Re: The splendor of true

Le 09/03/2025 à 03:05, Richard Hachel a écrit :

Le 09/03/2025 à 01:28, Python a écrit :

Le 09/03/2025 à 00:54, Richard Hachel a écrit :
...

Into what strange world would we fall, if, instead of using Z=aa'-bb'+i(ab'+a'b), we used the much more logical and natural equation Z=aa'+bb'+i(ab'+a'b).

 What makes you claim that ( aa' + bb', ab' + a'b ) is "more logical and natural" than ( aa' - bb', ab' + a'b) ?

 Several reasons.
First, the simplicity of writing, all in positive terms.

Not very convincing. bb' as well as -bb' can be (depending on the signs of b and b')
In arithmetic a + b is not "simpler" than a - b.

The concordance with statistical science.
The ease of natural understanding (Plougastel high school).

Quite a weak argument. What you noticed for integer values of real and imaginary parts of numbers i n the split-complex numbers does not provide any simpler way to compute the cardinals of cartesian products. You basically noticed what is well-known :
z = x + y*j = (x - y)*e + (x + y)*ee with e = (1-j)/2 and ee = (1+j)/2 then if you denote z1 = (a,b), z2 = (a',b') the coordinates of z in the (e, ee) base you got :
z1*z2 = (aa', bb') (in the same base)
see: https://en.wikipedia.org/wiki/Split-complex_number#The_diagonal_basis
Interesting, but not that much a big deal.

The perpetual monitoring of derivations using i=-1 as a simple numerical concordance check.

This sentence has no meaning.

The simplicity of the explanation and the generalization of i^x=-1

It is false. Hopefully! As R(j) is consistent, and j^2 = 1 NOT -1, the existence of an item i such as i^x = -1 for all x leads to a direct contradiction.
In other words: R(j) that you somewhat "rediscovered" is totally unrelated to this other (silly) idea of yours of i^x = -1 for all x.

The visualization of the associated curves symmetrical in $(0,y).
Roots often much simpler to calculate and write...

Again, this is 100% unrelated with R(j).
What you are calculating is NOT supplemental roots of the function you start with. You are calculating the roots of *another* function : g:g(x) = 2*f(0) - f(-x) (i.e. the graph of g is the point symmetric graph of f with respect to (0, f(0)) for a polynomial the relation between coefficient of f and g is very simple :
f(x) = a_n*x^n + a_(n-1)x^n-1 + a_(n-2)x^n-2 + ... + a_2 x^2 + a_1 x + a_0
g(x) = 2f(0) - f(-x)
g(x) = 2*a_0 - a_n*(-1)^n*x^n - a_(n-1)*(-1)^(n-2)*x^(n-1) - ... - a_2 (-1)^2 x^2 - a_1*(-1)*x - a_0
g(x) = (-1)^(n+1)*a_n*x^n + (-1)^n*a_(n)*x^(n-1) +  ... + a_2 (-1)^3 x^2 + a_1*(-1)^2 + a_0
(for instance for f(x) = x^3 + 2x^2 + 3x + 4, g(x) = 8 - (-x)^3 - 2*(-x)^2 - 3*(-x) - 4  = x^3 - 2x^2 + 3x + 4)
See: https://www.wolframalpha.com/input?i=x%5E3+%2B+2x%5E2+%2B+3x+%2B+4%2C+x%5E3+-+2x%5E2+%2B+3x+%2B+4 for the graphs of f and g.
There is no "need" for an item i such as i^x = -1 to get these coefficient (hopefully, as i^x = -1 is inconsistent as a property.
Anyway g this has nothing to do with the roots of f. By definition of "root": a root of function or polynomial f is a value a such as f(a) = 0. PERIOD).

So basically, it's simpler and more beautiful.

The only argument you have, at the end, is that this is something you proposed. This is egotism, pathological egotism.
Moreover the *three* "ideas" you proposed (R(j) i.e. "split-complex numbers" as it is usually called, i^x = -1 and "mirror curves") are 100% unrelated: the first one is consistent and already known, the second is contradictory, and the last ones does not address the point of roots of a function f).

Now what does that change for fractals, for Gauss's representation, for experimentally validated equations, I don't know.

When it comes to representation of R(j) I, and other, posted several links. The natural representation is R^2 (like for complex numbers C), and there are surprising stuff. For instance the set of points at a constant distance from the origin, is no more a circle put an hyperbola).
Anyway, nothing you can write could change what the set of complex numbers are - by definition - and the properties it has that R(j) does not have. Mainly: being a field, making true the fundamental theorem of Algebra and allowing to extend analysis from R to R^2 (i.e. derivative and integrals).