• Revilo30703

    KASE pt 2

    January 23, 2016 by Revilo30703

    Original Blog:

    Last time I got up to Ø, where if you have aØb, each place with arrows equals ^(a,a,a,a....a,a,a,a) where the number of "dimensions" equals a and the amount of length/depth or whatever it is in that dimension equals a.

    To take this to the next step we'll have to create treat Ø as a single ^. Once we have done that we can do what we did before by creating Ø(2), Ø(2,1), Ø(3,6,34352,34) or whatever craziness you can dream up. To solve that amount of Ø's you would treat them as ^, just make sure not to confuse them with actual ^s. Then you would get a singular Ø the same way ^. Then you would turn that into arrows...

    Now if you remember when…

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  • Revilo30703

    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" (^^^..…

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