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This bigger than grahams number?

by fat f...
Tags: bigger, grahams, number
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Robert1986
#19
Sep30-12, 04:21 PM
P: 828
Quote Quote by acabus View Post
Stop it. You're absolutely unimaginably nowhere near Graham's number, and it's pointless to try and come up with a larger number.
Really? Surely you can't mean *exactly* what you said. :)

It seems trivial to come up with a number larger than Graham's Number. Of course, I understand what you mean, though.
qpwimblik
#20
Oct25-12, 07:28 PM
P: 22
I got some numbers way bigger than Grahams puny number done with Conway's arrows.

10^100→10^100→... a Googol times over which = 1 HyperGoogol and each HyperGoogol is like the number before it inside each bit of the chain that many times over on the Conway chain.

This my friends is the place where you find numbers which can only be described by us mere mortals as silly numbers.

A Googol Hypergoogol's Is like flashing up the digits of this number part by part with the whole visible universe with transistors working at the plank length until the whole visible Universe around this visible universe you've turned into a computer is one big elephant simply because of the probability that it could become that and still having more digits to count.

It's just plain silly.
mfb
#21
Oct29-12, 07:47 AM
Mentor
P: 11,617
Quote Quote by Robert1986 View Post
It seems trivial to come up with a number larger than Graham's Number. Of course, I understand what you mean, though.
It is easy if you use notations with a similar power (e. g. G65 is larger than G64, of course, and GG64 is even larger) or even more powerful notations (Conway chained arrows).

It is not easy if you try to come up with something similar yourself.
Every tower of exponents (without more powerful definitions inside) is tiny compared to G1, and it is completely pointless to use those towers once you have Gn or conway arrows. G64G64 < G65, and that holds even if you build some large tower of exponents on the left side.
Plasmacircle
#22
Jan12-13, 06:22 AM
P: 1
Quote Quote by Deedlit View Post
fuga(n) = (...((n^n)^n)...^n)^n with n n's = n^(n^(n-1)) < n^(n^n).

So fugagargantugoogolplex < gargantugoogolplex ^ (gargantugoogolplex ^ gargantugoogolplex)
= 10^10^10^10^100^(10^10^10^10^100^(10^10^10^10^100))
= 10^10^10^10^100^(10^10^(10^10^100 + 10^10^10^100))
= 10^10^(10^10^100 + 10^(10^10^100 + 10^10^10^100))
< 10^10^10^(1 + 10^10^100 + 10^10^10^100))
< 10^10^10^10^10^10^101
< 3^3^3^3^3^3^3^3^3
= 3^^9,
This is wrong because you assumed "a gargantugoogolplex raised to a gargantugoogolplex a gargantugoogolplex times" to be "a gargantugoogolplex raised to the power of a gargantugoogolplex, raised to the power of a gargantugoogolplex". A gargantugoogolplex raised to the power of a gargantugoogolplex a gargantugoogolplex times = gargantugoogolplex^^gargantugoogolplex, = (10^10^10^10^100)^^(10^10^10^10^100) = (10^10^10^10^100)^((10^10^10^10^100)^(10^10^10^10^100)). MUCH larger than your 3^^9. however, it is still absolutely incomprehensibly smaller than Graham's number.
Jheavner724
#23
Jun23-13, 01:38 AM
P: 33
I just want to say that the amazing thing about Graham's number is that it actually serves a use, other numbers of that immense magnitude are simply trivial. I just thought I would mention this because it seems to be largely ignored on this thread.
Jheavner724
#24
Jun26-13, 08:38 PM
P: 33
Also, for anyone that wants to see a bigger number, although it is arguably less useful, check out the Kruskal Tree Theorem. There are a few other numbers like this that are bigger than Graham's number and serve a purpose, but they aren't as possible and usually considered more trivial.


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