Re: New way of dealing with complex numbers

Le 09/03/2025 à 00:31, Moebius a écrit :

Am 09.03.2025 um 00:26 schrieb efji:

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)]

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

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

 Might be the case, yes. So what? :-P
 But -hint- you talked about *associativity*, not about *commutativity*. :-)

I just pointed out the fact that the notation x^n is never used in the case of non associative operators because it is ambiguous without further definition. Think about the vector product in R^3 for example, which is not associative, and not commutative too. Nobody would write x^3 for (x \wedge x)\wedge x.
In the case of Hachel's delirium, the product is obviously associative, thus i^2 = -1 and i^4 = -1 makes no sense.
And of course, even with your recursive definition, it makes no sense.

 Trying to use crank strategies?

fighting fire with fire :)
--
F.J.