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> In one model there is an object called Q that satisfies all of the properties in ZFC of being a natural number, but is infinitely large. In this model the Tur

> In one model there is an object called Q that satisfies all of the properties in ZFC of being a natural number, but is infinitely large. In this model the Turing Machine halts after Q steps. That doesn’t make any sense. A Turing machine can’t halt after a infinite number of steps. It either halts after a finite number of steps, or it never halts. I’m sure there are models of hypercomputation and corresponding “what’s the largest number of steps they can run?” functions that would admit infinities, but those would not be Turing machines and the function would not be the Busy Beaver.

Kranar 1y ago
There is a lot of nuance you are skipping over that needs to be fully appreciated if you wish to understand this topic.
- wat10000 1y ago
  I can accept that there is a lot of nuance on the math side that I’m completely missing, but the Turing machine side is really straightforward. A Turing machine either never stops, or it stops after a finite number of steps. If it stops, the number of steps that it runs is a finite whole number, no different from “three” in its relationship to infinity or its theoretical ability to be written down. This doesn’t depend on your mathematics, only on your Turing machine.
  - bo1024 1y ago
    The point is that when it "never stops", there are models of ZFC in which the "infinity" number of steps it runs for isn't considered infinity by the model, it's a made-up "nonstandard" number that is smaller than infinity but larger than any integer. And that model considers that to be "halting", so that model says the TM halts.
    - wat10000 1y ago
      That’s just a change of definition. That isn’t really saying that BB(748) is different under a different model, just that there’s a BB’ equivalent for that model and BB’(748) is equal to something else.
raincole 1y ago
It's not about hypercomputation. What the commenter above you said doesn't make sense in our daily life, but it makes perfect sense when in comes to non-standard models. You got confused because you're thinking natural numbers as something we can count in real physical world, which is a perfectly sane mental model, and that is why there was a comment above said: > People find that weird because they don't think about non-standard models, as arguably they shouldn't. Q is not a number you can actually count, so it doesn't fit into our intuition of natural number. The point is not that Q exists in some physical sense in real life, like "3" in "3 apples" (it doesn't). The point is that ZF itself isn't strong enough to prevent you from defining random shit like Q as a natural number.
- red75prime 1y ago
  > The point is not that Q exists in some physical sense in real life Ultrafinitism? If you'd run the Turing machine that performs BB(748) steps in a physical universe that admits it, you'd get a physical representation of BB(748). If you have a competing theory about which Turing machine computes BB(748), you can run them both alongside in this universe and see with your own eyes which one finishes first. I guess from ultrafinitist's point of view such universe has different mathematics, but isn't it a fringe viewpoint in mathematical circles?
  - raincole 1y ago
    > ultrafinitism I'm not sure what flavor of ultrafinitism you're referring here. If it's the "very big numbers, like TREE(3), are not natural numbers as they are far bigger than the number of atoms in this universe..." kind, then it has nothing to do with what this is about. > physical representation > your own eyes Non standard models of ZFC have nothing to do with our physical world. That's why no physicist or engineer cares about them (or cares about axiom systems at all). So we need to be very careful when connecting the idea of physical, running "stuff" to the discussion of ZFC. Anyway, back to > you can run them both alongside in this universe and see which one finishes first There are two Turing Machines, Foo and Bar. We build and run them in our physical universe. Foo halts at the standard BB(748) steps. Bar just keeps running and running. That's what we will see with our own eyes. The issue is that when we try to reason out whether Bar will ultimately halts, ZFC doesn't prevent us from defining a non-standard model where Bar halts after a non-standard number of steps. Note that the physical Bar will not halt in our universe. The "non-standard number of steps" is as nonsense as it sounds. It's just that ZFC doesn't prevent us from defining such a nonsense. The point of ZFC is it's compatible with almost all the useful, sane math. It's not necessarily incompatible with bullshit and insane math. That is it. The fact that Bar is still keeping running in our universe is completely irrelevant.

Kranar 1y ago

There is a lot of nuance you are skipping over that needs to be fully appreciated if you wish to understand this topic.

wat10000 1y ago
I can accept that there is a lot of nuance on the math side that I’m completely missing, but the Turing machine side is really straightforward. A Turing machine either never stops, or it stops after a finite number of steps. If it stops, the number of steps that it runs is a finite whole number, no different from “three” in its relationship to infinity or its theoretical ability to be written down. This doesn’t depend on your mathematics, only on your Turing machine.
- bo1024 1y ago
  The point is that when it "never stops", there are models of ZFC in which the "infinity" number of steps it runs for isn't considered infinity by the model, it's a made-up "nonstandard" number that is smaller than infinity but larger than any integer. And that model considers that to be "halting", so that model says the TM halts.
  - wat10000 1y ago
    That’s just a change of definition. That isn’t really saying that BB(748) is different under a different model, just that there’s a BB’ equivalent for that model and BB’(748) is equal to something else.

raincole 1y ago

It's not about hypercomputation. What the commenter above you said doesn't make sense in our daily life, but it makes perfect sense when in comes to non-standard models. You got confused because you're thinking natural numbers as something we can count in real physical world, which is a perfectly sane mental model, and that is why there was a comment above said: > People find that weird because they don't think about non-standard models, as arguably they shouldn't. Q is not a number you can actually count, so it doesn't fit into our intuition of natural number. The point is not that Q exists in some physical sense in real life, like "3" in "3 apples" (it doesn't). The point is that ZF itself isn't strong enough to prevent you from defining random shit like Q as a natural number.

red75prime 1y ago
> The point is not that Q exists in some physical sense in real life Ultrafinitism? If you'd run the Turing machine that performs BB(748) steps in a physical universe that admits it, you'd get a physical representation of BB(748). If you have a competing theory about which Turing machine computes BB(748), you can run them both alongside in this universe and see with your own eyes which one finishes first. I guess from ultrafinitist's point of view such universe has different mathematics, but isn't it a fringe viewpoint in mathematical circles?
- raincole 1y ago
  > ultrafinitism I'm not sure what flavor of ultrafinitism you're referring here. If it's the "very big numbers, like TREE(3), are not natural numbers as they are far bigger than the number of atoms in this universe..." kind, then it has nothing to do with what this is about. > physical representation > your own eyes Non standard models of ZFC have nothing to do with our physical world. That's why no physicist or engineer cares about them (or cares about axiom systems at all). So we need to be very careful when connecting the idea of physical, running "stuff" to the discussion of ZFC. Anyway, back to > you can run them both alongside in this universe and see which one finishes first There are two Turing Machines, Foo and Bar. We build and run them in our physical universe. Foo halts at the standard BB(748) steps. Bar just keeps running and running. That's what we will see with our own eyes. The issue is that when we try to reason out whether Bar will ultimately halts, ZFC doesn't prevent us from defining a non-standard model where Bar halts after a non-standard number of steps. Note that the physical Bar will not halt in our universe. The "non-standard number of steps" is as nonsense as it sounds. It's just that ZFC doesn't prevent us from defining such a nonsense. The point of ZFC is it's compatible with almost all the useful, sane math. It's not necessarily incompatible with bullshit and insane math. That is it. The fact that Bar is still keeping running in our universe is completely irrelevant.