How much bigger than Graham’s Number is TREE(3)?

March 11, 2025 12:23 PM

They say the latter “dwarfs” the former, but I don’t know how to interpret that. Would a nested G(G(…G(x))) function filling every Planck volume in the observable universe even be in the ballpark? On a logarithmic scale of increasing enormity, is Graham’s number closer to 1 than it is to TREE(3)? Is it practically indistinguishable from 1? Is TREE(3), like, a whole different strata of reality?
posted by Lemkin to Science & Nature (4 answers total)

[wiki:] Graham's number, for example, is much smaller than the lower bound ... approximately g_{3↑¹⁸⁷¹⁹⁶3}

[iflscience:] But that "is just some rubbish lower bound on it,” he explains. “We don’t have an upper bound”
posted by HearHere at 1:16 PM on March 11 [1 favorite]

TREE(3) is big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the Graham's Number, but that's just peanuts to TREE(3).
posted by jozxyqk at 1:26 PM on March 11 [7 favorites]

Best answer: According to this video from Numberphile, Graham's Number is effectively zero compared to Tree(3).
posted by justkevin at 1:35 PM on March 11 [3 favorites]

Best answer: > Would a nested G(G(…G(x))) function filling every Planck volume in the observable universe even be in the ballpark?

This article is a pretty good explanation, and also I think pretty much a specific answer to your question:

Graham’s Number is trivial, basically zero compared to TREE(3), and that is so much the case that if instead of going to G64 via the steps, you went to GGoogolplex, or even G(Graham’s Number), i.e., G(G64), you would still be at zero relative to TREE(3)

It's a bit like how multiplication is so much "faster" than addition, and then exponentiation is "faster" than multiplication, and then tetration is "faster" than exponentiation, and then the Graham sequence is even "faster" than that.

But TREE(3) is so much bigger, I don't think you can even start to approach it by going up this type of ladder at all, no matter how many rungs of this type you stack. It is like another universe of bigness that you just can't get to by stacking these exponential-type functions on top of each other the way Graham's Number does, even if you continue this type a stacking a bunch more times.

Here is another good answer that explains this concept:

[T]here is no known upper bound for TREE(3).

To have at least a slight idea how huge this gap is, you should study the "fast growing hierarchy":

* Graham's number has level ω+1.

* Conway chains far surpass Graham's number, already 4→4→4→4 is vastly larger. The level of Conway chains of length n is about ω2.

* Bowers arrays far surpass Conway chains. [3,3,3,3,3] is already inexpressible with Conway chains. Even Bowers planes only reach ωω2-level and with 3 dimensions ωω3.

* This is still far away from the ϵ0 level and this is FAR surpassed by the Γ0 level.

* And even this level is FAR too low for TREE(3).

I think you begin to understand that TREE(3) is not just another league, the difference in magnitude is barely comprehensible even in googology-standards.

posted by flug at 12:24 AM on March 12 [2 favorites]