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Author Topic: My number is bigger!  (Read 11684 times)

NAV

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Re: My number is bigger!
« Reply #45 on: November 16, 2014, 05:57:25 pm »

3
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alemagno12

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Re: My number is bigger!
« Reply #46 on: November 16, 2014, 07:31:13 pm »

3
lol.
But it's not valid.
I'm willing to dispute your statements that A(G,G) is smaller than your number.
Apparently you have no idea how large large numbers can actually be.

First of all, A(G) = A(G,G) = fω(G), and AG(G) = fωn(G) = fω+1(G), and your number is AAG(G)(G) < AAG(G)(AG(G)) = fω+1(fω+1(G)) = fω+12(G) < fω+1G(G) = fω+2(G). Note how incredibly unexact those inequalities are. And since ω+2 is so very very much smaller than ψ(ψI(0)), your number is so incredibly much smaller than fψ(ψI(0))(G) you can't even imagine how hard it would be to even come up with this shit in the first place.

EDIT: Suddenly I know so much about ordinals. Thanks, thread.

EDIT2: Aaaaaaactually, after reading through the definition of fast-growing hierarchies again, my previous statements are all not really true. Since alemagno's number is dependent on the choice of the fast-growing hierarchy (which is defined by the choice of fundamental sequences of limit ordinals) and since both ψ(ψI(0)) and ω are limit ordinals, fψ(ψI(0))(G) could very well be smaller than fω+1(G), namely if you choose the first G elements of ω's and ψ(ψI(0))'s limit sequences to both be the first G integers, then AAG(G)(G) > fω+1(G) > fω(G) = fω[n](G) = fG(G) = fψ(ψI(0))[n](G) = fψ(ψI(0))(G). So alemagno, you need to specify your fundamental sequences first before making such claims, because your number was not completely defined.
Noope.
Read again.
It's wrong.
ψ(ψI(0)) is FAR greater than ω.
And fω+1(G) beats your ''fψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.
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FallenAngel

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Re: My number is bigger!
« Reply #47 on: November 16, 2014, 08:00:29 pm »

Great, now I'm cross-eyed again.
I'm going to go back to my circles whose circumference is greater than πd and toot on my Gabriel's Horn.

Also, I realized a typo in what I said previously. I had many instances of fω+1, while I had meant to put fω+1.
Revising my previous statement...
I know how large numbers can get - I just wanted to pry some info out of the situation.
Going by the numbers that have been derived from the xkcd function, I have a new, larger number.
fω+1(fω+1(G)), my large ...thing made of nested Ackermann functions with the root of it all being A(G,G), is not close to your number, or fψ(ψI(0))(G).
Now, to take this information. Since fω+1(fω+1(G)) is approximately A(A(A(A(A(A(...(G,G)...), where the number of A() is equal to A(A(A(A(...(G,G)...), in which the number of A() is Graham's Number,  the two can technically be interchangeable - we're at large enough numbers where most differences between similar numbers can be ignored safely - A(G,G)-1 can be written as A(G,G) to no ill effect.
From this, my new number is fω+1(fω+1(fω+1(fω+1(fω+1(...(G)...), in which function the fω+1(...) are nested G taken to the xkcd hyperoperation taken to the xkcd hyperoperation times. Plus 1. Because one is my seventh-favorite number, and 7 is considered "lucky".
Also, you had previously stated that fω+1(G) is drastically lower than fψ(ψI(0))(G), but now you are saying that the latter is the smaller. Make up your mind.
And as a "I'm-tired-right-now" thing to put down, fω+(fw+G)(G). I'm admittedly sort of lost by now, but I'm rather confident that said number is a large number, and certain that it is larger than fω+1(G).

alemagno12

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Re: My number is bigger!
« Reply #48 on: November 16, 2014, 08:14:49 pm »

3
lol.
But it's not valid.
I'm willing to dispute your statements that A(G,G) is smaller than your number.
Apparently you have no idea how large large numbers can actually be.

First of all, A(G) = A(G,G) = fω(G), and AG(G) = fωn(G) = fω+1(G), and your number is AAG(G)(G) < AAG(G)(AG(G)) = fω+1(fω+1(G)) = fω+12(G) < fω+1G(G) = fω+2(G). Note how incredibly unexact those inequalities are. And since ω+2 is so very very much smaller than ψ(ψI(0)), your number is so incredibly much smaller than fψ(ψI(0))(G) you can't even imagine how hard it would be to even come up with this shit in the first place.

EDIT: Suddenly I know so much about ordinals. Thanks, thread.

EDIT2: Aaaaaaactually, after reading through the definition of fast-growing hierarchies again, my previous statements are all not really true. Since alemagno's number is dependent on the choice of the fast-growing hierarchy (which is defined by the choice of fundamental sequences of limit ordinals) and since both ψ(ψI(0)) and ω are limit ordinals, fψ(ψI(0))(G) could very well be smaller than fω+1(G), namely if you choose the first G elements of ω's and ψ(ψI(0))'s limit sequences to both be the first G integers, then AAG(G)(G) > fω+1(G) > fω(G) = fω[n](G) = fG(G) = fψ(ψI(0))[n](G) = fψ(ψI(0))(G). So alemagno, you need to specify your fundamental sequences first before making such claims, because your number was not completely defined.
Noope.
Read again.
It's wrong.
ψ(ψI(0)) is FAR greater than ω.
And fω+1(G) beats your ''fψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.

You were the one who said the latter, by the way. Does that mean I win?
I mean in the user's sense.
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FallenAngel

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Re: My number is bigger!
« Reply #49 on: November 16, 2014, 08:25:55 pm »

I agree.
...what?

I mean, I often make statements that make no real sense. Typically such statements are on purpose, but sometimes they aren't.
But that is one heck of a statement.
I'm going to assume you're discussing ontological inertia and continue on with my day.

alemagno12

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Re: My number is bigger!
« Reply #50 on: November 16, 2014, 08:29:58 pm »

So game over?
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FallenAngel

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Re: My number is bigger!
« Reply #51 on: November 16, 2014, 09:53:36 pm »

Technically no, if we stretch the rules a tad.
We can certainly get larger. Very easily, I might add.
Just the creativity would slowly boil away until we're doing the higher mathematical version of x+1.

Putnam

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Re: My number is bigger!
« Reply #52 on: November 16, 2014, 11:40:34 pm »

non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest
« Last Edit: November 16, 2014, 11:53:47 pm by Putnam »
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Putnam

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Re: My number is bigger!
« Reply #53 on: November 17, 2014, 01:38:07 am »

"The universe" is vague.

Putnam

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Re: My number is bigger!
« Reply #54 on: November 17, 2014, 02:05:58 am »

See, now it's not vague, because you gave a term :P And yeah, that's a very nice, all-encompassing definition. Trying to one-up it will just result in it getting "bigger".

MagmaMcFry

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Re: My number is bigger!
« Reply #55 on: November 17, 2014, 02:39:53 am »

-
Noope.
Read again.
It's wrong.
ψ(ψI(0)) is FAR greater than ω.
And fω+1(G) beats your ''fψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.
My point is that they don't necessarily have to, because the definition of f depends on the choice of all your fundamental sequences (which need not be related at all), and since the only requirement for a fundamental sequence for a limit ordinal is that it strictly increases and converges to that limit ordinal, you can very well choose such sequences for ω and ψ(ψI(0)) such that the first few elements are the same, e.g. ψ(ψI(0))[G] = G and ω[G] = G.

Now remember the definition of fast-growing hierarchy, which says that if α is a limit ordinal, fα(n) is defined as fα[n](n). So in this case, fω+1(G) = fωG(G) > fω(G) = fω[G](G) = fG(G) = fψ(ψI(0))[G](G) = fψ(ψI(0))(G).

Now as I said before, different choices of fundamental sequences lead to different fast-growing hierarchies, and it may very well be possible (for appropriate choice of the fundamental sequences) for your number to be larger than Fallen's. But until you provide a choice of fundamental sequence for all the limit ordinal numbers up to ψ(ψI(0)), your number is not even well-defined, as opposed to Fallen's perfectly well-defined AAG(G)(G).

My number is Ω_A(NiB), nested NiB times.
Well, I hereby define a function Z that takes an integer and adds the imaginary unit. Now your definition of A(x) is broken forever, because chained arrows don't operate on complex numbers. So there.
« Last Edit: November 17, 2014, 02:47:11 am by MagmaMcFry »
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alemagno12

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Re: My number is bigger!
« Reply #56 on: November 17, 2014, 08:53:12 am »

Ok, i'm going to see if your entries are valid.
Well, if we stretch the rules to include non-computable values, then...

I will start off with f1(G).
First off, ℵ1 is NOT an ordinal, it's a cardinal.
Also, if we consider ℵ1 to be ω1 (first uncountable ordinal), then we have a problem here, because ω1 does not have a fundamental sequence.
And i'm using ω1 because ω1 has cardinality ℵ1, if you were wondering.
So invalid.
non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest
Congratulations, you have just breaked the rules of the game!
non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest

Hehe. Well, let's get truly meta:

j: V -> V

There we go. Note- this is in ZK, not ZKC. j is a cardinal, and V is the universe. We don't know whether such a cardinal can exist, but if it does, it would be the supremum of cardinals- as in, there can not be any greater cardinal.
Is this finite?
Are finite numbers allowed?
Are cardinals allowed?
Are ordinals allowed?
Are cardinals infinite?
Are ordinals infinite?
To quote the Wikipedia page on why the number I just mentioned can't exist in ZKC (Kunen's Inconsistency):

"There is no non-trivial elementary embedding of the universe V into itself."

So, mathematical universe.
Uhh, i see.
...
Like there are no other posts, i present you the Q:
The Q is the largest number definable with every function posted in this thread as of the date of posting on this thread and n, where n is the largest possible computable number.
« Last Edit: November 17, 2014, 08:58:19 am by alemagno12 »
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FallenAngel

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Re: My number is bigger!
« Reply #57 on: November 17, 2014, 09:49:04 am »

I vaguely recall stating that, if you used functions defined prior in the thread, you must at the very least mention them by name.

alemagno12

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Re: My number is bigger!
« Reply #58 on: November 17, 2014, 10:28:37 am »

I vaguely recall stating that, if you used functions defined prior in the thread, you must at the very least mention them by name.
Then
non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest
is not valid.
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FallenAngel

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Re: My number is bigger!
« Reply #59 on: November 17, 2014, 12:04:15 pm »

I wasn't referring directly to you, I was just reminding everyone.
I have an unusually simple trump card that is crazy enough it could work.
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