5 ms·

Prime Factorization of the Real Numbers

2 points by goldenkey 9y ago

gus_massa 9y ago
This is a function defined in the real numbers (that are not 0? positive?), but it's not a prime factorization. (Never mind a "unique canonical" factorization as the original title says.) To define an extension of a well known function it's important that it preserve as much of the original properties as possible. This is an informal requirement, and one of the dark secrets of math is that the definitions are made up and are only a social construct. [1] But after a few years in math you get a good sense of what is a good extension and what is a bad extension. This is a bad extension. One of the most important properties of the prime factorization is that it behaves good with multiplication. i.e. if A = 2^x * 3^y * 5^z B = 2^p * 3^q * 5^r then A * B = 2^(x+p) * 3^(y+q) * 5^(z+r) and this extension doesn't have this property. For example: 2 = "2^1 ; 3^0 ; 5^0" 2.5 = "2^(1/2) ; 3^(1/2) ; 5^0" 2 * 2.5 = 5 = "2^0 ; 3^0 ; 5^1" ≠ "2^(1+1/2) ; 3^(1/2) ; 5^0" --- It is useful to compare it with a good extension. (This is also not standard.) You can define a "prime factorization" of the rational numbers if you allow negative exponents, so the "prime factorization" is the difference between the numerator and the denominator. (this is good, even for the multiple ways to write a fraction) 2 = 2^1 * 3^0 * 5^0 2.5 = 2^(-1) * 3^0 * 5^1 2 * 2.5 = 5 = 2^0 * 3^0 * 5^1 ≠ 2^(1-1) * 3^0 * 5^(0+1) IIRC this is not a standard construction, but there are some similar construction like the p-adic order https://en.wikipedia.org/wiki/P-adic_order https://en.wikipedia.org/wiki/P-adic_order I think that this can even be extended to some irrational numbers like sqrt(2) = 2^(1/2) * 3^0 * 5^0 but I should check the details. To extend this to all the (positive) real numbers I think it is necessary to use a non numerable base, or something as nasty as that, but this would be far away from a unique and canonical definition. [1] https://en.wikipedia.org/wiki/Natural_number#Notation https://en.wikipedia.org/wiki/Natural_number#Notation
- goldenkey 9y ago
  I thank you for the goodies that I will look into. I guess its just a pairing function using the primes .. a decomposition. Guess I oversold it. But be aware that your tuple for 2.5 is wrong. Sqrt(6) isn't even 2.5. In[389]:= CanonicalRealFactorization[2.5] Out[389]= <|2 -> 0.44966, 3 -> 0.55034|> What you probably aim for is a UFD domain. I'll definitely investigate your links. Thank you! [1] https://en.wikipedia.org/wiki/Unique_factorization_domain https://en.wikipedia.org/wiki/Unique_factorization_domain
  - gus_massa 9y ago
    [sorry for the delay] UPS. I totally missed the log in the definition. My example of 2.5 is wrong (slightly wrong :)) Anyway, the proposed factorization of 2.5 is something that involves only 2 and 3, but if you multiply it by 2 then magically a 5 appears. So the main problem is still there. I think that a UFD is not a structure similar to your idea, in particular it doesn't use fractional exponents. My idea was to use R as a Q-vector space and use the base {log(2); log(3); log(5); log(7); log(11); ...}, but to complete this base you must use the Axiom of Choice, so you get a lot of crazy values, not only the logarithm of the primes. Then the idea is to use this base to express the logarithm of the number. For example, if x=12 then log(12) = 2 * log(2) + 1 * log(3) that is equivalent to 12 = 2^2 * 3. This construction provides good values for sqrt(2), 1/6 and other simple algebraic numbers. The problem is that if you pick something like pi, you get a decomposition that is not very meaningful.
    - goldenkey 9y ago
      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.

2 points by goldenkey 9y ago

This is a function defined in the real numbers (that are not 0? positive?), but it's not a prime factorization. (Never mind a "unique canonical" factorization as the original title says.) To define an extension of a well known function it's important that it preserve as much of the original properties as possible. This is an informal requirement, and one of the dark secrets of math is that the definitions are made up and are only a social construct. [1] But after a few years in math you get a good sense of what is a good extension and what is a bad extension. This is a bad extension. One of the most important properties of the prime factorization is that it behaves good with multiplication. i.e. if A = 2^x * 3^y * 5^z B = 2^p * 3^q * 5^r then A * B = 2^(x+p) * 3^(y+q) * 5^(z+r) and this extension doesn't have this property. For example: 2 = "2^1 ; 3^0 ; 5^0" 2.5 = "2^(1/2) ; 3^(1/2) ; 5^0" 2 * 2.5 = 5 = "2^0 ; 3^0 ; 5^1" ≠ "2^(1+1/2) ; 3^(1/2) ; 5^0" --- It is useful to compare it with a good extension. (This is also not standard.) You can define a "prime factorization" of the rational numbers if you allow negative exponents, so the "prime factorization" is the difference between the numerator and the denominator. (this is good, even for the multiple ways to write a fraction) 2 = 2^1 * 3^0 * 5^0 2.5 = 2^(-1) * 3^0 * 5^1 2 * 2.5 = 5 = 2^0 * 3^0 * 5^1 ≠ 2^(1-1) * 3^0 * 5^(0+1) IIRC this is not a standard construction, but there are some similar construction like the p-adic order https://en.wikipedia.org/wiki/P-adic_order https://en.wikipedia.org/wiki/P-adic_order I think that this can even be extended to some irrational numbers like sqrt(2) = 2^(1/2) * 3^0 * 5^0 but I should check the details. To extend this to all the (positive) real numbers I think it is necessary to use a non numerable base, or something as nasty as that, but this would be far away from a unique and canonical definition. [1] https://en.wikipedia.org/wiki/Natural_number#Notation https://en.wikipedia.org/wiki/Natural_number#Notation

goldenkey 9y ago
I thank you for the goodies that I will look into. I guess its just a pairing function using the primes .. a decomposition. Guess I oversold it. But be aware that your tuple for 2.5 is wrong. Sqrt(6) isn't even 2.5. In[389]:= CanonicalRealFactorization[2.5] Out[389]= <|2 -> 0.44966, 3 -> 0.55034|> What you probably aim for is a UFD domain. I'll definitely investigate your links. Thank you! [1] https://en.wikipedia.org/wiki/Unique_factorization_domain https://en.wikipedia.org/wiki/Unique_factorization_domain
- gus_massa 9y ago
  [sorry for the delay] UPS. I totally missed the log in the definition. My example of 2.5 is wrong (slightly wrong :)) Anyway, the proposed factorization of 2.5 is something that involves only 2 and 3, but if you multiply it by 2 then magically a 5 appears. So the main problem is still there. I think that a UFD is not a structure similar to your idea, in particular it doesn't use fractional exponents. My idea was to use R as a Q-vector space and use the base {log(2); log(3); log(5); log(7); log(11); ...}, but to complete this base you must use the Axiom of Choice, so you get a lot of crazy values, not only the logarithm of the primes. Then the idea is to use this base to express the logarithm of the number. For example, if x=12 then log(12) = 2 * log(2) + 1 * log(3) that is equivalent to 12 = 2^2 * 3. This construction provides good values for sqrt(2), 1/6 and other simple algebraic numbers. The problem is that if you pick something like pi, you get a decomposition that is not very meaningful.
  - goldenkey 9y ago
    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.