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Right - the error is still present despite your mistake. I understand that I wrote a poor extension. I see what you are doing with the logs, but how do we know

Right - the error is still present despite your mistake. I understand that I wrote a poor extension. I see what you are doing with the logs, but how do we know that all irrationals and reals are targeted by the log basis - and I think you mention it with regard to pi -- won't we end up with an infinite number of basis elements / factors? I don't want an infinite factorization for reals with a finite komogorolov complexity - there must be a way to get a basis that is finite for reals that have finite descriptions (yet infinite digits.) Because nomatter what, the reals that have infinite descriptions will only be provably factorized by AC, not actually factorized in a tangible sense. And those cases are less interesting to me because I want the utility more for practical investigation of number theory, not the 10th lemma in a non constructive proof.

gus_massa 9y ago
Just to clarify, each number will have a finite "factorization" using the decomposition in the basis of the Vector Space. The problem is that the base is so big that it is not very meaningful. To allow an infinite decomposition you must add a topology or a metric to the space, so there is possible to define a convergence of the infinite sum. This are for example https://en.wikipedia.org/wiki/Schauder_basis https://en.wikipedia.org/wiki/Schauder_basis and https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_base... . They are not basis in the Vector Space sense, they are an extension. :) The usual metric doesn't play nice with factorization, so the problem is to choose the right metric. There is s standard trick to use the p-adic order in rationals, but I'm not sure if something similar is defined for multiple primes (It should be possible.) But if you try to complete the rationals with the new metric, you don't get the real numbers, you get a different set. -- One recommendation is to think first were are you going to use the new definition/extension. There are just to many form to extend a definition, but it's very difficult to choose the right one (whatever it means). If you know where you are going to use it, you can try to make a tradeoff to get something similar enough to the old definition and something big enough to include more interesting cases.
- goldenkey 9y ago
  I'm thinking the only reason this is difficult is because we don't have prime-based formulas for all the interesting reals we care about enough to name. That makes this a lifelong quest to solve but probably other things need to be tackled first in order to make way for it. Thoughts?
  - gus_massa 9y ago
    Probably you need 3 or 4 semesters of Algebra classes in the university to understand the details. I'll try to explain the main ideas, but this is not my specialization so I don't remember the details :(. Too much Analysis in my life... A possible way to extend the integer is to consider the Gaussian Integers, that are the numbers a+bi, where a and b are integers and i^2=-1. https://en.wikipedia.org/wiki/Gaussian_integer https://en.wikipedia.org/wiki/Gaussian_integer They have an unique factorization and primes. Some of the normal primes are still primes, for example 3, 7, ... and some of the normal primes are split, foe example 5 = (2+i) * (2-i) . Now 2+i and 2-i are the new primes. Anyway, you still have primes and unique factorization and can live happily there. Another extensions are a+b.i.sqrt(d) aka Z[sqrt(-d)]. In the case d=2 and d=3 you still have primes and unique factorization. But if you consider d=5 or more, you don't have primes or unique factorization (see for example https://math.stackexchange.com/questions/1141412/mathbb-z-sqrt-5-is-not-a-ufd https://math.stackexchange.com/questions/1141412/mathbb-z-sq... , d=14 is apparently also a good value to get easy examples) But this examples are extensions of the integer numbers in the complex numbers. You can make similar examples with the real version, i.e. the numbers a+b.sqrt(d) aka Z[sqrt(d)] I tried to use google to find an example of where there is no unique factorization, but I failed. I suspect that perhaps there is an easy way to proof that all of them but I didn't find it now and I'm too lazy to try to prove it. I'm not sure. Anyway, when you add more roots, you have more problems. You can use the numbers a+b.cubicroot(d)+c.cubicroot(d^2) and try to analyze if they have unique factorization. This is related for example to the proof of the Last Fermat Theorem. The idea is to extend the integers to some more general set of numbers, and if you assume that there are primes and unique factorization the proof is not very difficult. For small exponents the extension use only square roots or cubic roots and it's not too difficult. But for big exponents the idea of unique factorization of the extended numbers fails. Try to look at the history part of https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem So ... I don't remember the details now, but I suspect that for a general (algebraic) number it is not possible to have a good generalization of primes.

gus_massa 9y ago

Just to clarify, each number will have a finite "factorization" using the decomposition in the basis of the Vector Space. The problem is that the base is so big that it is not very meaningful. To allow an infinite decomposition you must add a topology or a metric to the space, so there is possible to define a convergence of the infinite sum. This are for example https://en.wikipedia.org/wiki/Schauder_basis https://en.wikipedia.org/wiki/Schauder_basis and https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_base... . They are not basis in the Vector Space sense, they are an extension. :) The usual metric doesn't play nice with factorization, so the problem is to choose the right metric. There is s standard trick to use the p-adic order in rationals, but I'm not sure if something similar is defined for multiple primes (It should be possible.) But if you try to complete the rationals with the new metric, you don't get the real numbers, you get a different set. -- One recommendation is to think first were are you going to use the new definition/extension. There are just to many form to extend a definition, but it's very difficult to choose the right one (whatever it means). If you know where you are going to use it, you can try to make a tradeoff to get something similar enough to the old definition and something big enough to include more interesting cases.

goldenkey 9y ago
I'm thinking the only reason this is difficult is because we don't have prime-based formulas for all the interesting reals we care about enough to name. That makes this a lifelong quest to solve but probably other things need to be tackled first in order to make way for it. Thoughts?
- gus_massa 9y ago
  Probably you need 3 or 4 semesters of Algebra classes in the university to understand the details. I'll try to explain the main ideas, but this is not my specialization so I don't remember the details :(. Too much Analysis in my life... A possible way to extend the integer is to consider the Gaussian Integers, that are the numbers a+bi, where a and b are integers and i^2=-1. https://en.wikipedia.org/wiki/Gaussian_integer https://en.wikipedia.org/wiki/Gaussian_integer They have an unique factorization and primes. Some of the normal primes are still primes, for example 3, 7, ... and some of the normal primes are split, foe example 5 = (2+i) * (2-i) . Now 2+i and 2-i are the new primes. Anyway, you still have primes and unique factorization and can live happily there. Another extensions are a+b.i.sqrt(d) aka Z[sqrt(-d)]. In the case d=2 and d=3 you still have primes and unique factorization. But if you consider d=5 or more, you don't have primes or unique factorization (see for example https://math.stackexchange.com/questions/1141412/mathbb-z-sqrt-5-is-not-a-ufd https://math.stackexchange.com/questions/1141412/mathbb-z-sq... , d=14 is apparently also a good value to get easy examples) But this examples are extensions of the integer numbers in the complex numbers. You can make similar examples with the real version, i.e. the numbers a+b.sqrt(d) aka Z[sqrt(d)] I tried to use google to find an example of where there is no unique factorization, but I failed. I suspect that perhaps there is an easy way to proof that all of them but I didn't find it now and I'm too lazy to try to prove it. I'm not sure. Anyway, when you add more roots, you have more problems. You can use the numbers a+b.cubicroot(d)+c.cubicroot(d^2) and try to analyze if they have unique factorization. This is related for example to the proof of the Last Fermat Theorem. The idea is to extend the integers to some more general set of numbers, and if you assume that there are primes and unique factorization the proof is not very difficult. For small exponents the extension use only square roots or cubic roots and it's not too difficult. But for big exponents the idea of unique factorization of the extended numbers fails. Try to look at the history part of https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem So ... I don't remember the details now, but I suspect that for a general (algebraic) number it is not possible to have a good generalization of primes.