What is the largest number?

[KakeiTheWolf]

I was thinking about this question earlier, and I realized it's not simple to define in terms of the finite realm of numbers. We could easily say that infinity or a transfinite cardinal would be the largest, but what about the largest number using traditional finite means. That's what I wanted to find out.

 

In theory, the largest number that exists that can be enumerated within current scientific notation (without repetitive infinite notation) is Σ: (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) ↑^ (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b))), where the values for a, n, and b are all Graham's number. But let's broaden that out to more accessible laymen's terms.

 

First, you need to know what Σ is in this case. Σ is the value representing the Busy Beaver function. Per wikipedia: "In computability theory, a busy beaver is a Turing machine that attains the maximum number of steps performed or number of nonblank symbols finally on the tape among all Turing machines in a certain class. The Turing machines in this class must meet certain design specifications and are required to eventually halt after being started with a blank tape.".

 

In the Busy Beaver game, a class of Turing machines is present, each member having to fulfill each of the design specifications:

 

• The machine has n "operational" states plus a Halt state, where n is a positive integer, and one of the n states is distinguished as the starting state.
  (Typically, the states are labelled by 1, 2, ..., n, with state 1 as the starting state, or by A, B, C, ..., with state A as the starting state.)
• The machine uses a single two-way infinite (or unbounded) tape.
• The tape alphabet is {0, 1}, with 0 serving as the blank symbol.
• The machine's transition function takes two inputs:

1. the current non-Halt state,

2. the symbol in the current tape cell,

and produce three outputs:

1. a symbol to write over the one in the current tape cell (it may be the same symbol as the one overwritten),

2. a direction to move (left or right; that is, shift to the tape cell one place to the left or right of the current cell),

3. a state to transition into (which may be the Halt state).

The transition function may be seen as a finite table of 5-tuples, each of the form

(current state, current symbol, symbol to write, direction of shift, next state).

"Running" the machine consists of starting in the starting state, with the current tape cell being any cell of a blank (all-0) tape, and then iterating the transition function until the Halt state is entered (if ever). If, and only if, the machine eventually halts, then the number of 1s finally remaining on the tape is called the machine's score.

 

The n-state busy beaver (BB-n) game is a contest to find such an n-state Turing machine having the largest possible score — the largest number of 1s on its tape after halting. A machine that attains the largest possible score among all n-state Turing machines is called an n-state busy beaver, and a machine whose score is merely the highest so far attained (perhaps not the largest possible) is called a champion n-state machine.

 

Radó required that each machine entered in the contest be accompanied by a statement of the exact number of steps it takes to reach the Halt state, thus allowing the score of each entry to be verified (in principle) by running the machine for the stated number of steps. (If entries were to consist only of machine descriptions, then the problem of verifying every potential entry is undecidable, because it is equivalent to the well-known halting problem — there would be no effective way to decide whether an arbitrary machine eventually halts.).

 

In simpler terms, the Busy Beaver function is the second fastest growind level of notation that exists. The definition of the Busy Beaver is Σ: N → N, where Σ: N is the maximum attainable score (the maximum number of 1s finally on the tape) among all halting 2-symbol n-state Turing machines of the above-described type, when started on a blank tape.

 

N for the Busy Beaver notation is defined as (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b), where a, n, and b are Graham's number. To explain this, you first need to know what Graham's number is. But first, I will explain the gravity of the usage of the Busy Beaver function in this.

 

The Busy Beaver function grows faster than any known computation system, where Σ(1) = 1, Σ(2) = 4, Σ(3) = 6, Σ(4) = 13, Σ(5) = 4098, and Σ(6) is unknown but known to be above 3.515 × 10¹⁸²⁶⁷ in value. This is just for the first six valid numerals, and the numeral used as N here is (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b), where a, n, and b are Graham's number.

 

Graham's number was discovered by Ronald Graham, who, in 1971, along with Rothschild, answered a problem in Ramsey Theory, stating:

 

"Connect each pair of geometric vertices of an n-dimensional hypercube to obtain a complete graph on 2ⁿ vertices. Colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured complete subgraph on four coplanar vertices?"

 

The bounds for this number are complicated to define, and were stated as 6 ≤ N* ≤ N by Graham and Rothschild, with N being a large but explicitly defined number F⁷(12) = F(F(F(F(F(F(F(12))))))),where F(n) = 2↑ⁿ3 in Knuth's up-arrow notation.

 

The standardized definition of Graham's number is as follows:

 

G =

 

"3↑ⁿ3, where n = g₆₃

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3↑ⁿ3, where n = g₆₂

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3↑ⁿ3, where n = g₆₁

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3↑ⁿ3, where n = g₆₀

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3↑ⁿ3, where n = g₅₉

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3↑ⁿ3, where n = g₅₈

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3↑ⁿ3, where n = g₅₇

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3↑ⁿ3, where n = g₅₆

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3↑ⁿ3, where n = g₅₅

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3↑ⁿ3, where n = g₅₄

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3↑ⁿ3, where n = g₅₃

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3↑ⁿ3, where n = g₅₂

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3↑ⁿ3, where n = g₅₁

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3↑ⁿ3, where n = g₅₀

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3↑ⁿ3, where n = g₄₉

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3↑ⁿ3, where n = g₄₈

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3↑ⁿ3, where n = g₄₇

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3↑ⁿ3, where n = g₄₆

^╰‒\/^‒╯
3↑ⁿ3, where n = g₄₅

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3↑ⁿ3, where n = g₄₄

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3↑ⁿ3, where n = g₄₃

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3↑ⁿ3, where n = g₄₂

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3↑ⁿ3, where n = g₄₁

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3↑ⁿ3, where n = g₄₀

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3↑ⁿ3, where n = g₃₉

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3↑ⁿ3, where n = g₃₈

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3↑ⁿ3, where n = g₃₇

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3↑ⁿ3, where n = g₃₆

^╰‒\/^‒╯
3↑ⁿ3, where n = g₃₅

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3↑ⁿ3, where n = g₃₄

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3↑ⁿ3, where n = g₃₃

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3↑ⁿ3, where n = g₃₂

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3↑ⁿ3, where n = g₃₁

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3↑ⁿ3, where n = g₃₀

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3↑ⁿ3, where n = g₂₉

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3↑ⁿ3, where n = g₂₈

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3↑ⁿ3, where n = g₂₇

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3↑ⁿ3, where n = g₂₆

^╰‒\/^‒╯
3↑ⁿ3, where n = g₂₅

^╰‒\/^‒╯
3↑ⁿ3, where n = g₂₄

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3↑ⁿ3, where n = g₂₃

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3↑ⁿ3, where n = g₂₂

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3↑ⁿ3, where n = g₂₁

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3↑ⁿ3, where n = g₂₀

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3↑ⁿ3, where n = g₁₉

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3↑ⁿ3, where n = g₁₈

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3↑ⁿ3, where n = g₁₇

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3↑ⁿ3, where n = g₁₆

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3↑ⁿ3, where n = g₁₅

^╰‒\/^‒╯
3↑ⁿ3, where n = g₁₄

^╰‒\/^‒╯
3↑ⁿ3, where n = g₁₃

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3↑ⁿ3, where n = g₁₂

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3↑ⁿ3, where n = g₁₁

^╰‒\/^‒╯
3↑ⁿ3, where n = g₁₀

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3↑ⁿ3, where n = g₉

^╰‒\/^‒╯

3↑ⁿ3, where n = g₈

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3↑ⁿ3, where n = g₇

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3↑ⁿ3, where n = g₆

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3↑ⁿ3, where n = g₅

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3↑ⁿ3, where n = g₄

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3↑ⁿ3, where n = g₃

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3↑ⁿ3, where n = g₂

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3↑ⁿ3, where n = g₁

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3↑↑↑↑3

 

Where G₁ = 3↑↑↑↑3, and each layer is G_n, and n = x+1, where x is the number of values for G_n so far defined.".

 

For this definition to have significant, we must go into what Knuth's up-arrow notation is.

 

Knuth's up-arrow notation is the second fastest method of notation, using arrows to signify levels of exponents, instead of traditional power towers. It was created by Donald Knuth in 1976 as a method of iterative exponentiation. The best way to demonstrate it is by example.

 

In Knuth's up-arrow notation:

 

a↑b = a^b = a multiplied by a for b amount of times.

Exempli gratis, you have 3↑3 = 3³ = 27.

 

a↑↑b = ^ba = a exponentiated by a for b amount of times.

Exempli gratis, you have 3↑↑3 = ³3 = 3²⁷ = 7,625,597,484,987.

 

a↑↑↑b = a↑↑(a↑↑(...↑↑a)) = a with b levels of tetration of number a

Exempli gratis, you have 3↑↑↑3 = ^₃3₃ = 3 with 7,625,597,484,987 exponent levels each of value 3.

 

a↑↑↑↑b = a↑↑↑(a↑↑↑(...↑↑↑a)) = a with b layers of pentation of number a.

Exempli gratis, you have 3↑↑↑↑3 = 3↑↑↑(^₃3₃) = 3 with 3 layers of pentation = 3^3↑↑↑3.

 

a↑↑↑↑↑b = a↑↑↑↑(a↑↑↑↑(...↑↑↑↑a)) = a with b layers of hexation of number a.

Exempli gratis, you have 3↑↑↑↑↑3 = 3↑↑↑↑(^₃3₃) = 3 with 3 layers of hexation = ^3↑↑↑33.

 

And lastly, a↑ⁿb = a↑^n-1(a↑^n-1(...a↑^n-1)) = a with b layers of the (n+1)th hyperoperation of number a.

 

So, for ease:

 

a↑↑b = ^ba.

 

a↑↑↑b = a with b levels of tetration of number a, with a brace below to note that there are b levels of tetration.

 

a↑↑↑↑b = a with b levels of pentation, represented as b layers of a with b levels of tetration.

 

a↑↑↑↑↑b = a with b levels of hexation, represented as b stacks of b layers of a with b levels of tetration.

 

a↑ⁿb = a with b levels of the (n+1)th hyperoperation.

 

This makes even just the amount of towers in the exponentiation of a within the base formula for determining the value of g₁ far larger than the amount of Planck's volumes in the observable universe. That is to say nothing of the actual value of g₁, let alone the rest of the 64 layers of G, which have an amount of arrows that equals G_n-1. That means there are g₆₃ arrows in G₆₄, which means that the 64th layer is 3 with 3 levels of the 64th hyperoperation, which would be named tetrahexacontation.

 

That is merely for Graham's number. This is a far larger. This is the proper explanation of Σ: (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) ↑^ (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b))):

 

For the value a of a↑ⁿb, we have (Σ: (G ↑^G G) ↑(^{_G↑G _G}) (G ↑^G G)). That is to say, the value of a is Σ:N where the value for n in Σ(n) is a↑ⁿb, where the value of a is (G↑^G G), which is G with G levels of the (G+1)th hyperoperation, where the value of n is (^{_G↑G _G}), which is G with G levels of the (G+1)th hyperoperation, and where the value of b is (G ↑^G G), which is G with G levels of the (G+1)th hyperoperation.

 

That means that the a↑ⁿb value for the Σ(n) value that defines a is [G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation (the name of which, if a conventional system was used, would be a word of an amount of letters exceeding the amount of letters written in human history).

 

For the value of n, we have an amount of arrows that is (Σ: (G ↑^G G) ↑(^{_G↑G _G}) (G ↑^G G)). That is to say, the amount of arrows is the value of Σ:N where the value for n in Σ(n) is a↑ⁿb, where the value of a is (G↑^G G), which is G with G levels of the (G+1)th hyperoperation, where the value of n is (^{_G↑G _G}), which is G with G levels of the (G+1)th hyperoperation, and where the value of b is (G ↑^G G), which is G with G levels of the (G+1)th hyperoperation.

 

That means that amount of arrows, n, is the a↑ⁿb value for the Σ(n) value that defines n, which is [G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation (This is more arrows than could be area of a sphere with the diametre being the distance between a redshift of 0 and the redshift of the cosic radiation of the universe (which is a redshift of 140), assuming each arrow took up one Planck's length).

 

For the value of b, we have (Σ: (G ↑^G G) ↑(^{_G↑G _G}) (G ↑^G G)). That is to say, the amount of arrows is the value of Σ:N where the value for n in Σ(n) is a↑ⁿb, where the value of a is (G↑^G G), which is G with G levels of the (G+1)th hyperoperation, where the value of n is (^{_G↑G _G}), which is G with G levels of the (G+1)th hyperoperation, and where the value of b is (G ↑^G G), which is G with G levels of the (G+1)th hyperoperation.

 

That means the amount of levels in the formula is the a↑ⁿb value for the Σ(n) value that defines n, which is [G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation (Bear in mind there were only 64 levels in the calculation of G).

 

These three values make the a↑ⁿb value for the Σ(n) value of our original equation, Σ: (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) ↑^ (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b))). This means that the value of the n value for the original equation's Σ(n) is enumerated as being the following:

 

([G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation]) ↑^ ([G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation]) ([G with G levels of the (G+1)th hyperoperation] with [G with G levels of the (G+1)th hyperoperation] levels of the [G with G levels of the (G+1)th hyperoperation]th hyperoperation]). This cannot really be simplified, as even a graphical representation of the most compact form of the equation concievable would be larger than the observable universe if each part were compacted into as few Planck's volumes as possible whilst still retaining recognizable form.

 

That equation above is merely the n value for the original Σ(n) in original equation, Σ: (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) ↑^ (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b)) (Σ: (a ↑ⁿ b) ↑(ª↑ⁿ ᵇ) (a ↑ⁿ b))).

 

That is, to the best of my knowledge, the largest concievable number without repetitive infinite notation (endless addition to the base formula is easily possible.

 

But just what is the point of such a number? It cannot be described in relatable terms, nor represented on computer or physical space, nor could there be a computer that could compute such a calculation (of lack of power), nor can we even concieve the computation time of such a calculation. The mere name of this number would be so long that a lifetime could not realize its writing in full, nor perhaps the entirety of all lifetimes. For all intents and purposes, the end result of this equation is of such unspeakable magnitude that it is worthless, having taken a 2,650 word post just to describe it in broad terms. This may be the largest number, but, honestly, what is the point of such a number?

Edited October 25, 2013 by KakeiTheWolf

[Hyacathusarullistad]

....

 

....

 

.... I ate a bee once.

[Thor.]

----

 

----

 

----

 

???? ??? ???? ???? infinite and beyond :pinch: :geek: :teehee: :huh:

[KakeiTheWolf]

Such coherent comments...

[Thor.]

I have verdict, no such thing as the largest number, quantum computing will be the one to judge that :geek:

[billyro]

1 / 0? Because it doesn't work?

[Thor.]

Nope because quantum computer overlaps 1 and 0, they defy the laws of physics, and it literally taps into a alternative universe.

 

1 and 0 in the same space.

[billyro]

I just said something random because I have no freaking clue. I'm going to say the largest number ever is... 5.

Edited October 27, 2013 by billyro

[KakeiTheWolf]

1 / 0? Because it doesn't work?

 

Actually, that's technically incorrect. In the case of both ℝ ∪ {∞} (The real projective line) and ℂ ∪ {∞} (The Riemann Sphere), where both cases use an unsigned infinity, a/0 can be solved, iff a ≠ 0, as being a/0 = ∞.

[billyro]

... Ok.

 

I'll take your word for it.