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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-standa

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.
  - wat10000 1y ago
    > ZFC doesn't prevent us from defining a non-standard model where Bar halts after a non-standard number of steps. But it does prevent you from defining a non-standard model where Bar halts after a finite number of steps. Since BB is finite by definition, the non-standard number of steps after which Bar halts cannot be BB(748). I’m pretty sure you and the other commenter have this mixed up. The fact that BB(748) is independent of ZFC doesn’t mean there are different models that have different values of BB(748). It means that ZFC is insufficient to determine the value of BB(748). That value is still some finite integer, you just can’t prove which one it is. Equivalently, there is some 748-state Turing machine which never halts but ZFC cannot prove never halts. And no, you can’t change your model such that this Turing machine halts in some non-standard number of steps. Or rather, you can, but that doesn’t actually change anything. The machine still doesn’t halt for the purposes of defining BB(748).
    - raincole 1y ago
      > I’m pretty sure you and the other commenter have this mixed up. We really don't. > that BB(748) is independent of ZFC > there are different models that have different values of BB(748) > ZFC is insufficient to determine the value of BB(748) These three statements are equivalent. f(n)=X is independent of ZFC means there are different models of ZFC that have different values of f(n). It's a very trivial theorem[0]. If you don't like it, I can't convince you otherwise. > that doesn’t actually change anything Changing the model will not change how any machine works in our physical, mechanical universe. However, it does change the value of BB(748). I understand your line of thinking: There is only one mechanical universe, which is the one where we exist. We can build Turing machines in this universe. BB(n) depends on Turning machines. Since there is only one single universe, there is only one single value of BB(n). It's a perfectly fine mental model for most cases. This was exactly how I thought when the first time I heard about BB(n). But it's not the kind of math than Scott Aaronson et al. are doing. Bar keeps running in our mechanical universe. But it can also halt in some non-standard number of steps. This weird, absurd-sounding proposition works because non-standard numbers simply don't map to anything in mechanical universe. They're purely abstract objects living in ZFC+~Con(ZFC). [0]: Given f(n)=X is independent of ZFC. Which means f(n)=X and ~(f(n)=X) are both consistent relative to ZFC. Therefore, if there is any model of ZFC, there is a model M1 that entails ZFC+(f(n)=X), and a model M2 that entails ZFC+~(f(n)=X). The value of f(n) cannot be the same in M1 and M2.
wat10000 1y ago
But Q is a number you can actually count, for a definition of “actually” that includes unimaginably large space and time. That finiteness comes from the basic mechanics of the Turing machine, which don’t depend on your mathematical axioms. Sure, you can come up with a set of axioms where the natural numbers include infinities. You may be able to use it to prove interesting things. But all that does here it make it so that the set of numbers describing how many steps a Turing machine runs before it stops is no longer the “natural numbers.”

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.
- wat10000 1y ago
  > ZFC doesn't prevent us from defining a non-standard model where Bar halts after a non-standard number of steps. But it does prevent you from defining a non-standard model where Bar halts after a finite number of steps. Since BB is finite by definition, the non-standard number of steps after which Bar halts cannot be BB(748). I’m pretty sure you and the other commenter have this mixed up. The fact that BB(748) is independent of ZFC doesn’t mean there are different models that have different values of BB(748). It means that ZFC is insufficient to determine the value of BB(748). That value is still some finite integer, you just can’t prove which one it is. Equivalently, there is some 748-state Turing machine which never halts but ZFC cannot prove never halts. And no, you can’t change your model such that this Turing machine halts in some non-standard number of steps. Or rather, you can, but that doesn’t actually change anything. The machine still doesn’t halt for the purposes of defining BB(748).
  - raincole 1y ago
    > I’m pretty sure you and the other commenter have this mixed up. We really don't. > that BB(748) is independent of ZFC > there are different models that have different values of BB(748) > ZFC is insufficient to determine the value of BB(748) These three statements are equivalent. f(n)=X is independent of ZFC means there are different models of ZFC that have different values of f(n). It's a very trivial theorem[0]. If you don't like it, I can't convince you otherwise. > that doesn’t actually change anything Changing the model will not change how any machine works in our physical, mechanical universe. However, it does change the value of BB(748). I understand your line of thinking: There is only one mechanical universe, which is the one where we exist. We can build Turing machines in this universe. BB(n) depends on Turning machines. Since there is only one single universe, there is only one single value of BB(n). It's a perfectly fine mental model for most cases. This was exactly how I thought when the first time I heard about BB(n). But it's not the kind of math than Scott Aaronson et al. are doing. Bar keeps running in our mechanical universe. But it can also halt in some non-standard number of steps. This weird, absurd-sounding proposition works because non-standard numbers simply don't map to anything in mechanical universe. They're purely abstract objects living in ZFC+~Con(ZFC). [0]: Given f(n)=X is independent of ZFC. Which means f(n)=X and ~(f(n)=X) are both consistent relative to ZFC. Therefore, if there is any model of ZFC, there is a model M1 that entails ZFC+(f(n)=X), and a model M2 that entails ZFC+~(f(n)=X). The value of f(n) cannot be the same in M1 and M2.

wat10000 1y ago

But Q is a number you can actually count, for a definition of “actually” that includes unimaginably large space and time. That finiteness comes from the basic mechanics of the Turing machine, which don’t depend on your mathematical axioms. Sure, you can come up with a set of axioms where the natural numbers include infinities. You may be able to use it to prove interesting things. But all that does here it make it so that the set of numbers describing how many steps a Turing machine runs before it stops is no longer the “natural numbers.”