Could there be an "i" of the tetrative operations?
#1
Hello. One day I was browsing Youtube, and I noticed a video that sprung me into thinking about tetration somehow. The Youtube video was a low effort math video and it said "Solve -a = 1/a", it will take anyone some basic algebra to figure out that a = i, thus, -i = 1/i. This was interesting to me, because I found it interesting the inverse orders of operations of different orders of magnitude have a result for a value which is the same and is a "unit" with magnitude 1 so to speak. This got me thinking about every "unit of dimensionality" for each operation in math. Let me know if you see a pattern:


0+1 = 1
1*-1 = -1
-1^(1/2) = i


So the  next level in this step would be either the 1/2 tetration power of i or the 1/3 tetration power of i. This means that i = x^x or i = x^x^x.
The latter didn't net me any decent results,  but I plugged i=x^x into Wolfram Alpha but I couldn't interpret it.
What I am adding to this tetration board is a postulate! I am postulating that there is an 'i' of tetration, where, like

1 = -(-1)
-i = 1/i,
there is a unit where
(1/x) = tetration(x,(1/2)),

This unit would follow the former pattern in the first list.

I got my undergrad in CS, so I may be a little out of my league, but let me know what you guys think.
-000_Era
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Could there be an "i" of the tetrative operations? - by 000_Era - 07/06/2024, 02:59 AM

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