Re: New way of dealing with complex numbers

Le 08/03/2025 à 23:55, Moebius a écrit :

Am 08.03.2025 um 23:47 schrieb Moebius:

Am 08.03.2025 um 14:32 schrieb efji:

Le 08/03/2025 à 14:18, Richard Hachel a écrit :

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Associativity is MANDATORY to be able to write something like i^4 = i*i*i*i.
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For a non associative operator, i^4 means NOTHING.

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Oh, i^(n+1) just might mean (i^n) * i (with n e IN).
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[And i^0 = 1.]
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Then: i^4 = ((i*i)*i)*i.
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[Hint: recursive definition:
  x^0 = 1
  x^(n+1) = x^n * x   (for all n e IN)]

      x^0 = 1
     x^(n+1) = (x^n) * x   (for all n e IN)]
 ... if you like.

I don't like.
What if * is not commutative ?
(x^n) * x =/= x * (x^n)
--
F.J.