Re: y=f(x)=(x²)²+2x²+3

Liste des GroupesRevenir à s math 
Sujet : Re: y=f(x)=(x²)²+2x²+3
De : r.hachel (at) *nospam* tiscali.fr (Richard Hachel)
Groupes : sci.math
Date : 07. Feb 2025, 22:12:18
Autres entêtes
Organisation : Nemoweb
Message-ID : <cp3Ba1KrGTesnSdl-vQCmBomsbg@jntp>
References : 1 2 3 4 5 6 7 8
User-Agent : Nemo/1.0
Le 07/02/2025 à 14:27, Alan Mackenzie a écrit :
Richard Hachel <r.hachel@tiscali.fr> wrote:

(i²)²=-1, and not 1.
 Where do you get that garbage from?  i^4 = 1.
 
And there, the whole structure that we thought we had defined by a simple i²=-1, which was true, collapses for everything else.
 Garbage.  Nothing "collapses".  The theory of complex numbers is, as far
as mathematicians can determine, consistent.  It is vast and fascinating
in its own right.  It is also useful to scientists and engineers.
 I suggest you make more humble efforts to learn and understand it.
 
R.H.
No, it is not consistent.
When I speak in real terms, I set (-x)(-x)=x².
If x=-5, then x²=25.
So far, it is perfectly consistent.
But if I use another form of mathematics, and I set 1=-i², in order to get rid of both the (-) sign and the square root, I have to go to the end of the structure used. I can no longer do everything I want and anyhow. I have to use this structure in a consistent way (it is, and it is magnificent if we know how to use it well).
In an imaginary universe, it is a bit like practicing in a mirror, the real becomes imaginary, and the imaginary becomes real.
But once in the imaginary, you have to stay WITH the laws of imaginary operations, and the mistake you make is to say: if (-n)(-n)=n² in reality, then it is obvious that (-i)(-i)=i² in the imaginary, then that (i²)²=1 and so on.
This is not how to proceed, it is incoherent.
It is like hammering a nail with a screwdriver. Some can do it, but a simple little hammer blow is more useful.
THIS is what mathematicians do without paying attention to their blunder.
But it is not a mathematically correct notion, and we play with numbers without knowing what we are doing.
R.H.

Date Sujet#  Auteur
6 Feb 25 * y=f(x)=(x²)²+2x²+321Richard Hachel
6 Feb 25 `* Re: y=f(x)=(x²)²+2x²+320sobriquet
6 Feb 25  `* Re: y=f(x)=(x²)²+2x²+319Richard Hachel
6 Feb 25   `* Re: y=f(x)=(x²)²+2x²+318sobriquet
6 Feb 25    +* Re: y=f(x)=(x²)²+2x²+34Chris M. Thomasson
6 Feb 25    i+- Re: y=f(x)=(x²)²+2x²+31FromTheRafters
7 Feb 25    i`* Re: y=f(x)=(x²)²+2x²+32Richard Hachel
7 Feb 25    i `- Re: y=f(x)=(x²)²+2x²+31FromTheRafters
6 Feb 25    `* Re: y=f(x)=(x²)²+2x²+313Richard Hachel
6 Feb 25     +- Re: y=f(x)=(x²)²+2x²+31Python
7 Feb 25     `* Re: y=f(x)=(x²)²+2x²+311sobriquet
7 Feb 25      `* Re: y=f(x)=(x²)²+2x²+310Richard Hachel
7 Feb 25       +* Re: y=f(x)=(x²)²+2x²+38Alan Mackenzie
7 Feb 25       i+* Re: y=f(x)=(x²)²+2x²+34Richard Hachel
7 Feb 25       ii`* Re: y=f(x)=(x²)²+2x²+33Python
7 Feb 25       ii +- Re: y=f(x)=(x²)²+2x²+31Richard Hachel
22 Feb 25       ii `- Re: y=f(x)=(x²)²+2x²+31Moebius
7 Feb 25       i+- Re: y=f(x)=(x²)²+2x²+31Richard Hachel
22 Feb 25       i`* Re: y=f(x)=(x²)²+2x²+32Moebius
22 Feb 25       i `- Re: y=f(x)=(x²)²+2x²+31Alan Mackenzie
7 Feb 25       `- Re: y=f(x)=(x²)²+2x²+31sobriquet

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal