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 International Skeptics Forum Continuation Deeper than primes - Continuation 2

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 27th September 2017, 07:43 PM #2761 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,793 Originally Posted by doronshadmi Please observe 111...111. or 111111. The first is infinite... No, the first is nonsense as an infinite sequence. Once again, you are trying to present an infinite sequence that terminates. You really need to stop doing that. Infinite sequences do not terminate. __________________ 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
 27th September 2017, 08:02 PM #2762 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by jsfisher Infinite sequences do not terminate. 111...111. is not terminated. The details are given in http://www.internationalskeptics.com...postcount=2760. __________________ 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 )
 27th September 2017, 11:24 PM #2763 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by jsfisher Your own special view of Mathematics once again gets you the wrong answer. And once again, I will remind you that you don't get to redefine Mathematics to accommodate your misunderstandings. You binary tree has countably infinite levels. None of them correspond to an infinite level. These facts do not change just because you may disagree. jsfisher, by deducing in terms of wholeness instead of in terms of completeness (as already given in http://www.internationalskeptics.com...postcount=2753) one redefines Mathematics. If you disagree with this redefinition, then please demonstrate its inconsistency in terms of wholeness. In order to do it both your visual_spatial AND verbal_symbolic brain skills must be activated, in order to demonstrate its inconsistency in terms of wholeness. __________________ 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 September 2017 at 11:32 PM.
 28th September 2017, 03:00 AM #2764 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 In http://www.internationalskeptics.com...postcount=2733 I was talking against ℵ0 as a fixed size, which means that by the fixed size argument, for example, ℵ0 = ℵ0 + 1. In http://www.internationalskeptics.com...postcount=2740 I show why this argument does not hold, in case that The Infinite Binary Tree is taken as a whole (which is not the same as complete). __________________ 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 September 2017 at 03:05 AM.
 28th September 2017, 03:52 AM #2765 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Since the last discussion is based on the differences between Complete and Whole, please look at these links, for example: https://www.quora.com/What-is-the-di...e-and-complete https://english.stackexchange.com/qu...hole-vs-entire http://wikidiff.com/wholeness/completeness If the set of natural numbers is taken as a whole, it does not mean that new natural numbers cannot be added to it, since being a whole is being both variant AND invariant without getting into contradiction (as shown in http://www.internationalskeptics.com...postcount=2740). __________________ 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 September 2017 at 04:22 AM.
 28th September 2017, 04:57 AM #2766 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,793 Originally Posted by doronshadmi Originally Posted by jsfisher Infinite sequences do not terminate. 111...111. is not terminated. You clearly meant 111...111. to represent an infinite sequence, one that comes to an abrupt end just before that binary point. __________________ 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 September 2017, 05:29 AM #2767 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by jsfisher You clearly meant 111...111. to represent an infinite sequence, one that comes to an abrupt end just before that binary point. EDIT: 111...111. is a unique path of bits that permanently growing from within as a whole, so the radix point (which is not any one of the bits) is the level of The Infinite Binary Tree that defines the unique path of The Infinite Binary Tree as its infinite whole organ. __________________ 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 September 2017 at 05:47 AM.
 28th September 2017, 06:48 AM #2768 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,793 Originally Posted by doronshadmi EDIT: 111...111. is a unique path of bits that permanently growing from within as a whole, so the radix point (which is not any one of the bits) is the level of The Infinite Binary Tree that defines the unique path of The Infinite Binary Tree as its infinite whole organ. If at any point you have any Mathematics to present and discuss, let me know. Meanwhile, enjoy your fantasy play with words and malformed ideas. __________________ 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 September 2017, 07:05 AM #2769 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by jsfisher If at any point you have any Mathematics to present and discuss, let me know. Meanwhile, enjoy your fantasy play with words and malformed ideas. Thank you jsfisher for the discussion, it enabled me to express my non-standard notions by further details that are also taken as a whole. I wish you the best. __________________ 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 September 2017 at 07:06 AM.
 28th September 2017, 02:44 PM #2770 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Ok, let's continue. The first level along The Infinite Binary Tree, which is mapped with the smallest cardinal number that is greater than any finite cardinal number, defines the ℵ0 domain along the tree, where each cardinal number that is mapped with a given level in that domain, must be represented by at least ℵ0 bits. The first level along The Infinite Binary Tree, which is mapped with the smallest cardinal number that is greater than any infinite cardinal number of ℵ0 domain, defines the ℵ1 domain along the tree, where each cardinal number that is mapped with a given level in that domain, must be represented by at least ℵ1 bits. The first level along The Infinite Binary Tree, which is mapped with the smallest cardinal number that is greater than any infinite cardinal number of ℵ1 domain, defines the ℵ2 domain along the tree, where each cardinal number that is mapped with a given level in that domain, must be represented by at least ℵ2 bits. etc. ... .<== radix point So, 111...111. is used (without loss of generality) in order to represent infinite cardinal numbers that are mapped with any infinite level in The Infinite Binary Tree. __________________ 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 September 2017 at 02:58 PM.
 29th September 2017, 07:04 AM #2771 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Cardinality and positional notation If one orderly marks each level of The Infinite Binary by cardinal numbers that are represented by bits, then the amount of places that is needed in order to represent each cardinal number is, at least, equal to the finite or infinite level number. The first finite level (which includes bits) of The Infinite Binary, is covered by a cardinal number that is represented by at least 12 places. The second finite level (which includes bits) of The Infinite Binary, is covered by a cardinal number that is represented by at least 102 places. The third finite level (which includes bits) of The Infinite Binary, is covered by a cardinal number that is represented by at least 1002 places. ... The first infinite level (which includes bits) of The Infinite Binary, which is marked by the smallest cardinal number that is greater than any finite cardinal number, is represented by at least ℵ0 places. ... The first infinite level (which includes bits) of The Infinite Binary, which is marked by the smallest cardinal number that is greater than any infinite cardinal number in domain ℵ0, is represented by at least ℵ1 places. By this observation ℵ1 > 2ℵ0 since ℵ1 is greater than any cardinal number that is involved with ℵ0 (so CH is solved). ... The first infinite level (which includes bits) of The Infinite Binary, which is marked by the smallest cardinal number that is greater than any infinite cardinal number in domain ℵ1, is represented by at least ℵ2 places. ... __________________ 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 September 2017 at 08:13 AM.
 29th September 2017, 11:00 AM #2772 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Cardinality and positional notation - The corrected post Let's correct the previous post. It has to be written as follows: If one orderly marks each level of The Infinite Binary by cardinal numbers that are represented by bits, then the amount of places that is needed in order to represent each cardinal number is, at least, equal to the finite (> 0) or infinite level's place in the The Infinite Binary, as follows: The first finite level (which includes bits) of The Infinite Binary, is covered by a cardinal number (represented by bits) that is represented by at least 12 places. The second finite level of The Infinite Binary, is covered by a cardinal number that is represented by at least 102 places. The third finite level of The Infinite Binary, is covered by a cardinal number that is represented by at least 1002 places. ... The first infinite level of The Infinite Binary (which is marked by the smallest infinite cardinal number that is greater than any finite cardinal number) is covered by an infinite cardinal that is represented by at least ℵ0 places. ... The first infinite level of The Infinite Binary (which is marked by the smallest infinite cardinal number that is greater than any infinite cardinal number in domain ℵ0) is covered by an infinite cardinal that is represented by at least ℵ1 places. By this observation ℵ1 > 2ℵ0 since ℵ1 is greater than any infinite cardinal number (represented by bits) that is involved with ℵ0 (so CH is solved). ... The first infinite level (which includes bits) of The Infinite Binary (which is marked by the smallest infinite cardinal number that is greater than any infinite cardinal number in domain ℵ1) is covered by an infinite cardinal that is represented by at least ℵ2 places. ... __________________ 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 September 2017 at 12:52 PM.
 29th September 2017, 05:30 PM #2773 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Some CH observation Ok, I do not need anymore the radix point in order to show the following: By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, is covered by a cardinal number (represented by bits) that is represented by 12 places. The term “covered by” means that 21 numbers (represented by bits) can be represented by 12 places. Generally, the number of places is determined by number x , which is used as the power value of any expression of the form 2x . The second finite level of The Infinite Binary, is covered by a cardinal number that is represented by 102 places. The third finite level of The Infinite Binary, is covered by a cardinal number that is represented by 1002 places. ... The first infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ0 places (it means that ℵ0 places can represent any cardinal number from 0 up to 2ℵ0 ). ... The second infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ1 places (it means that ℵ1 places can represent any cardinal number from 0 up to 2ℵ1). By this observation ℵ1 > 2ℵ0 ... The third infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ2 places (it means that ℵ2 places can represent any cardinal number from 0 up to 2ℵ2). ... etc. __________________ 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 September 2017 at 05:33 PM.
 30th September 2017, 01:13 AM #2774 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Some GCH observation Since the observation in the previous post holds for any base (finite or infinite) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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 )
 30th September 2017, 06:56 AM #2775 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 GCH (important typo correction) By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, is covered by a cardinal number (represented by bits) that is represented by 1 places. The term “covered by” means that 21 numbers (represented by bits) can be represented by 1 places. Generally, the number of places is determined by number x , which is used as the power value of any expression of the form 2x . The second finite level of The Infinite Binary, is covered by a cardinal number that is represented by 2 places. The third finite level of The Infinite Binary, is covered by a cardinal number that is represented by 3 places. ... The first infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ0 places (it means that ℵ0 places can represent any cardinal number from 0 up to 2ℵ0 ). ... The second infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ1 places (it means that ℵ1 places can represent any cardinal number from 0 up to 2ℵ1). By this observation ℵ1 > 2ℵ0 ... The third infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ2 places (it means that ℵ2 places can represent any cardinal number from 0 up to 2ℵ2). ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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 September 2017 at 08:30 AM.
 30th September 2017, 09:59 AM #2776 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Some GCH observation, without the nonsense that I wrote before Some GCH observation, without the nonsense that I wrote before. By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, needs cardinal number with 1 bit in order to represent 21 cardinal numbers. The second finite level of The Infinite Binary, needs cardinal number with 2 bits in order to represent 22 cardinal numbers. The third finite level of The Infinite Binary, needs cardinal number with 3 bits in order to represent 23 cardinal numbers. ... The first infinite level of The Infinite Binary, needs cardinal number with ℵ0 bits in order to represent 2ℵ0 cardinal numbers. The second infinite level of The Infinite Binary, needs cardinal number with ℵ1 bits in order to represent 2ℵ1 cardinal numbers. By this observation ℵ1 > 2ℵ0 The third infinite level of The Infinite Binary, needs cardinal number with ℵ2 bits in order to represent 2ℵ2 cardinal numbers. ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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 )
 1st October 2017, 11:14 AM #2777 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 My previous post is wrong. It has to be as follows: By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, needs cardinal number with 1 place in order to represent 21 cardinal numbers. The second finite level of The Infinite Binary, needs cardinal number with 2 places in order to represent 22 cardinal numbers, such that 2 = 21 The third finite level of The Infinite Binary, needs cardinal number with 3 places in order to represent 23 cardinal numbers, such that 3 < 22 The forth finite level of The Infinite Binary, needs cardinal number with 3 places in order to represent 23 cardinal numbers, such that 4 < 23 ... The first infinite level of The Infinite Binary, needs cardinal number with ℵ0 places in order to represent 2ℵ0 cardinal numbers. The second infinite level of The Infinite Binary, needs cardinal number with ℵ1 places in order to represent 2ℵ1 cardinal numbers, such that ℵ1 = 2ℵ0 The third infinite level of The Infinite Binary, needs cardinal number with ℵ2 places in order to represent 2ℵ2 cardinal numbers, such that ℵ2 < 2ℵ1 The forth infinite level of The Infinite Binary, needs cardinal number with ℵ3 places in order to represent 2ℵ3 cardinal numbers, such that ℵ3 < 2ℵ2 ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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; 1st October 2017 at 12:34 PM.
 1st October 2017, 01:22 PM #2778 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Some correction of the previous post. The forth finite level of The Infinite Binary, needs cardinal number with 4 places in order to represent 24 cardinal numbers, such that 4 < 23 __________________ 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 )
 1st October 2017, 01:38 PM #2779 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 I am not comfortable with my last 3 posts about CH or GCH, so at this stage all I can show is that The Binary Tree is not limited by 2ℵ0 By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, needs cardinal number with 1 place in order to represent 21 cardinal numbers from 0 to 21-1 The second finite level of The Infinite Binary, needs cardinal number with 2 places in order to represent 22 cardinal numbers from 0 to 22-1 The third finite level of The Infinite Binary, needs cardinal number with 3 places in order to represent 23 cardinal numbers from 0 to 23-1 ... The first infinite level of The Infinite Binary, needs cardinal number with ℵ0 places in order to represent 2ℵ0 cardinal numbers from 0 to 2ℵ0-1 The second infinite level of The Infinite Binary, needs cardinal number with ℵ1 places in order to represent 2ℵ1 cardinal numbers from 0 to 2ℵ1-1 The third infinite level of The Infinite Binary, needs cardinal number with ℵ2 places in order to represent 2ℵ2 cardinal numbers from 0 to 2ℵ2-1 ... etc. --------------------- The observation above holds for any base, finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides. So The Infinite Binary Tree is some case without loss of generality. __________________ 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; 1st October 2017 at 02:20 PM.
 2nd October 2017, 10:45 AM #2780 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Let x be a placeholder for any cardinal number, finite or infinite. Because the sequences along The Binary Tree represent ordered cardinal numbers, any cardinal number of the form 2x-1 < 2x exactly by the finite cardinal 1. So one learns at least four novel things about infinite cardinals: 1) By using the binary tree (without loss of generality) one directly proves that x < 2x. 2) Finite cardinals > 0 can be added to and subtracted from infinite cardinals, such that the result is different from the considered infinite cardinal. 3) The bijection between an infinite set and its proper subset, is an illusion based on finite presentation of infinite sets. 4) The Axiom Of Infinity guarantees the accessibility of a given set of elements to the levels above it. __________________ 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; 2nd October 2017 at 12:32 PM.
 3rd October 2017, 09:37 AM #2781 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 The axiom of infinity (in words): Quote: there is a set I (the set which is postulated to be infinite), such that the empty set is in I and such that whenever any x is a member of I, the set formed by taking the union of x with its singleton {x} is also a member of I. Such a set is sometimes called an inductive set. Such an induction is limited to members with finite cardinality (i.e. the natural numbers). Let's use The Infinite Binary Tree (without loss of generality) as follows: The idea is to define x>0 of 2x as a placeholder for any given level with more than one node, such that the finite cardinal 1 is used as a successor in order to represent 2x cardinal numbers, by unique sequences with x bits for each one of them. By doing so x can be any cardinal number, finite or infinite, and we get a binary tree which is not bounded by any x. __________________ 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; 3rd October 2017 at 09:44 AM.
 4th October 2017, 03:45 AM #2782 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 The last 3 posts represent totally ordered cardinal numbers, such that cardinal number 1 is their successor, no matter what is the cardinal number of the placeholder, that is used to represent them. --------------------------------------------- Such observation is not accepted by mathematicians that define infinite cardinal numbers in terms of limit-cardinals, which means that given an infinite cardinal, no addition operation with some cardinal taken from any level below it, is used as a successor of this infinite cardinal, for example: ℵ0 + 2n = ℵ0, ℵ1 + 2ℵ0 = ℵ1, ℵ2 + 2ℵ1 = ℵ2, ℵ3 + 2ℵ2 = ℵ3, ... etc. In this case the following holds: By carefully observe this diagram I realized that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, is covered by a cardinal number (represented by bits) that is represented by 1 places. The term “covered by” means that 21 numbers (represented by bits) can be represented by 1 places. Generally, the number of places is determined by number x , which is used as the power value of any expression of the form 2x . The second finite level of The Infinite Binary, is covered by a cardinal number that is represented by 2 places. The third finite level of The Infinite Binary, is covered by a cardinal number that is represented by 3 places. ... The first infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ0 places (it means that ℵ0 places can represent up to 2ℵ0 cardinal numbers). ... The second infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ1 places (it means that ℵ1 places can represent up to 2ℵ1 cardinal numbers). By this observation ℵ1 > 2ℵ0, since ℵ1 is a limit-cardinal of anything that is involved with ℵ0 . ... The third infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ2 places (it means that ℵ2 places can represent can represent up to 2ℵ2 cardinal numbers). ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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; 4th October 2017 at 04:37 AM.
 4th October 2017, 01:24 PM #2783 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Strong limit cardinals, GCH and infinite trees This post is an improved version of the previous post. By carefully observe this diagram I realized (by exclude the root, which is not defined by any bit) that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, is covered by a cardinal number (represented by bits) that is represented by 1 places. The term “covered by” means that 21 cardinal numbers (represented by bits) can be represented by 1 places. Generally, the number of places is determined by number x , which is used as the power value of any expression of the form 2x, whether x is finite or infinite cardinal number. The second finite level of The Infinite Binary, is covered by a cardinal number that is represented by 2 places. The third finite level of The Infinite Binary, is covered by a cardinal number that is represented by 3 places. ... The first infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ0 places (it means that ℵ0 places can represent up to 2ℵ0 cardinal numbers). ... The second infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ1 places (it means that ℵ1 places can represent up to 2ℵ1 cardinal numbers). By this observation ℵ1 > 2ℵ0, since ℵ1 is a limit-cardinal of anything that is involved with ℵ0, exactly as ℵ0 is a strong limit-cardinal of anything that is involved with n (where n is any finite cardinal number). ... The third infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ2 places (it means that ℵ2 places can represent can represent up to 2ℵ2 cardinal numbers). ... etc. Some examples: ℵ0 + 2n = ℵ0, ℵ1 + 2ℵ0 = ℵ1, ℵ2 + 2ℵ1 = ℵ2, ℵ3 + 2ℵ2 = ℵ3, ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) GCH is solved. So The Infinite Binary Tree is some case without loss of generality. __________________ 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; 4th October 2017 at 02:31 PM.
 11th October 2017, 01:42 PM #2784 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Strong limit cardinal numbers and infinite trees By carefully observe this diagram I realized (by exclude the root, which is not defined by any bit) that every sequence of bits in its left side, has a complement in its right side and vise versa, such that no matter how many bits are involved, the complement property is invariant, which guarantees the uniqueness of each sequence along the tree. By carefully observe these notions I have found the following: The first finite level (the one that includes two bits) of The Infinite Binary, is covered by a finite cardinal number that is represented by 1 place (1 place is needed in order to represent the members of {0, 1}). Generally, the number of places is determined by number x , which is used as the power value of any expression of the form 2x, whether x is finite or infinite cardinal number. The second finite level of The Infinite Binary, is covered by a finite cardinal number that is represented by 2 places (2 places are needed in order to represent the members of {00, 01, 10, 11}). The third finite level of The Infinite Binary, is covered by a finite cardinal number that is represented by 3 places (3 places are needed in order to represent the members of {000, 001, 010, 011, 100, 101, 110, 111). ... The first infinite level of The Infinite Binary is covered by an infinite strong limit cardinal (please look at https://en.wikipedia.org/wiki/Limit_cardinal) that is represented by ℵ0 places (ℵ0 places are needed in order to represent the 2ℵ0 members of the power set of S0, where each one of its unique members has ℵ0 bits). ... The second infinite level of The Infinite Binary is covered by an infinite strong limit cardinal that is represented by ℵ1 places (ℵ1 places are needed in order to represent the 2ℵ1 members of the power set of S1, where each one of its unique members has ℵ1 bits). By this observation ℵ1 > 2ℵ0, since ℵ1 is a strong limit cardinal of anything that is involved with ℵ0, exactly as ℵ0 is a strong limit-cardinal of anything that is involved with n (where n is any finite cardinal number). ... The third infinite level of The Infinite Binary is covered by an infinite cardinal that is represented by ℵ2 places (ℵ2 places are needed in order to represent the 2ℵ2 members of the power set of S2, where each one of its unique members has ℵ2 bits). ... etc. Some examples: ℵ0 + 2n = ℵ0, ℵ1 + 2ℵ0 = ℵ1, ℵ2 + 2ℵ1 = ℵ2, ℵ3 + 2ℵ2 = ℵ3, ... etc. --------------------- Since the observation above holds for any base (finite or infinite, where the invariant complementary property is taken as an average between trees' left and right sides) The Infinite Binary Tree is some case without loss of generality. __________________ 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; 11th October 2017 at 02:05 PM.
 12th October 2017, 07:34 AM #2786 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 More details about the previous post are already given in http://www.internationalskeptics.com...postcount=2733 and http://www.internationalskeptics.com...postcount=2734. __________________ 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 )
 12th October 2017, 02:33 PM #2787 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 __________________ 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; 12th October 2017 at 03:08 PM.
 16th October 2017, 11:00 PM #2788 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 The usefulness of order and complementarity among two-valued logical oprators Please observe the example of the following 16 two-valued logical operators on propositions p and q. As can be seen they are ordered from contradiction to tautology or backwards. Moreover, each two-valued logical operator has a complement. Boolean algebra is a generalization of Power set algebra, but as can be seen in Wikipedia the order and complemntarity (as seen in the example of two-valued logical operators) are not defined. Are there formal mathematical researches that are focusing on the order and complementarity among uncountable two-valued logical operators (please provide concrete examples)? __________________ 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; 16th October 2017 at 11:04 PM.
 17th October 2017, 03:54 AM #2789 Hevneren Thinker   Join Date: Jul 2007 Posts: 136 Originally Posted by doronshadmi Yet pi is a proportion among the endless larger and the endless smaller non-composed circles. So a fixed value can be related to infinitely many things as long as it is not used to define their amount, their sum or any other fixed value that contradicts their property of being endless larger or endless smaller things. Ok, so circles come in all sizes, but all of them have the same proportion between area and squared radius, namely pi. That's fifth grade curriculum.
 17th October 2017, 04:28 AM #2790 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by Hevneren Ok, so circles come in all sizes, but all of them have the same proportion between area and squared radius, namely pi. That's fifth grade curriculum. There are no "all of them" if endless smaller or endless larger circles are considered, therefore it is definitely not fifth grade curriculum. --------------------------------- Please air your view about http://www.internationalskeptics.com...postcount=2788. __________________ 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; 17th October 2017 at 04:30 AM.
 17th October 2017, 05:29 AM #2791 Hevneren Thinker   Join Date: Jul 2007 Posts: 136 Originally Posted by doronshadmi There are no "all of them" if endless smaller or endless larger circles are considered, therefore it is definitely not fifth grade curriculum. Let me rephrase it without "all": A circle's area divided by the square of its radius equals pi. Relating a fixed value to infinitely many things isn't such a big deal: I hereby relate the word "fish" to the rational numbers. That was easy. Originally Posted by doronshadmi Please air your view about http://www.internationalskeptics.com...postcount=2788. Not easy to understand. Why did you mix boolean expressions with implications, and what were you thinking? And what does "p not implies q" mean? Is it "p doesn't imply q" or "p implies not-q"? Last edited by Hevneren; 17th October 2017 at 05:30 AM.
 17th October 2017, 02:26 PM #2792 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by Hevneren Let me rephrase it without "all" In that case you have no argument. Originally Posted by Hevneren Not easy to understand. https://en.wikipedia.org/wiki/Material_nonimplication and still you do not reply to http://www.internationalskeptics.com...postcount=2788 __________________ 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 )
 17th October 2017, 03:07 PM #2793 Hevneren Thinker   Join Date: Jul 2007 Posts: 136 Originally Posted by doronshadmi In that case you have no argument. My argument is that this is fifth grade stuff. The area of a circle equals pi times the radius squared. Regardless of circle size. Originally Posted by doronshadmi https://en.wikipedia.org/wiki/Material_nonimplication and still you do not reply to http://www.internationalskeptics.com...postcount=2788 I replied to it in my previous post, where I asked you for some clarification. You answered with another hyperlink to the post I had commented on, and a Wikipedia link that may or may not be relevant. No clarification.
 17th October 2017, 03:12 PM #2794 Hevneren Thinker   Join Date: Jul 2007 Posts: 136 Originally Posted by doronshadmi I guess "p not implies q" means "NOT (p implies q)" then?
 17th October 2017, 03:22 PM #2795 Hevneren Thinker   Join Date: Jul 2007 Posts: 136 Originally Posted by doronshadmi Given a non-composed endless straight line and a point not on that line, there are endless larger non-composed circles that are smaller than that line, and there are endless smaller non-composed circles that are larger than that point. I want to learn this properly, so I hope it's OK that I ask a couple of questions: Are there any composed endless straight lines, and what's the difference between them and the non-composed ones? Are there any composed circles, and what's the difference between them and the non-composed ones? What do you mean when you say that a circle is smaller than a line?
 18th October 2017, 05:15 AM #2796 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Originally Posted by Hevneren I want to learn this properly, so I hope it's OK that I ask a couple of questions: Questions and answers are like two legs for any abstract or non-abstract research. Originally Posted by Hevneren Are there any composed endless straight lines, and what's the difference between them and the non-composed ones? In my model there is one and only one non-composed endless line that is vibrating upon infinitely many smaller or larger scales. A composed line is defined by collection parts, where a non-compsed line does not have parts. Originally Posted by Hevneren Are there any composed circles, and what's the difference between them and the non-composed ones? The same as explained about lines. Originally Posted by Hevneren What do you mean when you say that a circle is smaller than a line? Any non-composed circle is smaller than the endless non-composed straight line. ------------------------ Still not even a single word from you about my question in http://www.internationalskeptics.com...postcount=2788. __________________ 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; 18th October 2017 at 05:22 AM.
 19th October 2017, 02:27 AM #2797 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Whole yet incomplete Ok, here is my answer to http://www.internationalskeptics.com...postcount=2788 First some wikipedia quote: 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 axiom of incompleteness: Let ... be the inaccessibility of sequence S to target T. By using an infinite tree of two-valued logical operators, we construct the following infinite sequences (where each sequence is a distinguished two-valued logical operator) such that they are ordered and complements of each other (like pairs of boots), yet they are inaccessible to a given target, as follows: ...000 ...001 ...010 ... ...101 ...110 ...111 As constructively seen, each infinite sequence has: a) An immediate predecessor or successor (where ...000 has no predecessor and ...111 has no successor). b) A complement in the other side of the considered tree of infinite two-valued logical operators. c) ... defines the inaccessibility of each sequence to the tree's root. d) ... defines the inaccessibility of each sequence to the tree's centre. So AC is not used, and unlike Russel's notion that is limited to ℵ0 bits, the inaccessibility a given S to a given T, is not limited by any particular infinite cardinality. ------------------ Please pay attention that an infinite tree of two-valued logical operators is taken as a whole from ...000 up to ...111, yet it is incomplete by its inaccessibility to its root an its centre. Actually we get a constructive proof of Godell's incompleteness theorems, which directly shows that given infinite logical operators, there are always logical operators (truth values) that are inaccessible to a given target. __________________ 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; 19th October 2017 at 02:47 AM.
 21st October 2017, 12:49 PM #2798 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Wholeness and Incompleteness, understand Mathematics in constructive terms Ok, let's put the last posts in one typo free (I hope) comprehensive post. --------------------------- I read Foundations of mathematics: an optimistic message. I wish to share with you my view on the issue at hand. Please observe the example of the following 16 two-valued logical operators on propositions p and q. As can be seen they are ordered from contradiction to tautology or backwards. Moreover, each two-valued logical operator has a complement. Boolean algebra is a generalization of Power set algebra, but as can be seen in Wikipedia the order and complemntarity (as seen in the example of two-valued logical operators) are not defined (by generalization I mean that x (which is used as the power value of the form 2x) is a placeholder for any cardinal number, whether it is finite or infinite). I did not find formal mathematical researches that use the order and complementarity among infinitely many two-valued logical operators (I think that it is important, since by doing so order and comlementarity enable to constructively distinguish between logical operators without the need of AC, even if uncountable logical operators are involved). My work on the issue at hand (in case of infinite sequences) is as follows: First some wikipedia quote: 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 axiom of incompleteness: Let ... be the inaccessibility of sequence S to target T. By using an infinite tree of two-valued logical operators, we construct the following infinite sequences (where each sequence is a distinguished two-valued logical operator) such that they are ordered and complements of each other (like pairs of boots) yet they are inaccessible to a given target, as follows: ...000 ...001 ...010 ... ...101 ...110 ...111 As constructively seen, each infinite sequence has: a) An immediate predecessor or successor (where ...000 has no predecessor and ...111 has no successor). b) A complement in the other side of the considered tree of infinite two-valued logical operators. c) ... defines the inaccessibility of each sequence to the tree's root. d) ... defines the inaccessibility of each sequence to the tree's centre. So AC is not used and, unlike Russel's notion that is limited to ℵ0 bits, the inaccessibility of a given S to a given T, is not limited by any particular infinite cardinality. ------------------ Please pay attention that an infinite tree of two-valued logical operators is taken as a whole from ...000 up to ...111 and it has a root (notated as "Unity") yet it is incomplete by its inaccessibility to its root and its centre. Actually we get a constructive proof of Godell's incompleteness theorems, which directly shows that given infinite logical operators, there are always logical operators (truth values) that are inaccessible to a given target, even if 2P bits (where P > ℵ0) are involved. ------------------ My optimistic view of Mathematics development is its wholeness in spite of its incompleteness. Furthermore, in my opinion, incompleteness is an essential signature of openness and endless creativity of Mathematics as a whole. __________________ 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; 21st October 2017 at 12:52 PM.
 27th December 2017, 07:11 AM #2799 doronshadmi Penultimate Amazing     Join Date: Mar 2008 Posts: 12,730 Actual Infinity and the Pythagorean Theorem By the Pythagorean Theorem a2+b2=c2 Let a2 be infinite and let b2 be finite and > 0. In that case c2 (which is infinite) > a2 (which is infinite) by finite b2 > 0, and we can't claim that finite values are inaccessible to infinite values (or in other words, finite values have impact on infinite values). In other words, the notion of actual infinity as used in Cantorean set theories, can't be used in case of The Pythagorean Theorem. Instead, potential infinity (an endless increased value) is used. So Cantorean set theories can't be considered as the foundations of any possible interesting mathematical framework. __________________ 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 December 2017 at 08:52 AM.
 27th December 2017, 07:30 AM #2800 jsfisher ETcorngods survivorModerator     Join Date: Dec 2005 Posts: 21,793 Originally Posted by doronshadmi By the Pythagorean Theorem a2+b2=c2 ...for the measures of the sides A and B of a right triangle with hypotenuse C, in an Euclidean plane. (Details can be such important things sometimes. Best to not exclude them unnecessarily.) Quote: Let a2 be infinite and let b2 be finite and > 0. Then it wouldn't be a triangle in an Euclidean plane, now would it. Quote: In that case c2 (which is infinite) > a2 (which is infinite) This does not follow. You've tried to apply a theorem outside its domain, and you've tried to exploit normal arithmetic outside its domain. __________________ 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

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