Could there be an "i" of the tetrative operations? - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: Could there be an "i" of the tetrative operations? (/showthread.php?tid=1788) |
Could there be an "i" of the tetrative operations? - 000_Era - 07/06/2024 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 RE: Could there be an "i" of the tetrative operations? - MphLee - 07/12/2024 Hi 000_era, welcome to the tetration forum. Sadly nowadays the forum is only just an archive of old discussion and is not very active anymore, read as "completely dead". Hopefully it will comeback, if enough new users get curious about these topics. To this to be a possibility we poster should do our best to favour engagement and questions: by being respectful of ppl's time and kind. This translates in putting effort in our questions by doing the homeworks, being as clear as we can, eg. by trying to define things and use consistent notation. Something that sometimes was missing in the past. Excuse me for my long welcome message. In the near future we will set up some mild moderation practices and set of guidelines for new users. Back to your question. I'm not sure of the analogy you see there. If I recall correctly one can argue that there are not equations containing tatration functions that are not solved by complex numbers or by limits of sequences of complex numbers... so no new sets of numbers I guess... anyways I never understood if there was a proof or just some heuristics. Can you make more precise what kind of analogy/scheme you see? For example I don't get why the role of the \(1/2\) doesn't break completely any scheme. Btw: here you can display math notation just by enclosing your equations by \ ( your_equation \ ) as follows: Code: \( (a+b)^n\) renders as \( (a+b)^n\). |