Which is larger, Graham's number or

  • Context: Graduate 
  • Thread starter Thread starter MathJakob
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the comparison between Graham's number and the factorial of the estimated number of atoms in the universe, denoted as (1080)!. It is established that (1080)! is significantly smaller than Graham's number, specifically g1, and that Graham's number itself is g64. The conversation also touches on the Arildno number, defined as Graham's number + 1, which is deemed meaningless in practical terms. Additionally, the discussion highlights TREE(3) as a number that surpasses both Graham's number and the lower bounds associated with it, emphasizing its position in the fast-growing hierarchy.

PREREQUISITES
  • Understanding of large numbers and their notation, specifically Graham's number and factorials.
  • Familiarity with the Ackermann function and its implications in number theory.
  • Knowledge of the fast-growing hierarchy and its significance in mathematical logic.
  • Basic comprehension of ordinals and their applications in set theory.
NEXT STEPS
  • Research the properties and implications of Graham's number in combinatorial mathematics.
  • Explore the Ackermann function and its role in defining large numbers.
  • Study TREE(n) and its position within the fast-growing hierarchy.
  • Investigate the concept of ordinals, particularly the Small Veblen Ordinal and Large Veblen Ordinal.
USEFUL FOR

Mathematicians, logicians, and enthusiasts of theoretical mathematics who are interested in large numbers, combinatorial theory, and the complexities of the fast-growing hierarchy.

MathJakob
Messages
161
Reaction score
5
This isn't a homework problem or anything just simply curious. Which is a larger number, Graham's number or the estimated number of atoms in the universe factorial?

So Graham's number or ##(10^{80})!##
 
Mathematics news on Phys.org
Well, as an upper bound, \left(10^{80}\right)! < \left(10^{80}\right)^{\left(10^{80}\right)} = 10^{80 * 10^{80}} = 10^{8*10^{81}}

which, if I am not mistaken, is far less than g_1.
 
So ##(10^{80})!## isn't even as large as g1... and Graham's number is g64... WOW
 
The Arildno number is even greater:

Arildno number =Graham's number+1.

Can I also get a WOW! from you?
:smile:
 
arildno said:
The Arildno number is even greater:

Arildno number =Graham's number+1.

Can I also get a WOW! from you?
:smile:

No. Your number is meaningless, it's never been used in any meaningful way.
 
MathJakob said:
No. Your number is meaningless, it's never been used in any meaningful way.

:cry:
 
Sorry to link to wikipedia, but you may be interested in TREE(3), and I can't find too much info on it.

an extremely weak lower bound for TREE(3), is A(A(...A(1)...)), where the number of As is A(187196)

Graham's number, for example, is approximately A^64(4) which is much smaller than the lower bound A^(A(187196))(1).
A is the Ackermann function.

http://en.wikipedia.org/wiki/Kruskal's_tree_theorem
 
MathJakob said:
This isn't a homework problem or anything just simply curious. Which is a larger number, Graham's number or the estimated number of atoms in the universe factorial?

So Graham's number or ##(10^{80})!##

Tip: if you can write out the number in any "normal" way, it's waaaaaaaaaaaaaaaaay smaller than Graham's number.
 
What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!
 
  • #10
Deedlit said:
What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!

It seems the Arildno number is a dwarf, after all..
 
  • #11
arildno said:
It seems the Arildno number is a dwarf, after all..

Poor thing, the Arildno number is coming under lots of fire in here!

Deedlit said:
What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!

I haven't really been able to find any sources that explain some of the larger operators in the fast growing hierarchy. For example,

http://googology.wikia.com/wiki/Fast-growing_hierarchy

gives a list of growth rates, but there's a lot of details being left out. Sadly, I only understand how to calculate up to and including f_{\epsilon_0}(n).

And to do so, I use a calculator of course :biggrin:
 
  • #12
Good to see you found the Googology Wiki!

Beyond \varepsilon_0 we have:

\varepsilon_1 = \varepsilon_0 ^ {\varepsilon_0 ^ {\varepsilon_0 ^\ddots}}

\varepsilon_2 = \varepsilon_1 ^ {\varepsilon_1 ^ {\varepsilon_1 ^\ddots}}

\varepsilon_\omega = \sup \lbrace \varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots \rbrace

and so on. Eventually we get to

\zeta_0 = \varepsilon_{\varepsilon_{\varepsilon_\cdots}}

We can set \varphi(0, \alpha) = \omega^\alpha, \varphi(1, \alpha) = \varepsilon_\alpha, \varphi(2, \alpha) = \zeta_\alpha. More generally,

\varphi(\alpha+1, \beta) = the \betath fixed point of f(\gamma) = \varphi(\alpha, \gamma).

When \alpha is a limit ordinal, \varphi(\alpha, \beta) is the \betath ordinal in the intersection in the ranges of f(\delta) = \varphi(\gamma, \delta) for all \gamma < \alpha.

So, for example,

\varphi(3, 0) = \varphi(2, \varphi(2, \varphi(2, \ldots)))

\varphi(\omega, 0) = \sup (\varphi(1, 0), \varphi (2, 0), \varphi (3, 0), \ldots)

and so on. This takes us up to

\Gamma_0 = \varphi(\varphi(\varphi(\ldots, 0),0),0)

or alternatively, \Gamma_0 is the smallest ordinal \alpha such that \alpha = \varphi(\alpha, 0).

We can continue the notation with \varphi(1, 0, 0) = \Gamma_0, and \varphi(1, 0, \alpha) is the \alphath fixed point of f(\beta) = \varphi(\beta, 0).

The ordinals \varphi (1, \alpha, \beta) are defined analogously to \varphi(\alpha, \beta), i.e. each function f(\beta) = \varphi (1, \alpha+1, \beta) enumerates the fixed points of g(\beta) = \varphi(1, \alpha, \beta), and at limit ordinals you enumerate the intersection of the ranges of previous ordinals. Then \varphi(2, 0, \alpha) is the \alphath fixed point of f(\beta) = \varphi(1, \beta, 0), and you can construct the hierarchy \varphi(2, \alpha, \beta) similarly. At limit ordinals you take the intersection of the ranges again, and so we define \varphi (\alpha, \beta, \gamma) for all ordinals \alpha, \beta, \gamma. Then we can define

\varphi (1, 0, 0, \alpha) to be the \alphath fixed point of f(\beta) = \varphi(\beta, 0, 0).

We can then define \varphi(\alpha, \beta, \gamma, \delta), \varphi(\alpha, \beta, \gamma, \delta, \epsilon), and so on. The general definition for the n-ary Veblen funciton is:

\varphi (\alpha) = \omega^{\alpha}

\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n + 1, 0, \ldots, 0, \beta) is the \betath fixed point of the function f(\gamma) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_n, \gamma, 0, \ldots, 0)

When \alpha_n is a limit ordinal, \varphi (\alpha_1, \alpha_2, \ldots, \alpha_n, 0, \ldots, 0, \beta) is the \betath ordinal in the intersection of the ranges of f(\delta) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}, \gamma, \delta, 0, \ldots, 0) for all \gamma < \alpha_n.

We define the Small Veblen Ordinal as

\sup (\varphi (1, 0), \varphi(1, 0, 0), \varphi(1, 0, 0), \ldots).

TREE(n) is larger than the fast-growing hierarchy at the level of the Small Veblen Ordinal. (how much further is not known.).

I'll go one step further: we can extend the n-ary Veblen function to transfinitely many places. Obviously, we can't write out transfinitely many variables, so we need to modify our notation: instead of

\varphi(\alpha, \beta, \gamma, \delta, \epsilon)

we write

\varphi(\alpha @ 4, \beta @ 3, \gamma @ 2, \delta @ 1, \epsilon @ 0).

So we append "@ n" to every variable, where n represents the index of the variable. This allows us to skip variables that are 0, and so we can notate things like \varphi (1 @ \omega, \alpha @ 0). \varphi(1 @ \omega, \alpha @ 0) is defined as the \alphath ordinal that is a fixed point of f(\beta) = \varphi (\beta @ n) for all n < \omega.

More generally, we define

\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n + 1, \gamma @ 0) is the \gammath fixed point of f(\delta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \delta @ \beta_n)

When \alpha_n is a limit ordinal, \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \gamma @ 0) is the \gammath ordinal in the intersection of the ranges of f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n + 1, \epsilon @ \beta_n) for all \delta < \alpha_n

When \beta_n is a limit ordinal, \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n, \gamma @ 0) is the \gammath ordinal in the intersection of the ranges of f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \epsilon @ \delta) for all \delta < \beta_n

When \alpha_n and \beta_n are limit ordinals, \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \gamma @ 0 is the \gammath ordinal in the intersection of the ranges of f(\zeta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n, \zeta @ \epsilon) for all \delta < \alpha_n, \epsilon < \beta_n

This defines \varphi(\alpha_1 @ \beta_1, \ldots, \alpha_n @ \beta_n) for all \alpha_i and \beta_i. This notation is known as Schutte's Klammersymbolen.

The smallest ordinal \alpha such that \alpha = \varphi (1 @ \alpha) is known as the Large Veblen Ordinal. I would think that TREE(n) would not reach the Large Veblen Ordinal in the fast-growing hierarchy, but I don't think this is known.

Phew! I hope this was at least somewhat comprehensible.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 43 ·
2
Replies
43
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K