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Subject: Re: New equation
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Date: Sun, 02 Mar 25 17:10:45 +0000
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Le 02/03/2025 à 17:32, Richard Hachel a écrit :
> Le 02/03/2025 à 06:06, Jim Burns a écrit :
>  ⎜
>> (a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖
> 
>  Yes.
> 
>> (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖
> 
>  No.
> 
>  (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖
> 
>
>  In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖
> 

This is true in R(j) not in C.

I posted about this yesterday. Did you even read my post?