Most people on here know Knuth's Up Arrow, if you don't page on this is here:

In this blog I will take several "jumps," where it gets progressively more huge. With the arrow system, Knuth has just made the "first dimension." Basically the number of arrows increases out in one direction. (When I use ^, I mean a singular up arrow) Normal exponentiation is a^b, then a^^b and so on.  I have simplified it to a^(c)b where c-2 is the number of arrows (a^(2)b is a*b, and a^(1)b is a +b). The next "jump" is going into the second dimension. What I have done for this is have an ^ on top of ^ (I've stopped using the normal up arrows because ^ is a lot easier to use). ^(d,c) is arrows with a "length" (^^^...^^^) c (unless d is equal to 1, then the length equals c-2) , and a "height" (the number of arrows on top of each other).



a^(2,1)b=a^(a)(a^(a)(....(a^(a)a)....) where the number of a's (not the one showing the length of the arrows) is equal to b

a^(2,c)b=a^(2,c-1)(a^(2,c-1)(...(a^(2,c-1)a)...) where the number of a's is equal to b

Then what we do is make a stack three high, where it simplifies to a stack two high and a long.At this point you'd keep making the stack higher and higher, but we've "reached an infinity"

Time for the next step!

The first time when we went to the second dimension, we had a^(2,1)b where the arrows (you will always have the b amount of a's) were a long. Basically we had a^(c)b, then I changed all the variables to a to get a^(a)a, to get a^(2,1)2. So now we have a^(d,c)b with arrows c long and d high. now plug in a for all the variables and get a^(a,a)a. This equals a^(2,1,1)2. The third dimension! Basically the rules are for the same for the length, then you'd make the "jump" to hight (always having the same amount of "depth"). To simplify you will try to get the rightmost number in the arrow, (c in ^(e,d,c)) then figure out what that number is and continue till you have a ridiculously huge number. At this point we could make a fourth dimention, fifth dimention, but we've reached another infinity! I initially was going to have a^(a,a,a,....a,a,a)a where each step to the right is a higher dimention, but then I realized this was just infinity. So I made a^(a,a,a...a,a,a)a, but the number of dimentions is equal to a! In wrighting I made an ^ with an e inside it but on here I'm going to use Ø. So 1Ø2 equals 1^(1)1=1. 2Ø2=2^(2,2)2=2^(2,1)2=2^(2)2=2*2=4 (cool note: whenever you have 2?2 where ? is any symbol in KASE (Knuth's arrow system expantion as I'm calling it) it will always equal 4) 3Ø2=3^(3,3,3)3 which is ridiculously huge. I'm not sure what numbers you could relate this too and if it's bigger or smaller and if you know please say in the comments. At this point I stop but if you really need even bigger numbers you could treat Ø as ^ and do the whole thing over again creating like a double Ø (but not ØØ) then you could do it again and have a tripple Ø then have something to reprezent the magnitude of Ø's....

The weird thing about this is that you think you have some an infinity but then you can make something to reprezent that and so on creating maybe a new infinity? Probably not but that'd cool.

Also if any of this is confusing or you have questions or mistakes that I made please can you comment them. Thanks

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