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From: Moebius <invalid@example.invalid>
Newsgroups: sci.math
Subject: Re: New way of dealing with complex numbers
Date: Sun, 9 Mar 2025 00:47:25 +0100
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Am 09.03.2025 um 00:43 schrieb efji:
> 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 :
>>>>>
>>>>>> Associativity is MANDATORY to be able to write something like i^4 
>>>>>> = i*i*i*i.
>>>>>>
>>>>>> For a non associative operator, i^4 means NOTHING.
>>>>>
>>>>> Oh, i^(n+1) just might mean (i^n) * i (with n e IN).
>>>>>
>>>>> [And i^0 = 1.]
>>>>>
>>>>> Then: i^4 = ((i*i)*i)*i.
>>>>>
>>>>> [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)
>>
>> 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 :)

:-P