallelsummation among finite number of values (countably finite cardinality 2). 0.9_{10}+0.09_{10}+0.009_{10}+0.001_{10} is some example of how to get value 1 by parallelsummation among finite number of values (countably finite cardinality 4). etc. ...  Now let's investigate [0,1] by using countably infinite cardinality N. Let's use, for example, 0.9_{10}+0.09_{10}+0.009_{10}+... that is a parallelsummation among infinite number of values (countably infinite cardinality N). It is obvious that no parallelsummation of countably finite cardinality n (where n is some natural number > 1) among 0.9_{10}+0.09_{10}+0.009_{10}+... gets value 1. Actually, we are using here (at the level of countably infinite cardinality N) the fact that N+1=N in order to conclude that 0.9_{10}+0.09_{10}+0.009_{10}+...=1  Now let's investigate [0,1] by using uncountably infinite cardinality R. Again, let's use, for example, 0.9_{10}+0.09_{10}+0.009_{10}... that is a parallelsummation among infinite number of values (countably infinite cardinality N). It is obvious that no parallelsummation of countably infinite cardinality N among 0.9_{10}+0.09_{10}+0.009_{10}... gets value 1 simply because of the fact that N+1=N<R=P(N).  So, the general conclusion is as follows: We get value 1 by parallelsummation only if we observe the values by some particular cardinality. If a greater cardinality is also used, value 1 is inaccessible to the parallelsummations that are based on smaller cardinalities.  Let's transcend the realline as follows: BY using the tower of powers line P(P(N)) < P(P(P(N))) < P(P(P(P(N)))) < ... there are obviously the following values: Between any two irrational values along the tower of powers line, there are P(P(N)) values. Between any two P(P(N)) values along the tower of powers line, there are P(P(P(N))) values. Between any two P(P(P(N))) values along the tower of powers line, there are P(P(P(P(N)))) values. ... etc. ... ad infinitum, where the inaccessible limit of the tower of powers line is simply the noncomposed 1dimesional space (which is not necessarily a metric space).  My theorem is false only if n>1 < N < P(N) < P(P(N)) < P(P(P(N))) < P(P(P(P(N)))) < ... is false. 
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17th November 2014, 09:13 AM  #74 
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Thank you for replying to my request ^_^
So, parallelsummation is... addition? But different because it adds an entire sequence in one operation, without intermediate steps? Short of representing a sequence of additions using some kind of equivalent identity (for example, and identity using multiplication ×, summation Σ, product Π, integral, etc), I don't really think a "parallelsummation" operator exists, intermediate additions are unavoidable: "a_{1} + a_{2} + a_{3} + . . . + a_{n}" involves at least n1 intermediate steps: ((((a_{1} + a_{2}) + a_{3}) + a_{3}) + . . . a_{n}) Even if not shown explicitly in a proof, intermediate steps  some kind of process which perform the additions  is implied. If the basis of your proof depends on the existence of a parallelsummation operator, how is it defined mathematically? I can't analyze your proof beyond this point without a meaningful description of a parallelsummation.
Quote:
For a start, the statement that you say is "obvious" is absolutely not "obvious" in any sense. Your statement requires justification, you must define your parallelsummation operator and prove that it has the property you say it has. I cannot infer your property from the diagram you posted earlier. Please be as explicit as you can by writing out your proof for the highlighted statement, stepbystep. 
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17th November 2014, 09:52 AM  #75 
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Doron is trying to resolve a criticism that his concepts rely on some form of stepbystep process (impossible processes at that) rather than being simply a result. An infinite series is not determined by an infinite sequence of additions, although Doron has in the past expressed that view. He is now trying to reduce the infinitely many additions down to just one by handwaving "parallel summation" into existence. It is still a process, albeit only one step, if he could ever formally define parallel summation, which he cannot.
Also, as you so correctly point out, the full set of intermediate additions is necessary, if for no other reason than the order of operations can be important. Consider the series: 1  1 + 1  1 + 1  1 + ... This is a nonconvergent series since the partial summations oscillate between 1 and 0. It does not have a limit. However, if I change the order of operations a bit to be: (1  1) + (1  1) + (1  1) + ... = 0 + 0 + 0 ... That would be 0, right? So the original series has a value of 0, right? Then again: 1 + (1 + 1) + (1 + 1) + (1 + 1) ... = 1 + 0 + 0 + 0 + ... And that would be 1. The point of all this being that order of operations matters. Doron's "parallel summation" is completely bogus (for this and other reasons, too).
Quote:

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17th November 2014, 10:45 AM  #76 
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Doron, before you comment on my lack of visual/spacial thinking, I want to show you a subset of my Amazon order history:
http://i346.photobucket.com/albums/p...ps8b63eee9.png I encourage you to enjoy some posts I've written on computer science, be amused by my implementation of the cons/car/cdr combinators in Ruby, be more amused by my AVL tree implementation using only lambda combinators, and enjoy my book on functional programming. Trust me, I am a goddamn nerd of the highest order. I'm confident my mathematical, visual, and spatial reasoning are more than adequate to analyze a proof that 0.999... < 1 as long as you provide the details. 
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17th November 2014, 10:51 AM  #77 
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17th November 2014, 11:35 AM  #78 
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The number of arranged levels that are involved have no impact on the result of a given parallelsummation because parallelsummation is done in one step on all arranged levels (again, no stepbystep is involved with parallelsummation).
Actually, for example, a value like 0.009 is also a parallelsummation among several arranged levels, for example: 0 (at the level of whole numbers) + (level of .1)*0 + (level of .01)*0 + (level of .001)*9 
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17th November 2014, 12:07 PM  #79 
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It at least an addition among two values (including negative values), and yes, parallelsummation is done on all arranged levels in one step.
If you wish to understand better the notion of parallelsummation that is not influenced by arranged levels, please think in terms of the cardinality of, for example: 
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17th November 2014, 12:20 PM  #80 
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Your proof relies on the existence of a "parallelsummation" operator which lacks a proper definition. I can't infer the correctness of your proof from undefined operators.
Would you mind replying to the rest of my post? 
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