07252014, 10:04 PM  #1 
Karyian
Join Date: Nov 2008
Location: Calthax's World
Posts: 403

My number is, in fact, Larger!
The goal is to come up with a computable, finite, welldefined number that's Larger than the last poster's number. You can only use already established numbers and notations (the xkcd number counts, as does Graham's number. But you can't just oneup that by saying g_65 without defining the recursion here), and you can't refer directly to the previous number. (If I post a description D of some number x, your number can't be defined as "D"+ 1". If it turns out to be D+1, fine. Just don't do anything like quoting the previous entry and putting +1 at the bottom.)
I'll start us off with one million. Because we all know that anything over one million is Larger Edit: Also, if someone questions the size of your number, you have to be able to prove that it's really Larger than the previous one. You can steal a notation from someone else, and the number will be valid. 
07262014, 10:06 AM  #2 
Not Evil, Just British
Join Date: Mar 2008
Location: Britain
Posts: 1,322

Re: My number is, in fact, Larger!
Define set F as { x!  0 < x < G, x is an integer} where G is Graham's number.
Let N be equal to the summation of all elements of F. I win. Last edited by hyperme; 07262014 at 10:10 AM. Reason: you can only ! an int. 
07262014, 02:55 PM  #3 
Karyian
Join Date: Nov 2008
Location: Calthax's World
Posts: 403

Re: My number is, in fact, Larger!
How does that work

07262014, 09:54 PM  #4 
Pinkie Pirate!
Join Date: Oct 2009
Location: Over There!
Posts: 1,007

Re: My number is, in fact, Larger!
Define set N as { x!  0 < x < G, x is an integer} where G is the product of all the terms in set F as { x!  0 < x < G, x is an integer} with G as Grahams number.
Let N be equal to the summation of all elements of F. I, infact, win. 
07272014, 12:40 AM  #5  
The Demagogue
Join Date: Jun 2007
Location: At home duhhh!
Posts: 2,416

Re: My number is, in fact, Larger!
Quote:
Define set F as the set of all complex numbers whose magnitudes are intergers and whose bounds on both the real and imaginary axis of +/ G(raham's number). Then define set E as {y!  y = x for all x}. Let N be the sum of all elements in E.
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Back to the roots: Lets put the Hamumu back into Hamumu Forums! 

07302014, 01:00 PM  #6 
Karyian
Join Date: Nov 2008
Location: Calthax's World
Posts: 403

Re: My number is, in fact, Larger!
Define set N as { x!  0 < x < (3>3>3>3), x is an integer} where R is the product of all the terms in set F as { x!  0 < x < (3>3>3>3), x is an integer} with 3>3>3>3 defined in conway chained arrow notation
Let N be equal to the summation of all elements of F. I win Last edited by calthax; 07302014 at 06:57 PM. 
07302014, 03:54 PM  #8  
Veteran Programmer
Join Date: Oct 2007
Posts: 3,076

Re: My number is, in fact, Larger!
Quote:
Quote:


07312014, 02:14 AM  #9 
Not Evil, Just British
Join Date: Mar 2008
Location: Britain
Posts: 1,322

Re: My number is, in fact, Larger!
hey 'googolgy' isn't an actual field of mathematics please find good sources please.

07312014, 11:43 AM  #10 
Veteran Programmer
Join Date: Oct 2007
Posts: 3,076

Re: My number is, in fact, Larger!
Sorry, that was the best source I could find quickly. After searching harder, here's something more reliable, with its own sources, but less detailed and doesn't mention computability.

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