Comments by "MC116" (@angelmendez-rivera351) on "TREE(Graham's Number) (extra) - Numberphile" video.

  1. He never claimed that was the largest number to store. He said this number expresses a cap on the data in terms of the area of the universe.
    6
  2. Mike Wagner TREE(n) for all n can be proven to be finite for all n in transfinite arithmetic, but not with finite arithmetic. However, for each individual value of n = k, a theorem stating that TREE(k) is finite exists in finite arithemetic, but this proof would be impossible to complete, it would take "too long" in a rigorous sense.
    4
  3. No, the amount of seconds will be even less than g(64). Actually, it's smaller than g(1). You are REALLY underestimating how stupidly large g(1). This number cannot be expressed using exponentiation alone.
    3
  4. kazedcat TM No. First of all, it has never been confirmed a multiverse exists. Second, that multiverse you describe is a type II multiverse, but again, there is no guarantee of its existence.
    3
  5. Jamie Den Adel Mmmmm... no, it is not.
    2
  6. Zathras Yes That is not how physical existence works.
    2
  7. Michoss9 TREE(3), not tree(3). tree(3) is a different but related quantity.
    2
  8. Also, if you really want to talk about physics here, the consensus is that it is not meaningful to talk about the universe beyond the observable universe. For scientific purposes, it does not exist. We can make hypothesis all the time, but hypothesis is not science.
    2
  9. Kelly Stratton 10^10^10^10^10 is an upper bound to the Poincaré recurrence time of what in physics is called an "empty, vast universe," which represents an universe bubble several orders of magnitude larger than our own observable universe. This is to say, if we had to consider the amount of time it would take for the universe to reset itself and traverse every single quantum microstate possible, then 10^10^10^10^10 planck units of time is an upper bound. This number must obviously be bigger than the number of possible microstates of the universe, which is in itself bigger than the biggest possible number that can be encoded in the universe.
    2
  10. erik2000 There i no such a thing as "the whole universe," that's ill defined.
    2
  11. Alazoral Also, we can almost be certain Everett's interpretation is not true. Sure, it makes sense as an interpretation of QM, but once you delve into the more accurate description of reality, QFT, the concept of the Everett multiverse ceases to be coherent. People need to abandon this whole notion of "interpretations of quantum mechanics," as they are extremely inaccurate and outdated.
    2
  12. Alazoral I'm not sure why you are making a philosophical argument on a topic that is strictly physical. Also, bold of you to claim that angles cannot be quantized without proof. If that is the game you want to play, then nobody has any reason to give proof to anything.
    2
  13. rotwang2000 Actually, the universe will just reset itself before this happens, making this never happen.
    1
  14. There is no faster speed of light, though.
    1
  15. Andrew Cunningham In, fact, even a number such as 10^10^10^10^10 is unfathomable. It is beyond the realm of comprehensibility of anyone, depending on how you comprehensibility.
    1
  16. Arnaz87 Because the idea of "the universe" is completely meaningless, it's like talking about 1/0. The only thing that is known to exist is the observable universe, so the only thing we are logically allowed to pretend exists is the observable universe.
    1
  17. Perplaxus 10^122 is the binary size of the biggest number that can be stored in the universe, not the number itself. But what is the biggest number we can represent? There is no such biggest number, since you can make arbitrary notation to represent arbitrarily large finite quantities, and that does not even account for transfinite quantities, of which there is no largest representable member. So it is not sensible question. Instead, the question that makes sense is the largest number that can be stored in the universe, which is what was addressed in the video.
    1
  18. A weak consensus is not a consensus at all.
    1
  19. Awwkaw Do you not understand the holographic principle?
    1
  20. Awwkaw You really are forming strong conclusions based on the first sentence of a Wikipedia article of all things? I'm sorry, but if you want to have a serious discussion, that's not going to cut it for anyone. You must be joking if you think I can call that researching a definition. You could at least put a bit more effort, don't you think? Going down too many dimensions just loses information. I can make a projection of a ball onto a plane to get a disk, but I cannot do the same and get a point or a line. Intuitively, this makes sense, and if you're skeptical, just try thinking about it yourself. You'll end up concluding this yourself. This alone is sufficient to show the amount of dimensions this lower-dimensional boundary has is not arbitrary and cannot be made as small as one wants it. You could do further maths to then show that for must situations, you only can go down 1 dimension, but the maths for that are rather complicated and too much to discuss on YouTube. However, the intuitive idea is there.
    1
  21. James If a person is going to be asking questions without putting the effort of doing the "research" that will allow them to understand the answer to said question, and when they are going to decide to form a conclusion off of the first sentence of Wikipedia, then that person deserves a harsh response. That's not genuine curiousity, that's called being disingenuous. It's like forming your opinion of what the answer should be before getting said answer and using that opinion to then discard the answer. If you want to do that, then don't ask at all, geez. And you can disagree and say a person has the right to be disingenuous. Fine by me. But that doesn't entitle anyone to polite responses. Welcome to the Internet. That's how it works. P.S: and no, I am not saying a person has to be an expert to be allowed to ask questions. But if I ask you "do you know what the holographic principle is?" just say you don't know and problem is solved instead of giving me a half-assed answer taken from a non-comprehensive sentence of a wiki. I'm just asking for people to not be disingenous when asking a question and projecting biases. That's all.
    1
  22. Josh Duewall Sure, but that is an abuse of language. These compressions are not what computer scientists actually call "storing a number."
    1
  23. Daniel V That is a completely different thing. You're confusing different notions of compressing. One thing has nothing to do with the other.
    1
  24. None of these combinatorial quantities exceed 10^10^10^10^10. So, no.
    1
  25. erik2000 It has everything to do with it, since which number is correct to answer the question very well might depend on the truth of said hypothesis.
    1
  26. Alazoral Mathematically, those combinations on a Rubik's cube exist, but physically, many of those are impossible to achieve from construction of the cube, and some are inaccessible with a finite algorithm. And that has literally nothing to do with the Everett multiverse interpretation, which also has nothing to do with the storage limit of the universe.
    1
  27. Head Librarian Because they are physically inaccessible. See, the problem with the calculation of the combinations of a Rubik's cube is that it ignores the way a Rubik's cube works. There nine slabs you can rotate, and there are three possible ways you can rotate each. You cannot, for instance, remove the corner cube out, permute its faces, and insert it back into the Rubik's cube. It's not part of how you are able to rotate these slabs. No finite combination of rotations will achieve a configuration which has only that one corner cube permuted while everything else stays identical. In fact, this very fact is very important when it comes to solving the Rubik's cube in the first place. The minimum number of moves required to solve the Rubik's is directly dependent on the constraints that are being placed on what can and cannot be moved.
    1
  28. Chris Sekely There is no such largest number, since arbitrary symbolic notation can be used to express arbitrarily large numbers. We can go to arbitarily high orders of logic to compress expressions as much we want. More concretely, I can define a function F(n) such that this is equal to the smallest number not representable in x-order logic with n symbols. Then let n be the number of Planck volumes in the universe and I have expressed a large number. Normally, you would want to use the label F_x to specify the order of logic of the function. With this method, there is such a largest number. However, I can circumvent this entirely by simply creating new symbols. Symbolic logic sets no restriction on what symbols I can use, so long as they are part of the language. I can arbitrarily expand the language and add new symbols arbitrarily, which allows me to not need to label the function, but rather just use a new symbol for a higher order logic function. Then any limitations would come from the limit of possible symbols I can use. As far as I understand, though, there is no limit. For any symbol that exists, I can make a new symbol from it. Okay, I suppose you may be able to come up with such a limitation symbols. But I'm already a step ahead of you: I can define a function such that there is a number not expressible in this type of notation, and I can do so in symbolic logic. In fact, I can define the function F(n) as the smallest number not expressible in n symbols in any lower order of logic with new symbols. And so on. You may have to consider transfinite orders of logic, and so, but you can always go one order higher and form a compression defined explicitly from the limits of the previous orders.
    1
  29. Benny Carpenter-Deason Something like 10^122.
    1
  30. Daniel Bamberger This number has already been calculated, and it is not bigger than e^e^10^10^10. Not even close to 3^^^3. So even by taking into account combinations, the amount is still tractable. Not 3^^^3.
    1
  31. Ξενοφώντας Σούλης Sure, but that the rules exist is not relevant. Computer scientists don't exactly care about whether the compressed version of a quantity can be compressed (it always can be if you go a high enough order of logic.) Calling that "expressible" means you don't understand the definition of "expressible."
    1
  32. Ξενοφώντας Σούλης Describing and expressing are not the same thing. But whatever. I'm not going to waste my time explaining such a basic difference to people on YouTube. It's not what degrees are for. Believe what you want. Have a nice day.
    1
  33. Gamesaucer Sure, but some functions are incompressible, meaning that for low values of n, they exceed R[R(2^10^122 + 1)], where R(n) is Rayo's function. Their second order logic formulas don't exist and are not expressible with all the storage space in the universe.
    1
  34. Gamesaucer Well, there is always a way to compress a function in a higher order logic. If it is impossible to compress a function to be expressible by a certain amount of symbols in a certain order logic, then you go to a higher order logic and define the number as the number such that it takes more symbols than available in the previous order. But in every order of logic, incompressible functions exist. But going beyond first order logic pretty much the concept meaningless and intractable.
    1