Your number is, in fact, not bigger!

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Vytron
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Your number is, in fact, not bigger!

Postby Vytron » Sat May 10, 2014 8:47 pm UTC

A successor to this thread, but with different rules so it doesn't reach a deadlock.

1. You must be able to describe your number using computable functions.
2. Your number must be finite.
3. Your number must be well-defined.
4. Your number should be easy to understand for the layman.
4b Your complain against a number should be easy to understand for the layman.
5. You canNOT use meta references like "all the posted numbers so far +1" or "the highest number that will be posted on page 20 + 1" and such, only actual numbers count.
6. You can, however, just use the highest number posted on the thread, or steal a notation from someone else, and add 1, and it'll be valid.

4 is there because people tend to make very long posts with hard to read notation that makes it pretty much impossible to know if the new number is, in fact, bigger.

5 is there because, if you want to come to the thread and make a metadefinition like that, you actually have to take a look at numbers posted and make calculations for a valid entry.

6 is there so the thread can always have an influx of entries, you can come here and post a number that is, in fact, bigger than all the previous posted.

So I'll start with:

0

Beatable by any non-negative number other than 0 - let's see how long it takes us to beat the other thread :)
Last edited by Vytron on Tue Mar 03, 2015 2:02 am UTC, edited 5 times in total.

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Re: My number is, in fact, bigger!

Postby faubiguy » Sat May 10, 2014 8:54 pm UTC

10-10

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 8:56 pm UTC

10-10 + 10-100

Nitrodon
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Re: My number is, in fact, bigger!

Postby Nitrodon » Sat May 10, 2014 9:00 pm UTC

10-10+10-99.9

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Re: My number is, in fact, bigger!

Postby faubiguy » Sat May 10, 2014 9:01 pm UTC

1/4

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 9:13 pm UTC

594/2375

Nitrodon
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Re: My number is, in fact, bigger!

Postby Nitrodon » Sat May 10, 2014 9:16 pm UTC

1/4 + 1/1168

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 9:20 pm UTC

Aha, that's in fact smaller than my last posted number, so I'm still champion with 594/2375 8-)

Fear those 8 symbols, the first number on the thread that failed to be beaten on first attempt!

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Re: My number is, in fact, bigger!

Postby Nitrodon » Sat May 10, 2014 9:24 pm UTC

594/2375 = 0.25010526... = 1/4 + 1/9500
1/4 + 1/1168 = 0.25085616...

I challenge your challenge.

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Re: My number is, in fact, bigger!

Postby brenok » Sat May 10, 2014 9:31 pm UTC

1

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sat May 10, 2014 9:47 pm UTC

1 + 1/9999999999999999999

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 9:49 pm UTC

Oh... god. Let it be known that I fail at math D:

1 + 1/999999999999999999

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Re: My number is, in fact, bigger!

Postby faubiguy » Sat May 10, 2014 9:58 pm UTC

(e+π)/5.859

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Re: My number is, in fact, bigger!

Postby Nitrodon » Sat May 10, 2014 10:12 pm UTC

eπ/20000

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sat May 10, 2014 10:25 pm UTC

1000/999

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sat May 10, 2014 10:44 pm UTC

So, how exactly is a layman defined?

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 11:07 pm UTC

Um, I guess it needs "on the fly definition", like, someone goes and say "Hello, I'm a layman and I don't understand your number" makes your number invalid (but you can make it valid by explaining in detail whatever you're doing), otherwise, it's fine.

100/99

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Re: My number is, in fact, bigger!

Postby Daggoth » Sat May 10, 2014 11:13 pm UTC

layman, in this case, i'd say is a man(or a person of arbitrary gender/genders) who does not earn money or aspire to earn money from doing math, for example: engineers, researchers, math professors etc..

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Re: My number is, in fact, bigger!

Postby Nitrodon » Sat May 10, 2014 11:14 pm UTC

1.1

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sat May 10, 2014 11:33 pm UTC

ln(π)

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Re: My number is, in fact, bigger!

Postby Vytron » Sat May 10, 2014 11:50 pm UTC

1.2
Go! Go! You can do it username5243!
Cheers Marsh'n!

Image
THANKS KARHELL!! :)

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 12:39 am UTC

Let's go crazy and all the way up to

2

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Re: My number is, in fact, bigger!

Postby brenok » Sun May 11, 2014 1:01 am UTC

10100

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faubiguy
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Re: My number is, in fact, bigger!

Postby faubiguy » Sun May 11, 2014 1:16 am UTC

10101

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Re: My number is, in fact, bigger!

Postby Vytron » Sun May 11, 2014 1:34 am UTC

Landmark! We have beaten 3 posts from the other thread!

10202

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Re: My number is, in fact, bigger!

Postby faubiguy » Sun May 11, 2014 1:36 am UTC

10^10^10

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Re: My number is, in fact, bigger!

Postby Vytron » Sun May 11, 2014 1:39 am UTC

10^(10^10)

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 1:41 am UTC

I think a tower of exponents is supposed to be solved from the top, so 10^10^10 = 10^(10^10)

Anyway, 10^^10

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Re: My number is, in fact, bigger!

Postby Nitrodon » Sun May 11, 2014 1:48 am UTC

11^(10^^9)

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Re: My number is, in fact, bigger!

Postby Daggoth » Sun May 11, 2014 3:28 am UTC

Nice try but what does ^^ mean? ;)

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 3:39 am UTC

It's using up-arrow notation.

But since we are only using two arrows so far, you don't really have to understand it all yet. =D

a^^b means a tower of exponents a^a^a^a^...^a with b instances of a.

So 10^^10 = 10^10^10^10^10^10^10^10^10^10

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Re: My number is, in fact, bigger!

Postby WaffleKirby » Sun May 11, 2014 5:18 am UTC

10↑googol10
O_o

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Re: My number is, in fact, bigger!

Postby Daggoth » Sun May 11, 2014 5:38 am UTC

what does ↑googol mean?, by the way, what is a googol?

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Re: My number is, in fact, bigger!

Postby Earthling on Mars » Sun May 11, 2014 12:57 pm UTC

We shall start with the number "Million"; that is 1,000,000. Now we shall multiply a Million by itself a Million times. This will be written 1000000^1000000 or 10000001000000, which is called "raising a Million to the Millionth power". Now we shall raise a Million to the Millionth power a Million times; that is 1000000^1000000^1000000... with a Million layers. This huge number shall be called Mega.

Now we shall multiply a Mega by itself a Mega times; that is Mega^Mega or MegaMega. And then we shall raise Mega to the Mega-th power Mega times; that is Mega^Mega^Mega... with a Mega layers. This number shall be called Mungo.

Now we shall raise a Mungo to the Mungo-th power Mungo times; that is Mungo^Mungo^Mungo... with a Mungo layers. This number shall be called Humungo.

Now imagine a series of numbers, where each number made by taking the previous one and raising it to its own power that many times. The first number in this series is a Million, the second is a Mega, the third is a Mungo, and the fourth is a Humungo. We could continue this series with ever larger numbers.

My number is the Humungo-th number in this series.

(I'm not sure if this will beat WaffleKirby's number, but it is more layman-level.)

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 1:46 pm UTC

Your number is dwarfed by WaffleKirby's, but since somebody has challenged it's understandability, your number is still the biggest that follows the rules.

Let's see if I can simplify the notation above...

f(n) = n^^n

Million = 1000000
Mega = f(1000000)
Mungo = f(Mega) = f(f(1000000)) = f2(1000000)
Humungo = f(Mungo) = f(f(f(1000000))) = f3(1000000)
n'th number in this series = fn-1(1000000)

g(n) = fn(1000000)

This means that your number is g(g(3)).


Now, let me introduce you to triple arrows. n^^^m is a tower of double arrows, n^^n^^n^^n^^n^^n^^...^^n of length m.

f(n) = n^^n = n^^^2
f(f(n)) = (n^^n)^^(n^^n) < n^^n^^n^^n = n^^^4
Each subsequent use of f doubles the number of n's in the tower, so
fk(n) < n^^^(2^k)
This means that g(n) < 100000^^^(2^n)
g(3) < 1000000^^^8
g(g(3)) < 1000000^^^1000000^^^8

I can easily beat this with 4^^^4^^^4^^^4. Note that this number is still beaten easily by WaffleKirby's if anybody explains that number well enough for a layman.

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Re: My number is, in fact, bigger!

Postby firesoul31 » Sun May 11, 2014 2:03 pm UTC

Daggoth wrote:what does ↑googol mean?, by the way, what is a googol?


As a layman, let me see if I understand it.

Firstly, a googol is 10100.

Secondly, Knuth arrows are an alternate way to write exponentiation, right? So ↑ just means "to the power of". Two up arrows, however, mean something different; they mean you iterate the power that many times. So 5↑↑3 would be 5↑5↑5, or just 5^5^5. Three up arrows follow the same pattern but with ↑↑ instead of ↑, so 5↑↑↑3 would be 5↑↑5↑↑5. This continues, so for any given n up arrows, for the formula a↑cb (the superscript indicates the number of up arrows), you get a↑c-1a↑c-1a↑c-1.... (continued for b times).

This means that Wafflekirby's number is 10 with 10100 up arrows connecting it to another 10, so following out all the calculations might be a bit time-consuming, but this is at least how you would do it.
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Re: My number is, in fact, bigger!

Postby Vytron » Sun May 11, 2014 2:27 pm UTC

I could. But my options are explaining WaffleKirby's number, and beating it, or just beating ODF's.

Also, please, let's keep the definitions of ^^... and ↑↑... separated so the latter one is much stronger. Since we have to define symbols on the thread for the layman, we can use non-standard definitions to make much larger numbers.

Now, I'll introduce, factorials.

Factorials are simple enough, you just, get some number, then, reduce 1 from this number to produce a second number, and multiply both numbers to get a third number. Then you reduce 2 from the first number and multiply it by the third number. Until the first number reaches 1.

5! = 5*4*3*2*1=120

My number:

(4^^^4^^^4^^^4)!

(4^^^4^^^4^^^4*((4^^^4^^^4^^^4)-1)*((4^^^4^^^4^^^4)-2)*((4^^^4^^^4^^^4)-3)...)

NINJA: Damn, I refuse to accept that 5↑5↑5 is just 5^5^5, I hope we can define ↑s to be successor arrows, like:

3+3 = 3 (sum)
3*3 = 3+3+3 (iterated sums)
3^3 = 3*3*3 (iterated multiplication)
3^^3 = 3^(3^3) (iterated exponentiation)
3^^^3 = 3^^(3^^3)
...
3^n3 = 3^^^n times^^^3

a↑b = a^bb
a↑↑b = aa↑bb
a↑↑↑b = aa↑↑bb
...
a↑nb = a↑↑↑a↑^(n-1)b times↑↑↑b

With this definition:

3↑644 is already higher than Graham's number :D

10↑googol10 goes higher than Graham(googol), but not higher than Graham(googol+1)

So, huh, I'll just go with:

(10↑googol10)!

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 2:39 pm UTC

In that case let's just start on the fast-growing sequence.

f1(n) = n+1
fk(n) = fk-1n(n)
fω(n) = fnn(n)
fω+k(n) = fω+k-1n(n)
fω2(n) = fω+nn(n)


Examples:
Spoiler:
f1 is the successor function. (n+1)
f2, which is iterated succession, rivals addition. (n+n)
f3, which is iterated addition, rivals multiplication. (n*)
f4, which is iterated multiplication, rivals exponentiation (^).
and so on, so we are basically developing a set of functions which are as powerful as up-arrow notation, with
fk rivaling ^k-3

fω(n) does pretty much the same as ↑, diagonalising over ^n
fω+1(n) does pretty much the same as ↑↑, iterating over ↑
fω+2(n) does pretty much the same as ↑↑↑, iterating over ↑↑
and so on, so we have
fω+k(n) does pretty much the same as ↑k+1

fω2(n) surpasses all of the ↑.


fω2(fω2(50))

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Re: My number is, in fact, bigger!

Postby Vytron » Sun May 11, 2014 2:57 pm UTC

Man, that's massive overkill. But still after looking at the other thread these numbers still look tiny.

I'm have to use my secret weapon: Conway arrows.

The thing with Conway arrows is that they have built in recursion, see, you had to add an extra "layer" of recursion for good measure, at the point you need to add an extra fω2(n) to your function to get to the next big number, Conway just adds 1 to the rightmost arrow.

Now, let's compare:

fω(n) will be dominated by 3 → 3 → n → 2
fω+k(n) will be dominated by 3 → 3 → 3 → 2+k
fω2(n) will be dominated by 3 → 3 → 3 → n → 2

Now, applying extra layers of fω2 will not do much as Conway does that automatically at → n+1, so I can beat your number with:

3 → 3 → 3 → 51 → 3

Conway arrows for the layman:

a → b → c = a^cb

a → b → c → d = a → b → (a → b → c-1 → d) → d (when c reaches 0 decrease d by 1, when d reaches 1, kill it).

Also: Landmark! we have beaten the first page of the other thread, yes! We're going faster!

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Re: My number is, in fact, bigger!

Postby orangedragonfire » Sun May 11, 2014 3:19 pm UTC

hmm, I guess I'll have to go straight to fω2.

fω2 (n) = fωn (n)

This should beat conway-arrow chains of pretty much any length.

fω2 (7)



/edit: also, we should wait for a layman to tell us when we stopped making sense =D


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