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{{Mathematical arguments subpage}}
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==Argument not work==
Trovatore deleted the following 1 minute after post:
The argument, as presented, (for reals, sets of bits, or sets of naturals) cannot work even if its conclusion is true because
For proof A to prove B to be FALSE it must allow B room to be true.
Consider someone asking you to count all 1000 three digit numbers, on 3 lines so only 3 numbers fit! Or they ask you to count all 1 digit numbers and after you count 1 number they say count all 2 digit numbers and after you count a second number they say count all 3 digit numbers …!
The height of the list HAS TO BE the exponential of the width to make room for all the sets to be counted per the statement of the proof 'count all the...' for it to be a PROOF. Saying that doesn't count with ∞ because with ∞ one can do magic has to be PROVEN for the rest to be a proof!
Example of proper counting:
In set theory Aleph infinity is infinite increase without end. Set the rate of increase of the height of the sets, the real numbers, to normal, have the digits of each real number produced by separate algorithms for each line, small to large algorithms, they can produce true random bits. Force the first digits to be a count sequence so each line is different. Set rate of increase of number of digets of nonrandom reals to normal, random reals to the (same base) logarithm of normal, this will force the diagonal to the same rate. The sort is by algorithm, like a program but programing language very complex so will always produce infinite list of digits, and in reasonable time as function of the number of digits yet produced by the algorithm. The width of the random numbers relative to the number of numbers counted is the same as natural numbers. The height keeps pace with the increase of possible numbers per the size of the algorithms and number of bits of randomness used to make any random real. Thus π is not ∞ly far thru the list. The height counts as Aleph0 not Aleph1 per analogy to the height : width relation of the set of natural numbers. π , for example, would be as precise in digits as the number of reals counted.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]])
00:28, 24 April 2017 (UTC)
==Simple counterexample ==
"In his 1891 article, 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 theorem:
If s1, s2, … , sn, … is any enumeration of elements from T, then there is always an element s of T which corresponds to no sn in the enumeration.
The proof starts with an enumeration of elements from T, for example:"
''Any'' enumeration, you say?
Let's fully specify the enumeration, as follows: the first digit of a sequence in the enumeration is given by the repeating pattern "10101010...", the second digit is given by pattern "110011001100...", the third digit by "1111000011110000...", the fourth by "1111111100000000..." and so on, or in other words the nth digit is "1" in the first 2^(n-1) sequences of the enumeration, "0" in the next 2^(n-1) sequences, and so on:
:{|
|-
| ''s''<sub>1</sub> || = || (<u>'''1'''</u>, || 1, || 1, || 1, || 1, || ...)
|-
| ''s''<sub>2</sub> || = || (0, || <u>'''1'''</u>, || 1, || 1, || 1, || ...)
|-
| ''s''<sub>3</sub> || = || (1, || 0, || <u>'''1'''</u>, || 1, || 1, || ...)
|-
| ''s''<sub>4</sub> || = || (0, || 0, || 1, || <u>'''1'''</u>, || 1, || ...)
|-
| ''s''<sub>5</sub> || = || (1, || 1, || 0, || 1, || <u>'''1'''</u>, || ...)
|-
| ''s''<sub>6</sub> || = || (0, || 1, || 0, || 1, || 1, || ...)
|-
| ''s''<sub>7</sub> || = || (1, || 0, || 0, || 1, || 1, || ...)
|-
| ''s''<sub>8</sub> || = || (0, || 0, || 0, || 1, || 1, || ...)
|-
| ''s''<sub>9</sub> || = || (1, || 1, || 1, || 0, || 1, || ...)
|-
| ''s''<sub>10</sub> || = || (0, || 1, || 1, || 0, || 1, || ...)
|-
| ''s''<sub>11</sub> || = || (1, || 0, || 1, || 0, || 1, || ...)
|-
| ''s''<sub>12</sub> || = || (0, || 0, || 1, || 0, || 1, || ...)
|-
| ''s''<sub>13</sub> || = || (1, || 1, || 0, || 0, || 1, || ...)
|-
| ''s''<sub>14</sub> || = || (0, || 1, || 0, || 0, || 1, || ...)
|-
| ''s''<sub>15</sub> || = || (1, || 0, || 0, || 0, || 1, || ...)
|-
| ''s''<sub>16</sub> || = || (0, || 0, || 0, || 0, || 1, || ...)
|-
| ''s''<sub>17</sub> || = || (1, || 1, || 1, || 1, || 0, || ...)
|-
| ''s''<sub>18</sub> || = || (0, || 1, || 1, || 1, || 0, || ...)
|-
| ''s''<sub>19</sub> || = || (1, || 0, || 1, || 1, || 0, || ...)
|-
| ''s''<sub>20</sub> || = || (0, || 0, || 1, || 1, || 0, || ...)
|-
| ''s''<sub>21</sub> || = || (1, || 1, || 0, || 1, || 0, || ...)
|-
| ''s''<sub>22</sub> || = || (0, || 1, || 0, || 1, || 0, || ...)
|-
| ''s''<sub>23</sub> || = || (1, || 0, || 0, || 1, || 0, || ...)
|-
| ''s''<sub>24</sub> || = || (0, || 0, || 0, || 1, || 0, || ...)
|-
| ''s''<sub>25</sub> || = || (1, || 1, || 1, || 0, || 0, || ...)
|-
| ''s''<sub>26</sub> || = || (0, || 1, || 1, || 0, || 0, || ...)
|-
| ''s''<sub>27</sub> || = || (1, || 0, || 1, || 0, || 0, || ...)
|-
| ''s''<sub>28</sub> || = || (0, || 0, || 1, || 0, || 0, || ...)
|-
| ''s''<sub>29</sub> || = || (1, || 1, || 0, || 0, || 0, || ...)
|-
| ''s''<sub>30</sub> || = || (0, || 1, || 0, || 0, || 0, || ...)
|-
| ''s''<sub>31</sub> || = || (1, || 0, || 0, || 0, || 0, || ...)
|-
| ''s''<sub>32</sub> || = || (0, || 0, || 0, || 0, || 0, || ...)
|-
| ...
|-
|
|-
| ''s'' || = || (<u>'''0'''</u>, || <u>'''0'''</u>, || <u>'''0'''</u>, || <u>'''0'''</u>, || <u>'''0'''</u>, || ...)
|}
So the complements of the first 5 digits are 0,0,0,0,0.
For the nth sequence, n>5, the digit is 1 if n <= 2^(n-1). When n=6: 6 <= 2^5 so the digit is 1. Since the series 6,7,8,... grows slower than the series 64,128,256,... n <= 2^(n-1) is true for all n>5 and as such the complements of the digits for all n>5 are 0.
This means that the sequence s is just all zeroes, which is in the set T and in the enumeration. But according to Cantor's diagonal argument s is not in the set T, which is a contradiction. Therefore set T cannot exist. Or does it just mean Cantor's diagonal argument is bullshit? [[Special:Contributions/37.223.145.160|37.223.145.160]] ([[User talk:37.223.145.160|talk]]) 17:06, 27 April 2020 (UTC)
: You do not say where 0,0,… is in your enumeration. To them the construction, that is an algorithm producing all 0s, would need to be in T, and your enumeration. [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 13:25, 28 April 2020 (UTC)
::The sequence 0,0,0,0,0,... is the last sequence in the enumeration, just like sequence 1,1,1,1,1,... is the first sequence in the enumeration. If you prefer 0,0,0,0,0,.... would be somewhere else, you can just apply a bijection f(x) from natural numbers to natural numbers to shuffle or reverse the order of the enumeration afterwards. [[Special:Contributions/37.223.145.160|37.223.145.160]] ([[User talk:37.223.145.160|talk]]) 11:35, 30 April 2020 (UTC)
::: The enumeration is a sequence - it assigns items of T to natural numbers. But there is no ''last'' natural number. If the element (0,0,0,...) of T belongs to the enumeration, it ''must'' be at some specific position. But by the definition you gave, at every possible position ''n'' there is a sequence whose ''n''-th digit is 1. Hence (0,0,0,...) is at no position in the enumeration. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 15:45, 30 April 2020 (UTC)
:::: By construction, for every sequence in the enumeration with a '1'-digit, there is another sequence in the enumeration where that '1'-digit is a '0'-digit and all other digits in the sequences are the same. So, for the 1,1,1,1,1,... sequence there is also the 0,1,1,1,1,... sequence, and for that 0,1,1,1,1,... sequence there is also the 0,0,1,1,1,... sequence, and for that 0,0,1,1,1,... sequence, there is also the 0,0,0,1,1,... sequence, and so on and so forth until the 0,0,0,0,0,... sequence which has the fewest of '1'-digits, that is, none. Therefore, the 0,0,0,0,0,... sequence is in the enumeration. We can call the position of that 0,0,0,0,0,.... sequence whatever we like. [[Special:Contributions/37.223.145.160|37.223.145.160]] ([[User talk:37.223.145.160|talk]]) 16:37, 7 May 2020 (UTC)
:::: But perhaps you are still not in agreement that the way I specify the enumeration includes all the sequences terminating with infinitely repeating 1's as well as those terminating with 0's, especially the 0,0,0,0,0,... sequence. If that's the case, then construct a second enumeration with only the 0,0,0,0,0,... sequence included and then, one by one, like you were doing Cantor's diagonal argument, add every sequence in turn from the first enumeration in-front of the 0,0,0,0,0,... sequence in the new enumeration. Now you have an enumeration which ''by definition'' includes the 0,0,0,0,0,... sequence in the last position and the rest proceeds as usual. [[Special:Contributions/37.223.145.160|37.223.145.160]] ([[User talk:37.223.145.160|talk]]) 20:23, 7 May 2020 (UTC)
::::: You made two mistakes. Mistake no. 1: there is no ''last position'' in any infinite sequence, speciffically there's no last position in your enumeration (a sequence of numeric sequences), because the last position would be a position numbered with the greatest natural number. And we know there is no greatest natural number.
::::: Mistake no.2: after inserting the zeros-only sequence in front of your enumeration, the list begins with:
:::::: 0 0 0 0 0...
:::::: 1 1 1 1 1...
:::::: 0 1 1 1 1...
:::::: 0 0 1 1 1...
:::::: 0 0 0 1 1...
::::: so its diagonal is {{nowrap|0 1 1 1 1...}} and the resulting sequence is {{nowrap|1 0 0 0 0...}}. Which again is absent in your (new) enumeration. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 21:15, 7 May 2020 (UTC)
==How is it a proof of any kind? ==
So, I enumerate
<pre>
S_0 = 0 0 1...
S_1 = 1 0 1...
S_2 = 1 1 1...
</pre>
Then,
<pre>
S_n = 1 1 0...
</pre>
And so, how does that prove anything at all? It's pretty obvious S_n will not be in my enumeration because I have not enumerated all the elements from the binary tree.
Am I missing something? :-)
[[Special:Contributions/197.79.24.8|197.79.24.8]] ([[User talk:197.79.24.8|talk]]) 09:09, 20 February 2014 (UTC)
:Well, that is kind of the point. You start off assuming that you can list all real numbers (in this case between 0 and 1) and then come up with a number that isn't on the list. That is a contradiction, so the assumption that you can list the real numbers must be wrong.[[Special:Contributions/155.95.80.243|155.95.80.243]] ([[User talk:155.95.80.243|talk]]) 21:20, 21 February 2014 (UTC)
== Simple disproof ==
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*Which integer does 0.01010101010101... correspond to? Your chart would say ...1010101010101 but this is not an integer! --[[User:Paul Laroque|Paul Laroque]] ([[User talk:Paul Laroque|talk]]) 20:37, 13 April 2009 (UTC)
** It would correspond to ...10101010101010. You simply flip the digits over. It's unclear why you would say it's not an integer. Granted, it is a large integer with an infinite number of base-2-representation digits. [[User:Frederick Bryan|Frederick Bryan]] ([[User talk:Frederick Bryan|talk]]) 17:48, 24 September 2013 (UTC)
* From the common sense, it doesn't seem to be true that ''in order to define an infinite number of integers one must have an infinite number of digits''. 'Infinite' means 'endless': there's no end to enumeration. But every other (enumerated) number does have a finite number of digits just like its predecessor: from time to time, the number of digits ''n'' increases by one, but it still has a successor ''n+1'' and is a valid number; the digits can be all enumerated, and the end to enumeration comes when ''n'' digits have been enumerated. So, no enumaration of natural numbers, no matter how long, would dig out a number that has infinitely many digits. In fact, such a natural number would be an absurd: called a number, but doesn't have a value. Besides, you couldn't do anything to such a 'number', because you'd have no tools (definition, addition, subtraction, …). The number of the natural numbers is infinite, but the numbers themselves are not (and the lengths of their digit strings are not either). ;) — [[Special:Contributions/178.71.141.48|178.71.141.48]] ([[User talk:178.71.141.48|talk]]) 00:00, 20 January 2013 (UTC)
** That's absurd. The largest integer you've counted to is also the number of integers you've counted. If the number of integers is infinite, then the number you've counted to, to get an infinite number of integers, is also infinite. Most integers therefore have an infinite number of digits. --[[User:Frederick Bryan|Frederick Bryan]] ([[User talk:Frederick Bryan|talk]]) 17:57, 24 September 2013 (UTC)
***That doesn't follow. Every integer has a finite number of digits. Sure, you can find an integer with an arbitrarily high number of digits, but no integer has an infinite number of digits. In addition, you don't count to "somewhere" to get an infinite number of integers. There's no point on the number line where it switches from finite to infinite. <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/73.189.35.161|73.189.35.161]] ([[User talk:73.189.35.161|talk]]) 02:32, 16 July 2014 (UTC)</span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->
== Other, "Rational" Cantor diagonalizations ==
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:Are you sure it was Cantor that came up with that one? -Dan 04:25, 19 December 2005 (UTC)
:It isn't a diagonalization proof in the same sense (though a common diagram for it uses diagonals); it doesn't belong here. There are other diagonalization proofs which share essential properties with the Cantor diagonal proof: they include the [[halting problem]] argument, standard proofs for Godel's [[incompleteness theorem]] and [[Tarski's theorem]] on the
== Would be Amusing ==
Line 71 ⟶ 221:
J.
==Apoorv1's arguments==
The following shows the vulnerability of cantor,s proof.
Line 503 ⟶ 653:
This to me, is an unqualified extrapolation from a sample to the entire population, and is more a conjecture. Another portion will show that same partial sequence.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 01:09, 29 January 2008 (UTC)
:No, it shows that it's distinct from all the sequences. Otherwise it would be the same as one of the sequences, but as has been shown, it can't be. That's really all there is to say about it -- it's ironclad and there is no subtletly remaining. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 02:28, 29 January 2008 (UTC)
:: Looks like the list ''S'' already contains all the sequences ''S(i)'', because every index ''i'' corresponds to a position inside the sequences, and no new positions can be added because we have agreed that the sequences are already presented in full (although it is impossible to imagine them being presented so). Still, this all is counter-intuitive (intuition tells we cannot do anything to infinite lists and sequences because infinite lists and sequences do not exist in the real world where it's possible to act).
:: Probably, the article could better go into some detail on why and for what the diagonal argument being applied to natural and real numbers is important in mathematics as well as in the "real world", so that a lay reader like me could really understand there is no life without this argument in this or that important field of human activity or imagination. A layman's view is that generic real numbers correspond to nothing in the world we see and experince, and in this sense do not exist, that they are only idealizations of our real and always finite knowledge, and that infinite lists of numbers (be they real or natural) do not exist in the same way, so any layman and not only me is naturally interested why mathematicians study different kinds of non-existence. [[Special:Contributions/91.122.7.114|91.122.7.114]] ([[User talk:91.122.7.114|talk]]) 00:51, 19 January 2013 (UTC)
::: Forget all that philosophy, what bothers me for real is this: how can one say he has picked ''all'' the natural numbers? One gave me a bucket of numbers and said: "That's all, there are no more natural numbers". I determined what number is the maximal one in the bucket (if that guy was able to pick them all, then I can inspect them all; and if the guy is not able to pick them all, then no infinite list can be built anyway) and then named a number that is greater by one, so that to add it into the bucket. Then I said, "Now all the natural numbers are in the bucket", and gave the bucket back to my friend, but he disagreed, on the same grounds. Something here must be wrong, but what exactly? - [[Special:Contributions/91.122.7.114|91.122.7.114]] ([[User talk:91.122.7.114|talk]]) 05:27, 19 January 2013 (UTC)
:::: Well, seems like the beaut is that the work is never done but the features of its state are the same at whatever point. I can imagine Cantor's devil as a machine that, first, declares it's going to do something abominable (names what and why), reserves ''B-1'' infinite-length registers (''B'' being the base of the numeral system to use), initializes the registers with empty sequences, and then proceeds: at each step, it picks a natural number, generates an (infinite: ???) sequence of digits, seeks a digit on the position that corresponds to the number just picked, makes a set of the other digits, and for each of them, adds a digit to the sequence in a reserved register. Of course, it never finishes its work, but whatever the finish can be, we'll see that when it has come, there are only so much natural numbers, but the sequences of digits outnumber the numbers anyway (whatever nonsense those sequences are).
:::: The problem (my problem and all the readers') is that the finish can't be (from the habitual point of view), and thinking of it is meaningless, unless there is such and such reason to accept this way of thinking in this exactly situation. The rest is to understand what are the reasons to accept these or such things, and I think ''this is what the article fails to demonstrate'', and that's why so many questions from us all. I guess it might be improved in this direction: i.e., show context. -- (the same user) [[Special:Contributions/178.71.141.48|178.71.141.48]] ([[User talk:178.71.141.48|talk]]) 18:59, 19 January 2013 (UTC)
==Cantor's Nonsense: Pseudo-mathematics at its best.==
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::::So: in one small thing you're right; Cantor was not the first person to consider infinite sets. He was just the first person to show that something of value came out of considering them (e.g. the existence of transcendental numbers). Gauss died too soon to see the fruits of Cantor's work (for that matter, so did Cantor -- the most important applications didn't really get moving until the early 20th century). Whether Gauss would have continued to reject the completed infinite, in the face of all the great mathematics that came out of it, is speculation. Certainly there are many important workers who continued to reject it; perhaps he would have been one.
::::But that has nothing whatsoever to do with these arguments about the same proof showing there are uncountably many naturals, or with the thing about adding the diagonal real back to the list. Those are just flat misunderstandings -- in the first case, misunderstanding of what a natural number is (can't have infinitely many nonzero digits), and in the second case, of the nature of a proof by contradiction. You will find good mathematicians that reject the completed infinite; you will not find ''any at all'' that think those two arguments are valid reasons to reject it. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 04:44, 1 June 2008 (UTC)
::::: Could you please give a brief overview of these applications? Why are they important? Thank you. - [[Special:Contributions/91.122.7.245|91.122.7.245]] ([[User talk:91.122.7.245|talk]]) 13:28, 27 April 2016 (UTC)
:: For a minute there, I thought you understood the significance of the "completed infinite" (CI) as applied to this argument. Apparently you don't. Those who reject CI can always counter Cantor's constructed real by simply adding a new natural number to correspond with it. It's really that simple. If you don't understand that, you don't understand the issue at all. And in this context, the diagonal argument yields no contradiction (and therefore no proof) because all infinite sets have the same cardinality. These are really very elementary notions that you should understand if you are going to defend Cantor. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/66.67.96.142|66.67.96.142]] ([[User talk:66.67.96.142|talk]]) 22:16, 1 June 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
:::No, you're the one who doesn't understand. Cantor's argument says "assume a completed list exists", and then derives a contradiction. If there were such a completed list, the contradiction would follow; therefore there is none. That finishes it, period. If you don't understand this then you are the one who is missing the very elementary notions.
Line 665 ⟶ 823:
--[[User:Vibritannia|Vibritannia]] ([[User talk:Vibritannia|talk]]) 14:42, 9 May 2009 (UTC)
:I have the exact same objection to Cantor's diagonal argument. I don't disagree with its result, that integers and reals can't be mapped to each other, but I don't see the fallacy in Vibratannia's point either.[[User:Mister Mormon|Mister Mormon]] ([[User talk:Mister Mormon|talk]]) 00:23, 11 September 2010 (UTC)
The visualization of a list screws up a lot of peoples intuition apparently. So I rewrote the proof any a more algebraic and precise way. I use function argument instead of subscript and g will denote the s_0 in the article:
Line 695 ⟶ 855:
Their equality would imply s(m)(m)=0.5, which is impossible. So g and s(m) differ at the mth term. Therefore g!=s(m) for any m, thus s is not surjective. QED [[User:Scineram|Scineram]] ([[User talk:Scineram|talk]]) 09:57, 30 October 2009 (UTC)
== proof or illusion? ==
A statement from 'Cantor's diagonal argument'':
1. "Therefore this new sequence s' is distinct from all the sequences in the list".
It should say "is distinct from all the sequences in the sample list".
There is no logical reasoning from 'some' to 'all'. You can't reach a conclusion for all, based on a comparison of some. For example, a doctor can't state with certainty that 100 people don't have disease A, based on examination of 10 from that group. Since the truth of statement 1 depends on the comparison of all sequences, it is based on information not available, i.e. speculation.
The problem with isolated experiments is the denial of information that would assist in making an accurate conclusion. If the list L is ordered, and divided into two subsets, L0, all sequences beginning with '0', and L1, all sequences beginning with '1', the first comparison and symbol swap would exclude the subset L0, and it is only necessary to compare s' to a member of the subsets L10 and L11 for the next position. The process becomes more efficient by eliminating redundant comparisons. A simple pattern emerges with the choice for each position always the same, '0' or '1'. If s' is compared to a member of one subset, its symbol swap will place it in the complementary subset. Continuing with any arbitrary 7-symbol sequence, the next comparison would be with one of the following subsets, each containing an unlimited number of sequences not compared, that match s' to the current position.
L10111010={10111010...} (all sequences beginning with s' and appended with 0)
L10111011={10111011...} (all sequences beginning with s' and appended with 1)
conclusion:
It's obvious that no matter which choice, there is an existing sequence matching s'.
The refined process is equivalent to forming any sequence.
The original process is not forming a 'new' sequence, it's just copying an existing one.
Assuming a complete list,
10111... and 11111... differ in one position and
010101... and 101010... differ in all positions.
If each sequence is unique, then each must differ from all others in at least one position.
The fact of being different does not imply exclusion.
Assuming in principle, if one complete sequence can be formed, then all possible sequences can be formed and presented as a complete ordered list L, otherwise there is no list.
In either case the diagonal proof fails.
The diagonal and partial list are instances of misdirection.
[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 18:59, 21 March 2013 (UTC)
: The sequence s' isn't being compared to 'some' of the sequences in the list, it's being compared to ALL of the sequences in the list. Your 'refined process' merely divides your list into several sub-lists. s' is not in any of these sub-lists because s' isn't in their union (the original list). [[User:Gazzawhite|Gazzawhite]] ([[User talk:Gazzawhite|talk]]) 02:20, 13 November 2013 (UTC)
You missed the part where the ordering of subsets eliminates redundant comparisons, and accounts for all possible combinations of 0 and 1. We could also just admit that even one infinite sequence cannot be constructed, and therefore no list.
On a more fundamental level, no human can conceptualize infinity, since there is no related experience. To model it using finite entities doesn't work. Even if humans became immortal, they still have a beginning. [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 20:26, 14 February 2014 (UTC)
: The ordering of the subsets (to "eliminate redundant comparisons", whatever that means) is irrelevant. If s' isn't in L, then it can't be in the subsets. Your change of format from a list to a collection of subsets does not prove that s' is in L any more so than our original assumption that L contains all binary sequences (and thus contains s'). Thus the contradiction is still there. [[User:Gazzawhite|Gazzawhite]] ([[User talk:Gazzawhite|talk]]) 04:52, 1 July 2014 (UTC)
== Moved from [[talk:continuum hypothesis]] ==
''Heading was "Proposed things to add"''
Note: A one-to-one correlation between the set of all natural numbers and all real numbers between zero and one, exclusive, can be conceived of. Each integer would pair with the real number obtained by flipping all of its digits across the decimal point. For example, 5.0 would coincide with 0.5, 27.000 would coincide with 000.72, and 271828182845904.00000000 would coincide with 00000000.409548281828172. Two arguments against this are that the correlation is not consistent between bases. While that is correct, and while 0.125 coincides with 521 in base ten, 0.125 in base twelve would coincide with 73 in base twelve, the existence of a correlation between the two sets is constant throughout all bases. The other argument against it is that recurring and irrational decimals (like 1/3 and e-2) would coincide with integers with infinitely many digits. However, a number can be infinitely large and still be an integer. Despite the presence of a counterargument against both arguments against this one-to-one correlation, the validity of this correlation is still controversial.
If the correlation is valid, then it would make the amount of all real numbers aleph-nought, making the Countinuum Hypothesis meaningless. <small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/98.195.88.33|98.195.88.33]] ([[User talk:98.195.88.33|talk]]) 04:49, 9 February 2016 (UTC)</small><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->
:What do you mean by "a number can be infinitely high and still be an integer"? You must have a meaning of "integer" in mind which is very different from the meaning generally accepted by mathematicians. More important, however, is that your extended concept of "integers" does not have cardinality aleph zero, which destroys the whole point of your suggestion. <small>''The editor who uses the pseudonym''</small> "[[User:JamesBWatson|JamesBWatson]]" ([[User talk:JamesBWatson#top|talk]]) 16:00, 9 February 2016 (UTC)
High meant large, and I fixed that. Let me explain what I mean to you. [[Graham's Number]] has a ton of digits, and it's still an integer. My definition of an integer is a number that has no non-zero digits after the decimal point. It can have as many non-zero digits as it wants, even an infinite amount, before the decimal point, as long as it doesn't have any after. Also both the set of [[natural number]]s and the set of [[integer]]s have cardinality zero.
== Wrong question ==
Folks, you're all asking the wrong question. The wrong question is: can every real number be enumerated? This question is too specific, it lacks generality, and detail. The real question is: is the situation conceivable, in which the creation of the set of all natural numbers has come to its end, the creation of the set of all real numbers has come to its end, the creation of all transitions from one natural number in the first set to one real number in the second set has come to its end, and no real number exists, such that it has no transition pointing to it from some natural number? Cantor says: no, one cannot conceive such situation, and here's why… Of course, one can argue that such situation is not conceivable simply because the completion of the set of all natural numbers is not conceivable, but that's another story, to which the argument does not apply. Sure, one needs to have finished the ''building actions'' about a set ''before'' saying it has the same cardinality as anything else. - [[Special:Contributions/91.122.4.3|91.122.4.3]] ([[User talk:91.122.4.3|talk]]) 23:07, 26 April 2016 (UTC)
: I think I can add some more to it.
: Let's just ask ourselves the question: what is the difference between finite and infinite sets? The difference is, we can conceive of every object in a finite set at the same time, while the same is not true of an infinite set. That ability of ours has consequences for all procedures of enumeration, as enumeration over only a finite set has a limitation: it only can find objects that can all be known and constructed beforehand. But one can freely get used to the idea that it does not matter whether this simultaneous conception of all objects in a set is possible. A set is good enough without that possibility, because what is substantial for a set is the kind of argument that we can do with whatever element in this set. The principle goes, what is true for an arbitrary number in the set is true for every number in it.
: So, what do we have? For a digital string whose digits are constructed in a certain way, it is true for an arbitrary string in the set of digital strings, on the condition that the string is put into a correspondence with a number in the set of integers, that that string is different than the string whose digits we can construct, because of the way those digits are constructed. Since it is true for an arbitrary string, it is true for every string that satisfies this condition. Therefore, the string that is different than all of them exists, and it satisfies all conditions that any member of the set of digital strings has to satisfy; that is, the set contains that string. It is not required that we actually construct all digits in the string; suffices that we may do so, because, again, what is true for an arbitrary digit in this string is true for all digits in the string.
: Sure, the construction and rememberance is possible only for a finite number of objects. But for a set, the rememberance of all objects is simply not necessary: it is just the same whether it is or is not allowed. I think there are sets for which it is not even known (as of yet) whether they are finite or (actually) infinite. - [[Special:Contributions/37.9.29.40|37.9.29.40]] ([[User talk:37.9.29.40|talk]]) 11:55, 24 November 2016 (UTC)
== Set theory isn't real ==
Suppose you construct the sets like this:
{{color|green|0}}000... (All zeros)
{{color|red|1}}{{color|green|0}}00... (1, then zeroes)
{{color|red|11}}{{color|green|0}}0... (11, then zeroes)
{{color|red|111}}{{color|green|0}}... (111, then zeroes)
.......
And so on. The 1s form a triangle. The diagonal pattern is the complement of a sequence of zeroes, so it's all 1s. Do we really believe that this triangle sequence, with a continuing number of 1s, is different from a constructed sequence that, so far, happens to be entirely 1s? [[Special:Contributions/23.121.191.18|23.121.191.18]] ([[User talk:23.121.191.18|talk]]) 04:14, 16 August 2017 (UTC)
: This question somehow resembles the one asked in the section [[#not a convincing argument]] above (although it's obviously not the same). --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 09:07, 16 August 2017 (UTC)
: The diagonal is an infinite sequence of zeros, so the constructed pattern is an infinite sequence of ones. However, ''none'' of the rows used in the construction is built of ones only. Please note that if one was, it would introduce a '1' into the diagonal, which by design contains zeros only! <br/> So the answer is:
:: YES, the resulting sequence of ones does NOT appear in the list of sequences used in the construction.
: --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 13:29, 17 August 2017 (UTC)
Yes, there is no single row with all ones. However, we can also show that the infinite set of all such rows does contain it, since for every column i, all (infinitely many) rows starting at i+1 indeed have ones there. <!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Zde71|Zde71]] ([[User talk:Zde71#top|talk]] • [[Special:Contributions/Zde71|contribs]]) 19:35, 4 October 2017 (UTC)</small> <!--Autosigned by SineBot-->
: {{re|Zde71}} Nope. The set, implied by the construction of the array above, does not contain a row with infinitely many ones. Certainly, there are sets of rows, which contain such rows. And those sets do not even have to be infinite – you can easily construct such finite sets. There is even a singleton (one-item set) with such a row:
{ '11111111...' }
: However, that doesn't imply such a row belongs in the set discussed above. It's true there are infinitely many rows in it with a one at position 7. There are infinitely many rows with a one at any chosen position. Anyway, none of them has ones only. What's more, each of them has '''more''' zeros than ones! (Because each ''n''-th row has only ''n''–1 ones-only leading part and infinitely long tail built of zeros only.) --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 07:46, 5 October 2017 (UTC)
That depends on your axioms.
At ∞ supposedly such exists, and whether any 0s follows depends on an additional axioms, and one can argue the number derived from the diagonal is at position ∞ that is line number ∞+1 having natural number ∞.
Your setting up your axioms to suit your tasts as is the other person. <!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko#top|talk]] • [[Special:Contributions/Victor Kosko|contribs]]) 23:45, 27 March 2018 (UTC)</small> <!--Autosigned by SineBot-->
: {{re|Victor Kosko}} No. [[Natural number]]s do not contain ∞. There is no such natural number. And our set is implicitly defined by a sequence, i.e. its items are numbered by natural numbers, hence there is no 'position ∞' in it. <br/>Let alone '∞+1', which is pure nonsense. Even if you insisted on building another arithmetics with something like 'the greatest natural number' denoted by ∞, adding 1 to it would immediately ruin the theory: either you have to admit '∞+1=∞' and your reasoning above becomes inconsistent, or you create an inner contradiction by proving there '''exists''' a 'number' ∞+1 '''greater than''' the number ∞, which was defined as the greatest one (which means '''no greater number exists'''). --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 07:30, 28 March 2018 (UTC)
I argued only that neither you nor the other(s) have sufficient axioms. I was using actual not potential ∞ since the diagonal argument is provably wrong with potential ∞ since it just keeps going on
If you want to be persnickety then per his argument the "last" row is "row number ∞" "indexed by ∞-1" having "∞-1" 1s followed by ONE 0 not ∞ 0s, the number derived from the diagonal on the disallowed next line. You are allowing transinfinitly many digits/rows in your counterargument. [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 00:32, 29 March 2018 (UTC)
== New simple arguments cardinal ∞ ==
1. Cardinal ∞ is pseudo or meta math:
∞ sets have no size and 2 unconnected ∞ sets have no relative size.
Proof; consider the set
1 2 3 4 5 6 7
The set has size 7, which is also the last element. Sets of this type have size = the last element. Consider the set
1 2 3 ………
It is the same type of set so has size = the last element. Set theorests would say
The set has the same size as the natural numbers
0 1 2 ………
The set has NO last member
But since it has no last member it has NO size, same as the natural numbers!
2 connected sets can have relative size without each having sizes such as the width and height of the set of all sets of 0s and 1s, the height is 2↑width. But the set of natural numbers is not connected to the set of even numbers, then the set of evens would be ½ the size of the natural numbers, or ∞+∞≠(the same)∞.
2. So why do they claim ∞ size sets have = > or < size? Because they think they have proof. The proofs are deceptive. There are 19 techniques directly in the same category as Injection and Surjection (Crudely and omissively):
One to one some matching
One to one matching
One to many matching
The one to one being one to one or one to one same
For ≤ < = > or ≥
2 sets or a set and itself, then < and > would be the same, = by one to one or one to one same.
Set theorists remove the < and > techniques, else 2 sets or a set and itself that are proven = in size would also prove ≤ < > and ≥ in size. This is not mentioned in Wikipedia!
They then axiomatically assume that which applies to finite cases applies to ∞ cases contrary to this unmentioned result!. So they claim ≤ means < or = rather than < or = or both. And they assume sets compared are < = or > in relative size but not a combination.
Contrary to this result they claim if ≤ and ≥ then =, bijection (techniqually tryjection including one to one all of each set to the other).
Since they dropped < and > they need to remove = from ≤, ≥. To attempt this they use the diagonal argument to disprove the = technique(s), however it only exhaustively disproves a proof, does not prove ≠. Disproving a proof is not proving the negative of that proof. Again axiomatically assuming that wich applies to finite cases applies to ∞ cases.
In fact = can be proved between sets they claim ≠, so >.
Proof:
Make a binary tree with an extra node at the top and all other nodes having 2 or 0 but not 1 node(s) connected directly below them. Call the nodes with 0 nodes connected below them bottom nodes. It is found in all cases the number of nodes is exactly 2× the number of bottom nodes. Apply this to ∞ cases. The branching to the bottom nodes left and right (and center from the top node) represents, a symbol from each node, 0.?????…, Thus all sequences of 0s and 1s, also thus all real numbers from 0 inclusive to 1 exclusive binary. Match the natural numbers 1 number to 2 nodes for all nodes. The natural numbers are thus 1:2 matched to nodes having 2:1 relation to all sequences of 0s and 1s.
This does not contradicting the diagonal argument proof since it only disproves a proof, not the negative of that proof.
3. Wikipedia falsly claims a proof that exponential of ∞ ≠ (the same) ∞ (vs = exponential of ∞ and =∞) rather than that being assumed thus an axiom, by the diagonal argument. Again exhaustive disproof of a proof, not a proof of the negative of that proof. In fact a proof of exponential of ∞ = ∞ is in the Wikipedia article on infinite hotel in the section using spaceships.
Proof:
Assume 2↑∞(Aleph0)≠∞(Aleph0) but = Aleph1
Consider two versions of the set of natural numbers;
One having ∞ many numbers
One having numbers with ∞ many digits (bits) so 2↑∞ many numbers
Consider two sets;
The set of "numerics" having subset the set of countable numerics described each by using the set of all math symbols having to do with numerics, so I, e, Euler's constant, the golden ratio, √π×e, etc. Now obviously the set of numerics has infinitesimal 1÷∞. So the set of numerics between 0 inclusive and 1 exclusive has size ∞ many numbers. That's the set being considered.
The set of numbers between 0 inclusive and 1 exclusive. Presumably these have ∞ many digits (bits) so size 2↑∞
Now are the two first sets the same set? Are the two later sets the same set? To be consistent with their writings set theorists would have to say yes to both
Likewise their are ∞ many points on a double ended line not Aleph1 as set theorists claim to have proved.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 23:58, 20 April 2018 (UTC)
Edit for outline, symbols, links, one diagram, please. [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 00:18, 25 April 2018 (UTC)
== Mainspace page for this talk page ==
I converted the mainspace one sentence "article" into a redirect pointing back here, but I'm not sure if that is even that is appropriate. Why was the mainspace page created? [[User:Elassint|<font color="Blue"> Elassint</font>]] <sup>[[User_talk:Elassint |<font color=#FF00FF>Hi</font>]]</sup> 02:42, 22 April 2018 (UTC)
:Should not exist at all. I'll tag it for speedy. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 17:00, 22 April 2018 (UTC)
:I've tagged it as G6, which is supposed to be "uncontroversial" maintenance. It is possible that there is someone who disagrees, since someone created it. I hope that person will reconsider and not block the deletion. The mainspace equivalent has no possible function. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 17:05, 22 April 2018 (UTC)
== Cantor's diagonal argument, float to integer 1-to-1 correspondence, proving the Continuum Hypothesis ==
[[User:Counting the Floats|Counting the Floats]] ([[User talk:Counting the Floats|talk]]) 16:29, 1 December 2018 (UTC)Hello,
[https://www.youtube.com/watch?v=i-QdlFj8sAc Pairing Floats and integers]
In the YouTube video above, I prove that there is a 1-to-1 correspondence between
positive integers and positive floats. This contradicts the conclusion arrived
from Cantor's diagonal argument. I do not attempt to disprove the diagonal argument
as it involves infinity which I am not qualified to perform calculations.
However, the very simple set of algorithms proving the 1-to-1 correspondence
actually proves the Continuum Hypothesis also unveiled by Georg Cantor.
I made an almost identical YouTube video which arrives at this conclusion.
[https://www.youtube.com/watch?v=bdQlNZr_b-Y&t=31s Proving the Continuum Hypothesis]
Tamas Varhegyi
secondcause@gmail.com
P.S. I am new to the rules and mechanics of trying to showcase some important
set theory algorithmic discoveries. I would appreciate positive, knowledgeable feedback
which would enable me to make my case according to established Wikipedia editing and authoring policies.
I might still have previous login or user info active on Wikipedia, please let me know if there is any conflict and if yes how do I clear it up.
[[User:Counting the Floats|Counting the Floats]] ([[User talk:Counting the Floats|talk]]) 16:29, 1 December 2018 (UTC)
:Have you been able to calculate to which integer the square root of 2 corresponds using your system? [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 17:26, 1 December 2018 (UTC)
:: {{Re|Counting the Floats}} Your 'unique '''one-to-one''' correspondence' doesn't even contain all ''rationals'', let alone <math>\sqrt 2</math> proposed by {{u|Dmcq}}. I'd bet you're unable to find an integer number corresponding to float <math>\frac 19</math> in your 'unique '''one-to-one''' correspondence'. And I'll be happy to tell you why 1/9 does not exist in your Table 6. You only need to ask. :) Will you dare to ask, {{u|Counting the Floats}}...? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 23:01, 1 December 2018 (UTC)
:<br>Hi {{u|Counting the Floats}}. Unfortunately there is no way to "make [your] case according to established Wikipedia editing and authoring policies". That is simply not what Wikipedia is for. If you believe you have an original insight, you will have to make your case out in the wide world, not here. Once original insights become accepted, or at least enough people pay attention to them to make them "[[WP:N|notable]]", then Wikipedia can talk about them. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 19:59, 1 December 2018 (UTC)
:{{re|Counting the Floats}} Your claim in the '''Step 5''' (timestamp 2:15) is patently false. I'm talking about '''Note the extreme simplicity of this navigational scheme compared to the one shown in Table 2'''
: I have deliberately removed the plural and the reference to the Table 3 from the sentence, as that part of the claim is actually true. What concerns Table 2, however, the difference is just in drawing: the Table 2 (timestamp 1:25) uses direct, straight returning arrows, while Table 5 uses semi-rectangular external paths, rounded at their corners. But the order of visiting nodes of the graph is essentially the same. Just flip Table 5 vertically to see this.
: As a result, you just re-invented the Cantor's scheme, which was '''convoluted, inconsistent and hard to follow''' (timestamp 1:30).
: Please note the difference in presence or absence of the zero label is out of scope here. I just comment on the claim about the 'navigational scheme' I quote above.
: The remaining comment at timestamp 1:30 results from a misunderstanding: the Cantor's pairing function was not invented for this particular diagonal argument and is not directly used in it. It is an example of <math>\mathbb N^2\leftrightarrow\mathbb N</math> bijection and if you want to apply it to pairs (numerator, denominator) of fraction, you just need to redefine the ___domain to get <math>\mathbb N\times\mathbb N^+ \leftrightarrow \mathbb N</math> . --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 00:38, 2 December 2018 (UTC)
== no reals ==
all depends on how you look at it...what Cantor does is show that real numbers aren't actually numbers since they can't be counted...[[Special:Contributions/108.20.114.62|108.20.114.62]] ([[User talk:108.20.114.62|talk]]) 14:42, 6 April 2019 (UTC)
: No, Cantor shows there's more binary strings (equivalently: more subsets of natural numbers) than natural numbers.
: This ''can'' be used to prove that there is more real numbers than integers, but that requires defining an additional function, which estabilishes a bijection between all binary sequences and real numbers (or at least a surjection from the former to the latter).
: Anyway, uncountability of real numbers is proven much simpler by showing they can't be enumerated with a [[sequence]] – see [[Cantor's first set theory article#The proofs]]. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 16:01, 6 April 2019 (UTC)
== The proof is (EDIT: NOT) incomplete ==
THIS IS WHAT I WROTE:
The current proof claims that the base-2 interpretation of infinite 0/1-sequences as reals into the unit interval [0;1] is an injection. The fact that it is an injection is used in an essential way to transfer the uncountability of the set of infinite bit sequences to the uncountability of reals in the unit interval. However, the interpretation is not an injection.
For instance, the sequences 011111.... and 100000.... both map to the same number 1/2, which happens to have two (periodical) base-2 representations: 0.011111... and 0.100000.... Thus, the only thing that is shown here is the uncountability of the set of infinite sequences, not of the reals. To properly extend the proof to the reals, the quotienting of bit sequences in the interpretation as reals has to be accounted for.
<!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Andreasabel|Andreasabel]] ([[User talk:Andreasabel#top|talk]] • [[Special:Contributions/Andreasabel|contribs]]) 18:50, 19 May 2019 (UTC)</small>
UPDATE: Sorry, I overlooked the box with the detailed argument that can be opened. So, the proof is indeed complete.
However, the text running up to this is slightly confusing.
Why mention the decimal interpretation of bit sequences? It does not help at all.
The assumption is that the infinite list of infinite bit streams is an enumeration of the reals in the unit interval. Which it is only in the base-2 interpretation! So what purpose does the decimal interpretation play here?
<!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Andreasabel|Andreasabel]] ([[User talk:Andreasabel#top|talk]] • [[Special:Contributions/Andreasabel|contribs]]) 19:01, 19 May 2019 (UTC)</small>
:For the proof of uncountability, it is sufficient to construct an ''injection'' ''T''→'''R'''. This is done by decimal interpretation. The remaining text is about constructing even a ''bijection'' ''T''→'''R''', using (a modification of) binary interpretation. I added an introductory sentence to make this more clear. - [[User:Jochen Burghardt|Jochen Burghardt]] ([[User talk:Jochen Burghardt|talk]]) 09:41, 20 May 2019 (UTC)
:PS: You can sign your comments by appending <nowiki>~~~~</nowiki> to them. This is strongly recommended on talk pages. - [[User:Jochen Burghardt|Jochen Burghardt]] ([[User talk:Jochen Burghardt|talk]]) 09:44, 20 May 2019 (UTC)
: {{re|Andreasabel}}
:# ''“Why mention the decimal interpretation of bit sequences? It does not help at all.”'' <br/> Because it's an encyclopedia, not a university lecture script, and its aim is to present facts and concepts to a widest audience. Whenever possible it's good to use a language and examples comprehensible for an 'ordinary reader', whose mathematical background may be as low as a mid-high school. A decimal notation is known to everyone, so this actually helps an 'ordinary reader' to grasp the idea of the injection we consider. If we talked about a binary interpretation, most of them would switch their thinking off, just because 'I have no idea what they're talking about'. Of course we may also include higher-level, more abstract and general examples – and this is in the box – but the basic description should be as simple as possible (but not simpler!).
:# ''“The assumption is that the infinite list of infinite bit streams is an enumeration of the reals in the unit interval. Which it is only in the base-2 interpretation!”'' <br/> No, and no. There is no such assumption, and that is not true for base-2 interpretation. The first ‘no’: we do not ''assume'' the list is an enumeration of anything (except the enumeration of binary sequences.) Instead, we ''prove'' (i.e. conclude at, not start from) it ''can be put into equivalence to'' an enumeration of numbers from some set. The second ‘no’: the equivalence is not direct or immediate even with binary representation of numbers, and that's because of tails of repeating digits: a repeating one in 0.101111... makes the representation equivalent to 0.110000... with repeating zero, hence the two ''different'' binary strings “101111...” and “110000...”, when simply transformed into binary representation of real numbers, become the same number. Hence the list of binary strings is ''not'' “an enumeration of the reals in the unit interval” by itself. We make a special trick with interleaving double representations, and only after that we get a desired bijection (“enumeration”).
: [[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 06:56, 21 May 2019 (UTC)
== Please fix the proof ==
In the current version, the proof of the [[special:permalink/953668864#Second theorem]] is incomplete. Can you fix it, please? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 16:24, 30 April 2020 (UTC)
:{{u|CiaPan}}, I think this would be more appropriate on the main talk page, together with a little more detail as to what you think is missing. I glanced at the section but haven't tried to figure out whether I think it's a complete proof or not (which is often a judgment call, since there's always a balance as to how much you expect the reader to fill in). --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 17:04, 1 May 2020 (UTC)
:<br>At another look, I notice that this complaint is not even about the [[Cantor's diagonal argument]] article, but about the (poorly named, but that's a separate issue) [[Cantor's first set theory article]] article. I think you should comment on the talk page there. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 18:49, 1 May 2020 (UTC)
== Victor Kosko's point of view ==
The proof provides one exception, to complete one proves it provides infinitely many. That is not, however, the major flaw. You Wikipedians claim it proves trans trans finite which it does not and Cantor knew that, and his proof is different from the articles in one point to bend it toward proving trans trans finite. He considers only hypothetical counts that are constructions. Presumably he came up with the diagonal argument to prove trans trans finite. The editors would not tolerate me fixing it, or commenting thus.
The recently added proof of his latter version was added to remove time from cantor's proof to supposedly make it prove trans trans finite, being a time based proof cannot, but the diagonal argument can be considered from a timeless perspective.
I do not wish to fix a small matter and ignore a major one. I want the new proof removed. [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 04:55, 1 May 2020 (UTC)
per Revision as of 19:17, 7 April 2020 of 'cantor's first set theory article' [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 05:07, 1 May 2020 (UTC)
== Set of all sets ==
In regards to the last two edits, you set theory editors are confused. The set of all sets to fit your argument would have to be 'size' ℵ1 Aleph1, but that would be the set of all sets of set theory constructions from finite length set theory sentences. The sets of the powerset of the powerset of the natural numbers are 'size' Aleph2 many sets for example, a subset of all sets.
If you take the powerset of the powerset etc of the empty set, for infinity that is aleph0 or ω many times, one gets all finite sets but only technically the set of natural numbers, which doesn't count same as the last natural number doesn't, so technically the powerset of the natural numbers is not included.
Instead one starts with all sets of finite sets the same 'size' cardinality of the natural numbers, or less, then repeatedly powersets that instead. The 'size' of that is Aleph(Aleph0). The statement it's powerset is larger is not correct. Aleph0 + 1 = Aleph0 and ω + 1 = ω , so the powerset has cardinality Aleph( (Aleph0 + 1) = Aleph0 ) so is the same 'size' and in fact the same set. In primitive set theory sets only consist of sets of sets, so if you think it through the set-all-sets contains itself (unless there are more than one of them which there are) and it's powerset and is it's own powerset.
Technically the set theory literature claims that set is not possible under the new axioms but is a class not containing itself.
Analogiesing to the Aleph0 Aleph1 diagonal argument situation isn't working.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 04:01, 31 May 2021 (UTC)
== diagonal counterexample ==
It would be an improvement if corrections are needed. Why would Wiki want to promote false ideas?
https://app.box.com/s/6u5ydjoo8f97dnsord7b49dff4quqlnz
This accesses a 3-page pdf showing Cantor's blind spot. Avoid extrapolating the binary system of symbols to any form of numbers.
Keep it simple, my lesson learned.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:54, 15 July 2021 (UTC)
I'm your only decent anti set theory person.
Per his argument, no.
The argument is correct, the rendition is garbage.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]])
Can't tell if that helps or not.
I updated the paper with the error (Cantor's blind spot).
[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 16:13, 23 July 2021 (UTC)
Your link is a specific link, not s general link. So the link needs to be updated.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 22:44, 1 August 2021 (UTC)
New link
https://app.box.com/s/bercujcbdvt3qdz544uwqb5w00gfhnjo
revised after a forum review.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:36, 11 August 2021 (UTC)
:"Since p differs from each S in the sample by construction, it will differ from all S in the list L, therefore a new sequence p will be formed not in the list L. The set of integers N is not sufficient to count the list L." This is correct and nothing that occurs later in the paper negates it. Fig.2 even illustrates how the numbers in the list L cannot be equivalent to the diagonal number. The rest of the post even contradicts itself about whether the diagonal number is included in the S. Whatever "blind spot" is here isn't Cantor's. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 18:38, 12 August 2021 (UTC)
: Main error number one.<br>The document linked begins with a claim: ''Assume a complete list L (...)''. It's not clearly defined what a 'list' is, but a sample presentation with symbols <math>S_1, S_2, S_3</math> etc. suggests it is a sequence. The term 'complete' in this context remains undefined as well, but the argument suggests it means 'containing all' (all possible binary sequences). So, the wannabe proof essentialy '''starts from assuming the intended conclusion'''. As such it proves nothing, it just makes a circle to present an assumption as a conclusion. <br>Error number two.<br>False premise allows arbitrary conclusion. The Cantor's proof shows precisely a ''sequence'' can not be 'complete' in the above sense - so the assumption quoted above is false. And false implies anything, hence the conclusion is meaningless. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 20:29, 12 August 2021 (UTC)
:: This article, linked from the article about the diagonal argument, should clear up some of your confusion: [[Proof by contradiction]]. See also [[Reductio ad absurdum]]. This has been a well-established form of logical argument for centuries. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 18:28, 13 August 2021 (UTC)
::: {{re|MartinPoulter}} You're clearly wrong. [[Proof by contradiction]] (read it, if you do not believe me) starts from assuming something ''opposite'' to what needs to be proven, and a contradiction following from that assumption ''disproves'' the assumption. It says: "Suppose NOT P. Now it follows that Q and NOT Q. Which is FALSE. Hence NOT (NOT P). As a result, P." On a contrary, the article in a PDF linked says: "Assume P. Then P. Hence P." Which is [[Begging the question]], an error known for more than two millenia. Nothing to do with a valid method of proving by contradiction. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 21:08, 20 August 2021 (UTC)
Cantor is attempting to show the need for 'transfinite numbers' by showing the infinite set N is not sufficient to form a 1 to 1 correspondence with an infinite set of sequences.
If he can form an additional sequence not in the list, the list is incomplete.The list of sequences is defined in section 1, the same as Cantor's, but using (0 and 1) in place of (m and w).
The binary tree (fig.1) is an overlay of all possible sequences, and by construction is complete if extended without limit. Its counterpart is the list L that contains all possible sequences itemized individually.
In fig.2, in swapping symbols, Cantor constructs the complement p of the diagonal d. The complement p is already in the list, but can't be detected with only one comparison.
Both are shown to be in the binary tree.
Cantor's conclusion:
''Since p differs from each S in the sample by construction'', it will differ from all S in the
list L, therefore a new sequence p will be formed not in the list L. The set of integers N is
not sufficient to count the list L.
Section 3 shows the basis (italics) for his conclusion is incorrect. Sequence p will differ from all minus one, itself, case 1. He uses case 2 as a basis.
It's a property of a set of unique elements, that each differ from the remainder of the set but not itself.
If p is new so is d, yet there it is in the list. Cantor contradicts himself.
It's simple but subtle and elusive. [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:36, 19 August 2021 (UTC)
: "The complement p is already in the list" Wrong: p is not in the list, and provably so, because p is different from every member of the list. "If p is new so is d" This does not logically follow. There is no necessity that d is not in the list, and the argument does not require that d is or is not in the list. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 19:55, 19 August 2021 (UTC)
Prove to yourself that the binary tree is complete. Form a short sequence of symbols that is not in the tree.
The tree is symmetrical, thus complementary S.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 16:19, 20 August 2021 (UTC)
His argument is error. Tree arguments do disprove set theory, however tree arguments and diagonal arguments (list, set) cannot correlate to one another, a common error of set theorists and anti set theorists.
[[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 03:33, 24 August 2021 (UTC)
The binary tree is complete by definition. After the first selection, 0 or 1, the process continues within the tree via a continuous loop. The set L contains itself at each position. You can follow any randomly selected sequence in L in the tree. There is a 1 to 1 correspondence for the tree and the list. The diagonal sequence is not special since orientation is not part of its definition.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 16:06, 24 August 2021 (UTC)
: {{re|Phyti}} You still try to prove nothing by nothing. You never defined what ''complete'' means, so 'being complete by definition' means exactly nothing. I have written ''what I think'' you mean by 'complete'. But that was just my guess, and explicitly emphasised as such. However, you never confirmed my guess nor fixed it, so your ''complete'' remains undefined, i.e. meaningless as 'a definition'. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 21:12, 26 August 2021 (UTC)
From Cantor's original 1891 paper:
"[Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements
E = (x1, x2, … , xv, …)
which depend on infinitely many coordinates x1, x2, … , xv, …, and where each of the coordinates is either m or w. Let M be the totality [Gesamtheit] of all elements E.
To the elements of M belong e.g. the following three:
EI = (m, m, m, m, … ),
EII = (w, w, w, w, … ),
EIII = (m, w, m, w, … ).
I maintain now that such a manifold [Mannigfaltigkeit] M does not have the power of the series 1, 2, 3, …, v, ….
This follows from the following 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."]"
The closest word to 'complete' is 'totality'.
His purpose is to show the set M cannot have a 1 to 1 correspondence with N, the set of integers, by forming an additional element E0 not in M. This would require his transfinite set aleph.
The binary tree is 'complete' by design for every position k accumulating 2^k branches.
The pdf shows why he can't detect the diagonal d and his construction p, and where his conclusion is incorrect. If his idea had truly been received from above, there would be no error.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 19:07, 29 August 2021 (UTC)
: I did not ask for your guesses about what Cantor could or wanted to prove. I did not ask what the word 'complete' is close to, either. I ask what the word 'complete' means. Specifically, and precisely, what does it mean 'the tree is complete'? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 16:52, 4 September 2021 (UTC)
It helps to read about the author to understand their ideas and their goal. They are not guesses.
The binary tree if extended without any boundary contains all possible sequences of 0 and 1.
That is why I ask anyone to attempt to form a short sequence that isn't contained in the tree. If the first position is 0 or 1, the sequence never leaves the tree![[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 17:09, 4 September 2021 (UTC)
After many revisions, my final answer. Posted on vixra as 2208.0022. No charge and no 'peer' review, but this is not rocket science!```` <!-- Template:Unsigned --><span class="autosigned" style="font-size:85%;">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Phyti|Phyti]] ([[User talk:Phyti#top|talk]] • [[Special:Contributions/Phyti|contribs]]) 16:08, 5 August 2022 (UTC)</span> <!--Autosigned by SineBot-->
== cantor complaint ==
In his 1891 article, Cantor considered the set T of all infinite sequences {sn} of
binary digits (i.e. consisting only of zeroes and ones). He then produced a
constructive proof of the following theorem:
If s1, s2, …, sn, … is ''any enumeration*'' of elements(which 1-1 corresponds to
the natural numbers or is a single thread as it were ) , then there always
exists some element s which corresponds to none of {sn } in that enumeration s.
To look at his proof, and then as a complaint associated to that proof: Consider
some “1-1 enumeration” consisting of an arbitrary collection C{of members sn},
and the dual of that collection (ie. a 1,0 mirror imaging C => D) ; where then
C and D may be bonded into a single-thread-initial-construction by imagining the
representations below of C,D,s,cs embedded in parallel planes, ie. for example
( C as viewed immediately over D over s-cs ) and as such any constructive thread of C may be looped
or pulled down into D for-and-at each so defined-corresponding-(1,0)site, and into s and cs only on the diagonal sites,
and thus for any-and-all C it’s dual D is and will automatically associate to it as also a single enumeration* C/D-s,cs.
C D S Cs
s1 =(0,1,0,0,0,1,0,...)______ d1 =(1,0,1,1,1,0,1,...)______s=(1,0,1,1,1,0,1,...)______cs = (0,1,0,0,0,1,0,...)
s2 =(1,1,1,1,1,1,1,...)______ d2 =(0,0,0,0,0,0,0,...)
s3 =(1,1,0,1,0,1,0,...)______ d3 =(0,0,1,0,1,0,1,...)
s4 =(1,1,1,0,1,0,1,...)______ d4 =(0,0,0,1,0,1,0,...)
s5 =(1,1,1,1,0,1,1,...)______ d5 =(0,0,0,0,1,0,0,...)
s6 =(1,1,1,1,1,1,1,...)______ d6 =(0,0,0,0,0,0,0,...)
s7 =(1,1,1,1,1,1,0,...)______ d7 =(0,0,0,0,0,0,1,...)
...
Cantor then "constructs" the sequence s just from C by choosing its nth digit as equal to the complement
to the nth digit on the diagonal of C. That is:
s=(1,0,1,1,1,0,1,...)
He next argues by construction that, s differs from each sn, in C since their nth digits differ, and
Hence, s cannot occur in the enumeration C.
Based on this theorem, Cantor then uses an indirect argument to show that:
. The set T is uncountable.
However, as a D' may also be seen as automatically bonded to any C' then by construction a derived initial C/D-s,cs always exists
such that
(s is contained in C/D-s,cs) but not C. and thus a complaint is posed that it seems the proof is maybe somehow higher dimensionally saturated dual flawed;
as claim:: there is no cantor-type-proof acceptable anti diagonalization of the set C/D-s,cs ,and
where s,cs are actually a visualization stepping stone and are unnecessary: As any further "construction" on or thru any C/D already implies an
associated matching construction on or thru its dual D/C which still produces a counter-example and complaint to cantors proof. (eg. repeat this
for any case s' , all that is really needed is C/D) as a countable set, and then it exists as a counter example
to the above statement '['any enumeration*']' .
- Crap
[[User:Pwaaben|Pwaaben]] ([[User talk:Pwaaben|talk]]) 13:51, 1 May 2022 (UTC)
Your argument technically does not deal with the diagonal, you say mirror instead of NOT, anti diagonal instead of NOT of the diagonal, and technically are changing his argument from EXACTLY his argument by yours which also has no actual argument in it's conclusion.
Clarify [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 20:56, 3 May 2022 (UTC)
:i hope this provides you a conclusion [[User:Pwaaben|Pwaaben]] ([[User talk:Pwaaben|talk]]) 09:00, 4 May 2022 (UTC)
== diagonal = contradiction ==
Cantor’s diagonal argument leads to a contradiction. Assume one wishes to list all the real numbers. Cantor’s diagonal real D cannot exist until the list is completed. But the list is not completed if there exists a real that isn’t on it. Therefore the existence of D entails that both the list is completed and the list in not completed. [[Special:Contributions/71.184.87.187|71.184.87.187]] ([[User talk:71.184.87.187|talk]]) 19:18, 21 May 2022 (UTC)
:Yes, that is exactly the structure of the argument. Are you posting it just to confirm that you've understood it? --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 20:54, 21 May 2022 (UTC)
Well, I understand that the existance of D means that both a proposition and its negation are true. Thus D does not exist. Do you understand that? [[Special:Contributions/71.184.87.187|71.184.87.187]] ([[User talk:71.184.87.187|talk]]) 13:52, 1 June 2022 (UTC)
:My previous posts failed to disprove the CDA, primarily for not recognizing the real flaw. Persistence won.
:It's non of the factors debated, but so subtle, it's almost invisible. Following the policy of not submitting original work here, I won't. Just informing the moderators and those who were deceived, and those who are free thinkers, it exists.[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]])phyti. [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 16:26, 13 June 2023 (UTC)
::I have returned to one of the original objections to Cantor's diagonal argument. Despite the fact that finite lists are not square, using his (v, u) coordinate system in his 1891 paper for all v, he assumes v=u for an infinite list. The progression of the geometric form of finite lists is exponential with a trend opposite to square.
::No one has ever formed an infinite list.
::Here is an example of a finite list with {0,1} substituted for {m,w} showing no missing sequence.
::2^3 = 8
:: position
::List 1 2 3
::000 x
::001 x x
::010 x x
::011 x x x
::100
::101 x
::110 x
::111 x x
::Since the list is random order, the 2nd diagonal is arbitrarily selected.
::D= 011.
::Its negation E (=not D) excludes 0 from position 1, and 1 from positions 2 and
::3.
::E=100, as the only sequence NOT excluded.
::A sequence cannot differ from itself [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 16:52, 4 January 2024 (UTC)
:::Freedom of expression opposes censorship. Here is the link to the latest paper,'Cantor's Illusion".
:::[https://drive.google.com/file/d/17Zz39YzAEEvD5pWB0f4NihEucQfZ3RHU/view?usp=sharing] [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:12, 7 January 2024 (UTC)
::[https://drive.google.com/file/d/17Zz39YzAEEvD5pWB0f4NihEucQfZ3RHU/view?usp=sharing] [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:17, 7 January 2024 (UTC)
:::Do I detect censorship here? [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:23, 7 January 2024 (UTC)
:::: It is possible to match each of the three-element strings from {0,1} with a number from 1 through 8. Here is an example such a matching:
:::: 1 '''0'''11
:::: 2 1'''0'''1
:::: 3 11'''0'''
:::: 4 111
:::: 5 010
:::: 6 100
:::: 7 000
:::: 8 001
:::: The sequence appearing along the diagonal is '''000'''. After adding one to each entry, you obtain '''111''', but this already appears on the list, at spot number 4. Does this refute Cantor's diagonal argument?
:::: The answer is that it does not. There is a key difference that separates this from Cantor's diagonal argument, which is that there are more rows in the list then there are columns. In Cantor's argument, the diagonal drawn through the list meets each row one time. This is key to the argument, because the reason the number obtained from the diagonal is not on the list is because it differs from the 1st number on the list at the 1st decimal place, it differs from the 2nd number on the list at the 2nd decimal place, etc. However, in this example, the diagonal misses some rows altogether. '''111''' is indeed different from the 1st listed string (011) because it differs in the 1st character, and different from the 2nd listed string at the 2nd character, and the 3rd listed string... but then, there is no reason that '''111''' should different from the 4th, 5th, etc. strings on the list, since we have run out of characters to compare. So it is fine for there to be a matching from the three-element strings from {0,1} with numbers from 1 through 8, since the diagonal argument needs the diagonal to meet each row once before Cantor's proof by contradiction can kick in.
:::: However, this loophole never happens in Cantor's argument about real numbers. Given any proposed matching from the natural numbers to the real numbers, once you get the number from the diagonal and change each digit, even the trillionth number on the list must differ from this one since they both have trillionth decimal places, and they are different. There is no "running out of digits to compare" that happens with real numbers. [[User:C7XWiki|C7XWiki]] ([[User talk:C7XWiki|talk]]) 06:55, 5 March 2024 (UTC)
:::::I’m the only decent anti set theory person you have
:::::the exponential argument is valid, if only aplied to the proof exponential cardinal infinity does not equal the same infinity (aleph 0)
:::::one infinity is exponential of the other to cause the set of reals be ‘all sequences of 1s and 0s’ qualified the same infinite size as the diagonal, yet one infinity must be at least the same size as the other to be able to complete the diagonal.
:::::if you say both are simultaneously so that disproves the final conjecture.
:::::your assuming something absolutely true that is philosophical debatable instead [[User:Victor Kosko|Victor Kosko]] ([[User talk:Victor Kosko|talk]]) 01:18, 7 March 2024 (UTC)
:::::: > if you say both are simultaneously so that disproves the final conjecture.
:::::: Exactly, that is where the contradiction is.
:::::: Cantor's argument is just a plain proof by contradiction, no more special than the proof of irrationality of the square root of 2. In the proof of irrationality of <math>\sqrt{2}</math>, you start out by making the totally wrong assumption that assume that <math>\sqrt{2}</math> can be written as a ratio of integers <math>\frac{a}{b}</math> reduced to lowest terms. Then over the course of the proof you show that the assumption is self-refuting, by producing proof that <math>\frac{a}{b}</math> was not really reduced to begin with. Even though something totally wrong (the claim that <math>\sqrt{2}=\frac{a}{b}</math> for integers <math>a,b</math>) is assumed at the beginning, this does not make the proof wrong. Instead it is what makes the proof work at all, since it works by asking "what if <math>\sqrt{2}</math> were rational?" and then showing that assumption is nonsensical, therefore showing that the only option that makes sense is for <math>\sqrt{2}</math> to be irrational.
:::::: Similarly, when proving that there are infinitely many primes by contradiction, you start the proof by making the totally wrong assumption that some finite set <math>\{p_1,\ldots,p_n\}</math> is an exhaustive list of the primes. Then the proof proceeds by showing that this assumption is absurd, by producing another prime that has to be outside of the set <math>\{p_1,\ldots,p_n\}</math>. Again, this proof starts by making the assumption "what if there were finitely many primes?", then showing that this assumption is nonsense, therefore showing the only correct option is that there are infinitely many primes.
:::::: However, with Cantor's diagonal argument, often people seem to reject the argument based on the claim that the assumption is nonsense. But this is the entire point of how proof by contradiction works, you start with the assumption that there is a complete list of the real numbers indexed by natural numbers, where when written out in decimal there are equally many columns as rows, and then show that this assumption is nonsense. Then the conclusion is that the only option that makes sense is for there to be '''no''' complete list of the real numbers with as many columns as rows, and this is what Cantor proved. [[User:C7XWiki|C7XWiki]] ([[User talk:C7XWiki|talk]]) 07:48, 7 March 2024 (UTC)
:::::::I'm tired and bored revising my paper to refute Cantor's diagonal argument 1891. The current version which I consider sufficient is available on google drive, if you or anyone else is interested. Remember it is not about any number system but the method itself. [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 15:13, 12 March 2024 (UTC)
::::::::Not about numbers, symbols, etc.
::::::::The diagonal is fake/imitation/counterfeit.
::::::::Latest revision only requires being conscious
::::::::https://drive.google.com/file/d/1J-5jNHE2raDD9xRwgwS6cTtrOBrTw7_j/view?usp=sharing [[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 18:46, 4 April 2024 (UTC)
:::::::::Just so we're clear, [[User:Phyti|Phyti]], you're [[preaching to the choir]] here: whatever you write here will do nothing to get your argument accepted or taken seriously by the mathematics community, and "[[I'm entitled to my opinion]]" is not an argument. Please also read [[WP:FREESPEECH]], [[WP:SOAPBOX]], and [[WP:NOR]].--[[User:Jasper Deng|Jasper Deng]] [[User talk:Jasper Deng|(talk)]] 20:50, 5 April 2024 (UTC)
[[Category:Wikipedia mathematical arguments]]
== Simple proof that this is bullshit ==
When you put the numbers from 0 to (exclusively) 1 on the right, but read them from right to left, what do you see? They become the natural numbers! But according to Cantor's argument this means that the set of natural numbers must be bigger than itself! [[Special:Contributions/2A02:8109:40B:EF00:D9BC:2CDE:BE30:15E8|2A02:8109:40B:EF00:D9BC:2CDE:BE30:15E8]] ([[User talk:2A02:8109:40B:EF00:D9BC:2CDE:BE30:15E8|talk]]) 07:43, 3 April 2025 (UTC)
:They don't become the natural numbers. Almost all of them will be infinitely long, and there are no infinitely long natural numbers. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 08:48, 3 April 2025 (UTC)
::That is only true if you don't allow natural numbers to be [[Actual infinity|actual infinite]], which does not make any sense in an infinite universe. [[Special:Contributions/2A02:8109:40B:EF00:F805:A5D1:158:660B|2A02:8109:40B:EF00:F805:A5D1:158:660B]] ([[User talk:2A02:8109:40B:EF00:F805:A5D1:158:660B|talk]]) 14:37, 3 April 2025 (UTC)
:::That all natural numbers are finite comes from any straightforward definition of the natural numbers, such as the [[Peano axioms]]. It's not a question of what I allow or don't allow, and it's nothing to do with the size of the universe. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 14:50, 4 April 2025 (UTC)
: Could you, please, clarify, where exactly you'd like to start "reading from the right" the number {{math|1/{{radical|2}}}}...? <br>And what natural number do you expect to get as a result? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 15:32, 4 April 2025 (UTC)
:: Or: where would you like to start reading {{math|1/7}} (possibly easier than {{math|1/{{radical|2}}}})...? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 13:57, 7 April 2025 (UTC)
:: Hello... Anybody out there...? --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 08:36, 16 April 2025 (UTC)
== Contradiction. ==
If you create a list of numbers contained in the matrix below, for example as in the link [[:File:Diagonal argument.svg]] then the diagonal method will not create a new number that is not on the list. The diagonal method creates a rational number in the next steps, so the question will probably be where on the list are irrational numbers. The question about the rational ___location of an irrational number in the matrix is a logical contradiction. To make it more interesting, let's assume that the matrix contains irrational numbers, but I only wrote down the initial digits, and you can add the missing digits yourself, creating e.g. 1/7 or any other. These additional digits that you add do not affect the fact that the diagonal method will not create a number that is not in the matrix. In the first step, writing the first digit after the decimal point creates the number that is in column A. In the second step, writing the second digit after the decimal point creates the number that is in column B. In the third step, writing the third digit after the decimal point creates the number that is in column C. In the "n" step, writing the "n" digit after the decimal point creates the number that is in column N.
A. B. C. D. ... n.
0. 0 0 0 0 0 0
1. 0.1 0.01 0.001 0.0001 ... ...
2. 0.2 0.02 0.002 0.0002 ... ...
3. 0.3 0.03 0.003 0.0003 ... ...
... ... ... ... ... ... ...
9. 0.9 0.09 0.009 0.0009 ... ...
10. 1.0 0.10 0.010 0.0010 ... ...
... ... ... ... ... ...
99. 0.99 0.099 0.0099 ... ...
100. 1.00 0.100 0.0100 ... ...
101. 0.101 0.0101 ... ...
... ... ... ... ...
999. 0.999 0.0999 ... ...
1000. 1.000 0.1000 ... ...
1001. 0.1001 ... ...
... ... ... ...
n. ... ... ...
[[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 14:18, 20 April 2025 (UTC)
:{{re|Krzysztof1137}} Unfortunately, your table contains only numbers with finite decimal representation. Arbitrarily long, but finite. So it does not even contain 1/3 or 3/7. As such it has nothing to do with a proof of uncountability of a set of all possible decimal strings. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 19:40, 20 April 2025 (UTC)
::The matrix was created only to contradict the diagonal method. It arranges numbers with all possible combinations of digits after the decimal point. You write: "Your table contains only numbers with a finite decimal representation." - Yes, but if you analyze the process of creating numbers in the diagonal method step by step, it turns out that each subsequent digit added creates a sequence of rational numbers, the same sequence of rational numbers as in any row of the matrix. You write: "Any length, but finite." - No. There is an infinite number of columns in the matrix, so in each row there is an infinite sequence of rational numbers, which means the matrix is not finite. You write: "So it does not even contain 1/3 or 3/7." - I wrote earlier that if you need irrational numbers, we can assume that the matrix contains the initial digits, and the remaining ones can be added, e.g.; in row 3, column A is 0.333... , the diagonal method does not check if all the numbers are there, it only creates a new number that is not on the list. A question for you; if in the following columns A, B, C... the number 0.3 , 0.33 , 0.333 ... etc. appears, does it mean that it approaches 0.333...? Can we say that it is 1/3? If so, then 1/3 exists, if not, the diagonal method will not create an irrational number. You write; "As such, it has nothing to do with the proof of the uncountability of the set of all possible decimal sequences." - I did not write anything about countability, I only mean the diagonal method itself. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 21:18, 20 April 2025 (UTC)
:::Can you please answer the question as to at what point 1/3 and 3/7 appear in your infinite list of lists? It seems clear that they don't. Your response goes straight to talking about irrational numbers. Please answer the question for rational numbers first, then answer it for irrational numbers. By telling us to "add the missing digits yourself", you're admitting that the irrational numbers don't appear on your lists. If your lists don't contain simple rational numbers like 1/3, nor any irrational number, then they obviously don't contain all the numbers from 0 to 1, or even most of them.
:::The diagonalisation process requires a list in which there is a sequence of digits for every natural number. Your process has multiple digit sequences for each natural number; in fact an apparently infinite number. So there isn't a diagonalisation process until you define how to combine all your lists into one list. [[User:MartinPoulter|MartinPoulter]] ([[User talk:MartinPoulter|talk]]) 09:54, 21 April 2025 (UTC)
::::I have already answered each of your questions earlier. The question about the ___location of irrational numbers written in the decimal positional system is impossible. The matrix contains an infinite number of infinite sequences of rational numbers. If you write down in a sequence of numbers, the numbers that arise step by step during the creation of a new number, which the diagonal method offers, then a sequence of rational numbers will be created, i.e. such a sequence as are in the matrix. You are asking about creating a list, I have previously indicated a link on how to write down the numbers from the matrix into a single list. You can also do this; A1 B1 A2 C1 B2 A3 D1 C2 B3 A4 E1 D2 ... However, if you learn how a matrix is built, and how a new number is created in the diagonal method, you will understand that this method is pointless, i.e. it will not create a new number that is not in the matrix. Now a question for you; 1. Can the diagonal method create an irrational number or an infinite sequence of rational numbers? 2. If the list does not contain all the numbers, will the diagonal method find a new number that is not on the list? I apologize if something was misunderstood, maybe it's because of Google Translate. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 17:21, 21 April 2025 (UTC)
::::I will describe it again to make everything clear. The matrix contains infinite sequences of rational numbers in the rows. The columns contain all possible numbers for any number of digits after the decimal point. Column "A" contains all numbers with one digit after the decimal point, column "B" contains all numbers with two digits after the decimal point, column "C" contains all numbers with three digits after the decimal point, etc... The number of columns is infinite. If you write the numbers in a list, e.g.; A1 B1 A2 C1 B2 A3 D1 C2 B3 A4 E1 D2 ... it turns out that the diagonal method will not create a number that is not on the list. The diagonal method only works on numbers written in the decimal positional system, in the next steps it creates subsequent rational numbers, if you write these numbers, an infinite sequence of rational numbers will be created. The diagonal method does not require the list to be complete, it does not check which number is missing, it creates a new number. -In the first step, we write down the first digit after the decimal point, the number that is in column "A" is created. Since column "A" contains all possible numbers with one digit after the decimal point, no new number is created that is not in column "A". -In the second step, we write down the second digit after the decimal point, the number that is in column "B" is created. Since column "B" contains all possible numbers with two digits after the decimal point, no new number is created that is not in column "B". -In the third step, we write down the third digit after the decimal point, the number that is in column "C" is created. Since column "C" contains all possible numbers with three digits after the decimal point, no new number is created that is not in column "C". -In the nth step, we write down the nth digit after the decimal point, the number that is in column "n" is created. Since column "n" contains all possible numbers with "n" digits after the decimal point, no new number is created that is not in column "n". I answered your questions, now you answer mine: - does the diagonal method create an irrational number or an infinite sequence of rational numbers? - if the list of numbers from the matrix is incomplete, why doesn't the diagonal method create a new number that is not on the list? [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 19:12, 25 April 2025 (UTC)
:::{{Re|Krzysztof1137}} ''"The matrix was created only to contradict the diagonal method."'' So it failed.<br>Apparently you confuse 'a possibility of expanding the decimal number infinitely' with 'having an infinite decimal number'. You certainly can find an arbitrarily long sequence of 3's after a decimal point in your table. But each such sequence is located in some specific row of the table and the number of the row determines how many threes there is in the sequence. You may have an infinite set of rows, but the ordinal number of each specific row is some natural number, which is certainly finite. As a result none of those sequences represents {{math|1/3}}. You definitely can find arbitrarily accurate aporoximations of one-third in your table, but not ''the'' {{math|1/3}} (let alone {{math|1/{{sqrt|3}}}}). --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 20:50, 21 April 2025 (UTC)
::::Now a question for you; 1. Can the diagonal method create an irrational number or an infinite sequence of rational numbers? 2. If the list does not contain all the numbers, will the diagonal method find a new number that is not on the list? [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 20:59, 21 April 2025 (UTC)
::::Please answer the questions. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 20:40, 25 April 2025 (UTC)
:::::I am not CiaPan, but:
:::::If you are referring to specifically the method from the article page, then my answers
:::::are both No: That creates/finds a sequence of binary digits, not a real number.
:::::(_That part_ is only to show that the set of
:::::infinite sequences of binary digits is uncountable.)
:::::If you are instead referring to
:::::" the real number such that its part before the decimal point is "0" and for all positions n after the decimal point, the digit in that position is the element of {1,2} with the opposite [[parity]] to the [[parity]] of the digit in position n after the decimal point of the [[Non-uniqueness of decimal representation and notational conventions|standard decimal representation]] of the n-th entry in the list "
:::::, then my answers are both yes: If you get different answers for this case, then:
:::::3. Do you think there is no real number
:::::"such that its part ... entry in the list"
:::::, like how there is no such integer for
:::::"the integer that is greater than all other integers" ?
:::::4. If you indeed think there is no real number
:::::"such that its part ... entry in the list"
:::::, then what about
:::::" the real number such that its part before the decimal point is "0" and for all positions n after the decimal point, the digit in that position is 3? "
:::::?
:::::[[User:JumpDiscont|JumpDiscont]] ([[User talk:JumpDiscont|talk]]) 23:06, 29 April 2025 (UTC)
::::::The diagonal method in the next steps creates the next numbers with a finite notation, if you write down all the possible numbers that can be created in the columns, then an infinite series of columns with numbers with a finite notation will be created. The columns contain all the possible numbers for any number of digits after the decimal point. Column "A" contains all the numbers with one digit after the decimal point. Column "B" contains all the numbers with two digits after the decimal point. Column "C" contains all the numbers with three digits after the decimal point. Column "n" contains all the numbers with "n" digits after the decimal point. The number of columns is infinite. If you write the numbers in a list, e.g.; A1 B1 A2 C1 B2 A3 D1 C2 B3 A4 E1 D2 ... etc. Only now can you use the diagonal method on the list, but we know that this list already contains all the numbers. There is a feedback loop that makes this list a trap for the diagonal method. There is no way to get beyond the numbers contained in the matrix. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 01:40, 30 April 2025 (UTC)
:::::::For the orderings you gave, the column number from the matrix will be [[Big O Notation|Big O]] of the square root of the position within the list - since each column is first reached at a position within the list that approximately the corresponding [[triangular number]] - so for all sufficienctly large n, the n-th number in that list will have a 0 for its n-th digit after the decimal point, which means the diagonal number will have a non-zero digit for its n-th digit after the decimal point.
:::::::More generally, for all lists of numbers, if [for all sufficiently large n, the n-th number in the list has a non-zero n-th digit after the decimal place] then for all sufficiently large n, at most n numbers in the list have only n digits after the decimal point, even though there are 10^n numbers (not necessarily all in the list, and indeed for such n they can't all be in the list) with only n digits after the decimal point.
:::::::[[User:JumpDiscont|JumpDiscont]] ([[User talk:JumpDiscont|talk]]) 04:18, 30 April 2025 (UTC)
::::::::A thought experiment and an analogy to a matrix. A drum counter like the ones used to have on car odometers, i.e. each column of the matrix is a drum with digits from zero to nine. That is, after the decimal point there is an infinite number of drums, the digits on the drums arranged in a line create a number with an infinite decimal expansion. The first drum after the decimal point rotates changing the digit, e.g. every second, the second drum rotates ten times faster than the first drum, the third drum rotates ten times faster than the second drum, the fourth drum rotates ten times faster than the third drum, etc. each subsequent drum rotates ten times faster than the previous one. After a full rotation of the first drum, i.e. after ten seconds, the counter will create all possible combinations of digits in the line, which will ultimately create all possible numbers from zero to one written in the decimal positional system. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 21:53, 29 June 2025 (UTC)
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