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From: efji <efji@efi.efji>
Newsgroups: sci.math
Subject: Re: New way of dealing with complex numbers
Date: Sun, 9 Mar 2025 00:43:53 +0100
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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 :)


-- 
F.J.