 Before you read this, you'll need to understand the notation of Graham's number, which can be found here. You'll also need to need to read up a little on the up-arrow notation done by Don Knuth Here .

In order to understand the world's largest finite number, Ed, you'll need to understand the notation involved at arriving at that number.Ā If, indeed, Graham's number is the largest defined number at the moment, it only seems fitting to create yet another notation to encompass and surpass it:

Don Knuth constructed Graham's number as follows:

63. Graham's number G = 3^^...^^3, where there are G62 up-arrows.

Similarly, I will define a function called Ed1, which uses the same method of construction:

Ed1(x) is defined as Gx = 3^^...^^3, where there are Gx-1 up-arrows, or in other words, defined as the number of times you follow the process. Ā Graham's number is Ed1(63).

Now imagine x was Graham's number. Ā This produces a number of significant size, but why stop there?Ā That number is simply Ed1(Ed1(63)).

why not go for Ed1(Ed1(Ed1(63)))? or Ed1(Ed1(Ed1(Ed1(63))))? Ā Well, now we have to come up with another function, Ed2(x), where x is the number of Ed1s in the number, as well as the x of the last Ed1, i.e. Ed2(3) = Ed1(Ed1(Ed1(3))), and Ed2(63) = Ed1(Ed1(Ed1...(Ed1(Ed1(63)))...))). Ā Naturally we'll want Ed2(Ed2(63)) and Ed2(Ed2(Ed2(63))), which means we'll have to have Ed3, then Ed4.Ā Edx(x) is defined, then, as the number of Edx-1s in the number, as well as the x of the last Edx-1. Ā To simplify things, we're going to define ED1(x) as Edx(x).Ā This makes ED1(1) 3^^^^3, but ED1(2) is already too large to express in the older notations.

We have already written the definition of Ed1(x). Ā If we wrote it out at the Ed1(x) level, those same words would be in the definition of ED1(x), but there would be other words added.Ā Those "other words" separating any two logical functions are now defined as the Difference in Logic Structure (DLS).

We're going to define Ed1(1) as the definition of Ed1(1) plus the DLS of Ed1(1) and ED1(1). Ā This makes Ed1(2) the definition of Ed2(2) plus the DLS of Ed2(2) and ED2(2), plus the DLS of ED2(2). and Ed2(2), where Ed1(1) is defined as EDx(x).Ā Since we started with Edx, Edx is defined as logic level 1.Ā EDx is logic level 2, and Edx is logic level 3.

Ed1(x), then, is the definition of Edx(x) plus the DLS of every logic level from 1 (Ed x(x) to x. Ā This makes Ed1(4) the definition of Ed1(4) plus the DLS of every logic level from 1 (Ed4(4)) to 4 [ED4(4), where (ED1(x) is defined as Edx(x).]. Ā

Ā

ĀEDx(y) is defined as the number whose definition is Edx(y), which makes ED1(1) equal to 3^^^^3, and ED1(2) equal to a far greater number.

Now, of course, there'll be ED1(ED1(ED1(ED1(4)))), which can be rewritten as ED2(4), by essentially the same rules that Ed, ED, ED, and all the other logic levels follow in their subscripts.Ā Now, we're going to define the DLS of Ed and ED as logiclevel(1) and the DLS of Ed and ED as logiclevel(2).Ā By defining ED1(x) as EDx(x), their DLS is logiclevel(1), and the DLS of Ed and ED is logiclevel(3).Ā Now, we'll define ED 1(x) as the function whose DLS from Ed is logiclevel(x), and whose subscript and independent variable is x.Ā thus ED 1(1) is ED1(1), and ED 1(3) is ED3(3).

Ā

Since I followed basically the same process to get from Ed1 to Ed2 to EDx to EDx to ED x, by making a new definition each time, we're going to define Edward(x) as the following this process x times.Ā Our last function, Tivursky(x), is defined as the function arrived at by Edward(x), with subscript and independent variable of x.Ā The number of bytes manipulated by every computer in history (an ever-growing and yet finite number) will be defined here as allbytes.