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 International Skeptics Forum Continuation Deeper than primes - Continuation 2

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 27th January 2018, 09:37 PM #2921 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by phiwum but it is nonsense to think that there are two different numbers 2. Unless redundancy is not ignored, for example: By the ,so called, "established mathematics" [A] and [A,A] are addressed by the distinct pairs {(A,1),(A,2)} exactly because each pair defines different redundancy degree. A = {1,2,3,...} (the set of all natural numbers). If redundancy degree > 1, in case of, for example [A,A] there can be more than 1 degrees of freedom of the mapping between the As copies, as addressed in http://www.internationalskeptics.com...postcount=2864. A = A is simply the particular case of the pair {(A,1)}, so nothing is fundamental here. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 27th January 2018 at 10:09 PM.
 27th January 2018, 09:48 PM #2922 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 See? 2 may not be a member of the set containing 2. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 02:25 AM #2923 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher See? 2 may not be a member of the set containing 2. Wrong, [2,2] is different than [2] exactly as the members of {(2,2),(2,1)} are different of each other by their redundancy degrees. In all the considered multisets above (whether the redundancy degree = or > 1) the copies of 2 are members of a given multiset. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 28th January 2018 at 03:51 AM.
 28th January 2018, 07:57 AM #2924 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi Wrong, [2,2] is different than [2] Why do you continue to respond to arguments no one has made? Of course the multiset [2, 2] is different from the multiset [2]. No one has suggested otherwise, but you seem compelled to dwell on it. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 08:06 AM #2925 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher Why do you continue to respond to arguments no one has made? Of course the multiset [2, 2] is different from the multiset [2]. No one has suggested otherwise, but you seem compelled to dwell on it. jsfisher, why do you take only a part of my reply to your wrong argument in http://www.internationalskeptics.com...postcount=2922 about 2 as if it is not a member of a given multiset ? __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 28th January 2018 at 08:11 AM.
 28th January 2018, 08:40 AM #2927 Little 10 Toes Graduate Poster     Join Date: Nov 2006 Posts: 1,866 Originally Posted by doronshadmi jsfisher, why do you take only a part of my reply to your wrong argument in http://www.internationalskeptics.com...postcount=2922 about 2 as if it is not a member of a given multiset ? Doronshadmi, why do you keep making up your own terms and why aren't you using the correct terms? Why do you continue to edit messages after people have replied to it? Why do you only take part of people's reply and use it wrong? __________________ I'm an "intellectual giant, with access to wilkipedia [sic]" "I believe in some ways; communicating with afterlife is easier than communicating with me." -Tim4848 who said he would no longer post here, twice in fact, but he did.
 28th January 2018, 08:52 AM #2928 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by jsfisher See? 2 may not be a member of the set containing 2. My phrasing was ambiguous and probably misleading for some, especially those for whom English was not their first language. I apologize. The meaning I meant to convey, the meaning as originally expressed by Doronshadmi when he made the claim was: 2 is not necessarily a member of {2}. It could be a different 2, or maybe 2 was busy off doing something else. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 02:03 PM #2929 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher My statement about 2 may not being a member of {2} was a bit of mockery. In other words, you wrote a reply that has nothing to do with http://www.internationalskeptics.com...postcount=2921, or in other words, another red herring of yours. Originally Posted by jsfisher You continue this conceptual failure with your understanding of multisets, but that aside for a moment, the multisets [2, 2], [2, 2, 2], and [2] are all different Exactly, they have different redundancy degrees among members' copies. Originally Posted by jsfisher By definition, distinct members may appear more than once in a multiset. And this is exactly the reason of why they have more than one degree of freedom, in case that at least two copies of infinite sets are mapped, as addressed in http://www.internationalskeptics.com...postcount=2864. Originally Posted by jsfisher It has nothing to do with the tortured vocabulary invention you keep trying to force on us. Worng jsfisher, you are the one that forces the notion of A=A, which is the particular case of {(A,1)}, as if it the only possibility in order to do Math. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 28th January 2018 at 02:10 PM.
 28th January 2018, 02:31 PM #2930 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher The meaning I meant to convey, the meaning as originally expressed by Doronshadmi when he made the claim was: 2 is not necessarily a member of {2}. It could be a different 2, or maybe 2 was busy off doing something else. No, 2 is an element. If this element is a member of some set, then this case may be written as {2} , {1,2,3} , {2, 4, 6, ...} , {{2}} etc. ad infinitum. The simple fact in this case is that nothing forces 2 to be a member of some non-empty set, since 2 can also be taken as its own mathematical universe even if it is taken in terms of a "pure" set like {{},{{}}} (Von Neumann constucation) or {{{}}} (Zeremalo costruction). Moreover, if 2 is not taken in terms of a "pure" set, then 2 is not a member of any framework that is defined in terms of "pure" sets. ------------------------------ So, jsfisher your red herring is not going to save you from your own artificial restrictions. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 28th January 2018 at 03:01 PM.
 28th January 2018, 02:39 PM #2931 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by Little 10 Toes Doronshadmi, why do you keep making up your own terms and why aren't you using the correct terms? I do not use the conventional terms, and by doing so I discover the different degrees of freedom of the mapping between at least two copies of a given infinite set under a given multiset, as done here. Generally, Mathematics is much more richer than any particular view, including my non-conventional view of the considered subject. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 28th January 2018 at 02:47 PM.
 28th January 2018, 03:14 PM #2932 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi I do not use the conventional terms No, you certainly don't. You also don't use the conventional concepts. So, you aren't talking about Mathematics at all when you are trying to show us how wrong it is. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 03:32 PM #2933 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi In other words, you wrote a reply that has nothing to do with http://www.internationalskeptics.com...postcount=2921, or in other words, another red herring of yours. If you really think that, then you are truly missing the meaning of your own words. You see, phiwum wrote: Originally Posted by phiwum it is nonsense to think that there are two different numbers 2. This is a perfectly correct statement. Mathematics doesn't need a Xerox machine. No matter how it is expressed, a single 2 is sufficient for all its needs. You then contradicted phiwum's statement with: Originally Posted by doronshadmi Unless redundancy is not ignored (By the way, you might want to look up the meaning of that word. Its really meaning is rather amusing in light of how you continue to abuse it.) You continue to labor under a conceptual block that prevents you from accepting that a multiset can have, for example, 2 as a member of it 17 times without the need for any copies. And this all ties quite nicely to comments you have made in the past about 2 and its possible non-membership in {2}. My post was completely topical and stood as a reminder to phiwum that you think copies are necessary. So, no, no red herrings were used in the making of my posts. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 03:35 PM #2934 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Out of curiosity, Doronshadmi, have you figured out yet you do not need to appeal to multisets to define mappings from N to N? If you wanted to discuss mappings, you could have gone directly there. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 28th January 2018, 11:01 PM #2935 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher You also don't use the conventional concepts. Generally, Mathematics is much more richer than any particular view, including my non-conventional view of the considered subject. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com )
 28th January 2018, 11:38 PM #2936 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher This is a perfectly correct statement. Mathematics doesn't need a Xerox machine. No matter how it is expressed, a single 2 is sufficient for all its needs. You jsfisher, and phiwum are simply closed under the notion of A=A that is simply the particular case of {(A,1)}. Originally Posted by jsfisher (By the way, you might want to look up the meaning of that word. Its really meaning is rather amusing in light of how you continue to abuse it.) It has more than one meaning, so you have no case. Originally Posted by jsfisher You continue to labor under a conceptual block that prevents you from accepting that a multiset can have, for example, 2 as a member of it 17 times without the need for any copies. It has 17 copies, but since your notion is restricted to {(A,1)}, you are not aware of it. Originally Posted by jsfisher And this all ties quite nicely to comments you have made in the past about 2 and its possible non-membership in {2}. My post was completely topical and stood as a reminder to phiwum that you think copies are necessary. Also http://www.internationalskeptics.com...postcount=2930 is not closed under your convectional notions. Originally Posted by jsfisher So, no, no red herrings were used in the making of my posts. It is a red herring, and in both paths your restricted notions can't catch the red herring nor the fox. -------------------------------------- Generally, Mathematics is much more richer than any particular view, including my non-conventional view of the considered subjects. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 29th January 2018 at 12:57 AM.
 29th January 2018, 01:07 AM #2937 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher Out of curiosity, Doronshadmi, have you figured out yet you do not need to appeal to multisets to define mappings from N to N? If you wanted to discuss mappings, you could have gone directly there. Out of curiosity, jsfisher, have you figured out that multiset is generalization of the concept of set, such that the mappings from N to N (such that redundancy is ignored) is only the particular case of redundancy = 1 ({(N,1)}). If you wanted to discuss mappings you can't cover the considered subject by {(N,1)} particular case. Jsfisher, under {(N,2)} ([N,N]) bijection is not the only possible mapping, but you are unaware of it since your notions of the considered subject are restricted to {(N,1)}. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 29th January 2018 at 01:16 AM.
 29th January 2018, 03:52 AM #2938 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do. In other words, their notion is restricted only to (A,1). __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com )
 29th January 2018, 05:18 AM #2939 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do. Phiwum and I understand multisets in terms of how the concept is defined. You, Doronshadmi, don't bother with such formalities. By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics? __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 29th January 2018, 05:47 AM #2940 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher Phiwum and I understand multisets in terms of how the concept is defined. Your understanding of, for example, [A,A] is understood in terms of {(A,1),(A,1)}. Originally Posted by jsfisher You, Doronshadmi, don't bother with such formalities. Wrong, you and Phiwum don't bother to comprehend the formality of [A,A] in terms of {(A,2)}. Originally Posted by jsfisher By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics? No mystery. (A,2) has redundancy degree > 1, where (A,1) has redundancy degree = 1. Under redundancy degree > 1 two copies of infinite sets have more than 1 degrees of freedom of the mapping between them, as very simply addressed in http://www.internationalskeptics.com...postcount=2864. {(A,2)} is a mystery if observed in terms of {(A,1),(A,1)}. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 29th January 2018 at 06:02 AM.
 29th January 2018, 06:03 AM #2941 Hevneren Thinker   Join Date: Jul 2007 Posts: 134 Originally Posted by jsfisher Phiwum and I understand multisets in terms of how the concept is defined. You, Doronshadmi, don't bother with such formalities. By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics? I think (A,1) represents the first A in the (apparently ordered) (multi)set, while (A,2) represents the second one. But I don't know which one of them is the original A and which one is the copy, or if the original A is stored in a safe place somewhere else, so nobody can put it in a set where it doesn't belong.
 29th January 2018, 06:54 AM #2942 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by Hevneren I think (A,1) represents the first A in the (apparently ordered) (multi)set, while (A,2) represents the second one. No, the second parameter, for example, in (A,1), (A,2), (A,3) etc. ... , defines the number of copies of a given member (the first parameter, which in this case is A) in a given multiset. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 29th January 2018 at 06:56 AM.
 29th January 2018, 07:31 AM #2943 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 If (x,y) is meant as an ordered pair where x is an element and y is whole number representing multiplicity, then, sure, that is a way to embed the multiset concept into set theory. (It isn't perfect, but it is workable.) E.g. { (2,3), (3,1), (5,1) } would be the multiset of the prime factors of 120 under this model. Doronshadmi, since no one has argued otherwise, why do you continue to argue as if we had? You have been doing that a lot of late, arguing against positions nobody has taken. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 29th January 2018, 09:42 AM #2945 phiwum Philosopher     Join Date: Aug 2010 Posts: 9,168 Originally Posted by doronshadmi To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do. In other words, their notion is restricted only to (A,1). Nonsense. The multiset [A,A] is obviously represented in set theory as {(A,2)}. Duh. But A had no referent here, unless you mean that for all sets A this is the case. Which is, of course, true. Thus, in [A,A], the single set A is represented twice, but it is nonetheless the same A.
 29th January 2018, 10:24 AM #2946 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi jsfisher, you are still missing the notion of redundancy as defined, for example, in {(2,3)}, because your notion is restricted only to redundancy = 1 (defined in this case as {(2,1),(2,1),(2,1)}). Why do you say that? Other than your insistence on using the wrong term (multiplicity is the correct one), [2, 2, 2] and { (2,3) } are identical constructs (in terms of the context of this thread). It is not clear why you brought up { (2,1), (2,1), (2,1) }. Just two more instances of you arguing against things no one else claimed. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 29th January 2018, 09:19 PM #2947 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 More of the same is not just the same. A = {1,2,3,...} (the set of all natural numbers) [A,A] is more of the same, such that bijection is only one case among the As under [A,A], as very simply addressed in http://www.internationalskeptics.com...postcount=2864. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 29th January 2018 at 09:21 PM.
 30th January 2018, 05:57 AM #2948 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi More of the same is not just the same. A = {1,2,3,...} (the set of all natural numbers) [A,A] is more of the same That is an odd way to express it, but no one has argued to the contrary. Will all of your posts be striking out against strawmen? Quote: ...such that bijection is only one case among the As under [A,A] The multiset [A, A] does not include nor imply any sort of mapping from A to A, bijection or otherwise. It is just a multiset with A as a member with multiplicity of 2. Mappings, on the other hand, require three things: a domain, a co-domain, and a functional relationship from domain to co-domain. Please stop conflating the two. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 30th January 2018, 08:08 AM #2949 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher The multiset [A, A] does not include nor imply any sort of mapping from A to A, bijection or otherwise. I agree with you. Originally Posted by jsfisher It is just a multiset with A as a member with multiplicity of 2. [A, A] is an example of more of the same, such that redundancy > 1. Since by convectional mathematics multiplicity is not understood in terms of redundancy, one wrongly understands, for example, {(A,2)} in terms of {(A,1),(A,1)} without being aware of his\her failure. Here is a concrete example: Since jsfisher is unaware of the redundancy in [2,2,2] he wrongly understands {(2,3)} (which is actually an expression that does not ignore the redundancy), in terms of {(2,1),(2,1),(2,1)} which ignores the redundancy in [2,2,2], where this reply Originally Posted by jsfisher It is not clear why you brought up { (2,1), (2,1), (2,1) }. very simply demonstrates jsfisher's non-awareness of the considered subject. Originally Posted by jsfisher Mappings, on the other hand, require three things: a domain, a co-domain, and a functional relationship from domain to co-domain. If mappings is considered among the As under [A, A] such that A is an infinite set, then since [A, A] is more of the same (where redundancy is not ignored) bijection is only one case among the As under [A,A], as very simply addressed in http://www.internationalskeptics.com...postcount=2864. In other words, jsfisher, you continue your off topic replies, exactly because you understand {(A,2)} (redundancy is not ignored) in terms of {(A,1),(A,1)} (redundancy is ignored). __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 30th January 2018 at 09:08 AM.
 30th January 2018, 09:02 AM #2950 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 As for phiwum, he wrongly thinks that I claim that [2,2] has two different members: Originally Posted by phiwum but it is nonsense to think that there are two different numbers 2. But this is not my argument. My argument simply does not ignore the redundancy > 1 in a given multiset, and in case that the redundancy > 1 is not ignored among the mappings between redundant infinite members, bijection is not necessarily the one and only one mapping, as very simply addressed in http://www.internationalskeptics.com...postcount=2864. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 30th January 2018 at 09:04 AM.
 30th January 2018, 10:12 AM #2951 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi Originally Posted by jsfisher The multiset [A, A] does not include nor imply any sort of mapping from A to A, bijection or otherwise. I agree with you. Originally Posted by jsfisher It is just a multiset with A as a member with multiplicity of 2. [A, A] is an example of more of the same, such that redundancy > 1. No, you do not get to redefine Mathematics in your desperate quest to disprove it. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 30th January 2018, 02:02 PM #2952 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher No, you do not get to redefine Mathematics in your desperate quest to disprove it. Mathematics is not a property of anyone or any group of people not in the past, not now and not in the future. Your desperate quest to close it under dogmas actually transforms it into a religion. A more fruitful way is to reply in details to http://www.internationalskeptics.com...hp?p=12165528& , but currently it seems that your best reply is "you do not get to redefine Mathematics". __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 30th January 2018 at 02:53 PM.
 30th January 2018, 04:39 PM #2953 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,513 Originally Posted by doronshadmi Mathematics is not a property of anyone or any group of people not in the past, not now and not in the future. Property rights have nothing to do with your misrepresentation of Mathematics. __________________ A proud member of the Simpson 15+7, named in the suit, Simpson v. Zwinge, et al., and founder of the ET Corn Gods Survivors Group. "He's the greatest mod that never was!" -- Monketey Ghost
 30th January 2018, 07:17 PM #2954 Little 10 Toes Graduate Poster     Join Date: Nov 2006 Posts: 1,866 Originally Posted by doronshadmi Mathematics is not a property of anyone or any group of people not in the past, not now and not in the future. Your desperate quest to close it under dogmas actually transforms it into a religion. A more fruitful way is to reply in details to http://www.internationalskeptics.com...hp?p=12165528& , but currently it seems that your best reply is "you do not get to redefine Mathematics". Doronshadmi, can I get my orange juice metric please? Extra 50 weight. __________________ I'm an "intellectual giant, with access to wilkipedia [sic]" "I believe in some ways; communicating with afterlife is easier than communicating with me." -Tim4848 who said he would no longer post here, twice in fact, but he did.
 31st January 2018, 07:22 AM #2955 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Originally Posted by jsfisher Property rights have nothing to do with your misrepresentation of Mathematics. You get Mathematics as if it is the non-redefined property of a group of people that you are one of its members, and this is exactly your misrepresentation of Mathematics. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 31st January 2018 at 08:16 AM.
 31st January 2018, 09:45 PM #2956 Little 10 Toes Graduate Poster     Join Date: Nov 2006 Posts: 1,866 How about instead of making up words with your own private definitions, you work on using the correct terms? __________________ I'm an "intellectual giant, with access to wilkipedia [sic]" "I believe in some ways; communicating with afterlife is easier than communicating with me." -Tim4848 who said he would no longer post here, twice in fact, but he did.
1st February 2018, 01:48 AM   #2957
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Join Date: Mar 2008
Posts: 12,721
Let's generalize the mathematical notion of strict membership (which is a particular case of Fuzzy logic).

Let E be a placeholder of any entity.

Let # be a placeholder of any strict membership.

Let (E,#) be a generalization of any strict membership of any entity.

Multiset is a generalization of the concept of set.

Any entity with strict membership 0 ( notated as (E,0) ), can't be defined but as E (it is strictly not a member of any multiset).

Let [] (empty multiset) be a multiset of any entity with strict membership 0 ( notated as (E,0) ).

Any entity with strict membership 1 ( notated as (E,1), can be defined as E = E (it is a strict and unique member of any multiset).

Convectional mathematics is the mathematical framework of strict membership of the form (E,1), and it needs the axiomatic method in order to define [] in terms of (E,1), instead of simply deduce it in terms of strict membership 0 ( notated as (E,0) ).

Let A = [1,2,3,… etc.]

Let S = [A,A]

S is a (E,2) mathematical framework, which enables mapping among infinite multisets that is not restricted to (E,1) mathematical framework, such that bijection is only a particular case among As under S, exactly because A is a strict but non-unique member of S (as very simply addressed in http://www.internationalskeptics.com...postcount=2864).

Exactly as one can't claim that strict membership is the one and only one possible membership (since it is simply a particular case of fuzzy logic), one also can't claim that (E,1) is the one and only possible strict membership.

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 Edited by Agatha: Edited to remove rule 12 breach

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Mathematics is indeed deeper than primes.
__________________
As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help.
----
If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com )

Last edited by Agatha; 3rd February 2018 at 01:41 PM.

 6th February 2018, 08:54 AM #2958 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,721 Here is some quote taken from Russell's article about the axiom of choice : Quote: The boots-and-socks metaphor was given in 1919 by Russell 1993, pp. 125–127. He suggested that a millionaire might have ℵ0 pairs of boots and ℵ0 pairs of socks. Among boots we can distinguish right and left, and therefore we can make a selection of one out of each pair, namely, we can choose all the right boots or all the left boots; but with socks no such principle of selection suggests itself, and we cannot be sure, unless we assume the multiplicative axiom, that there is any class consisting of one sock out of each pair. Russell generally used the term "multiplicative axiom" for the axiom of choice. Referring to the ordering of a countably infinite set of pairs of objects, he wrote: There is no difficulty in doing this with the boots. The pairs are given as forming an ℵ0, and therefore as the field of a progression. Within each pair, take the left boot first and the right second, keeping the order of the pair unchanged; in this way we obtain a progression of all the boots. But with the socks we shall have to choose arbitrarily, with each pair, which to put first; and an infinite number of arbitrary choices is an impossibility. Unless we can find a rule for selecting, i.e. a relation which is a selector, we do not know that a selection is even theoretically possible. Russell then suggests using the location of the centre of mass of each sock as a selector. The notion that defines a set as a collection of distinct members, is actually based on an act of selection. In order to be aware of it, all one needs is to use the notion of membership 0, such that exactly one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given set (in this case the used notion is membership 1, such that, for example, {2} , {3,2} , {1,2,3,...} etc. are all cases of membership 1). {2,2} , {2,3,2} , {1,2,2,2,2,2,2,2,3,...} etc. are all examples of membership > 1, such that more than one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) are chosen as members of a given multiset. If multiset is defined as a generalization of the concept of set, then: [] is the case that no element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given multiset. [2] is the case that one element (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) is chosen as a member of a given multiset. [2,2] is the case that two elements (out of infinitely many totally isolated and identical elements (for example: infinitely many identical socks that are not members of any set)) are chosen as members of a given multiset. etc. ... ad infinitum. Convectional mathematics is simply the particular case of choosing exactly one element (out of infinitely many totally isolated and identical elements (for example: infinitely many yellow identical socks that are not members of any multiset, infinitely many rad identical socks that are not members of any multiset etc. ad infinitum)) such that infinitely many membership 1 multisets have no more than one yellow sock and one rad sock. __________________ As long as notion is impossible because of partial usage of one's brain skills, new glasses will not help. ---- If a tree falls in the forest, and no one’s there to see it, the tree and ground still measure each other. ( http://www.askamathematician.com ) Last edited by doronshadmi; 6th February 2018 at 09:32 AM.

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