6 ms·

We don't know that - some people posit that but it's far from being provable. All physically measured numbers have uncertainty, meaning the value obtained is n

We don't know that - some people posit that but it's far from being provable. All physically measured numbers have uncertainty, meaning the value obtained is not an actual number, but is a range, perhaps with some associated probability spread. This is not measure zero. We are also very capable "directly" using various things that may be continuums, such as energy, or time, or velocities, or many other physical quantities.

bobbylarrybobby 4y ago
All the numbers we can use are computable (how can you use a number if you can't actually talk about it?) and there are only countably many of those.
- ChrisLomont 4y ago
  >All the numbers we can use are computable That's not true :) A nice example is Chaitin's constant [1], which I can use in proof and books and define and on and on.... And it's explicitly and most definitely NOT computable :) There are lots of numbers in lots of areas of mathematics, even symbolic mathematics that are not computable in the sense you want them to be computable. Chaitin's constant is the tip of a very big iceberg. You're using circular logic by claiming the only numbers I can are are the computable ones then claiming all numbers I can use are computable. That's not true. It's a circular argument. [1] https://en.wikipedia.org/wiki/Chaitin%27s_constant https://en.wikipedia.org/wiki/Chaitin%27s_constant
chowells 4y ago
No, I mean something simpler. The subset of R that can be identified is countable, because language is countable. No matter what system you devise for describing numbers, it will be countable. And that means that it cannot describe approximately 100% of the numbers in R. They can't be uniquely described; they can't be used in computations. They're phantoms, at best. You know they're out there, but they will never be usable the way numbers you can actually describe are. The only thing they give you is the ability to declare R to be Cauchy complete in some proofs. They're a polite fiction.
- ChrisLomont 4y ago
  Yeah, I'm well aware of these style arguments ala Gregory Chaitin..... >The subset of R that can be identified is countable, because language is countable Define "identified" in mathematically precise terms without using circular reasoning - and there is the flaw in this line of claims. You will find such definitions miss common uses of real numbers in the same way no finite set of axioms catch all true statements about integers. You will maybe get a nice consistent math subset of the reals with measure zero, but it will not cover all the cases you want it to. Thanks Godel For example, using your "proof" of "because language is countable" would imply there can be no infinity, yet we use it all the time. Time, for example, may be a physical continuum, so a finite interval of time contains infinitely many time steps, and if your math cannot even represent physical reality then it's a pretty weak system. >You know they're out there, but they will never be usable the way numbers you can actually describe are Plenty of formal computational systems are capable of using the same set of numbers I can describe. You're conflating being able to list every number in use one at a time, and operating on sets of numbers, like computation has done almost since day 1 using interval arithmetic. >they can't be used in computations Interval arithmetic, formal proof system - both capable of using all the numbers as sets that I can use - well beyond (Lebesgue) measure zero. And to be pedantic, even countable numbers have infinite measure for certain measures :)
  - chowells 4y ago
    Interval arithmetic doesn't use those numbers. It works just fine in the computable reals, for instance. (We can of course uniquely refer to elements of R beyond the computable reals, but it suffices for the purpose of demonstration.) If there's no difference between the number being part of the set vs being absent, is it contributing anything? Well, they contribute "this sequence has a limit" for all Cauchy sequences. That's it. Some proofs that depend on the existence of those limits work. But you never do anything else with those numbers. Their "existence" is about faith more than it is utility. That's a very different kind of "existence" than a number like 2, or ⅓, or π, or Chaitin's Ω, or anything else that be communicated uniquely in finite space.
    - 4y ago
voldacar 4y ago
The Bekenstein limit would seem to imply that the numbers / probability spreads we deal with in the real world are not infinite precision and therefore form a countable subset of R

bobbylarrybobby 4y ago

All the numbers we can use are computable (how can you use a number if you can't actually talk about it?) and there are only countably many of those.

ChrisLomont 4y ago
>All the numbers we can use are computable That's not true :) A nice example is Chaitin's constant [1], which I can use in proof and books and define and on and on.... And it's explicitly and most definitely NOT computable :) There are lots of numbers in lots of areas of mathematics, even symbolic mathematics that are not computable in the sense you want them to be computable. Chaitin's constant is the tip of a very big iceberg. You're using circular logic by claiming the only numbers I can are are the computable ones then claiming all numbers I can use are computable. That's not true. It's a circular argument. [1] https://en.wikipedia.org/wiki/Chaitin%27s_constant https://en.wikipedia.org/wiki/Chaitin%27s_constant

chowells 4y ago

No, I mean something simpler. The subset of R that can be identified is countable, because language is countable. No matter what system you devise for describing numbers, it will be countable. And that means that it cannot describe approximately 100% of the numbers in R. They can't be uniquely described; they can't be used in computations. They're phantoms, at best. You know they're out there, but they will never be usable the way numbers you can actually describe are. The only thing they give you is the ability to declare R to be Cauchy complete in some proofs. They're a polite fiction.

ChrisLomont 4y ago
Yeah, I'm well aware of these style arguments ala Gregory Chaitin..... >The subset of R that can be identified is countable, because language is countable Define "identified" in mathematically precise terms without using circular reasoning - and there is the flaw in this line of claims. You will find such definitions miss common uses of real numbers in the same way no finite set of axioms catch all true statements about integers. You will maybe get a nice consistent math subset of the reals with measure zero, but it will not cover all the cases you want it to. Thanks Godel For example, using your "proof" of "because language is countable" would imply there can be no infinity, yet we use it all the time. Time, for example, may be a physical continuum, so a finite interval of time contains infinitely many time steps, and if your math cannot even represent physical reality then it's a pretty weak system. >You know they're out there, but they will never be usable the way numbers you can actually describe are Plenty of formal computational systems are capable of using the same set of numbers I can describe. You're conflating being able to list every number in use one at a time, and operating on sets of numbers, like computation has done almost since day 1 using interval arithmetic. >they can't be used in computations Interval arithmetic, formal proof system - both capable of using all the numbers as sets that I can use - well beyond (Lebesgue) measure zero. And to be pedantic, even countable numbers have infinite measure for certain measures :)
- chowells 4y ago
  Interval arithmetic doesn't use those numbers. It works just fine in the computable reals, for instance. (We can of course uniquely refer to elements of R beyond the computable reals, but it suffices for the purpose of demonstration.) If there's no difference between the number being part of the set vs being absent, is it contributing anything? Well, they contribute "this sequence has a limit" for all Cauchy sequences. That's it. Some proofs that depend on the existence of those limits work. But you never do anything else with those numbers. Their "existence" is about faith more than it is utility. That's a very different kind of "existence" than a number like 2, or ⅓, or π, or Chaitin's Ω, or anything else that be communicated uniquely in finite space.
  - 4y ago

voldacar 4y ago

The Bekenstein limit would seem to imply that the numbers / probability spreads we deal with in the real world are not infinite precision and therefore form a countable subset of R