NTuft Posted April 19, 2022 Share Posted April 19, 2022 So, if we take the set of square roots of prime numbers, each number in that set can match the set of "Natural" numbers by extending the decimal points. As the "Natural" numbers get large a further decimal 0-9 matches each n, n+1, ...; 1:1. The ordinality of i' is the same as the ordinality of "infinity", and there is a degree of aleph=2 to the subset i' before it reaches anything including all the "Reals". So, between the sets "infinity", i', various Real sets, and the Natural set, there are at least two degrees, hence the aleph=2, hence Cantor's conjecture is towards the negative. Does that make sense? P.s. yada, yada, cardinality. Link to comment Share on other sites More sharing options...

uncool Posted April 19, 2022 Share Posted April 19, 2022 18 hours ago, NTuft said: Does that make sense? Sorry, but no. It reads as word salad. Link to comment Share on other sites More sharing options...

NTuft Posted April 19, 2022 Author Share Posted April 19, 2022 (edited) Let aleph_0 = N, the set of Natural numbers If we take the set of square roots of prime numbers as a subset of the real numbers, we can say that this subset is smaller than all reals, R. This subset, although smaller than all Reals, can have each member of the set equate to the set N, 1:1, as the decimals keep extending infinitely to match the increasing Naturals. Hence, this set is of aleph (cardinality) greater than N, aleph_0, but less than the cardinality of all reals, R, because it is a subset therefrom; hence, the cardinality of R is at least aleph=2. Edited April 19, 2022 by NTuft Link to comment Share on other sites More sharing options...

uncool Posted April 19, 2022 Share Posted April 19, 2022 that’s less salad, at least. it’s clear enough to be simply wrong, rather than incomprehensible. the set of square roots of prime numbers has cardinality equal to that of the natural numbers. 2 Link to comment Share on other sites More sharing options...

NTuft Posted April 19, 2022 Author Share Posted April 19, 2022 (edited) Edited April 19, 2022 by NTuft Link to comment Share on other sites More sharing options...

NTuft Posted April 20, 2022 Author Share Posted April 20, 2022 (edited) How Many Numbers Exist? Infinity proof moves math closer to an answer https://www.quantamagazine.org/how-many-numbers-exist-infinity-proof-moves-math-closer-to-an-answer-20210715/ What I am alluding to is akin to a forcing procedure that only opens a further decimal in each member of the irrational subset (for the sake of the conjecture here the set of square roots of primes; I see no reason now why not to use all irrationals) that can then each by itself match the natural numbers as countably infinite. I think it follows that this set is of a higher cardinality than the natural numbers... By what sort of bijection do you suppose to map the naturals, which are infinite but countable, to what is the larger subset of the real, the irrationals, when the irrationals are considered larger than the rationals (of which the naturals are a subset), and the part that makes the real numbers infinitely uncountable? Then in a separate comparison take the set of sq. rt.'s of prime as being fully enumerated as compared to the full set of reals. The set of reals will have other irrationals that can be used in a similar forcing procedure to create the set of sq. rt.'s of primes as subordinate in cardinality to all reals but greater than naturals. Edited April 20, 2022 by NTuft Link to comment Share on other sites More sharing options...

NTuft Posted April 20, 2022 Author Share Posted April 20, 2022 3 hours ago, NTuft said: Then in a separate comparison take the set of sq. rt.'s of prime as being fully enumerated as compared to the full set of reals. The set of reals will have other irrationals that can be used in a similar forcing procedure to create the set of sq. rt.'s of primes as subordinate in cardinality to all reals but greater than naturals. Rather, going back to the full set of ℝ, we know that the set of square roots of prime numbers (here we'll call them i', a subset of R\Q or P, the irrationals) are enumerated and matched, that these members are then at least doubled by the corresponding negative values of the same non-terminating decimal values, and further outnumbered by all other Reals, then the set ℝ clearly has more members than the set i', thus is of a higher cardinality. Therefore, presuming the forcing procedure described above is valid, 0א = ℕ ; 1א = i' ; 2א ≤ ℝ Link to comment Share on other sites More sharing options...

uncool Posted April 20, 2022 Share Posted April 20, 2022 You’ve returned to word salad; I can’t determine even what statements you are proposing to be true beyond the supposed conclusion. The pop-sci article is…poorly informed. This statement: Asperó and Schindler showed that one of these axioms implies the other, raising the likelihood that both axioms — and all they intimate about infinity — are true. makes no sense - one axiom implying another is not a way to gauge whether they are “more likely”. And the first half of the article keeps implying there’s a “battle” over whether the continuum hypothesis is “true”, before the second half reveals the actual state of affairs: it is independent of our current axiom set, meaning it could freely be true or false, depending on which model you choose. 1 Link to comment Share on other sites More sharing options...

uncool Posted April 20, 2022 Share Posted April 20, 2022 A better article would be able to articulate that axiom choice is much less about what’s true and more about what’s convenient. These mathematicians have come up with some axioms which work together in an apparently surprising way, and therefore may be more convenient. That doesn’t make them more true, merely more likely to be used in the future. 1 Link to comment Share on other sites More sharing options...

NTuft Posted April 20, 2022 Author Share Posted April 20, 2022 21 hours ago, uncool said: the set of square roots of prime numbers has cardinality equal to that of the natural numbers. When both sets are increasing "vertically" infintely, but only one has the potential to expand infinitely "horizontally", you think they're equatable? Please, defend your assertion. Also perhaps see the guidelines re: "salad". 2 hours ago, uncool said: You’ve returned to word salad; I can’t determine even what statements you are proposing This conjecture was presented before the current ZF axioms, axiom of choice, axiom of determinancy, MM++ (*), etc. Is it necessary to write it out in set theorem logic symbols? Logic via language ought to be able to suffice for discussion. Perhaps be specific about what it is you can't determine? Link to comment Share on other sites More sharing options...

uncool Posted April 20, 2022 Share Posted April 20, 2022 (edited) 19 minutes ago, NTuft said: When both sets are increasing "vertically" infintely, but only one has the potential to expand infinitely "horizontally", you think they're equatable? Please, defend your assertion. It’s trivial to construct a bijection between the set of prime numbers and the set of square roots of prime numbers. The set of prime numbers is obviously countable, so the set of square roots of prime numbers is countable, too. It sounds like you are trying to talk about the set of digits of the square roots of prime numbers. This is still countable, though slightly less obviously, but it is not the same as the set of square roots of prime numbers. 19 minutes ago, NTuft said: Logic via language ought to be able to suffice for discussion. It should be, if you can make clear statements. My point is that you have not done so. 19 minutes ago, NTuft said: Perhaps be specific about what it is you can't determine? I have been: I can’t determine what statements you are proposing to be true. Edited April 20, 2022 by uncool 1 Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 5 hours ago, uncool said: It’s trivial to construct a bijection between the set of prime numbers and the set of square roots of prime numbers. I am taking a subset from the irrational numbers | i' ⊆ {ℝ\ℚ}. Quote The set of prime numbers is obviously countable, so the set of square roots of prime numbers is countable, too. The set of prime numbers is a subset of the natural numbers. The set of the square roots of prime numbers is not. Quote It sounds like you are trying to talk about the set of digits of the square roots of prime numbers. No, I did mention that, at one point. Quote This is still countable, though slightly less obviously, but it is not the same as the set of square roots of prime numbers. No, it is obvious, and yes, it is not. 5 hours ago, uncool said: I have been: I can’t determine what statements you are proposing to be true. I state that there is a subset of ℝ, from {ℝ \ ℚ}, that is of cardinality between that of ℕ and ℝ, "...determine whether there were any infinite sets of real numbers that were of intermediate size, that is, whether there was an infinite set of real numbers that could not be put into one-to-one correspondence with the natural numbers and could not be put into one-to-one correspondence with the real numbers."^{1} and that is the set of square roots of primes. ^{1 }https://plato.stanford.edu/entries/continuum-hypothesis/ Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 (edited) @uncool My apologies, that may not be satisfactory. If we have the set of numbers, making use of: Quote Statement If a and b are algebraic numbers with a ≠ 0, 1, and b irrational, then any value of a^{b} is a transcendental number.^{1} where a is a prime number, and b is a square root of a prime number. Since this set would be transcendentals, it could be of cardinality greater than the natural set, but would it be of intermediate cardinality under the reals? Thank you ^{1 }https://handwiki.org/wiki/Gelfond–Schneider_theorem Edited April 21, 2022 by NTuft repeated; a^b Link to comment Share on other sites More sharing options...

uncool Posted April 21, 2022 Share Posted April 21, 2022 The set of all transcendentals has cardinality equal to that of the reals. The set of all square roots of primes has cardinality equal to that of the natural numbers, by rather obvious bijection. Neither of those have “intermediate” cardinality. If you are claiming to have constructed a set of intermediate cardinality, what is it? 11 hours ago, NTuft said: The set of the square roots of prime numbers is not. And your point is? 9 hours ago, NTuft said: Since this set would be transcendentals, it could be of cardinality greater than the natural set Cardinality is not about underlying members. Just because the elements of a set are transcendentals and there are lots of transcendentals doesn’t mean that set is large. The cardinality of the set containing just pi is still 1. 1 Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 (edited) 2 hours ago, uncool said: [...] a set of intermediate cardinality, what is it? {ℕ^{i'}}, where 1∉ℕ... ? Edited April 21, 2022 by NTuft Link to comment Share on other sites More sharing options...

uncool Posted April 21, 2022 Share Posted April 21, 2022 What do you mean by {ℕ^{i'}}? If you mean the set of all i’-indexed sequences of natural numbers, that set has cardinality equal to that of the continuum. It clearly has the same cardinality as ℕ^{ℕ}, and it’s not hard to establish that that has the same cardinality as the reals. 1 Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 I mean the set of numbers from each member of ℕ being exponentiated by the set of square roots of primes. Do you see anything about using the set of square roots of primes as the b component here: 12 hours ago, NTuft said: Statement If a and b are algebraic numbers with a ≠ 0, 1, and b irrational, then any value of a^{b} is a transcendental number.^{1} ,cited above? Link to comment Share on other sites More sharing options...

uncool Posted April 21, 2022 Share Posted April 21, 2022 (edited) So it is a set that can be indexed by ordered pairs of integers, namely the base and the prime whose square root is the exponent. It will again be countable, as the set of ordered pairs of natural numbers has a bijection with the set of natural numbers. You keep trying to use the fact that these numbers are transcendental as if that gives you a natural connection to uncountability. It doesn’t. The fact that the set of all transcendentals is large doesn’t mean that any particular subset will be large. Edited April 21, 2022 by uncool 1 Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 (edited) Non-integers from ℚ for the base? I can understand your point from the example you gave with pi. What I'm trying to do now is find a set composed of transcendentals because they supposedly vastly outcount the naturals, but it hasn't been well defined. So if there were such a thing don't you suppose it would have countability differing from the naturals? Edited April 21, 2022 by NTuft Link to comment Share on other sites More sharing options...

uncool Posted April 21, 2022 Share Posted April 21, 2022 Still countable. Can be indexed by a triple of integers - numerator, denominator, and exponent - and the set of triples of integers is countable (for largely the same reason that the set of pairs of integers is countable). Again, there is nothing “inherently” uncountable about transcendental numbers. 16 minutes ago, NTuft said: So if there were such a thing don't you suppose it would have countability differing from the naturals? I see no specific reason to assume it would, unless you assume the negation of the continuum hypothesis to begin with. Which is your right, but you would then only have a circular argument. Link to comment Share on other sites More sharing options...

NTuft Posted April 21, 2022 Author Share Posted April 21, 2022 (edited) "[...] every square root of a prime number is a distinct quadratic irrational [...] Quadratic irrationals have useful properties, especially in relation to continued fractions, where we have the result that all real quadratic irrationals, and only real quadratic irrationals, have periodic continued fraction forms." "The periodic continued fractions can be placed in one-to-one correspondence with the rational numbers." and hence also with the natural numbers. Is there a bijection of an index that places {i'^{i'}} in with the cardinality of ℚ & ℕ? Or can we find Quote Cardinality Definition 3: |A| < |B| A has cardinality strictly less than the cardinality of B, if there is an injective function, but no bijective function, from A to B. This is again looking to use the Gelfond-Schneider theorem, as I don't see a restriction that the base a can not be a real irrational quadratic, only that it is algebraic and ≠ 0, 1. Edited April 21, 2022 by NTuft Link to comment Share on other sites More sharing options...

uncool Posted April 21, 2022 Share Posted April 21, 2022 again: if a set can be indexed by pairs of natural numbers, then it is countable. i’ is countable, so it can be indexed by the set of countable numbers, so the set of all powers of the form x^{y} where x and y are in i’ can be indexed by pairs of natural numbers, so that set is also countable. Link to comment Share on other sites More sharing options...

NTuft Posted April 25, 2022 Author Share Posted April 25, 2022 (edited) What about using the Gelfond-Schneider theorem to generate a set of relatively uncountable transcendental numbers? Taking all algebraic numbers as the base and exponentiating to a (sub)set of the irrationals. So the intermediate cardinality would be comprised of not all reals, but at least all algebraics relative to the naturals, plus a portion of the reals that makes them uncountable. Thanks for the instruction over the board. P.s. Ordinality. Edited April 25, 2022 by NTuft word salad Link to comment Share on other sites More sharing options...

uncool Posted April 25, 2022 Share Posted April 25, 2022 (edited) The set of algebraic numbers is countable; therefore, if the set of powers you use is also countable, then the set you get as a result will again be countable. In fact, the output set will be of the same cardinality as the exponent set, for whatever exponent set you choose (possibly subject to some rare exceptions). Edited April 25, 2022 by uncool Link to comment Share on other sites More sharing options...

uncool Posted April 27, 2022 Share Posted April 27, 2022 Part of the point of “forcing” is that it is not constructing new sets within the same model - which is your approach - but instead constructing a new model of the real numbers. Link to comment Share on other sites More sharing options...

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