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8^413 is the largest number, there is no number larger. This number is so large that if you add 1 to it, it will absorb the additional number because if it incremented then 8^413 would not be the largest number, which it clearly is therefor you cannot add to it. Now what about negative numbers? simple. The span can not exceed 8^413. The V-Limit, Limits of Limits, The maximum sum of all numbers.
Now since I don’t want to have to compute the v-limit we will talk in pseudo limits. So the mathematicians can follow along.
Set 8=8^413
If you include negative numbers then the span needs to equal 8.
-4 through 4 = 8
If you have fractions then the sum of the component parts cannot exceed 8.
Set 1000 to 8 // 8 was set to 8^413
Introduce .01 Then if counting from 0 then the v-limit cannot exceed 10. If you allow negative numbers the range would be -5 to 5.

The GooglePlex and the V-Limit.

Since the 10^(10100) May appear larger than the v-limit in reality it is not. Since there are only 8⁽⁴¹³⁾ numbers allowed the incremental space between the number is greater than 1. The increments of the Googleplex must be arranged in such a way that there are only  8⁽⁴¹³⁾ iterations in the defined span.

I cannot stress the importance of the defined span. In order to operate in a finite space name space rules must be written ahead of time in order to make the mathematical problem valid. Too often mathematicians fall into the trap of adding apples and oranges. When deliberating on a mathematical problem there must be a header included in the equations establishing what portions of the occupying reality will be used and how those portions will operate.

In pseudo mathematics where infinity is discussed a popular paradox would be the statement that there are as many numbers between 0 to 1 as there are between 0 to 2. Even under Vince’s Law one can argue that this is true. If we define VL(temporary number until I can find a better one) to occupy the fractions between 0 to 1 and then the same between 0 to 2 disseminating a value twice the size of the latter then it might appear as proof that there are as many numbers between 0 to 1 as there are between 0 to 2. This statement is both true and false. Binary mathematicians hold on and bare with me.

The VL is constant, but contingency based

We will set two name-spaces.

Set n1 to 0-1

Set n2 to 0-2

If one was applying the VL to 0-1 is not the same as apply it to 0-2. In 0-1-2 the VL would mandate that half of the numbers be allocated to 0-1 as there is from 1-2. The amount of numbers between 0-1 is half of what it is from 0-2. In another separate instance in which the name-space starts at 0 and ends at 1 the distance between the numbers is 1/2 that of the previous name-space. Therefor there are as many numbers in n1 as there are in n2. However you cannot double n1 to equal n2 because the two sets occupy two different name-space and cannot coexist in the same area of understanding because of incompatible name-spaces.