My number is bigger than yours!, Redux

[Holo]

Welcome to the second "My number is bigger than yours!", after the first was derailed by people who didn't understand what "bigger" meant.

The objective of this game is simple: post a number bigger than the previous number.

Here are the rules you need to know:

1. It must be a bigger number.

1a. It must not be a smaller number.

1b. It must be possible to prove that the number is bigger.

1c. If it is not trivial to prove that your number is bigger, you are expected to prove it.

2. No infinities

2a. No transfinite numbers

3. NO MORE THAN 10 DIGITS PER NUMBER!

3a. No walls of exponents

3b. No numbers where you copy and paste zeroes, or any other digit

3c. Functions are certainly allowed, as long as the total number of digits in all of the inputs is not greater than 10.

3d. You can recursively define functions to beat this limit, as long as the function itself has no more than 10 digits of input and none of the functions used in defining the main function have more than 10 digits of input as well.

4. No referencing the number before yours.

5. Your number should be sufficiently well defined that, given infinite time and space, one would be able to write it out in full.

6. Bonus points if you name the number

7. You must be able to define the number in terms of existing notation

7a. Constructing new notation to define your number is also allowed as long as the new notation is defined based on existing notation

7b. You may also define new notation with more new notation, and so on, all in one post, as long as the final number doesn't break any rules.

8. Functions (inc. computer programs) without inputs are allowed will be judged by their location on the fast-growing hierarchy.

To ensure that you are not breaking rule 1, here is the scoreboard:

1. (10^100)!, by Palladium Corp
   
2. 10¹⁰¹⁰⁰, by Holo
   
3. (10^10^100)^10!+1, by Palladium Corp
   
4. 10¹⁰¹¹⁰+1, by Holo
   
5. ((10^100^100)^100)!+1, by Palladium Corp
   
6. 10^1010500+1, by Holo
   
7. ((10^100^100^100^100^100)!)!+1, by Palladium Corp
   
8. ⁸100 + 1, by Holo
   
9. (^100!100)!+1, by Palladium Corp
   
10. 100↑↑↑3, by Holo
    
11. (100↑↑↑3)!+1, by Palladium Corp
    
12. (100↑↑↑100)!+1, by Holo
    

Past Scoreboards (For Historical Interest)

Round 1

1. 3 (three), by Holo
   
2. tan(90-99^-999999), by SunJumper
   
3. Graham's Number, by rasheed
   
4. 4 <<64>> 4 (duseedu), by Holo
   
5. 999<<9999>>999, by SunJumper
   
6. 4 <<<64>>> 4 (terseedu), by Holo
   
7. Terseedu<<(Terseedu)^(Grahm's number)>>Terseedu, by zekes
   
8. (GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP), by SunJumper
   

Round 2

1. -1, by theattackcorgi
   
2. 100000, by 11JRidding
   
3. 100001, by theattackcorgi
   
4. S6, by Holo
   
5. A(g64, g64), by jaidens111
   

Now that those are out of the way, let's begin

Edited January 21, 2014 by Holo

[rasheed]

Graham's number

[Holo]

Graham's number

That reminds me of a Rule 7 I needed. Apologies for missing it out . All you need to do is define Graham's number and the spot on the leaderboard shall be yours.

[The Destroyer]

100 Sextillion.

[SunJumper]

Until he proves his number,

tan(90-99^-999999)

Proof: As theta approaches 90, tan(theta) approaches infinity. arctan(4) is 76 degrees, and as 90-99^-999999 is less than 90 (otherwise the number would be infinite or negative), and greater than 76, tan(90-99^-999999) is greater than 4.

I don't have a numerical approximation.

[Goodle...]

99999 googolplex

[rasheed]

In response to my number.

or . This is the number we will call g1. After that, g2 is equal to ; the number of arrows in this number is g1.

g3 is equal to , where the number of arrows is g2.

We keep going in this way. We stop when we define g64 to be , where the number of arrows is g63.

This is Graham's number.

Here you go

[Holo]

Excellent, I'll update the leaderboard with your numbers. 99999 googolplex and 100 sextillion were just names, not explained. I'm working on my own Googological notation, so I'll be able to join the race soon.

EDIT:

x {4} y = x^x...^x, with x being repeated "y" times

x {5} y = x{4}x...{4}x, with x being repeated "y" times

Now, imagine x {y{y}y} x. This will be written as x {{2}} y

This can be nested any number of layers deep. So, for example, x {{3}} y is x {y{y{y}y}y} x and so on.

Expressed like this, g1 would be 3 {6} 3, g2 would be 3 {g1+2} 3, and so on. Graham's Number would be 3 {g64+2} 3. To simplify the writing of Graham's number, we can say that "x {y} z" equals "x <y-2> z".

Like this, g64 would be 3 <3 <3... <4> ... 3> 3> 3, with 64 layers in total.

Remember the double brackets I showed you earlier? Well, g64 would be between 3 <<64>> 3 and 3 <<64>> 4.

Therefore, my next number shall be 4 <<64>> 4, which I shall name the Duseedu (derived from Waterbien, a conlang I sometimes work on).

Have fun. There are more numbers bigger than Duseedu than there are smaller.

Edited November 29, 2013 by Holo

[SunJumper]

999<<9999>>999

[Wrench Head]

3,141,589,996.85841

Die Regel von Pi

In der Dunkelheit, in der Nacht, ich werde da sein!

[Holo]

999<<9999>>999

Excellently done. Adding you to the OP.

3,141,589,996.85841

Die Regel von Pi

In der Dunkelheit, in der Nacht, ich werde da sein!

This is smaller than the previous number, and it has more than 10 digits. I put that up in red!

My next number:

Imagine x <<y<<y>>y>> x. This will be notated as x <<<2>>> y. x <<y<<y<<y>>y>>y>> x is x <<<3>>> y, and so on.

Picture 4 <<<64>>> 4. This number is the huge Terseedu (from Waterbien, translates as "three-six-two").

Edited November 30, 2013 by Holo

[zekes]

999<<Grahm's number>>9999

[Holo]

And now we have the challenge of working out whether Terseedu or 999<<Grahm's number>>9999 is larger. I suspect Terseedu is, but Graham's Number is well into deutero-hyperion territory so it's not clear cut

Edited November 30, 2013 by Holo

[NQMT]

99^99.

Bam.

[Holo]

99^99.

Bam.

This is a horrific violation of Rule 1. In fact, it's so horrific, I will make a

Googology PSA

Exponentiation will not work, and in fact stopped working after Graham's Number. No matter how large the coefficients, it is impossible to get any higher with exponentiation without violating both the physical size of the universe and Rule 3.

[zekes]

And now we have the challenge of working out whether Terseedu or 999<<Grahm's number>>9999 is larger. I suspect Terseedu is, but Graham's Number is well into deutero-hyperion territory so it's not clear cut

What about...

Terseedu<<(Terseedu)^(Grahm's number)>>Terseedu

woah.

[Holo]

What about...

Terseedu<<(Terseedu)^(Grahm's number)>>Terseedu

woah.

Well, that definitely beats mine. Have fun on the leaderboard.

I'm reading up on the latest in Googology so I can continue making bigger numbers.

Edited November 30, 2013 by Holo

[SunJumper]

Well then:

If I use {x}y{x}, that represents x amount of '<'s before and after the y number.

Also, with a googolplex being 10^a googol, a googoldiplex being (10^a googol)^a googol, and so on, and witrh Zeke's Number being represented by 'Zeke', I give you:

GoogolyottaZekeplex (GYZP)

So, (GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP)

[Holo]

Well then:

If I use {x}y{x}, that represents x amount of '<'s before and after the y number.

Also, with a googolplex being 10^a googol, a googoldiplex being (10^a googol)^a googol, and so on, and witrh Zeke's Number being represented by 'Zeke', I give you:

GoogolyottaZekeplex (GYZP)

So, (GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP){(GYZP^GYZP)}(GYZP^GYZP)

So, if I'm not mistaken, a GoogolyottaZekeplex is (...(10^goolol)^googol)...^googol, with a septillion Zekes of googols? Nonetheless, this is glorious and there's no doubt it's huge.

[SunJumper]

Until he proves his number,

tan(90-99^-999999)

Proof: As theta approaches 90, tan(theta) approaches infinity. arctan(4) is 76 degrees, and as 90-99^-999999 is less than 90 (otherwise the number would be infinite or negative), and greater than 76, tan(90-99^-999999) is greater than 4.

I don't have a numerical approximation.

I think I have an estimate on this number.

So, I have found that tan(90-99^(-x)) is approximately 57.29 (or 180/pi) times 99^x.

So, my number there is 57.29X99^999999. Hence, according to Wolfram|Alpha, the number is about 9X10^1995634.

[minerman30]

∞^∞10chars

[theattackcorgi]

-1. this is because since the distance from your number to -1 is more then it is currently my number is bigger :3.

[11JRidding]

100000 Your number is in negatives, mine is positives. Mine's bigger

[SunJumper]

Here are the rules you need to know:

1. It must be a bigger number.

1a. It must not be a smaller number.

1b. It must be possible to prove that the number is bigger.

1c. If it is not trivial to prove that your number is bigger, you are expected to prove it.

2. No infinities

2a. No transfinite numbers

3. NO MORE THAN 10 DIGITS PER NUMBER!

3a. No walls of exponents

3b. No numbers where you copy and paste zeroes, or any other digit

3c. Functions are certainly allowed, as long as the total number of digits in all of the inputs is not greater than 10.

3d. You can recursively define functions to beat this limit, as long as the function itself has no more than 10 digits of input and none of the functions used in defining the main function have more than 10 digits of input as well.

4. No referencing the number before yours.

5. Your number should be sufficiently well defined that, given infinite time and space, one would be able to write it out in full.

6. Bonus points if you name the number

7. You must be able to define the number in terms of existing notation

7a. Constructing new notation to define your number is also allowed as long as the new notation is defined based on existing notation

7b. You may also define new notation with more new notation, and so on.

Your numbers must be bigger than mine.

Also, it mustn't be infinity (as that would be boring).

So, please read the forum game rules beforde you post

[theattackcorgi]

Error no number

100001