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Subject: Re: Division of two complex numbers
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From: Python <jp@python.invalid>

Le 20/01/2025 à 23:22, Tom Bola a écrit :
> Am 20.01.2025 23:08:44 Python schrieb:
>> Le 20/01/2025 à 23:00, Tom Bola a écrit :
>>> Am 20.01.2025 21:20:51 Python schrieb:
>>>> Le 20/01/2025 à 21:09, Tom Bola a écrit :
>>>>> Am 20.01.2025 20:33:12 Moebius schrieb:
>>>>>> Am 20.01.2025 um 19:27 schrieb Python:
>>>>>>> Le 20/01/2025 à 19:23, Richard Hachel  a écrit :
>>>>>>>> Le 20/01/2025 à 19:10, Python a écrit :
>>>>>>>>> Le 20/01/2025 à 18:58, Richard Hachel  a écrit :
>>>>>>>>>>>> Mathematicians give:
>>>>>>>>>>>>
>>>>>>>>>>>> z1/z2=[(aa'+bb')/(a'²+b'²)]+i[(ba'-ab')/(a'²+b'²)]
>>>>>>>>>>>>
>>>>>>>>>>>> It was necessary to write:
>>>>>>>>>>>> z1/z2=[(aa'-bb')/(a'²-b'²)]+i[(ba'-ab')/(a'²-b'²)]
>>>>>>>>
>>>>>>>>> I've explained how i is defined in a positive way in modern algebra. 
>>>>>>>>> i^2 = -1 is not a definition. It is a *property* that can be deduced 
>>>>>>>>> from a definition of i.
>>>>>>>>
>>>>>>>>  That is what I saw.
>>>>>>>>
>>>>>>>>  Is not a definition.
>>>>>>>>  It doesn't explain why.
>>>>>>>>
>>>>>>>> We have the same thing with Einstein and relativity.
>>>>>>>>
>>>>>>>> [snip unrelated nonsense about your idiotic views on Relativity]
>>>>>>> 
>>>>>>>> It is clear that i²=-1, but we don't say WHY. It is clear however that 
>>>>>>>> if i is both 1 and -1 (which gives two possible solutions) we can 
>>>>>>>> consider its square as the product of itself by its opposite, and vice 
>>>>>>>> versa.
>>>>>>> 
>>>>>>> I've posted a definition of i (which is NOT i^2 = -1) numerous times. A 
>>>>>>> "positive" definition as you asked for.
>>>>>> 
>>>>>> I've already told this idiot:
>>>>>> 
>>>>>> Complex numbers can be defined as (ordered) pairs of real numbers.
>>>>>> 
>>>>>> Then we may define (in this context):
>>>>>> 
>>>>>>           i := (0, 1) .
>>>>>> 
>>>>>>  From this we get: i^2 = -1.
>>>>> 
>>>>> For R.H.
>>>>>   By the binominal formulas we have: (a, b)^2 = a^2 + 2ab + b^2
>>>> 
>>>> Huh? This is not the binomial formula which is (a + b)^2 = a^2 + 2ab + b^2
>>>> 
>>>> (a, b)^2 does not mean anything without any additional definition/context.
>>>> 
>>>>>   So we get: (0, 1)^2 ) 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1 
>>>> 
>>>> you meant  (0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1 ?
>>>> 
>>>> This does not make sense without additional context.
>>>> 
>>>> In R(epsilon) = R[X]/X^2 (dual numbers a + b*epsilon where epsilon is such 
>>>> as
>>>> epsilon =/= 0 and epsilon^2 0) we do have :
>>> 
>>> So your epsilon or i squared is 0 --- but the complex i sqared is -1.
>> 
>> Sure. I didn't pretend that R(epsilon) was C. My point is that 
>> multiplication in this ring makes perfect sense even if it is not 
>> isomorphic to C.
>> 
>>>> (0, 1) ^ 2 = 0
>>> 
>>> But with your vector (a,b), b^2=-1 that vector square really does calculate 
>>> with the first binominal formula and you don't get " > (0, 1) ^ 2 = 0"... 
>> 
>> The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about 
>> (0, 1)^2 or (a, b)^2.
> 
> Sure,

This was my only objection.

> but remember the definition of a complex number: Z = a+bi, i^2 = -1.
> 
> See?

But the point of the discussion is about the 
validity/usefulness/consistency of defining i^2 = -1 and C = { a + bi }

To assume that it is so necessarily is not about to address the issue.