Пікірлер

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  @kephalopod305421 күн бұрын

  I always felt that CDA was iffy. For the new sequence created from the diagonal, we could always look further down the list for the first sequence that matches the first n digits, and I don't see why there would be any n where we couln't find a match.

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  @karenhartl21152 ай бұрын

  Interesting

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  @_strangelet__3 ай бұрын

  Thank you so much sir! ❤

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  @mohib50803 ай бұрын

  Sir thank you so much I really appreciate this

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  @colinjava84474 ай бұрын

  Your board says the "digits are finite or countably infinite" This is true. Then it says "The digits are also a natural number" Which would imply when x = 0.3333.... that 333.... is a natural number. But you recently said: "You are correct. 333… is not a natural number. It designates a hypothetical unique limit of natural numbers." So you were lying then or lying now, which is it? You won't respond, but this will help other readers to see what garbage you are spewing.

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  @wernerhartl20694 ай бұрын

  The infinite digital sequence 333…. Is a natural number for any n digits. The complete infinite sequence is not a natural number.

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  @colinjava84474 ай бұрын

  In fact, a simple way to show you are wrong about infinite length natural numbers. Starting at 1, can you count to 333... or ...333 (assuming you won't die or get bored)? The answer is no, since at each step the number you are on will always be finite in length, and so will the next number. So how is ...3333 or 3333... a counting number if you can't count to it? So its not, and you are taking absolute bollox as usual.

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  @wernerhartl20694 ай бұрын

  That’s why it’s called countably infinite.

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 So starting at 1 can you count to ...333 or 333... (with same assumptions) ? Yes or no.

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  @wernerhartl20694 ай бұрын

  @@colinjava8447 Yes. I can count to 333….3_n for all n because the number of digits is countably infinite. en.m.wikipedia.org/wiki/Countable_set

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 What on earth is 333...3_n ? Do you mean (10^n - 1)/3 ? Cause that's trivial. That wasn't what I asked was it, I said can you count to 333... or ...333 The ... means it goes on forever. Please try again.

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  @wernerhartl20694 ай бұрын

  @@colinjava8447 333…3_ n n 3 digits, which you can count to for all n. For ex, if n=1, count one, two, three. If n=2, count one, two, three, …thirty-three.

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  @colinjava84474 ай бұрын

  So you can't provide a link to a single document saying natural numbers can have infinitely many digits? Just say if you can't do it.

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  @wernerhartl20694 ай бұрын

  Does the set of natural numbers have a largest member? Yes or no.

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 No. So you can't provide documentation I see. Is 3333.... infinite or finite? Is 3333.... infinite or finite in its number of digits?

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  @wernerhartl20694 ай бұрын

  @@colinjava8447 333…. Is either infinite or finite. It is not finite. Therefore it is (countably) infinite, ie, doesn’t end.

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 You can't seem to distinguish value from length even though I asked specifically. So show me a document that says natural numbers can be infinite in value/length. What is the problem? Question, let F = subset of natural numbers that have finitely many digits. Is F a finite or infinite set?

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  @wernerhartl20694 ай бұрын

  @@colinjava8447 .333….. represents a countably infinite (unending!) string of 3’s. therefore its value doesn’t exist. Its limit exists.

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  @colinjava84474 ай бұрын

  How is it you can understand vectors but not natural numbers, sets, infinities, proof by contradiction, induction, rationals, irrationals, fractions? I'd have to go through this in detail though to see if it was wrong.

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  @wernerhartl20694 ай бұрын

  Thanks for watching. Glad to answer any questions.

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 Still waiting to see a single document from this planet showing natural numbers that are infinitely long. Can you provide a link?

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  @colinjava84474 ай бұрын

  @@wernerhartl2069 So you can't provide a link to a single document saying natural numbers can have infinitely many digits? Just say if you can't do it. So not glad to answer this question then, but you said any questions, so more lying.

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  @wernerhartl20694 ай бұрын

  @@colinjava8447 Off topic. Raise your “question” here. kzread.info/dash/bejne/Y2dp1MaSZ82ueaQ.htmlsi=YXKFWRCrly0kK5_i

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  @colinjava84474 ай бұрын

  ​@@wernerhartl2069Topic raised but you ignored it. If you can't give evidence of infinitely long natural numbers just say, why so afraid?

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  @rineric32144 ай бұрын

  The aether was retained, thanks to Sagnac's effect calculation. There goes relativity!

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  @wernerhartl20694 ай бұрын

  Actually, it says the speed of light is constant wrt source. Which also explains the Michelson Morley experiment. Relativity says the speed of light is independent of the source.

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  @rineric32144 ай бұрын

  @@wernerhartl2069What does "wrt" mean?

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  @wernerhartl20694 ай бұрын

  @@rineric3214 wrt: with respect to. This is the emission or ballistic theory of light, still alive outside the mainstream. Einstein postulated that the speed of light was independent of the source. A correct Sagnac calculation (this video) confirms the ballistic emission theory as does the Michelson-Morley experiment.

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  @rineric32144 ай бұрын

  @@wernerhartl2069Thank you for your PROMPT reply! I am unfamiliar with "ballistic emission theory". Is that an attempt to believe the Double Slit Experiment? Light is not a bullet. Ballistic Emission theory is not trying to say that when you turn the flashlight on, the light beam takes awhile to reach the double slit is it? It would not do that. The beam would immediately extend the full distance. Light is not a bullet traveling toward a target. I was more focused on the retention of the ether (aether).

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  @wernerhartl20694 ай бұрын

  @@rineric3214 A bullet is emitted from a gun with the same speed relative to the gun independent of the speed of the gun. For example: A bullet is fired from a train moving through a station. Speed relative to gun is 1000m/s and the train is traveling 200m/s. Relativity says an observer standing on the station platform would measure the speed of the bullet as 1000m/s. Ballistic (emission) theory says he would measure the speed as 1200m/s.

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  @livdamnit69985 ай бұрын

  That was cool but can you show me how to measure so that it fits my art? It seems so easy but it is not. I'm using the same corner molding you are but when i measure i can't figure out how to do it very accurately and how to line it up with the saw. I don't know how to explain. I do a lot of measuring and cutting flat things but this is dimensional and i need to be so precise.

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  @wernerhartl20695 ай бұрын

  For 1/2” border on floating frame, add 1” to frame size to get corner to corner size. Important thing is that opposite pieces are exactly the same length. That’s why a stop helps (for hand saw, don’t use stop on power saw).

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  @madmathematician44585 ай бұрын

  Sir, i would very much like to discuss your work with you. I am also working on the different sizes of infinite cardnality argument and I like your work but I've been tackling the argument from a different perspective proving that the really is no such thing as a real number and that these supposed "real numbers" have an infinite cardnality which makes the argument an oxymoron. Id very much like to discuss my work with you since you have a strong interest in the subject.

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  @wernerhartl20695 ай бұрын

  Yes, real numbers don’t exist anymore than points on a line, which have zero size. If I have a dot in a plane of finite dia, and shrink it to zero dia, does the dot still exist? And yet a line ends at a point on the line, or rather a location. Hmm. So a point on the line is really a location? On the other hand, if I can write .333…. and define mathematical operations on (countably) infinite series, they exist as mathematically defined objects. Anyhow, “degrees of infinity” really don’t interest me. I just wanted to show CDA was wrong. I only believe countable infinity and induction, the real numbers (as limits of countably infinite series) are countable, and CDA is wrong.

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  @wernerhartl20695 ай бұрын

  By the way, my argument on Countability of countably infinite decimal numbers in [0,1) was based on the Countability of all n-place digits for all n, and including limits at most doubles the count.

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  @madmathematician44585 ай бұрын

  @@wernerhartl2069 Even if your not interested in sizes of infinity your countability argument still applies to degrees of infinity. I'd very much like to share my work with you which proves that real numbers don't exist. I have different standards for going public with my work & would rather not deal with the ridicule that you are enduring. I will only publish my work if it's within the proofing standards which are only taught at the PhD level unfortunately. Even the best mathematician Ramanujan needed Hardy to proof his work. I hate that the skills for formal mathematical proofing are confined to PhD programs in which very few can afford.

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  @wernerhartl20695 ай бұрын

  @@madmathematician4458 Are terms precisely defined? Does it make sense? Don’t need a PhD for that. Who cares about ridicule and insults unless you are a Professor with a job to Iose. Do a video. Haven’t yet seen anyone who defines “ infinity” when giving CDA. First take care that you have a source of income.

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  @madmathematician44585 ай бұрын

  @@wernerhartl2069I care about proofing my work within the rigorous standards but I actually want my work to be published and reviewed by the best mathematicians in the world. I have actually had a conversion with Kip Thorne regarding another mathematical proofing that I submitted for publication and I understand the requirements that these elite individuals seek. I don't care about convicing your average KZreadr that my math work is correct. If you want your work reviewed at a college level you should be thinking the same thing. Now to address your terms, no they are not adequately defined & that's one aspect that I think I can improve upon your work. Mathematicians can often benefit from each other's peer reviews and maybe you can help me with my work also. Such terms like ciuntabke infinity and uncountable infinity are not only poorly defined but also complete nonsense. If you don't want my help or at least review my work that's fine, but finding mathematicians that want to dispute Cantor's diagonal are rare.

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  @seneca9836 ай бұрын

  Description: "n digits give rise to 2^n sequences for all n. If infinity means greater than countable infinity, CDA fails because you are assuming what you are trying to prove." But it doesn't assume what one is trying to prove. 2^n > n for both all natural numbers and also for all cardinal numbers n, including infinite ones.

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  @seneca9836 ай бұрын

  10:15 Not all proofs of uncountability use the diagonal argument. Cantor's original proof that the set of reals is uncountable used a different argument which relies on the completeness of reals. It goes as follows. Take an interval [x, y] on the reals. We prove that it's uncountable (and therefore the full set of reals is too) by getting to a contradiction by assuming it's not. Assume there's a sequence (a0, a1, a2...) containing all reals on the interval. Now let's construct two other sequences (b0, b1, b2...) and (c0, c1, c2...) from it. We choose them such that b0=a0, c0 is the first entry in the sequence a such that c0>b0, a1 is the first entry such that a0<a1<b0, b1 is the first entry such that a1<b1<b0, etc. The idea is that we always pick numbers that are between the numbers in the sequences so far. Now sequence b is strictly increasing and c is strictly decreasing and it's also true that all entries in the sequence b are less than those in the sequence c and both sequences are also bounded. Now take the supremum of the sequence b which must exist in [x,y] as per the completeness axiom of real numbers and call that. Is that number somewhere in the original sequence a? It is not. If it were, it would be either in the sequence a or b because it's necessarily be between the numbers chosen so far. After that any entry in the sequence c would have to be smaller than this number which would make it smaller than than some numbers in the sequence b because we picked the supremum of b. This contradicts our original assumptions because we chose the sequences b and c such that all entries in b are smaller than those in c. Thus we have a contradiction and the interval [x,y] (and thus also reals) is uncountable.

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  @wernerhartl20696 ай бұрын

  The set of all irrational fractions is countable. Therefore so are their limits. Pi is not an irrational fraction, it is the limit of an irrational fraction. The limits “plug up the holes.”

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  @seneca9836 ай бұрын

  @@wernerhartl2069 I'm not sure what you mean by "irrational fractions" but the set of reals is not countable. I just gave a proof, or at least a rough outline thereof (which is different from the diagonalization one), that reals are not countable because of their completeness property.

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  @wernerhartl20696 ай бұрын

  @@seneca983 I can’t follow your proof. Where is Cantor’s proof you refer to?

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  @seneca9836 ай бұрын

  @@wernerhartl2069 I'll give you a link in a separate comment. Hopefully KZread won't remove it.

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  @seneca9836 ай бұрын

  @@wernerhartl2069 Here's a link to the proof. This isn't Cantor's original text but it's basically the same proof so it shouldn't matter. personal.math.ubc.ca/~PLP/book/sec_cantor_first.html

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  @jeffjo87327 ай бұрын

  Yet another Math lesson for Werner, in the hope that he will learn some of what he gets wrong here. 1) A _bijection_ between two sets *_A_* and *_B_* is a matching relationship that pairs each member of either set with exactly one member in the other. 1A) If you can make a bijection between finite sets *_A_* and *_B_* then any attempt to match every member of either (let's say *_A_* for example) with exactly one member of the other will result in a bijection. 1B) A good example of how this works is a square table. Its length and width are equal (L=W), meaning, and the only way to number the rows and columns consecutively from 1 will always end at L=W. 2) If *_A_* can be put in a bijection with both *_B_* and *_C,_* then these two bijections can be used to construct a bijection between *_B_* and *_C._* 3) A set is countably infinite if it can be put into a bijection with the set infinite *_N_* = {1,2,3,...}. (I'll call this set the natural numbers. Some systems say that 0 is a natural number, but that system can be confusing to work with, and I don't want to give Werner any way to find more things to deliberately misunderstand.) 3A) Any two sets that are countably infinite can be put into a bijection with each other 3B) But (1A) does not hold for countably sets. You can pair the even natural numbers and the odd naturals numbers by matching 1 with 2, 3 with 4, 5, with 6, etc. Or you can make a non-bijection by matching 1 with 2, 3 with 2, 5 with 4, 7 with 4, 9 with 6, etc. This will map every member of both sets, but two evens are associated with every odd. 3C) This is why you can't call the "sizes" of a countably infinite array its "length" and "width." You can always create a diagonal that crosses every row and column, but you can also create a diagonal that misses some. 3D) This bears repeating: YOU CAN ALWAYS CREATE A DIAGONAL THAT CROSSES EVERY ROW. What Werner calls Cantor's Diagonal Construction (CDC) starts with countably infinite list of countably infinite strings. Werner and Wikipedia call the strings s1, s2, s3, ..., indicating a known bijection with *_N._* Each string is a countably infinite list of 0s and 1s that, by (3A), can be put into a bijection with each other. This means that we can use the members of *_N_* as an index to find every string sn in this list, and by (2) as an index to a character position. What CDC does is create a new string s0 that, by definition, differs from every string sn in Werner's list in character position n. There is nothing about this that "fails." Many people learn Cantor's Diagonal Argument (CDA) incorrectly They think that the list s1, s2, s3, ... is supposed to include every possible such string. This in incorrect. It is supposed to include a countably infinite number of strings. The fact that any countably infinite list must be missing at least one is what proves that a complete list is impossible.

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  @wernerhartl20697 ай бұрын

  You can’t do arithmetic with countably infinite. It is not a number. You can only do induction. If I have two sets that each contain ten elements, can I establish a bijection? Yes. If I have two sets that each contain countably infinite elements, can I establish a bijection? No.

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  @jeffjo87327 ай бұрын

  @@wernerhartl2069 "You can’t do arithmetic with countably infinite. It is not a number." BZZZT. Wrong again. 1) While not every system of arithmetic does, some do define arithmetic with transfinite numbers. Specifically, set theory does. Addition is equivalent to the union of two disjoint sets. So Alpeh0+(Finite)=Aleph0+Aleph0=Aleph0. So your claim here is wrong. 2) I don't use transfinite numbers in any of this. I use countably infinite sets, but all the numbers I use are finite. So your claim, even if it were right, is vacuous. 3) With one trivial exception, I don't do arithmetic in any of this. That exception is that when inverting a string, I replace each character X with 1-X. I could just as easily say "swap 0s and 1s." Again, your claim is vacuous. "If I have two sets that each contain countably infinite elements, can I establish a bijection? No." Since a countably infinite set is defined to be one that can be put into a bijection with the set of all natural numbers *_N,_* this is blatantly false. (And before you make a fool of yourself again, any set can be put into a bijection with itself, which is why *_N_* is countably infinite.) Please, next time try to understand what is being said before you throw words you do not understand into a nonsensical sentence that you think disproves something.

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  @servicebavaria75497 ай бұрын

  Excellent.

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  @wernerhartl20697 ай бұрын

  Bottom line: Cantor’s Diagonal Construction is meaningless because infinity (countable or uncountable) is not a positive integer.

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  @jeffjo87327 ай бұрын

  Werner's objection is meaningless because CDA (what he calls CDC) never uses the word "infinity," nor anything like the number he thinks he means. CDA proves that if you think you have a function that maps *_N,_* the set of of natural numbers, to the set that Cantor called *_M_* or the equivalent set that Wikipedia calls *_T,_* then there is an element of that other set that is not mapped by this function. It follows immediately that you can't have a function that maps to every member of that other set. Otherwise, the element that CDA finds both is, and is not, in that set.

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  @wernerhartl20697 ай бұрын

  @@jeffjo8732 if I have two countably infinite series of elements, which one is longer?

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  @wernerhartl20697 ай бұрын

  A set of 2^n elements is larger than the set {1,2,3. n} for all n. Yet both are countably infinite. That’s the trick underlying CDA.

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  @jeffjo87327 ай бұрын

  @@wernerhartl2069 Um, no. A set of 2^n elements has, well, 2^n elements. This is a finite number. Even you should understand that. If we let n increase without bound; that is, consider the similarly constructed infinite set? Then the form that, when finite, had 2^n elements is larger. That's essentially what Cantor's Theorem says, and CDA is a specific example of it. A _finite_ set of 2*n elements, such as {1, 3/2, 2, 5/2, ..., n, (2n+1)/2} is larger than a _finite_ set of n elements, such as {1, 2, 3, ... , n}. Yet if we do the same, {1, 3/2, 2, 5/2, ..., n, (2n+1)/2, ...} can be put into a 1:1 correlation with the set of all natural number starting with 1; that is, {1, 2, 3, ..., n, ...}. This is easily seen by expressing each each member of the first set as (n+1)/2, where n is any member of the second. This is no "trick." It is a necessary property of infinite sets. In fact, one definition of an infinite set used in axiomatic systems, is a set that can be put into a bijection with a strict subset of itself.

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  @jeffjo87327 ай бұрын

  @@wernerhartl2069 By definition, both have cardinality Aleph0. "Longer" isn't the right word, because you can also form injections that are not surjections, and surjections that are not injections. But if you want to know whether you can form the diagonal when the rows and columns are countably infinite, the answer is "yes." Since your claim in the video is essentially that you can't form a diagonal, your video is wrong.

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  @jeffjo87327 ай бұрын

  1) Axiom, of Infinity: There is a set *_N_* that (A) contains the natural number 1 and, (B) if it contains the natural number n, it also contains the natural number n+1. #1B means both that there is no "last number" in *_N,_* and also no "last element" in any list enumerated by *_N._* This means the list is infinite. It also means that the cardinal number representing the cardinality of *_N_* is (C) does not correspond to any natural number and (D) is infinite. 2) A Cantor String is any function s(n) that maps *_N_* to the set {0,1}. Actually, Cantor used the est {'m','w'}, but the what the two elements in the set are makes no practical difference. 3) Assume that S(n) is any list of Cantor Strings {s1(n), s2(n), s3(n),...}. Note that this is an infinite list, so it has no "last element." The one error Cantor made, was that he needed to demonstrate that such lists exist. But examples are trivial. One defines S(n)(m) - that is, the nth string's mth number from {0.1} - to be 0 unless n=m, in which case S(n)(n)-1. 4) The function s0(n) = 1-S(n)(n) is a Cantor String that is not in the list S(n). 5) Assume the list T(n) exists and includes all possible Cantor Strings. From that enumeration we can construct a Cantor String s0(n) that both is listed by T(n) (since it lists all) and is not listed in T(n) (from #4). This contradiction disproves the assumption that such a T(n) exists. Nothing that Werner has said invalidates this proof, which is the one Georg Cantor published.

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  @wernerhartl20697 ай бұрын

  Unintelligible. I assume you are trying to mimic Cantors published proof: www.logicmuseum.com/cantor/diagarg.htm which labels the strings in his set with subscripts, clearly denoting a countably infinite set for which CDA fails, and presumably that’s why it is impossible to find. If you Google “Cantors Diagonal Argument” you get the wiki version of the proof, which doesn’t subscript the strings.

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  @jeffjo87327 ай бұрын

  @@wernerhartl2069 "Unintelligible." Only if you don't understand High-School level math. That, or you don't want to face facts, so you choose to misunderstand it. "[Cantor] labels the strings in his set with subscripts, clearly denoting a countably infinite set." You probably missed the fact that this countably-infinite set of strings is not assumed to contain every such string. "For which CDA fails." Vacuous. CDA takes this set of strings, and defines a new string from it. And proves that this new string is not in the table. What do you think is the "failure" here? Do you think the new string is not properly defined? Do you think it is in the set that CDA operated on? "If you Google "Cantors Diagonal Argument' you get the wiki version of the proof, which doesn’t subscript the strings." You seem to forget that I am the one who pointed both sites out to you. And you don't seem to understand that Wikipedia's s1, s2, s3, ... are subscripted stgrings. This is why you are ridiculed. because you refuse to see what is plainly placed before you;.

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  @wernerhartl20697 ай бұрын

  From www.logicmuseum.com/cantor/diagarg.htm proposition: "If E1, E2, …, Ev, … is any simply infinite [einfach unendliche] series of elements of the manifold M, then there always exists an element E0­ of M, which cannot be connected with any element Ev." What is a “simply infinite” series of elements. He’s dealing with infinities as if they were numbers. Suppose M is {00,10,01,11}, what is a subset of M in Cantors sense? Obviously, if you take all subsets consisting of two elements and apply Cantor’s construction you will come up with an element not in the subset. For example, 1) 01 2) 11 CDA gives 10, which is not in this subset but in M. Obviously. Repeating the argument for “infinite”strings is gibberish. Suppose M consists of binary strings with number equivalents {1,2,3…} which is an “infinite” set which contains all possible infinite strings of {0,1}. In that case T is the set {1,2,3…} and CDA says there is an element in {1,2,3…} which is not in {1,2,3…}. On the other hand {2,4,6…..} is an “infinite” subset which has elements not in M. Of course. So what?

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  @wernerhartl20697 ай бұрын

  More colloquially. Theorem: The set T of all countably infinite strings of 0 and 1 is uncountable. Proof: Assume T is countably infinite. Then by Cantors Diagonal Construction (CDC) there is an element of T not in T. The contradiction implies T is not countable. Cantor’s Diagonal Construction (CDC) Consider any countably infinite subset of T. By CDC there is an element of the subset not in T. Failure Countably Infinite is not a number. The construction CDC is meaningless. Comment. The generalization of applying CDC to any countably infinite subset of T is an unnecessary complication serving only to confuse the proof. It is sufficient to show that, by CDC, T is countably infinite leads to a contradiction, hence T is not countable. But CDC is meaningless.

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  @wernerhartl20697 ай бұрын

  Infinity, countable or otherwise, is not a positive integer so Cantor’s Diagonal Construction is meaningless and CDA fails.

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  @servicebavaria75497 ай бұрын

  Werner, your many videos are inspirational. I share your views. Most physicists, who are at an age, when they are still employed, don't have time to question their textbooks. And most have neither guts nor wits, to rise against a complicated, mathematically involved scheme like SRT or GR anyhow. It is easiest to go with the flow, particularly in science. My respect to you.

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  @wernerhartl20697 ай бұрын

  Glad to hear from someone who shares my conclusions. Thanks, Service.

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  @user-qx1og1qn7i7 ай бұрын

  kzread.info/dash/bejne/mYd1zsN9gLnIk6g.htmlsi=DO4We_WzHdWBSHa7 New analysis of special relativity

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  @a.hardin6208 ай бұрын

  I think the major flaw(ignoring minor flaws) is at the end where he mixes sets and elements. The list going down has single elements. This is important: Each infinite binary string is a single element in the list. He switches suddenly and treats the 1 and zeroes of each infinite binary element as sets themselves and then “counts them” using that curving method treating the ones and zeroes as elements themselves. Also putting the new number in the list does nothing because you can just derive a new number not in that new list with the same diagonal argument.

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  @wernerhartl20698 ай бұрын

  So if you list all the countable numbers in the list there is a countable number not in the list.

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  @wernerhartl20698 ай бұрын

  Apply CDA to the set of all natural numbers in binary form. You find that there is a natural number not in the set of all natural numbers. CDA fails.

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  @wernerhartl20698 ай бұрын

  First things first. Define infinite.

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  @colinjava84477 ай бұрын

  @@wernerhartl2069 Numbers are not countable, sets are countable, but not the reals obviously.

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  @colinjava84477 ай бұрын

  @@wernerhartl2069 How can it fail if its not even designed to be used with natural numbers? Its designed to be used with real numbers. Again for the billionth time, you don't know what natural numbers are, that's why you think it fails. Lets assume your "natural numbers" that are infinitely long are used in the list, then you would get a "natural number" not in the list, which shows your set of "natural numbers" is not countable. But natural numbers are countable, proving your "natural numbers" are not actually natural numbers. Well done, you just proved yourself wrong about natural numbers.

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  @wernerhartl20699 ай бұрын

  The general case of an arbitrary ring or mirror configuration is given here: kzread.info/dash/bejne/eH-DxMyFcseoXdI.htmlsi=uVC-R7Hok0yRc0FO

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  @JNorth879 ай бұрын

  Absolutely fascinating insights on the Sagnac calculation and its implications for our understanding of the speed of light. The idea that the speed of light is constant with respect to the source, as opposed to a fixed inertial coordinate system, is truly groundbreaking. It's intriguing to see how the Sagnac effect demonstrates the rotation of the track of photons when observed from a rotating coordinate system. The formula EF = R Omega L / C for calculating the rotation angle of a circuit is a testament to the mathematical elegance of this phenomenon. Moreover, the revelation that there's a difference in distance traveled depending on the direction of rotation adds another layer of complexity to our understanding of Earth's motion. This challenges not only our traditional notions of time and space but also emphasizes the generality and applicability of the findings to any arbitrary rotating loop. My question is, how do these findings reconcile with the results of the Michelson-Morley experiment, and what implications might this have for our broader understanding of relativity?

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  @wernerhartl20699 ай бұрын

  Thanks. For over one hundred years the belief by most physicists has been that Mitchelson-Morley refutes the Aether theory and Sagnac confirms it, and the only resolution was relativity. But the ballistic theory explains both results. From en.m.wikipedia.org/wiki/Emission_theory_(relativity) “Emission theory, also called emitter theory or ballistic theory of light, was a competing theory for the special theory of relativity, explaining the results of the Michelson-Morley experiment of 1887. Emission theories obey the principle of relativity by having no preferred frame for light transmission, but say that light is emitted at speed "c" relative to its source instead of applying the invariance postulate.”

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  @debrayoung53110 ай бұрын

  8:23 good. detailed n clear. thanks 🎉

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  @lee546310 ай бұрын

  i’m doing my mathematics thesis based around Strangs article so thank you for this video!

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  @photosphotos10 ай бұрын

  The earth is stationary

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  @I_dreamed_my_name_was_Brandon9 ай бұрын

  hey, someone gets it!

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  @Professor-taboo5 ай бұрын

  And flat friend .

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  @bookofsounds0144 күн бұрын

  False. 1mph motion.

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  @photosphotos4 күн бұрын

  @@bookofsounds014 hahaha.. its not budging. Its stationary

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  @wernerhartl206911 ай бұрын

  The problem is the mirror is rotating, so the velocity of the incident light and reflected light relative to the velocity of the rotating mirror has to be calculated. I feel this will give the same result as the calculation in the rotating frame* thus there is no need for Aether or SRT. * kzread.info/dash/bejne/hK1smpeSfauwpLA.html

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  @wernerhartl206911 ай бұрын

  The expressions for ds=.. and ds’= are wrong. Theta is a constant. Don’t know why it gives correct answer. A clean and transparent calculation comes from using a reference frame attached to the ring. This is given in kzread.info/dash/bejne/hK1smpeSfauwpLA.html I will delete this video after vacation.

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  @wernerhartl206911 ай бұрын

  I thank colinjava for correctly pointing out my approximation for cos(theta) was wrong. It should be 1-theta^2/2. However, result is the same. For the vector triangle on the left with alpha and theta small (5mins into video): csinα=vsin(θ-α) sinx ≈ x -> α=vθ/(c+v) as in the video. The calculation from a reference frame in the ring is much cleaner and transparent, and gives same result: kzread.info/dash/bejne/hK1smpeSfauwpLA.html

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  @Chris_531811 ай бұрын

  Now all you have to do is realise that he has correctly pointed out that almost everything you have said in your videos and comments is wrong.

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  @deimosthewizard Жыл бұрын

  I think it's fine if people disagree with your theory, I do for one, but I think it's really not justifiable to be as mean and insulting to you as some people in the comment section are. Really quite disrespectful considering how much effort you put into making the video.

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  @colinjava84477 ай бұрын

  Well, we all started off being nice, but after weeks or months of trying to help him out of his delusions he still says we are all wrong. At times I think he enjoys pretending to be a moron to piss people off, but now I tend to just think he has stupid ideas about what natural numbers are, which means any theorem using natural numbers is then wrong. Try exchanging messages with him for a week and see if you can go without ridiculing him. The point is he should stop posting bullshit videos that lie to the viewers, that's the main reason we ridicule.

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  @jeffjo87327 ай бұрын

  I try not to ridicule, but I do try to examine what he says, and point out where it strays from valid mathematical theory. Bit here's the thing: HE WON'T DO THE SAME. Here are some basic mathematical concepts he refuses to recognize: The empty set exists. A finite set is one that can be put into a bijection with a set of consecutive natural numbers of the form {1,2,3,...,N}. And yes, this includes the empty set. An infinite set is a set that is not finite. While you can define a "number" that is infinite, such as Aleph0, it does not belong to the set of natural numbers (as he tries to do with his c00, and with ∞, as the horizontal index in his table). And most importantly, he does not present CDA correctly. And the biggest problem, is that he tries to twist the explanations of why he is wrong to fit how he misrepresents these concepts.

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  @wernerhartl20697 ай бұрын

  @jeffjo8732 Define empty set. From the usual definition, a collection of objects, no objects, no set. I agree the set of all natural numbers is not finite, so satisfies your definition of an infinite set. You are confusing infinite set with infinite number. There is no such thing as an infinite number. I use oo in the video as an adjective, not as a number. It’s explained in the video. Apparently you are just looking at the pictures. Basically, CDA finds the last entry in an enumerated list, which doesn’t exist.

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  @colinjava84477 ай бұрын

  @@wernerhartl2069 You say there's no such thing as an infinite number, but claim 3333.... Is a "natural number". So is 3333.... Infinite or finite? It doesn't find the last entry, there isn't a last entry, it finds one of many numbers that are not in the supposed list.

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  @jeffjo87327 ай бұрын

  See? This is why he receives criticism. "Define empty set." A set that has no members. Duh. "From the usual definition, a collection of objects, no objects, no set." Nope. From all definitions, it does exist. Since youtube has stopped allowing links to other domains, just look for "Axiom of empty set" or simply "Empty set" on wikipedia. From the latter, "Some axiomatic set theories ensure that the empty set exists by including an axiom of empty set, while in other theories, its existence can be deduced." In case you missed the implication, every set theory allows the empty set to be a set. A set is not defined to be a set by what it contains, it is defined to be set by the fact that it defines what objects it does, and does not, contain. "You are confusing infinite set with infinite number." And you are confusing different definitions of numbers. Specifically, there is no "infinity," or "infinite number," in either the natural number system (which you try to use in your video) or real number system. But there are infinite cardinal numbers and infinite ordinal numbers. These kinds of numbers have different definition than the one you use. They are valid definitions, but you refuse to recognize those definitions. For example, the cardinal number representing the cardinality of the set of all natural numbers is the infinite cardinal number Aleph0. And the fact that you don't even try to understand the differences between these definitions is why you can't understand that CDA is correct. "There is no such thing as an infinite number." There most certainly is. The cardinality of the set of all natural numbers is one. "I use oo in the video as an adjective, not as a number." You use it, and c00, as indices for the columns in your table. You compare their values, to each other and to other (finite) numbers. This makes them "numbers" in your treatment. "Basically, CDA finds the last entry in an enumerated list." Nope. CDA demonstrates that there must be an element that missing from a list that has no end. You are the only one trying to locate its end.

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  @deimosthewizard Жыл бұрын

  As someone learning linear algebra for the first time currently, mainly in hopes of later being able to go into abstract algebra without lacking any prerequisites, these videos are greatly appreciated. Thanks!

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  @wernerhartl2069 Жыл бұрын

  detEr’=detEr -> detA’=detA detAB=detAdetB. detA’=det(E1E2)’=det(E2’E1’)=detE2’detE1’=detE2detE1=detA. If nxn A is singular, row rank =col rank <n and detA’=detA=0. (ErEs)’=Er’Es’ from definition of matrix multiplication.

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  @abietester9037 Жыл бұрын

  Thanks for illuminsting a tough topic Dr. Hartl.

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  @wernerhartl2069 Жыл бұрын

  My pleasure!

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  @warrenhaney6038 Жыл бұрын

  Exactly what I needed to watch... thanks so much .. well done

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  @jeffjo8732 Жыл бұрын

  In the Zermelo-Fraenkel set theory, there is an axiom called the Axiom of the Empty Set. In words, it says: "There is a set such that no element is a member of it." So the empty set is a set that exists in Z-F set theory. Sorry, you are wrong. But you still get other things wrong in your rush to find mistakes made by people who know more than you do. The definition of set equality is called the Axiom of Extensionality. And it is not quite what you wrote (and misinterpreted). It says: "Given any set A and any set B, if for every element X, X is a member of A if and only if X is a member of B, then A is equal to B." The difference is that you only look at elements that are members of A and B, while the correct definition looks at every possible element. Then you misinterpret your definition when, for example, you imply that looking for every member of A "fails" when you find none. "Every" actually does mean "none" if there are none. Falling back on your fallacious argument, that an argument is "vacuous" if you don't understand it is, well, an actually vacuous argument. But your vacuous argument is based on an incorrect definition. The actual definition looks at everything that could be an element of a set in your mathematics. If that element is in A, you then look to see if it is in B. And vice versa. This search will test many, many elements, all of them successfully. Which in the case means none are found to be in A or B. The definition is satisfied. Besides, even if {} did fail to satisfy the definition, that won't make it "not a set." It means the definition is faulty, which may be why it is worded as it is. And yes, {} is a subset of {1,2,3}. The rest of your argument is gibberish. You haven't defined "you can count to." But I claim that I can count to any member of the set of all natural numbers, *_N_* . Every member of *_N_* is a finite natural number, call it n, and I can count to it by 1,2,3,...,n. But the set itself is infinite (endless). I'd go on, but what is the point if you refuse to consider any point but your own.

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  @wernerhartl2069 Жыл бұрын

  What is your definition of a set?

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 Something that the axioms of set theory says is a set. What is yours, that you think allows you to ignore an axiom? Then, what makes you think you can re-write an axiom to get the result you want?

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 Where in “the axioms of set theory” does it say that? Prove that something exists which satisfies these axioms. For example, saying something is an empty set because it is an axiom that it is an empty set is a circular definition.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 "Where in 'the axioms of set theory' does it say that?" Did you read what I wrote, or do you ignore anything you don't want to have to accept? It's called "The Axiom of the Empty Set." And it says the empty set exists. It is usually listed as the second Axiom of ZF set theory, right after the one you butchered to claim you proved that the empty set does not exist. There is absolutely no way that you can deny this. The empty set exists is set theory. "Prove that something exists which satisfies these axioms." The "Axiom of the Empty Set" says "There is a set such that no element is a member of it." QED. "For example, saying something is an empty set because it is an axiom that it is an empty set is a circular definition." So, are you admitting that you have no understanding of what "axiomatic mathematics" means? Mathematics has to begin with statements that are accepted to be true without proof, otherwise you do get circular reasoning. There has to be a starting point. These statements are called Axioms, and they are accepted as true statements. There is nothing circular here, Axioms are the starting point of all proof with nothing before them.. Now, do you care to give use your definition of a set? This is the second time I've asked. And please, tell us where you get it.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 I asked: “What is your definition of a set?” You responded: “Something that the axioms of set theory says is a set.” Where in the axioms of set theory does it say that?

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  @colinjava8447 Жыл бұрын

  Just watched a bit more, its too painful to watch in one go... What is this N.N nonsense? The real number 77.33333.... cannot be represented in the form N.N This goes all the way back to werners misunderstaing of Natural numbers, he claims 3333.... is a natural number, which is just retarded. The natural (or counting) numbers can literally be counted to starting with 1,2,3,... Consider the number 734982749873284728957430204340023423877562. Eventually a human can count to that number provided they could live long enough and not get bored. But 33333.... (even if it made sense as number) cannot be counted to, you'd never reach it. So its not a natural number. This destroys werners attempt to map the reals to the natural numbers. Doesn't matter how many times he tries it, he will fail every time since its an impossible task.

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  @jeffjo8732 Жыл бұрын

  Literally, every natural number is defined by one of these two axioms: 1) 0 is a natural number. 2) If n is a natural number, then n+1 is a natural number. Among the many things that can be deduced from these axioms, is that if n is a non-zero natural number, then n-1 is also a natural number. So, if n=33333... is a natural number, what is n+1? Or n-1?

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  @colinjava8447 Жыл бұрын

  @@jeffjo8732 I guess n-1 = 3333.....33332 But its nonsense, cause you have something infinitely long with 2 "ends". Sometimes werner does ...33333 so you can add 1 more easily. But then if x = ...33333, then 10x + 3 = x, and x = - 1/3 These numbers are interesting in their own way, but its ludicrous to call them natural numbers.

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  @jeffjo8732 Жыл бұрын

  An outline of a general form of CDA, with notes on each step: 1) Let T be the set of all binary strings that have cardinality C. 1A) "Cardinality" means the length of the strings, if they are finite. And C is a *_cardinal_* number. 1B) Finite cardinal numbers have the same values as the natural numbers, and can be compared. Literally, a natural number represents the *_position_* of a character, while a cardinal number represents the *_length_* of the string up to that position. I really don't think Werner understands this difference, but it isn't at all important for finite strings. 1C) If C is not finite, it is very important. In this proof, because it deals with lists, it can only mean the smallest transfinite cardinal number Aleph0. This is the "size" of the set of all natural numbers, but its value is not the same as that of any natural number. Sometimes this is what Werner means by "Coo," but sometimes he uses that in other contexts. Or he uses "ꚙ". He often asks what "ꚙ" means, or what it means relative to "Coo." The issue here, is that "ꚙ" is never used in CDA, or set theory in general. "Coo" is never used by anybody except Werner, and he can't keep his usage consistent so it is never clear what he means. It would be so easy for him to use the right words, but he either doesn't understand them or intentionally refuses to. 1D) Werner shows an example of what he calls T3, where C=3, in the upper-right corner of his blackboard. But he calls the set S, when that name is supposed to mean something else. As in... 2) Let S be any ordered subset of T with the same cardinality as the strings in T. 2A) This is where Werner starts to go wrong. He does not make a subset of T, he tries to use the entire set T. I'm not sure if he thinks calling it "S" makes it appropriate for this part of the proof, but it doesn't. S HAS TO HAVE THE SAME CARDINALITY AS THE STRINGS IT CONTAINS. 2B) A correct example of S is the *_first_* *_three_* elements of his T3, the ones he has drawn a diagonal line through. Werner calls them s1, s2, and s3. His mistake is that s4 thru s8 DO NOT BELONG TO SET S used in this part of the proof. 2C) By definition, such a set can't "fail" in the manner Werner claims. 3) Construct a string whose ith character is the opposite of the ith character the ith row of S, for all i from 1 to C. 3A) It should now be incredibly obvious why the cardinality restrictions on S exist. Werner can't see the obvious when it contradicts what he wants to be true. When he can't avoid it in a criticism, he changes the subject of the criticism and responds to that. 3B) If C is Aleph0, this means all natural numbers. The index i will never equal C. 3C) Werner shows i equaling, and exceeding, "Coo" in his table on the left. Here, his "Coo" does seem to mean Aleph0, which is just one of his inconsistencies. 4) By definition, this constructed string is in T. By construction, it is different from every string in S. 4A) This is the correct form of the process that Werner says "fails." 4B) Done correctly, it does not "fail." 4C) Done as Werner tries, it actually illustrates how CDA is correct. Werner will always ignore this fact. 5) This proves the lemma "If S is a set of binary strings that have cardinality C, and S also has cardinality C, then there is at least one binary string with cardinality C (call it s0) that is not in S." 6) From this lemma, it follows immediately that T cannot have the same cardinality as the strings it contains. Because otherwise, there would be a string s0 that both is, and is not, in T. This is Cantor's Diagonal Argument in a form that applies to both finite and infinite sets. It isn't very interesting for finite sets. One of the few things Werner gets right is when C is finite. He is correct that the cardinality of T is 2^C, and that must be greater than C. He even uses this result to claim that step 4 above "fails." His argument actually demonstrates the lemma in step 5. But Werner's biggest mistake is thinking that the finite case can be extended to the infinite simply by saying something like "taken to the limit." This is why the difference between cardinal numbers and natural numbers is important - when "taken to the limit" you can get different results than what applies to every finite case leading to it. For example, the set {2,4,6,...,2n} is always smaller than the set {1,2,3,...,2n}, for all n>=1. The size difference even gets bigger as n grows. But {2,4,6,...} and {1,2,3,4,5,6,...} have the same cardinality. Werner is just convinced that he sees an error overlooked by all great Mathematicians since 1890,. He is so desperate to prove this to the world that he refuses to see his own trivial mistakes.

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  @colinjava8447 Жыл бұрын

  He has another couple of videos on CDA too, its all the same bullshit though. One of them he uses a grid, so (7,2) will represent the rational 7/2 or (2/7), then he says something like there's a sequence of rationals converging to pi, so pi is at infinity on the graph and is rational. I think he calls that induction as well. I thought it was a joke at first, but who keeps a joke going for years on end?

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  @aleksandertorken8202 Жыл бұрын

  Hi and thank you

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  @colinjava8447 Жыл бұрын

  Thank you for what? You realise almost everything on that board is complete garbage? Any reasonably mathematically competent person who's argued with Werner knows he is a fool. Don't take anything he says seriously, cause it will be wrong.

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  @colinjava8447 Жыл бұрын

  This is nonsense, the infinity written at the top right corner of square is countable infinity, its just referring to the digits in a sequence, no need to invoke uncountable infinities at this stage. The number of rows is also countable infinity by assumption, so each row can correspond to each digit going across horizontally. This is really the ramblings of a mad man, its just gobbledegook, pure and utter drivel. 4:42, this is drivel, YOUR ASSUMPTION WAS THAT YOU COULD LIST THE SEQUENCES, SO THERE ARE COUNTABLE MANY, NOT 2^infinity MANY (Based on the supposition, which we then show is wrong, proving there are 2^infinity sequences). You can't even remember you made the assumption at the start. The right side of the board is just following on, and is nonsense. But interestingly, he correctly says there's 2^n sequences, for sequences of n digits, so is he saying the cardinality of all binary sequences is 2^infinity ? Which is what the theorem tells us? Who knows.

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  @wernerhartl2069 Жыл бұрын

  If the digits in a sequence are countably infinite, the number of sequences is countably infinite and CDA doesn’t work. If you invoke uncountable infinity, you are assuming what you are trying to prove and CDA doesn’t work.

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  @colinjava8447 Жыл бұрын

  ​@@wernerhartl2069 How are the number of sequences countably infinite? If binary sequences have n terms, the number of sequences is 2^n, that applies to n as a natural number or countable infinity.

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  @wernerhartl2069 Жыл бұрын

  @@colinjava8447 Yes, for all n.

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  @colinjava8447 Жыл бұрын

  @@wernerhartl2069 You didn't answer how its countable. Show how you can list all the infinite binary sequences (showing they are countable) and I'll admit defeat. But you can't do it, but have a go and I'll correct you.

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  @wernerhartl2069 Жыл бұрын

  @@colinjava8447 A list of all n binary digital places (2^n) is countable for all n.

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  @annmarie2839 Жыл бұрын

  Could you provide the make / model of the saw and box? How many tpi on the saw?

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  @wernerhartl2069 Жыл бұрын

  Have a Huskey and Stanley 14” mitre saw. 13 TPI. Don’t know model number but Stanley is still available at Home Depot for around $15. You can get the saw and mitre box as a set.

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  @Domi2gud Жыл бұрын

  Werner, would you say light behaves similar to a sound wave, where regardless of how fast you spin a speaker, the sound waves will propagate at the same speed through a given medium? Is a "photon" then the same thing as a "phonon?"

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  @wernerhartl2069 Жыл бұрын

  It was concluded from the Michelson-Morley experiment that light did not require a medium (the aether) to support it. It propagates in a vacuum. A photon is a packet of electrical energy generated by electrons dropping to lower orbits. If the frequency is correct, a photon will be absorbed and excite an electron to a higher energy orbit. Continuous motion of an electron in an electric field generates a light wave. A phonon is the mechanical equivalent of a photon. I am not familiar with it.

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  @jaykay92864 ай бұрын

  ​@@wernerhartl2069 And what about propagating NOT in a vacuum??

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  @wernerhartl20694 ай бұрын

  @@jaykay9286 It slows down. Like in a prism.

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  @kennyearthling7965 Жыл бұрын

  This guy talks more sense than Cantor

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  @colinjava84477 ай бұрын

  I really hope that was a joke, this guy is the most arrogant clueless person on earth with anything to do with sets, induction, infinities, decimals, fractions and even more. He doesn't even know what natural numbers are, even after being told a thousand times, I think he just likes being insulted, so that's what we do.

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  @rootm3b8204 ай бұрын

  @@colinjava8447 you are simply a bunch of incompetent and closed-minded precious people who think that mathematics is absolute, you are completely mistaken

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  @wernerhartl2069 Жыл бұрын

  In his original article, Cantor indexes the elements of his string, which means they are Coo in length, and CDA fails. Wiki, and everybody else, presumably realizing the problem, use the undefined term “infinite” sequence (string). When clarified, CDA fails, the point of this video. I note that Cantor’s article doesn’t show up when you google “Cantor’s Diagonal Argument.” Thanks to Jeff Jo for his reference* to the article. * www.logicmuseum.com/cantor/diagarg.htm en.m.wikipedia.org/wiki/Cantor%27s_diagonal_argument Note: Indexing implies Countability.

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  @jeffjo8732 Жыл бұрын

  And this explains what Werner does not understand, and actively tries to misunderstand. "In his original article, Cantor indexes the elements of his string, which means they are Coo in length, ..." "Length" has no defined meaning for countably-infinite lists like this. There are "countably-infinite many" elements (characters) in each string, but not " Coo" elements. You can't put a number, in the sense Werner understands numbers, on it. It is not "Coo in length." And there definitely is no column under "Coo" like he labels in his table. Cantor also indexes the sequences (rows) in the table the same way. So it "implies countability" the same way it does for the columns. Werner somehow keeps missing this, probably because he doesn't like what it means. Any concept that can be associated with the "length" he applies to columns, applies the exact same way to the rows. This makes it impossible for CDA to "fail." Because the failure mode Werner implies requires the indexes to be treated differently for the rows and columns. And the term "countably infinite" is well-defined in the math that Werner refuses to understand. It means that the set can be put into a bijection (en.wikipedia.org/wiki/Bijection,_injection_and_surjection) with the set of all natural numbers. But this definition doesn't support the conclusions that Werner wants to reach, so he claims that it doesn't exist and hasn't been explained to him before. This is CDA: 1) Let *_T_* be the set of all countably-infinite binary stings. That is, a string where there is a '0' or a '1' associated with every natural number n. 1A) Symbolically, we can call the characters c1, c2, c3, ..., cn, ... . This is exactly what Cantor does in the third paragraph of the paper Werner linked to, that he describes as "Cantor indexes the elements of his string..." 2) Let *_S_* be any subset of *_T_* that can be enumerated s1, s2, s3, ..., sn, ... . That is, where every string in *_S_* can be associated with a different natural number n, with no leftover strings or natural numbers. 2A) If a subset of *_T_* can't be enumerated this way, it is not an *_S_* that is subjected to step #3. 2B) This is the exact same thing that Cantor does in his fourth paragraph, that I do in 1A) above, and that Werner describes as "indexes the elements" but in this case, of set *_S_* . 3) Create a new string, s, where the character associated with each natural number n is the opposite of the nth character of the nth string in the enumeration of *_S_* . 4) Since s is not in *_S_* we have proven that any subset *_S_* of *_T_* that can be enumerated must be missing at least one string s that is in *_T_* . 5) It follows immediately from step #4 (not step #3) that *_T_* cannot be enumerated. Because if it could, we would have the contradiction that there is a string s that both is in *_T_* (because all strings are) and also not in *_T_* (by step #4).

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  @colinjava8447 Жыл бұрын

  @@jeffjo8732 He uses infinitely long natural numbers (that obviously aren't natural numbers) in a desperate attempt to pair off the natural numbers with the elements in [0,1], so 0.33333... would pair off with 3333.... He does that to conclude the reals or [0,1] are countable since the natural numbers are (his version of the natural numbers would be uncountable though, so its still retarded). Besides the fact of it being total drivel, it raises a myriad of problems, like what about 0.00003333..., what does that pair off with off?

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  @isadoradaniella5123 Жыл бұрын

  This was very helpful! Thank u so much

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  @TomasAragorn Жыл бұрын

  Btw, you're listing a finite sequence by ending it with a S∞. There's no such last sequence in the sequence and there's also no last digit in D∞. The fact that you're even using the symbol ∞ shows how nonsensical your proof is. I'd suggest you delete this video to spare you the embarrassment

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  @jeffjo8732 Жыл бұрын

  Well, it's unclear what he is doing. I see what you are saying, but the only thing that is clear is that he is very unclear on the concepts. So trying to describe what he does is hard. Look at his columns. He lists what _should_ be a countably infinite sequence of characters - and an informal, but perfectly acceptable, definition of that is that it has no end - as if it has an end. Maybe even two. The full sequence "ends" at a position he calls "∞". But there is a point in the middle that divides it in two - one the "ends" at "C∞" and another that continues beyond "C∞" and goes to to "∞". Maybe this will help him - I doubt it, but I believe in trying to help people to understand. I can define a kind of set I will call a "counting set." A counting set *_may_* contain the natural number 1. If a counting set contains the natural number n, it *_may_* also contain the natural number n+1. And finally, if a counting set contains the natural number n but *_not_* n+1, we say it has size n and call it CS(n). The important parts here are that, for each finite counting set, its size is a natural number in the set. In fact, it is the largest natural number in the set. We even use this as the definition of a different kind of number, called a cardinal number. It means, informally, the size of a set. And technically, we don't call it "size," we call it "cardinality." So far, the cardinal numbers seem to be redundant, since every natural number is also a cardinal number. The way we define the infinite is through the Axiom of Infinity: There is a set N that *_does_* contain the natural number 1 and, if it contains the natural number n, it *_does_* contain the natural number n+1. This satisfies the definition of a counting set, but there are two problems naming it. There is no natural number n that is in the set while n+1 is not. So there is no largest natural number in it, and (as yet) no cardinal number representing its "size" or cardinality. So we *_define_* a new kind of number, called a transfinite cardinal number. Alpeh0, or C∞ as Werner calls it, is the smallest. It means the size/cardinality of the set N that I defined with the Axiom of Infinity. It is unique in that it is the only cardinality of a counting set that is not also a natural number in the set. AND THIS IS THE POINT WERNER KEEPS CONFUSING. Each logical bit in a sequence is associated with a finite natural number, but the size/length/cardinality of that string is not. It is Aleph0 (or C∞). Most of what Werner claims about ∞ is confused because it does not recognize this. The last bit he gets wrong, is that "the construction" does not try to find a "square" table, or use every sequence in T. It only needs a set of sequences that associates a sequences with each n in N. The size/cardinality of this set of sequences is Aleph0 (or C∞), but there is no sequences associated with Aleph0 (or ∞, or C∞).

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 It’s quite simple. wiki begins with: “If s1, s2, ... , sn, ... is ANY enumeration of elements from T,….” Emphasis added. The video gives a case for which this is not true. CDA fails.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 It's really far simpler than you make it out to be. If we add the parts back in, that you deliberately omitted (btw, in most parts of the world this is called "lying"), the wiki "starts out": "Cantor considered the set T of all infinite sequences of binary digits..." It then says he begins the proof by first proving a lemma. It is that lemma that starts out: ""If s1, s2, ... , sn, ... is ANY enumeration of elements from T,[note 2]….”. Please notice that this does not say "enumeration of T." Nor does it say "enumeration of *_every_* element in T." In fact, there is a notation, between the "T," and the "..." that you cut from the quote. Apparently so you could make claims about what it prevents. It says that this does *_NOT_* mean every element in T. The "case for which this is not true" that the "video gives" is the exact one that the note says in not meant. If you could STOP LYING TO YOURSELF about this, we could convince you of what the proof says. This lemma applies to any subset S of T that can be put into such an enumeration. We know that some do exist; you even described one. Cantor does not assume that it does, or does not, apply to all of T. It applies to ANY such enumeration THAT ACTUALLY EXISTS. Is the enumeration given in the video an example of the form s1, s2, ... , sn, ... ? No? Then it isn't what the premise of the lemma applies to. Understand? This isn't hard. Your video then uses invalid logic to make a correct claim: that there are subsets that it doesn't apply to. What you can't seem to understand, is THAT IS THE POINT OF CDA. Your claim is that the contradiction Cantor uses is wrong because it actually is a contradiction.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 “[note 2] Cantor does not assume that every element of T is in this enumeration.” All elements is not excluded. If you disprove one case, ANY is invalid and CDA fails.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 BZZZZT. Wrong. Blatantly wrong. "Makes one wonder where you learned Math" wrong. "If XXX then YYY" makes no claim that XXX is true, only that when it is true YYY must also be true. It doesn't even need XXX to ever be true. That is, "If pigs can fly, then bluebirds like french pastry" is a true statement. Just not a proof. But in the lemma, XXX is "s1, s2, ... , sn, ... is any enumeration of <not necessarily all> elements from T." This makes no claim that there is such an enumeration of >>>SOME<<< elements, but it proves that the YYY part - "an element s of T can be constructed that doesn't correspond to any sn in the enumeration" is true for >>>ANY<<< that do. Not "EVERY SUBSET", which is what you are misinterpreting it to mean. Nowhere, in any part of this so far, is the full set considered. Nowhere, in any part of this so far, is there a failure point based on the full set not being enumerable. But the part you keep ignoring, is that you are assuming that CDA is true and correct when say it is a failure point.

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  @wernerhartl2069 Жыл бұрын

  There is an update to this video: kzread.info/dash/bejne/Y2dp1MaSZ82ueaQ.html

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  @jeffjo8732 Жыл бұрын

  Cantor's Diagonal Argument, simplified and based on Wikipedia terminology Werner uses: 1) Let string s be any countably-infinitely-long strings of the logical (not binary) characters '0' and '1', which can stand for 'false' and 'true'. 2) Let T be the infinite (how infinite is not specified, or needed) set of all such strings. 3) Let S be a subset of T with the additional property that it has countably-infinite members; that is, that they can be enumerated s1, s2, s3, ..., sn, ... for all natural numbers n. NOTE A: Nowhere in CDA is it assumed that S=T. NOTE B: For those with limited attention span, this means that the table you can make from S has countably-infinite rows. There is no need to ask what "infinite" means here, or to incorrectly use "oo" to mean "infinite." NOTE C: This assumes you can identify an actual set with this property. Examples are so trivial, that this is often overlooked. But an example is that sn has all 0's except a 1 at position n. 4) Create a new string, s0, whose nth character is the logical inverse of the nth character of sn. 5) The string s0 is different from any string sn in at least one position. So it is not in S. 6) This proves the proposition "If s1, s2, ... , sn, ... is any enumeration of elements from T, then an element s0 of T can be constructed that doesn't correspond to any sn in the enumeration." NOTE D: "Enumeration" means "countably infinite," which seems to be what Werner means by "Coo." But he can't seem to acknowledge this explicit definition, or what correct terminology is. This is CDA. It only attempts to prove the proposition is #6. It does not "fail" due to not defining what "oo" means, because it does not use "oo." If Werner's "Coo" is supposed to mean "countably infinite," one does not need to ask "“If oo means Coo, ..." cor two reasons. First, it doesn't use "oo" or "infinity." Second, where it uses "infinite" it clearly means countably infinite". But it doesn't "fail" as Werner claims, because he keeps assuming that S=T in step 3. It doesn't "fail" because all "success" means is that the proposition is step 6 has been proven. Itg doesn't "fail" due to strawman arguments about 2^n rows since it never claims or implies that many rows. +++++ But Cantor has a second proof, that is not CDA but uses the proposition proven by it. 1) What would happen if we assume that T can be enumerated? That is, that S=T? NOTE E: This does not assume such an enumeration actually exists. It explores what might be necessary consequences if one did. Werner can't seem to distinguish between the two. But this is how proof-by-contraction works. 2) But then we'd get the contradiction, that there must be string s0 that both is, and is not, in T. 3) Since the existence of such a sequence would lead to a contradiction, an enumeration of T as supposed in step 1 must be impossible. NOTE F: This does not "fail if 00 means C00" because the no actual enumeration of T is assumed. Only the necessary implications of one. +++++ And there is yet a third proof Werner seems to know nothing about. If P(A) is the power set of A - that is, the set of all subsets of A - then the cardinality of P(A) is strictly greater than the cardinality of A. In fact, |P(A)| = 2^|A|, which is where Werner gets his objection about needing 2^n rows for the set of length-n strings. The proof does not mention anything to do with "infinity," "oo,", the infinite (countable or not), or "Coo." It applies to any set A. It cannot "fail" for any of the strawman arguments Werner uses. This was the point of Cantor's paper, not either of the previous two proofs. They represent an example of this one, for the set of natural numbers. The set T can be considered to represent the power set of the natural numbers, P(N), by using each logical character to indicate whether the corresponding natural number is, or is not, in the subset. +++++ 0.43 "Assume they are countable..." Bzzzt. Wrong. This assumption is not made here. It is the subset S that is assumed to be countable. 0:47 "then construct an infinite binary sequence s not in t." Bzzzt. Wrong. The string is in T, but not in S. Read the Wikipedia article if you doubt me: "By construction, s is a member of T that differs from each sn." 0:52 "contradiction, T is not countable" Bzzzt. Wrong place in the proof. The only conclusion possible here is about S. 1:30 "assuming the sequences are listable then we list them as one two down to ... countably infinite." Bzzzt. Wrong. You keep implying an end; there is no end. That is what :infinite" can mean here. 3:05 "now there's a problem. Infinity isn't really defined." Bzzzt. Strawman argument. "Infinity" is never used. "Infinite" in this context is defined to be countably infinite. 3:15 "if infinite/Infinity means greater than countable Infinity...." Bzzzt. Strawman argument. Why do you keep asking this? The length of the strings is explicitly defined to be countable infinite. 3:36 "if you assume Infinity means countably Infinity ..." Bzzzt. You don't understand what infinite means. You can't have the strings "end" and Coo. There is no natural number C00. 3:48 " ... are more sequences than there are digits..." In T, yes. That is the point of the *_SECOND_* proof above. In S? No. Why? Because we require that any S we use can be enumerated. Again, S is not all of T. But if we accept the "all T" part? Then yes, there are more sequences than digits. THIS IS WHAT CANTOR IS TRYING TO PROVE. The contradiction you sense, when you misinterpret what set gets used, IS THE CONTRADICTION CANTOR IS HEADED FORE. In short, even though you do it incorrectly, you are proving that Cantor is right. Not that he is wrong. 4:42 "n digits lead to two to the N sequences for all n." This is about the first thing you get right, but you say it all wrong. First, n is not a natural number in CDA. It can be in the Power Set proof. But the proper statement is: "The cardinality of set A, C=|A|, leads to 2^C possible subsets of A." What you can't seem to accept, is that the rows in "the construction" do not correspond to the set T, they correspond to any way *_some_* members of T can be arranged so that the row as and columns *_do_* match up. The fact that you can't do it with *_all_* members of T is what Cantor is trying to prove with CDA

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  @wernerhartl2069 Жыл бұрын

  From wiki article on CDA: “Cantor considered the set T of all infinite sequences of binary digits (i.e. each digit is zero or one).[note 1] He begins with a constructive proof of the following lemma: If s1, s2, ... , sn, ... is any enumeration of elements from T,[note 2] then an element s of T can be constructed that doesn't correspond to any sn in the enumeration.” But “infinite” is undefined so his construction proof is meaningless. However, for the assumed case that all elements of T are enumerable, the construction fails, and so CDA fails, as shown in this video.

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  @jeffjo8732 Жыл бұрын

  What part of "If s1, s2, ... , sn, ... is any enumeration of elements from T, [note 2: Cantor does not assume that every element of T is in this enumeration]..." are you having difficulty understanding? Nothing in this says "the assumed case is that all elements of T are enumerable" is meant. In fact, it is *_EXPLICITLY_* stated that it is not. Where is the word "all" that you use found in what the article says? Or, if we go back to Cantor's paper, we find: "For proof, let there be E1 = (a1.1, a1.2, … , a1,v, …), E2 = (a2.1, a2.2, … , a2,v, …), Eu = (au.1, au.2, … , au,v, …), ………………………….". Do you notice that *_EVERY_* sequence Eu is indexed with a natural number? Do you understand that this means it is what you call "Coo"? And that no part of that paper implies anyhting like what you claim, even with your intentional misinterpretation? Next, how is it that you cannot grasp that this is a two-part proof? And that the construction you keep talking about exists only in part 1, and that the assumption you insist goes with it applies only to the second part? The one that *_DOES_* *_NOT_* *_USE_* the construction? Finally, the failure mode you imply here is that you cannot put the set of all sequences into a bijection with the columns (character positions). Since possibility is *_exactly_* what CDA is trying to prove, why do you keep insisting that it fails? Even if you were correct in this (you aren't), the theorem is still true BY YOUR OWN ADMISSION.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 The part that says: “the set T of all infinite sequences”. “Infinite” is undefined so construction is meaningless. Any discussion of CDA which doesn’t begin with a definition of “infinite” is random noise. I thought precise definitions were the foundation of mathematics.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 Cantors indexed construction fails for the same reason the diagonal construction fails when infinity means Coo. In a_i,j there are 2^n indices i for n indices j. So no matter what j is there are still more i, so you never exhaust the list. Thanks for bringing the Cantor article to my attention. www.logicmuseum.com/cantor/diagarg.htm

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  @jeffjo8732 Жыл бұрын

  Cantor doesn't use the word "infinity." "Infinite" is defined here. It means a set that can be - strike that, *_is_* *_actually_* - put into a bijection with the set of natural numbers. This is literally an explicit part of both Cantor's paper, and the Wikipedia article, for the set that "the construction" applies to. You keep ignoring this definition because it invalidates your claims.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 Yes, there are 2^n pairs for each element of a (A) square, (B) finite, table that (C) contains every possible row. Since none of these qualifications apply to the set of sequences that "the construction" applies to, yours is a straw man argument. Aside: "square" could apply if you extend its definition to a countable-infinite set of rows and columns. But then the reason you dismiss CDA assumes the proposition that CDA is supposed to prove.

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  @wernerhartl2069 Жыл бұрын

  From wiki on CDA: “Cantor considered the set T of all infinite sequences of binary digits (i.e. each digit is zero or one). He begins with a constructive proof of the following lemma: If s1, s2, ... , sn, ... is any enumeration of elements from T,[note 2] then an element s of T can be constructed that doesn't correspond to any sn in the enumeration.” However, if ALL elements of T are assumed to be enumerable, the construction fails, and hence CDA fails, as shown in this video.

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  @jeffjo8732 Жыл бұрын

  Try this informal version of the proof: 1) Make a table with a countably-infinite number of rows, and a countably-infinite number of columns. 2) Put a "0" or a "1" in every spot in this table. A) The number of such bits is also countably-infinite. B) Each row is one of Cantor's strings. Call each s_r, where r is the row number (starting with r=1). C) We do not know if it is possible to put every possible string into these rows. D) The diagonal, D, is also one of Cantor's strings. 3) Make a new string, s_0 by inverting each bit in D. Again, this is one of Cantor's strings. 4) The string s_0 is not represented in any row. In fact, any table that can be made using steps 1 and 2 has to be missing a string. 5) Since any table is missing a string, it is impossible to make such a table with every one of these strings. 6) Let T be the set of all such strings. T has infinitely many strings, but not a countably-infinite many because it cannot be put into the table. This is CDA, just phrased a little differently, and less formally. But every step is obviously true. What seems to confuse Werner about CDA, is that the table created in step 2 is the one used in CDA. What really confuses him, is that the formal logic, that proves the obvious statement in step 5, is where Cantor assumes he is using the complete set T. "Any table has to be missing a string" obviously means the entire set can't be there, but to formally prove it Cantor uses proof-by-contradiction. He assumes it can, and gets the contradiction that there is a string that is, and is not, in the table.

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  @wernerhartl2069 Жыл бұрын

  But there are more rows than columns and CDA fails. n column places give rise to 2^n rows, for all n, if you are consistent with Cantors assumption that you are listing all the members of T. Of course you can pick out a square array and find a member not in the array. So what. Irrelevant to CDA. Let T = {00,10,01,11} 11 is not in the array: 10 01 Cantors array in this case is 00 10 01 11 and CDA fails.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 What part of "a countably-infinite number of rows, and a countably-infinite number of columns" did you have trouble understanding? Even if you continue to intentionally misunderstand what "countably infinite" means, it means the same thing in both places. The "construction" only applies to a table with the same number of rows and columns.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 Coo is not a number. A CooXCoo array is not square, unless it’s nXn for all n. But it’s not, it is nX2^n for all n. CDA doesn’t say list a square array from T, it says list them all.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 "Coo is not a number." Aleph0 is a number, just not a natural number. It is a cardinal number. That seems to be what you want "Coo" to means. "A CooXCoo array is not square, unless it’s nXn for all n." I have no idea what you mean by "unless it's nxn for all n." That is a vacuous statement. "But it’s not, it is nX2^n for all n." If you are trying to say that the countably infinite rows cannot be put into a bijection with the countably infinite columns, you are wrong. "CDA doesn’t say list a square array from T, it says list them all." It says "If E1, E2, …, Ev, … is any simply infinite series of [sequeces]..." You keep reading from the wrong part of the proof. BBBUUUTTT AAAGGGAAAIIINNN: If you maintain that number of rows must exceed the number of columns, then you are maintaining that the proposition Cantor claims to prove with C DA is correct. That is exactly what it says. That "there is an infinite manifold, which cannot be put into a one-one correlation with the totality of all finite whole numbers 1, 2, 3, …, v, … " So, which is it: The fact this this proposition is true validates the conclusion but invalidates Cantor's proof of it, or the proof is valid?

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 Cantors construction applies to any enumeration of T. Nothing mathematically says you can’t test the construction on an assumed enumeration.

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  @jeffjo8732 Жыл бұрын

  "If infinite means countably infinite, CDA fails because Cantors array is longer than it is wide. n digits give rise to 2^n sequences for all n." This first part of this is wrong, CDA does not fail. The claim is based on a naive definition of "long" and "wide." The one you learned in Kindergarten. That you count the rows or columns to get length or width, and STOP COUNTING WHEN YOU REACH THE END. Cantor's array has no end (which can be what "infinite" means) to the rows or columns. In fact, you can label rows with natural numbers, and columns with even natural numbers, and still get an array that is just as long as it is wide. Even though "n digits give rise to 2*n sequences for all n." Just not with every possible sequence. So please, advance past pre-teen definitions before you try to understand CDA. The list Cantor uses in the diagonalization process (which is not _every_ binary sequence) is just as "long" as it is "wide" by any definition that allows for countably-infinite length and width. We know that, because it has one sequence for each natural number in the set {1,2,3,...}, and each sequence has a bit for each natural number in that same set. There is always an element on the diagonal. But the conclusion is right, and we know that because CDA does not "fail." There indeed are more possible sequences than natural numbers. Cantor never tries to put them all in a list. +++++ "If infinity means greater than countable infinity, CDA fails because you are assuming what you are trying to prove." I hesitate to mention this, because I know you will take it the wrong way. But CDA is not a formal proof, and was not the point of that paper. It lacks one (trivial) step. The first part assumes an actual list that is known to exist. Technically, it should demonstrate that it is possible. But that is trivial: sn=1/n gives us such a list. The second part is hypothetical. And it assumes the opposite of what it wants to prove. That's how proof-by-contradiction works. CDA itself (just the first part) proves that there must be a sequence missing from any countably-infinite list of binary sequences that actually exists. The second parts says "How do we know that we can't list them all?", and shows that in the hypothetical case where we could, we would get an absurd result: that a sequence we know exists, the one that can't be in the list, must be in the list. It is Cantor's Theorem that was the point of that paper: the Power Set of set A, called P(A), is the set of all possible subsets of A. You show one on your blackboard. A subset of the set {1,2,3} can be represented by a length-3 binary string; for example, {1,3} is represented by "101." Your eight rows show all eight subsets of {1,2,3} this way. If n is the size/cardinality of set A, then the cardinality of P(A) is 2^n. The cardinality of any finite set of natural numbers {1,2,3,...,n} is the natural number n. The cardinality the set of all natural numbers is not itself a natural number, it is called Aleph0 (not d_inf or "infinity."). The cardinality of P(N) is 2^Aleph0, which as you conclude is "bigger" than Alpeh0. +++++ "Countability (abbreviated) Any real number is of the form N1.N2, where N1 and N2 are natural numbers." I think you mean that, for example, when N1=123 and N2=456, that R=123.456 is a real number. That is true, but this statement is expressed backwards. A better definition is that N1.N2 is always a rational number, not that any rational number is always N1.N2. For example, 22/7 is 3.142857 142857 142857 ... , with this same six digits repeated endlessly. There is no natural number N2 for this in your notation. Let alone for a real that doesn't repeat, like pi=3.1415926535897932384.... .

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  @wernerhartl2069 Жыл бұрын

  Cantors construction fails for all n. See the example in my video for n=3. Countably infinite digits, as after . In 3-pi, represent a natural number.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 Rebuttal #1: Your construction for n=3 assumed that Cantor uses all possible combinations of the specified length. He does not. Rebuttal #2: That was for finite sets. The conclusion requires finding an end to the rows and/or columns. If there is no end, the conclusion fails. Rebuttal #3: A variation of your logic is that there is a row whose index is greater than the index of every column. With infinite sets, that is not true. Rebuttal #4: If your logic is correct, what it shows (you even say this!) is that there must be more rows than columns. Since the number of rows is the number of sequences, and the number of columns is the number of natural numbers, your claim is that you proved, without using CDA, that there are larger infinities. Is that what you want to claim? +++++ "Countably infinite digits, as after . In 3-pi, represent a natural number." Um, no. Elementary-school math "no." Try it this way: instead of thinking of the first digit past the decimal point as the "leading" digit of the natural number N2, think of it as the 1's place. The second digit is the 10's place. Etc. The decimal representation (do you understand the difference between a number, and its decimal representation?) of any natural number can be reversed this way, and will also represent a natural number. If you can't understand that, you don't understand math.

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 1) wiki- “The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, ... , sn, ... . Applying the previous lemma (Cantors Construction)* to this enumeration produces a sequence s that is a member of T, but is not in the enumeration.”- *added Actually, if the number of digits is Coo, then so is the list of all Coo sequences and CDA fails right off the bat, and you don’t even need the construction. But I wanted to show the construction fails. 2) Is {1,2,3….} larger than {2,4,6,…}? They are both Coo. 3) Video example illustrates all 3-Place binary numbers. It takes n digit places to represent 2^n natural numbers. It takes Coo digit places to represent all the natural numbers. 4) The Coo sequence 3.14159… IS, IS, the number pi, which has a geometrical meaning. There is no other real number equal to pi.

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  @jeffjo8732 Жыл бұрын

  @@wernerhartl2069 We can go 'round and 'round this merry-go-round as often as you deny the truth: 1) There are two propositions that are proved in that section of Wikipedia. The first is "If s1, s2, ... , sn, ... is any enumeration of elements from T[note 2], then an element s of T can be constructed that doesn't correspond to any sn in the enumeration." For it, the article says "The proof starts with an enumeration of elements from T." And, if you read "note 2," it says that this proof does not assume that T is countable. This is the one that contains what you call "Cantor's construction," but it does not claim to use a contradiction. Or all of T. The second is "The set T is uncountable'." It is the one where the article says ""Based on this lemma, Cantor then uses a proof by contradiction ...". This is the one that does not use "Cantor's construction." It does use contradiction, but not the one you claim. But you are correct, it assumes T is countable. It never uses any property of a countable T, like the first part does, it just invokes what that part proved. And that is how proof-by-contradiction works. You assume what you want to prove is untrue, and find contradictory implications of that assumption. Example: 2^(1/2) cannot equal the reduced fraction N/M, because that would imply both that M is odd and N is even, and that M is even and N is odd. We don't actually derive an M and an N - the equivalent of you applying "Cantor's Construction" to all of T - we use known results of arithemetic that uses natural numbers. 2) You need to define what you mean by "Set A is Coo." Because your video uses Coo as an index, and that is wrong. One property of two sets A and B that are Coo, but what it seems you want that to mean, is that you can make, between them, a surjection that is not an injection, a bijection, and an injection that is not a surjection. So what is your point about them "being Coo"? 3) See the rebuttal to your point #1 above. The video example illustrates the construction used in the article's first proof, but by "illustrat[ing] all 3-Place binary numbers," it is using the assumption from the second.. As such, it is a vacuous illustration. 4) The "Coo number" that you claim IS, IS the number Pi is not represented by a natural number N2 like you claim. ("Any real number is of the form N1.N2, where N1 and N2 are natural numbers.") Where is the one's digit?

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  @wernerhartl2069 Жыл бұрын

  @@jeffjo8732 What is your definition of infinity, as used in CDA. For n-place decimals, the 1’s digit is .00…1_n for all n. Why can’t you do Cantors construction for the particular set of all Coo digits?