Re: Hello!

Le 19/01/2025 à 18:16, guido wugi a écrit :

 Not "on", but "only" you ;)
 The definition of "i" is simple: it's a solution of the equation i^2=-1.
 So it's clear from the start that it's not a member of the real number line, and thus
neither a doubting case, or Schrödinger cat limbo state case, between [+1 and -1],
neither a case where it should be alternatively +1 and -1, to have i*i=-1 work...
 It's simply some "other number" outside the real number line, one creating its own number line.
What's there more to brood on?

Nothing prevents me from thinking that i is a special unit, existing in an imaginary world, and which is neither 1 nor -1.
If it were -1 or 1, its square would be 1.
This is also what it becomes once the equation is posed, when it is presented under the visible face of the moon, we immediately have i²=1 whatever the observed value (-1 or 1).
As long as it is not observed, that we do not know what it is, we can only imagine two solutions, and in the imagination, it is worth both 1 and -1.
At this precise moment of imagination, i having both values, we have (i)(i)=(-1)(1)=-1=i².
The immense error of mathematicians is then to consider that i remains forever unknown. This is false. We can then give TWO solutions to z (which is an imaginary double number) z=a+ib
But if we give the first solution, with i=-1, we get
as a product (a+ib)(a'+ib')=aa+bb+i(ab'+a'b). And exactly the same thing for i=1. Because each time, i being known, we can simplify like this.
We then just have to reintroduce the uncertainty, in the equation, which now contains only one ito have the two solutions.
But in each solution, the real part remains the same, and it is aa+bb', and certainly not aa'-bb".
R.H.