4 ms·

is Graham's Number a galactic number? https://en.m.wikipedia.org/wiki/Graham's_number https://en.m.wikipedia.org/wiki/Graham's_number

saagarjha 7y ago
You could say so: > As with these, it is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number—and so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe.
- function_seven 7y ago
  > for a number of times far exceeding the total number of Planck volumes in the observable universe. Just so I'm clear. They're saying that not only can I not fit Graham's Number in all the Planck volumes of the universe; and I can't even count the digits of GN and write that in the Planck volume of the universe (and so on), but the number of "indirections" is itself so large as to not fit in the universe? Like: 1. GN (can't fit). 2. Number of digits in GN (can't fit). 3. Number of digits in #2 (can't fit). 4. Number of digits in #3 (can't fit). ... N. <-- The numbered list item itself won't fit. Am I understanding that right?
  - Gunax 7y ago
    Yes. log(log(log... GN))) applied X times (where X is the number of planck volumes in the universe) is still greater than X. Where log = base 10 logarithm. Hofstadter talks a bit about this abstraction in his article 'On Number Numbness' > If, perchance, you were to start dealing with numbers having millions or billions of digits, the numerals themselves (the colossal strings of digits) would cease to be visualizable, and your perceptual reality would be forced to take another leap upward in abstraction-to the number that counts the digits in the number that counts the digits in the number that counts the objects concerned.
    - saalweachter 7y ago
      And despite that, Graham's Number is still in the countable set. :-)
  - marcosdumay 7y ago
    It shouldn't be surprising, because the Graham's Number is all about applying those indirections recursively into more indirections.
theemathas 7y ago
But have you heard of Rayo's number?

saagarjha 7y ago

You could say so: > As with these, it is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number—and so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe.

function_seven 7y ago
> for a number of times far exceeding the total number of Planck volumes in the observable universe. Just so I'm clear. They're saying that not only can I not fit Graham's Number in all the Planck volumes of the universe; and I can't even count the digits of GN and write that in the Planck volume of the universe (and so on), but the number of "indirections" is itself so large as to not fit in the universe? Like: 1. GN (can't fit). 2. Number of digits in GN (can't fit). 3. Number of digits in #2 (can't fit). 4. Number of digits in #3 (can't fit). ... N. <-- The numbered list item itself won't fit. Am I understanding that right?
- Gunax 7y ago
  Yes. log(log(log... GN))) applied X times (where X is the number of planck volumes in the universe) is still greater than X. Where log = base 10 logarithm. Hofstadter talks a bit about this abstraction in his article 'On Number Numbness' > If, perchance, you were to start dealing with numbers having millions or billions of digits, the numerals themselves (the colossal strings of digits) would cease to be visualizable, and your perceptual reality would be forced to take another leap upward in abstraction-to the number that counts the digits in the number that counts the digits in the number that counts the objects concerned.
  - saalweachter 7y ago
    And despite that, Graham's Number is still in the countable set. :-)
- marcosdumay 7y ago
  It shouldn't be surprising, because the Graham's Number is all about applying those indirections recursively into more indirections.

theemathas 7y ago

But have you heard of Rayo's number?