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Old 07-25-2014, 10:04 PM   #1
calthax
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Default My number is, in fact, Larger!

The goal is to come up with a computable, finite, well-defined 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 one-up 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.
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Old 07-26-2014, 10:06 AM   #2
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Default 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.
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Last edited by hyperme; 07-26-2014 at 10:10 AM. Reason: you can only ! an int.
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Old 07-26-2014, 02:55 PM   #3
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Default Re: My number is, in fact, Larger!

How does that work
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Old 07-26-2014, 09:54 PM   #4
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Default 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.
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Old 07-27-2014, 12:40 AM   #5
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Default Re: My number is, in fact, Larger!

Quote:
Originally Posted by Gigacat View Post
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.
Nope, you forgot to edit it fully. Also, you were told you can't mention the previous number.

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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Old 07-30-2014, 01:00 PM   #6
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Default 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
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Old 07-30-2014, 02:30 PM   #7
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Default Re: My number is, in fact, Larger!

I have no idea what any of these numbers mean, but from what I've gathered, Grahams number automatically wins.
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Old 07-30-2014, 03:54 PM   #8
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Default Re: My number is, in fact, Larger!

Quote:
Originally Posted by calthax View Post
The goal is to come up with a computable, finite, well-defined number that's Larger than the last poster's number.
Quote:
Originally Posted by calthax View Post
Define set N as { x! | 0 < x < R, x is an integer} where R is the product of all the terms in set F as { x! | 0 < x < R, x is an integer} with R as Rayos number.

Let N be equal to the summation of all elements of F.
I win
You do not, in fact, win, because you have named an uncomputable number.
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Old 07-31-2014, 02:14 AM   #9
hyperme
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Default Re: My number is, in fact, Larger!

hey 'googolgy' isn't an actual field of mathematics please find good sources please.
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Old 07-31-2014, 11:43 AM   #10
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Default 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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