User blog:Revilo30703/KASE pt 2

Original Blog: http://googology.wikia.com/wiki/User_blog:Revilo30703/Knuth%27s_Up_Arrow_Infinite_Expantion

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 we created Ø the first time we had arrows expanding out a times in a dimensions. Now if you have Ø expanding out a times in a dimensions, we get Ø{3}, a sort of "double" Ø. Now if you had aØ{2}b that would become a?(a?(...(a?a) with b a's where ? equals the "block" of Ø's. Now we'd do the same thing and make Ø{4}, Ø{5}.....

Next jump time! Now what we have is aØ{z}(i,h,g,...e,d,c)b where a is the number we're dealing with, b is the number of a's once we simplify it one time, c-i are dimensions, and z is the "level" of Ø. One is a normal ^, two is Ø, three is what we created in the paragraph above and so on. So this time we will plug in a for everything: aØ{a}(a,a,a....a,a,a)a and we make ¥! aØ{a}(a,a,a....a,a,a)a=a¥2. So... a¥b=aØ{a}(a,a,a....a,a,a)(aØ{a}(a,a,a....a,a,a)(....(aØ{a}(a,a,a....a,a,a)a) where the number of a's (just the once that are before and after Ø) is equal to b. So what can we do with this to make it bigger? Well lets treat it as an ^. Then we will redo everything and eventually have to make some sort of Ø-¥ hybrid. How about we forget the ¥ (I'm going to have to use it later) and rewrite it as if they were dimensions. I'm calling it Ø{2,2} where a¥b equals aØ{2,1}b (Looking familiar?) So every time we want some way to represent all the letters we would increase the rightmost number in the {} (if your brackets were {b,a}, it would be a) by one. Then the to do the same sort of thing but including the rightmost number by one in the {} you would increase b by one in the example above. But don't forget each Ø (even if it has different bracket numbers) can be made a ØØ so they all still have parentheses. ØØ equals two in the parentheses, not in the curly brackets. So now what we can do is make {c,b,a}. If you are having trouble following the rules all you have to do is make the numbers in the parentheses one then treat the curly brackets as parentheses and then simplify them normally. From this point onwards what we can do is keep adding more numbers to the curly brackets, kind of like dimensions. So now what we have is aØ{q,p,o...m,l,k}(i,h,g...e,d,c)b.

Wait a minute, what if we plug in a for everything? We get aØ{a,a,a...a,a,a}(a,a,a...a,a,a)a. Now I'm going to use more brackets. This thing above will equal aØ2 with a 2 in the first part of the new brackets and everything else in the brackets and parentheses will equal one. (and of course, a¥b equals aØ{a,a,a...a,a,a}(a,a,a...a,a,a)(aØ{a,a,a...a,a,a}(a,a,a...a,a,a)(aØ{a,a,a...a,a,a}(a,a,a...a,a,a)(...(aØ{a,a,a...a,a,a}(a,a,a...a,a,a)a) with b a's). Also the number of a's in each curly bracket or parentheses equals a.

Hopefully you can see where this is going to end up going, we will probably use [] and do the same thing as we did with {}, then make another set... Now I'm going to bring ¥. a¥2 will equal a?b where the question mark equals aØa, but the numbers in the brackets and parenthesis will all equal a a times and a different brackets/parenthesis! Consider this as the "second dimension" of this part. This might be a bit confusing at first but just wait. Because we can treat ¥ as a ^, we could recreate everything above using ¥, where the rightmost number in the rightmost brackets(the parentheses) would equal ¥¥. Where the second dimension comes in: imagine all the brackets "stacked on top of each other" Yes that's right, we have gone back to the very original parentheses.

From here I'm not going to explain how it works because I hope it's kind of self makes sense how it would work. Where we could make another layer then go back into 3-D then make the whole bracket idea again... I'm not 100% how that would look or even how you could even notate so yeah

Again if you have questions or comments or you really want to break my mind and make me make it bigger please say so, but to be completely honest I might not be able to answer the questions because I'm not 100% sure how it works but I'll do my best.