Division of two complex numbers.

Now let's set Z=(a+ib)/(a'+ib')
with
z1=a+ib
and
z2=a'+ib'

What becomes of Z=A+iB?

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

It seems to me that you are asking for the complex numbers A and B such
that Z = A + iB.

See here: https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:

https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Merci beaucoup.

I saw this.

<http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1>

Merci, je vais donc pouvoir répondre aux mathématiciens.

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Am 20.01.2025 um 17:23 schrieb Richard Hachel:

Le 20/01/2025 à 16:22, Richard Hachel  a écrit :

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:
https://en.wikipedia.org/wiki/
Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

 Merci beaucoup.

 I saw this.

 <http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1>

 Merci, je vais donc pouvoir répondre aux mathématiciens.

 R.H.

As I expected, it is completely wrong.

If you say so. :-)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 16:22, Richard Hachel a écrit :

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:

https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Merci beaucoup.

I saw this.

<http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1>

Merci, je vais donc pouvoir répondre aux mathématiciens.

R.H.

As I expected, it is completely wrong.

The same sign error.

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)]

Three sign errors (which is the same error) because each time we put
i²=-1 where b and b' are already defined.

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 17:23, Richard Hachel a écrit :

Le 20/01/2025 à 16:22, Richard Hachel a écrit :

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:

https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Merci beaucoup.

I saw this.

<http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1>

Merci, je vais donc pouvoir répondre aux mathématiciens.

R.H.

As I expected, it is completely wrong.

The same sign error.

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)]

Three sign errors (which is the same error) because each time we put i²=-1 where b and b' are already defined.

R.H.

It is not an "error". Complex numbers are defined in such a way that this relation is true. They are what they are.

You cannot object to a "definition", except if it is not consistent.
Definition of complex numbers is consistent, and they do have purposes.
Quite a LOT of useful purposes, from geometry to integral calculus,
electricity and quantum mechanics.

You can, nevertheless, propose that other rules for multiplication (so division) may be useful. But then you're not talking about complex numbers
but another kind of numbers.

There are already other kinds of numbers build from pairs of real numbers,
like dual numbers that are interesting. Dual numbers to name one.

I'm not convinced that *your* proposition is useful. Maybe is is.

You are ridiculing yourself when you pretend that you "fix" a error in the definition of complex numbers, in an even more pathetic way than when you pretend to redefine Relativity.

But that is the story of your life, right? Making a fool of yourself and
drown yourself in your pathetic mix of hubris and stupidity.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 18:19, Python a écrit :

Le 20/01/2025 à 17:23, Richard Hachel a écrit :

Le 20/01/2025 à 16:22, Richard Hachel a écrit :

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:

https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Merci beaucoup.

I saw this.

<http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1> >>>
Merci, je vais donc pouvoir répondre aux mathématiciens.

R.H.

As I expected, it is completely wrong.

The same sign error.

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)]

Three sign errors (which is the same error) because each time we put i²=-1 >> where b and b' are already defined.

R.H.

It is not an "error". Complex numbers are defined in such a way that this relation is true. They are what they are.

You cannot object to a "definition", except if it is not consistent. Definition
of complex numbers is consistent, and they do have purposes. Quite a LOT of useful
purposes, from geometry to integral calculus, electricity and quantum mechanics.

You can, nevertheless, propose that other rules for multiplication (so division)
may be useful. But then you're not talking about complex numbers but another kind
of numbers.

There are already other kinds of numbers build from pairs of real numbers, like
dual numbers that are interesting. Dual numbers to name one.

I'm not convinced that *your* proposition is useful. Maybe is is.

You are ridiculing yourself when you pretend that you "fix" a error in the definition of complex numbers, in an even more pathetic way than when you pretend
to redefine Relativity.

But that is the story of your life, right? Making a fool of yourself and drown
yourself in your pathetic mix of hubris and stupidity.

Si tu vérifies avec honnêteté tout ce que j'ai dit, et les équations
que j'ai corrigées, tu verras que tout se tient.

Maintenant, on peut se poser la question : oui, mais est-ce des nombres complexes qu'il parle?

Là je suis d'accord, posons nous la question.

Revenons à la base (comme dans la théorie de la relativité) et
progressons grain par grain, comme font les petits oiseaux.

Sur la notion des nombres complexes, posons nous la question : qu'est ce
que i?

Ce n'est pas 1, ce n'est pas moins 1, mais semble-t-il "quelque chose
d'autre" qui peut donner ce qui n'existe pas dans le réel, un carré
négatif, et plus précisément égal à -1.

Ne sachant pas ce que c'est que i, j'ai proposé l'idée qu'il soit à la
fois 1 et -1.

Cela induit que z, qui n'est autre qu'un multiplicateur de cette unité bizarre, a lui aussi une dualité,
et qu'il peut être à la fois 4 et 9, 4 et 12, 4 et 45, 3 et 27, etc...

Bref, z est lui aussi un nombre imaginaire qui est une dualité.

On revient au problème, et toi qui es très féru de définition
précise, qu'est ce que i?

Dire que i²=-1, c'est dire qu'une hirondelle vole quand l'hirondelle
vole.

Ca n'explique par ce que c'est qu'une hirondelle, ni pourquoi ça vole.

En tout ça, ce que j'ai proposé ici est quelque chose de très cohérent (comme ce que j'ai proposé en RR qui n'a jamais pu être attaqué sérieusement).

Les équations sont cohérentes, les réciprocités évidentes, les lois mathématiques respectées.

Est-ce un bonne façon de voir les nombres complexes, est-ce une MEILLEURE
et plus concrète façon? je ne sais pas.

Que devient cette façon appliquée à la trigonométrie, je ne sais pas.
Et pourquoi faut-il l'appliquer à la trigonométrie? Que devient z? Une hypoténuse entre la composante imaginaire et la composante réelle?
Pourquoi? Que viens faire le point M? L'argument? Le module? Où sont les
deux nombres Z? Où sont les deux racines imaginaires d'une équation sans racine réelle?

Il est clair que si les additions de complexes sont les mêmes chez moi et
chez les mathématiciens, les produits et les divisions ne le sont pas.

Les deux systèmes ont leur cohérence, mais parle-t-on de la même chose?

Si l'on prend une vérification statistique, on se rend compte en deux
minutes que mes équations sont correctes, et pas celles des
mathématiciens (problème du collège de Plougastel).

Je te laisse trouver des définitions plus appropriés que celle que j'ai données, ou que celles que les mathématiciens donnent...

Je rappelle que i²=-1, c'est très joli, mais ça n'explique pas pourquoi l'hirondelle vole.

On en est au même point en relativité. On nous explique que le temps
passe réciproquement moins vite sur les horloges opposées, mais n'ayant
rien compris au phénomène qui est d'apparence absurde, on est obligé de s'inventer un time-gap à la con et des vitesses apparentes non
réciproques, ce qui bafoue la loi de réciprocité et de covariance de
tous les phénomènes relativistes.

Mais je te l'ai déjà expliqué tout ça.

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 18:58, Richard Hachel a écrit :

Le 20/01/2025 à 18:19, Python a écrit :

Le 20/01/2025 à 17:23, Richard Hachel a écrit :

Le 20/01/2025 à 16:22, Richard Hachel a écrit :

Le 20/01/2025 à 15:46, Moebius a écrit :

Am 20.01.2025 um 12:02 schrieb Richard Hachel:

Division of two complex numbers.

Now let's set Z = (a + ib)/(a' + ib')

I guess you meant: Z = z1/z2

with
z1 = a + ib
and
z2 = a' + ib'

What becomes of Z = A + iB?

See here:

https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Merci beaucoup.

I saw this.

<http://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1> >>>>
Merci, je vais donc pouvoir répondre aux mathématiciens.

R.H.

As I expected, it is completely wrong.

The same sign error.

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)]

Three sign errors (which is the same error) because each time we put i²=-1 >>> where b and b' are already defined.

R.H.

It is not an "error". Complex numbers are defined in such a way that this
relation is true. They are what they are.

You cannot object to a "definition", except if it is not consistent. Definition
of complex numbers is consistent, and they do have purposes. Quite a LOT of useful
purposes, from geometry to integral calculus, electricity and quantum mechanics.

You can, nevertheless, propose that other rules for multiplication (so division)
may be useful. But then you're not talking about complex numbers but another kind
of numbers.

There are already other kinds of numbers build from pairs of real numbers, like
dual numbers that are interesting. Dual numbers to name one.

I'm not convinced that *your* proposition is useful. Maybe is is.

You are ridiculing yourself when you pretend that you "fix" a error in the >> definition of complex numbers, in an even more pathetic way than when you pretend
to redefine Relativity.

But that is the story of your life, right? Making a fool of yourself and drown
yourself in your pathetic mix of hubris and stupidity.

Si tu vérifies avec honnêteté tout ce que j'ai dit, et les équations que j'ai corrigées, tu verras que tout se tient.

Maintenant, on peut se poser la question : oui, mais est-ce des nombres complexes qu'il parle?

Là je suis d'accord, posons nous la question.

Revenons à la base (comme dans la théorie de la relativité) et progressons grain par grain, comme font les petits oiseaux.

Sur la notion des nombres complexes, posons nous la question : qu'est ce que i?

Ce n'est pas 1, ce n'est pas moins 1, mais semble-t-il "quelque chose d'autre"
qui peut donner ce qui n'existe pas dans le réel, un carré négatif, et plus
précisément égal à -1.

Ne sachant pas ce que c'est que i, j'ai proposé l'idée qu'il soit à la fois 1
et -1.

Then your "i" is NOT the "i" of complex numbers. End of Story. Moreover
"being at the same time 1 and -1" is MEANINGLESS.

Cela induit que z, qui n'est autre qu'un multiplicateur de cette unité bizarre,
a lui aussi une dualité,
et qu'il peut être à la fois 4 et 9, 4 et 12, 4 et 45, 3 et 27, etc...

Bref, z est lui aussi un nombre imaginaire qui est une dualité.

On revient au problème, et toi qui es très féru de définition précise, qu'est ce que i?

Dire que i²=-1, c'est dire qu'une hirondelle vole quand l'hirondelle vole.

Ca n'explique par ce que c'est qu'une hirondelle, ni pourquoi ça vole.

En tout ça, ce que j'ai proposé ici est quelque chose de très cohérent (comme ce que j'ai proposé en RR qui n'a jamais pu être attaqué sérieusement).

Sorry, your claims on SR have been show illogical, wrong and
contradictory.

Les équations sont cohérentes, les réciprocités évidentes, les lois mathématiques respectées.

Est-ce un bonne façon de voir les nombres complexes, est-ce une MEILLEURE et plus concrète façon? je ne sais pas.

Que devient cette façon appliquée à la trigonométrie, je ne sais pas.

Then ask, or study by yourself.

Et pourquoi faut-il l'appliquer à la trigonométrie? Que devient z? Une hypoténuse entre la composante imaginaire et la composante réelle?

Then ask, or study by yourself.

Pourquoi? Que viens faire le point M? L'argument? Le module? Où sont les deux
nombres Z? Où sont les deux racines imaginaires d'une équation sans racine réelle?

Then ask, or study by yourself.

Il est clair que si les additions de complexes sont les mêmes chez moi et chez
les mathématiciens, les produits et les divisions ne le sont pas.

So you are not talking about complex numbers. End of story.

Les deux systèmes ont leur cohérence, mais parle-t-on de la même chose?

Definitely NOT. Moreover, complex numbers are consistent. Your proposal,
so far, is NOT.

Si l'on prend une vérification statistique, on se rend compte en deux minutes
que mes équations sont correctes, et pas celles des mathématiciens (problème du
collège de Plougastel).

"correct" on what ground? ?

Je te laisse trouver des définitions plus appropriés que celle que j'ai données, ou que celles que les mathématiciens donnent...

Je rappelle que i²=-1, c'est très joli, mais ça n'explique pas pourquoi l'hirondelle vole.

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.

On en est au même point en relativité. On nous explique que le temps passe réciproquement moins vite sur les horloges opposées, mais n'ayant rien compris
au phénomène qui est d'apparence absurde, on est obligé de s'inventer un time-gap à la con et des vitesses apparentes non réciproques, ce qui bafoue la
loi de réciprocité et de covariance de tous les phénomènes relativistes.

Mais je te l'ai déjà expliqué tout ça.

And what you've "explained" is a pile of garbage.

R.H.

1. Richard, posting in French in an English-speaking group demonstrates a complete lack of respect to people

2. When it comes to complex numbers you are making a fool of yourself,
exposing your pathetic arrogance and stupidity, as usual

If you have questions about complex numbers, the ones you've asked above
in French, post them on fr.sci.maths. If you want to post them here, post
them in English.

Act as a man, not as a clown.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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.

You really think that what you decide to ignore deliberately does not
exist?

Act like a decent human being, Richard, not as a clown.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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.

Einstein, he's a nice guy, a tender guy, he says "The speed of light is constant by change of frame of reference".

But he doesn't explain why.

He gives a quality, but without specifying the cause.

Reread my chapters on the notions of simultaneity, on the notion of anosochrony, of synchronization of watches, you will see, if you make the effort, why the invariance of c is logical,
and why it is like that.

For i, it's the same.

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.

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/20/2025 6:02 AM, Richard Hachel wrote:

Division of two complex numbers.

Now let's set Z=(a+ib)/(a'+ib')
with
z1=a+ib
and
z2=a'+ib'

What becomes of Z=A+iB?

x + iy = (1 + i0)/(c + id)

(c + id)(x + iy) = 1 + i0

(cx + i²dy) + (cy + dx)i = 1 + i0

cx + i²dy = 1
dx + cy = 0

c²x + i²cdy = c
i²d²x + i²cdy = 0

(c²-i²d²)x = c

cdx + i²d²y = d
cdx + c²y = 0

(c²-i²d²)y = -d

x + iy = (1 + i0)/(c + id)
x + iy = (c - id)/(c²-i²d²)

If i² ≠ 0 then
then some c + id ≠ 0 + i0
do not have inverses.

Above, I have asserted that i² is real (is east.west).
Real i² can be proven from the assumption (w⋅z)⃰ = w⃰⋅z⃰
[1]

There might be yet another argument showing that
i² is specifically -1
Or there might not be. I'm not sure right now.

Even if there isn't an argument for i²=-1,
the north.south units can be re.defined
so that it is so.

That leaves us with essentially
only one complex multiplication.

[1]
(w⋅z)⃰ = w⃰⋅z⃰
w = a + ib
z = c + id

i² = sₑ + i⋅sₙ

(w⋅z)⃰ =
((a + ib)(c + id))⃰ =
(ac + i(ad+bc) + i²bd)⃰ =
(ac + i(ad+bc) + (sₑ+i⋅sₙ)bd)⃰ =
((ac+sₑbd) + i(ad+bc+sₙbd))⃰ =
(ac+sₑbd) - i(ad+bc+sₙbd)

w⃰⋅z⃰ =
(a + ib)⃰⋅(c + id)⃰ =
(a - ib)⋅(c - id) =
ac - i(ad+bc) + i²bd =
ac - i(ad+bc) + (sₑ+i⋅sₙ)bd =
(ac+sₑbd) - i(ad+bc-sₙbd)

(w⋅z)⃰ - w⃰⋅z⃰ = -2i⋅sₙbd
(w⋅z)⃰ = w⃰⋅z⃰ ⇒ sₙ = 0
i² = sₑ

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/20/2025 1:33 PM, Jim Burns wrote:

On 1/20/2025 6:02 AM, Richard Hachel wrote:

[...]

[...]

x + iy  =  (1 + i0)/(c + id)
x + iy  =  (c - id)/(c²-i²d²)

If i² ≠ 0 then
then some c + id  ≠  0 + i0
do not have inverses.

Better:

If i² ≥ 0 then
then some c + id ≠ 0 + i0
do not have inverses.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Am 20.01.2025 um 20:21 schrieb Richard Hachel:

Le 20/01/2025 à 19:33, Jim Burns a écrit :

((a + ib)(c + id))⃰ =
(ac + i(ad+bc) + i²bd)⃰ =

No. You can't.

Yes, we can. :-)

You don't know what i is but

We know that i is i, i e C and that i² = -1, idiot.

If it is 1 then 1² = 1.
If it is -1 then -1² = 1

Yeah, but since i ISN'T 1 or -1, this is immaterial. Go that, idiot?

Again, i is i, i e C and i² = -1.

<facepalm>

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/01/2025 à 19:33, Jim Burns a écrit :

On 1/20/2025 6:02 AM, Richard Hachel wrote:

Division of two complex numbers.

Now let's set Z=(a+ib)/(a'+ib')

((a + ib)(c + id))⃰ =
(ac + i(ad+bc) + i²bd)⃰ =

No. You can't.

=(ac + i(ad+bc) + (ib*id)⃰

You don't know what i is but at this moment, you know that it is both i=1
and i=-1.
If it is 1 then 1²=1.
If it is -1 then -1²=1
In any case, its square will be 1 because in your two solutions, you will
have to give your choice, but not both at the same time.

(ac + i(ad+bc) + (sₑ+i⋅sₙ)bd)⃰ =
((ac+sₑbd) + i(ad+bc+sₙbd))⃰ =
(ac+sₑbd) - i(ad+bc+sₙbd)

ac+bd+i(ad+bc) with i=(+/-)1 at final choise.

R.H.

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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.

You really think that what you decide to ignore deliberately does not
exist?

Act like a decent human being, Richard, not as a clown.

--- SoupGate-Win32 v1.05
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Le 20/01/2025 à 20:33, Moebius a écrit :

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.

The concept of context is far out of Hachel/Lengrand's ability. He's a
complete idiot as you noticed, and a stubborn crank.

When it comes to definition of C your definition stands, anyway I do
prefer the one in term of equivalence classes of R[X]. As we all know
(except Richard which is absolutely incapable of learning anything and out
of rational thinking) they are equivalent i.e. defines isomorphic sets and structures.

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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 :

(0, 1) ^ 2 = 0

--- SoupGate-Win32 v1.05
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Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:20 PM, Python wrote:

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 :

(0, 1) ^ 2 = 0

vec2 ct_cmul(in vec2 p0, in vec2 p1)
{
return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * p1.x);
}

So what? This is not an application of the binomial formula...

What's you point?

--- SoupGate-Win32 v1.05
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Le 20/01/2025 à 21:59, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:51 PM, Python wrote:

Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:20 PM, Python wrote:

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 :

(0, 1) ^ 2 = 0

vec2 ct_cmul(in vec2 p0, in vec2 p1)
{
     return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * p1.x); >>> }

So what? This is not an application of the binomial formula...

What's you point?

It's a way I multiply two vectors together as if they are complex numbers.

Another one:

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

I can pass in normal vectors to this in GLSL. vec2's

Good! You know how to write a C program. :-) (pun intended)

This is quite off-topic to point out that multiplication of complex
numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel would
admit this. It is *why* it makes sense to define multiplication *that
way*.

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Le 20/01/2025 à 22:06, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:04 PM, Python wrote:

Le 20/01/2025 à 21:59, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:51 PM, Python wrote:

Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:20 PM, Python wrote:

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 :

(0, 1) ^ 2 = 0

vec2 ct_cmul(in vec2 p0, in vec2 p1)
{
     return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * p1.x);
}

So what? This is not an application of the binomial formula...

What's you point?

It's a way I multiply two vectors together as if they are complex
numbers.

Another one:

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

I can pass in normal vectors to this in GLSL. vec2's

Good! You know how to write a C program. :-) (pun intended)

Fwiw, that is not is C, it's from one of my GLSL shaders. ;^)

It is also C.

Again what's *your* point? Your posts makes absolutely no sense in the
context of this thread!

This is quite off-topic to point out that multiplication of complex
numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel would
admit this. It is *why* it makes sense to define multiplication *that way*.

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Am 20.01.2025 um 21:07 schrieb Chris M. Thomasson:

On 1/20/2025 11:33 AM, Moebius wrote:

Complex numbers can be defined as (ordered) pairs of real numbers.

Then we may define (in this context):

          i := (0, 1) .

Indeed. 0+1i = (0, 1)

In some of my vector code I allow for traditional 2-ary vector
multiplication and complex multiplication.

Same in math. Even though we may use the same symbol, say "*", it
denotes two different operations (in this case).

A complex number can be a simple 2-ary vector.

Exactly.

[...] Multiplying two 2-ary vectors together is
different than multiply two complex number together,

Right. It's the context that "decides" here; at least in math.

however they can both be stored in the same vector representation.

z = 2 + 8i = (2, 8)

Where 2 = Re(z), and 8 = Im(z)

Right. In math we call such an object an ordered pair (or 2-tuple).

See: https://en.wikipedia.org/wiki/Ordered_pair
and: https://en.wikipedia.org/wiki/Tuple

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Am 20.01.2025 um 22:28 schrieb Chris M. Thomasson:

On 1/20/2025 1:09 PM, Python wrote:

Again what's *your* point? Your posts makes absolutely no sense in the
context of this thread!

Just a way to multiply two 2-ary vectors as if they were complex
numbers. Now, here is a little C99 program I just typed in the
newsreader. It should compile.
_____________________________
#include <stdio.h>

struct vec2
{
    float x;
    float y;
};

struct vec2
ct_cmul(
    struct vec2 p0,
    struct vec2 p1
){
    struct vec2 result = {
        p0.x * p1.x - p0.y * p1.y,
        p0.x * p1.y + p0.y * p1.x
    };

    return result;
}

int main()
{
    struct vec2 z = { 0, 1 };
    struct vec2 zmul = ct_cmul(z, z);

    printf("z = (%f, %f)\n", z.x, z.y);
    printf("zmul = (%f, %f)\n", zmul.x, zmul.y);

    return 0;
}
_____________________________

Let me run it on a C99 compiler... Ok, it works:

z = (0.000000, 1.000000)
zmul = (-1.000000, 0.000000)

I thought it might help out the OP.

Mückenheim? :-P

Note though:

This is quite off-topic to point out that multiplication of complex
numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel
would admit this. It is *why* it makes sense to define
multiplication *that way*.

See?!

You C code doesn't answer the latter "question".

.
.
.

--- SoupGate-Win32 v1.05
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Le 20/01/2025 à 22:28, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:09 PM, Python wrote:

Le 20/01/2025 à 22:06, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:04 PM, Python wrote:

Le 20/01/2025 à 21:59, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:51 PM, Python wrote:

Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:20 PM, Python wrote:

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 :

(0, 1) ^ 2 = 0

vec2 ct_cmul(in vec2 p0, in vec2 p1)
{
     return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * >>>>>>> p1.x);
}

So what? This is not an application of the binomial formula...

What's you point?

It's a way I multiply two vectors together as if they are complex
numbers.

Another one:

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

I can pass in normal vectors to this in GLSL. vec2's

Good! You know how to write a C program. :-) (pun intended)

Fwiw, that is not is C, it's from one of my GLSL shaders. ;^)

It is also C.

No. GLSL is not C at all, it has a similar style, but is different for sure.

It is exactly the same syntax. *facepalm*.

Ok, let's say so, if you wish, so you can implement complex multiplication
in a GLSL shader.

Again: SO WHAT? ? ? This is NOT THE POINT of the discussion.

Again what's *your* point? Your posts makes absolutely no sense in the
context of this thread!

Just a way to multiply two 2-ary vectors as if they were complex
numbers. Now, here is a little C99 program I just typed in the
newsreader. It should compile.
_____________________________
[snip irrelevant triviality]

So what? ? ?

I thought it might help out the OP.

In which way? ? ? Hachel didn't write that it cannot be done (he's not
that silly), he claimed (wrongly) that it is the wrong way to define multiplication between complex numbers.

This is quite off-topic to point out that multiplication of complex
numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel
would admit this. It is *why* it makes sense to define multiplication
*that way*.

--- SoupGate-Win32 v1.05
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Am 20.01.2025 um 22:47 schrieb Chris M. Thomasson:

On 1/20/2025 1:36 PM, Moebius wrote:

Am 20.01.2025 um 22:28 schrieb Chris M. Thomasson:

On 1/20/2025 1:09 PM, Python wrote:

Note though:

This is quite off-topic to point out that multiplication of
complex numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel >>>>>> would admit this. It is *why* it makes sense to define
multiplication *that way*.

See?!

Your C code doesn't answer the latter "question".

Sorry for missing something here. The division of complex numbers?

You are missing what Python told you:

The discussion is not about that it can be done, even crank Hachel
would admit this. It is *why* it makes sense to define
multiplication *that way* [i.e the usualy way --Moebius].

Again:

Hachel didn't write that it cannot be done (he's not that silly),
he claimed (wrongly) that it is the wrong way to define
multiplication between complex numbers.

See?!

.
.
.

--- SoupGate-Win32 v1.05
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Le 20/01/2025 à 22:45, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:35 PM, Python wrote:

Le 20/01/2025 à 22:28, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:09 PM, Python wrote:

Le 20/01/2025 à 22:06, "Chris M. Thomasson" a écrit :

On 1/20/2025 1:04 PM, Python wrote:

Le 20/01/2025 à 21:59, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:51 PM, Python wrote:

Le 20/01/2025 à 21:44, "Chris M. Thomasson" a écrit :

On 1/20/2025 12:20 PM, Python wrote:

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 :

(0, 1) ^ 2 = 0

vec2 ct_cmul(in vec2 p0, in vec2 p1)
{
     return vec2(p0.x * p1.x - p0.y * p1.y, p0.x * p1.y + p0.y * >>>>>>>>> p1.x);
}

So what? This is not an application of the binomial formula... >>>>>>>>
What's you point?

It's a way I multiply two vectors together as if they are complex >>>>>>> numbers.

Another one:

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

I can pass in normal vectors to this in GLSL. vec2's

Good! You know how to write a C program. :-) (pun intended)

Fwiw, that is not is C, it's from one of my GLSL shaders. ;^)

It is also C.

No. GLSL is not C at all, it has a similar style, but is different for
sure.

It is exactly the same syntax. *facepalm*.
Ok, let's say so, if you wish, so you can implement complex
multiplication in a GLSL shader.

No. C and GLSL are completely different languages. Have you ever even
used GLSL? You can do fun things in GLSL that C cannot do at all.

This is ridiculous nitpicking.

#define cx_mul(a, b) vec2(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x)

may compile in C, also can your function ct_cmul above.

I didn't wrote that GLSL was C, I wrote that the code you wrote was C.

Anyway, this is not the point. Either in C or GLSL the fact that you can implement complex multiplication (or in ANY language) is NOT THE POINT it
is IRRELEVANT!

Again: SO WHAT? ? ? This is NOT THE POINT of the discussion.

I thought it might help the OP.

In what manner? ? Nobody, not even the OP pretended that it cannot be implemented.

Seriously Chris, what's wrong with you?

Again what's *your* point? Your posts makes absolutely no sense in
the context of this thread!

Just a way to multiply two 2-ary vectors as if they were complex
numbers. Now, here is a little C99 program I just typed in the
newsreader. It should compile.
_____________________________
[snip irrelevant triviality]

So what? ? ?

I thought it might help out the OP.

In which way? ? ? Hachel didn't write that it cannot be done (he's not
that silly), he claimed (wrongly) that it is the wrong way to define
multiplication between complex numbers.

This is quite off-topic to point out that multiplication of complex >>>>>> numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel >>>>>> would admit this. It is *why* it makes sense to define
multiplication *that way*.

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Am 20.01.2025 um 22:50 schrieb Python:

Le 20/01/2025 à 22:45, "Chris M. Thomasson" a écrit :

Seriously Chris, what's wrong with you?

Have mercy, Python. He's a programmer, not a mathematician (or physicist).

Keep cool.

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Le 20/01/2025 à 22:51, Moebius a écrit :

Am 20.01.2025 um 22:47 schrieb Chris M. Thomasson:

On 1/20/2025 1:36 PM, Moebius wrote:

Am 20.01.2025 um 22:28 schrieb Chris M. Thomasson:

On 1/20/2025 1:09 PM, Python wrote:

Note though:

This is quite off-topic to point out that multiplication of
complex numbers in C/C++ can be done.

The discussion is not about that it can be done, even crank Hachel >>>>>>> would admit this. It is *why* it makes sense to define
multiplication *that way*.

See?!

Your C code doesn't answer the latter "question".

Sorry for missing something here. The division of complex numbers?

You are missing what Python told you:

The discussion is not about that it can be done, even crank Hachel
would admit this. It is *why* it makes sense to define
multiplication *that way* [i.e the usualy way --Moebius].

Again:

Hachel didn't write that it cannot be done (he's not that silly),
he claimed (wrongly) that it is the wrong way to define
multiplication between complex numbers.

See?!

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.

Maybe autistic syndrome?

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Am 20.01.2025 um 22:52 schrieb Python:

Le 20/01/2025 à 22:51, Moebius a écrit :

Am 20.01.2025 um 22:47 schrieb Chris M. Thomasson:

Sorry for missing something here. The division of complex numbers?

You are missing what Python told you:

      The  discussion is not about that it can be done, even crank Hachel
      would admit this. It is *why* it makes sense to define
      multiplication *that way* [i.e the usualy way --Moebius].

Again:

       Hachel didn't write that it cannot be done (he's not that silly),
       he claimed (wrongly) that it is the wrong way to define
       multiplication between complex numbers.

See?!

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.
.

Maybe autistic syndrome?

Well, just a a person that is not identical with you or me. :-P

He's thinking "like a programmer". ;-)

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Am 20.01.2025 um 22:56 schrieb Chris M. Thomasson:

On 1/20/2025 1:52 PM, Python wrote:

Le 20/01/2025 à 22:51, Moebius a écrit :

Am 20.01.2025 um 22:47 schrieb Chris M. Thomasson:

Sorry for missing something here. The division of complex numbers?

You are missing what Python told you:

      The  discussion is not about that it can be done, even crank Hachel
      would admit this. It is *why* it makes sense to define
      multiplication *that way* [i.e the usualy way --Moebius].

Again:

       Hachel didn't write that it cannot be done (he's not that silly),
       he claimed (wrongly) that it is the wrong way to define
       multiplication between complex numbers.

See?!

Yikes! I totally missed that underlying point. Damn!

Sorry. Did Hachel present a new formula for it?

EXACTLY! :-)

Nuff said.

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Am 20.01.2025 um 22:57 schrieb Moebius:

Am 20.01.2025 um 22:56 schrieb Chris M. Thomasson:

On 1/20/2025 1:52 PM, Python wrote:

Le 20/01/2025 à 22:51, Moebius a écrit :

Am 20.01.2025 um 22:47 schrieb Chris M. Thomasson:

Sorry for missing something here. The division of complex numbers?

You are missing what Python told you:

      The  discussion is not about that it can be done, even crank >>>> Hachel
      would admit this. It is *why* it makes sense to define
      multiplication *that way* [i.e the usualy way --Moebius].

Again:

       Hachel didn't write that it cannot be done (he's not that >>>> silly),
       he claimed (wrongly) that it is the wrong way to define
       multiplication between complex numbers.

See?!

Yikes! I totally missed that underlying point. Damn!

Sorry. Did Hachel present a new formula for it?

EXACTLY! :-)

Moreover, he's questioning the correctness of i^2 = -1. :-)

Nuff said.

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Am 20.01.2025 um 23:02 schrieb Chris M. Thomasson:

On 1/20/2025 1:59 PM, Moebius wrote:

Moreover, he's questioning the correctness of i^2 = -1. :-)

Strange indeed.

A typical crank, just like Mückenheim.

See: https://en.wikipedia.org/wiki/Mathematical_Cranks

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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.

When you wrote:

(0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1
----------------------^^^

where does that -1 comes from?

The binomial formula cannot be used to justify complex multiplication
without assuming i^2 (i.e. (0,1)) in the first place.

So pointless here.

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Am 20.01.2025 um 23:11 schrieb Tom Bola:

BTW - does one know if the plain set if imaginary numbers
[...] has a name so one can google it?

The set of /purely imaginary numbers/ (/rein imaginäre Zahlen/).

See: https://mathworld.wolfram.com/PurelyImaginaryNumber.html

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Le 20/01/2025 à 23:16, 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.

Why not! ( Ok, you could also write (a + bi)^2 = -1,i^2 := -1 as well... )

Why not? Because it is what it is.

When you wrote:

(0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1
----------------------^^^

where does that -1 comes from?

It does come from -2 + 1 = -1 as the solution of the line above...

'nuff said...

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Am 20.01.2025 um 23:15 schrieb Tom Bola:

Am 20.01.2025 23:08:44 Python schrieb:

The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about (0,
1)^2 or (a, b)^2.

Why not!

WEIL ES SO IST.

(Ok, you could <bla>

*gähn*

When you wrote:

(0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1
----------------------^^^

where does that -1 comes from?

Das "- 1" in "2*(0 - 1)" war/ist gemeint.

It does come from

Ja?

Hinweis: "The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not
about (0, 1)^2 or (a, b)^2."

<facepalm>

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Am 20.01.2025 um 23:15 schrieb Tom Bola:

Am 20.01.2025 23:08:44 Python schrieb:

The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about (0,
1)^2 or (a, b)^2.

Why not?

WEIL ES SO IST.

Ok, you could <bla>

*gähn*

When you wrote:

(0, 1)^2 = 0^2 + 2*(0 - 1) + 1 = 0 + (-2) + 1 = -1
----------------------^^^

where does that -1 comes from?

Das "- 1" in "2*(0 - 1)" war/ist gemeint.

It does come from

Ja?

Hinweis: "The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not
about (0, 1)^2 or (a, b)^2."

<facepalm>

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Am 20.01.2025 um 23:21 schrieb Tom Bola:

The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about
(0, 1)^2 or (a, b)^2.

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

See?

See WHAT?!

Was wir hier brauchen, ist die Definition der MULTIPLIKATION (*) auf C =
{(x, y) : x,y e IR}.

DARAUS folgt dann

i * i = (0, 1) * (0, 1) = ... = (-1, 0)

See?

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Am 20.01.2025 um 23:27 schrieb Tom Bola:

Am 20.01.2025 23:20:38 Moebius schrieb:

Hinweis: "The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not
about (0, 1)^2 or (a, b)^2."

As I told in the parallel post:
... remember the definition of a complex number: Z = a+bi, i^2 = -1

Dummes Gerede a la Mückenheim.

EOD

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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.

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Am 20.01.2025 um 23:40 schrieb Chris M. Thomasson:

[...] Fwiw, a fun part of GLSL is doing stuff like:

vec3 a = vec3(.25, 1, .75);
vec2 b = a.xz;
vec2 c = b + vec2(.75, .25);

c now equals (1, 1)

Nice.


In math:

a := (.25, 1, .75) ,

b := (a_1, a_3) ,

c := b + (.75, .25) .

Then c = (1, 1) .

:-P


Question. What if

vec4 a = vec4(.25, 1, .75, .999);

vec3 b = a.xz<?>;

I'd like to get b == (.25, .75, .999). :-P

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Am 21.01.2025 um 00:03 schrieb Chris M. Thomasson:

On 1/20/2025 2:48 PM, Moebius wrote:

Am 20.01.2025 um 23:40 schrieb Chris M. Thomasson:

[...] Fwiw, a fun part of GLSL is doing stuff like:

vec3 a = vec3(.25, 1, .75);
vec2 b = a.xz;
vec2 c = b + vec2(.75, .25);

c now equals (1, 1)

Nice.


In math:

a := (.25, 1, .75) ,

b := (a_1, a_3) ,

c := b + (.75, .25) .

Then c = (1, 1) .

:-P


Question. What if

vec4 a = vec4(.25, 1, .75, .999);

vec3 b = a.xz<?>;

I'd like to get b == (.25, .75, .999). :-P

That would be:

vec4 a = vec4(.25, 1, .75, .999);
vec3 b = a.xzw;

;^)

C'mon... lol

And then?

xyzw ... uv ...

I usually use the "series"

x, y, z, u, v, w

in math/physics.

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Le 21/01/2025 à 00:08, Tom Bola a écrit :

Am 20.01.2025 23:26:19 Moebius schrieb:

Am 20.01.2025 um 23:21 schrieb Tom Bola:

The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about
(0, 1)^2 or (a, b)^2.

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

See WHAT?!

Was wir hier brauchen, ist die Definition der MULTIPLIKATION (*) auf C =
{(x, y) : x,y e IR}.

DARAUS folgt dann

i * i = (0, 1) * (0, 1) = ... = (-1, 0)

See?

Ja! This is fully straight plus absolutely correct.

I was just looking for some idea. Aside from that please have a look here:

(english readers just look for "binomisch")

https://www.math.kit.edu/iana2/~mandel/seite/schnupperkurs/media/komplex.pdf

See?

*facepalm*

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Am 21.01.2025 um 00:08 schrieb Tom Bola:

Am 20.01.2025 23:26:19 Moebius schrieb:

Am 20.01.2025 um 23:21 schrieb Tom Bola:

The binomial formula is (a + b)^2 = a^2 + 2ab + b^2 it is not about
(0, 1)^2 or (a, b)^2.

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

See WHAT?!

Was wir hier brauchen, ist die Definition der MULTIPLIKATION (*) auf C =
{(x, y) : x,y e IR}.

DARAUS folgt dann

i * i = (0, 1) * (0, 1) = ... = (-1, 0)

See?

Ja! This is fully straight plus absolutely correct.

I was just looking for some idea. Aside from that please have a look here:

(english readers just look for "binomisch")

https://www.math.kit.edu/iana2/~mandel/seite/schnupperkurs/media/komplex.pdf

Gutes pdf.

Das steht:

"Der Division liegt die _dritte Binomische Formel_ (a + b)(a − b) = a^2
- b^2 zugrunde."

Nicht: (a + b)^2 = a^2 + 2ab + b^2

Got it?

Und es geht dabei um die Division, nicht um (a, b)^2. (*seufz*)

Hint: You wrote (the nonsense): "By the binominal formulas we have: (a,
b)^2 = a^2 + 2ab + b^2"

.
.
.

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Am 21.01.2025 um 00:50 schrieb Tom Bola:

Am 21.01.2025 00:25:01 Moebius schrieb:

Am 21.01.2025 um 00:08 schrieb Tom Bola:

https://www.math.kit.edu/iana2/~mandel/seite/schnupperkurs/media/komplex.pdf

Gutes pdf.

Das steht:

"Der Division liegt die _dritte Binomische Formel_ (a + b)(a − b) = a^2 - b^2 zugrunde."

Ja, mir ging es ganz allein um ...

Ja, ja. Verstehst Du auch, warum es da heißt: "Der Division liegt die
_dritte Binomische Formel_ (a + b)(a − b) = a^2 - b^2 zugrunde"?

Kannst Du das mal etwas genauer ausführen/zeigen?

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Am 21.01.2025 um 03:50 schrieb Chris M. Thomasson:

GLSL has x, y, z, w or [0], [1], [2], [3]

x, y, z, t ? :-)

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Am 21.01.2025 um 03:50 schrieb Chris M. Thomasson:

On 1/20/2025 3:15 PM, Moebius wrote:

Am 21.01.2025 um 00:03 schrieb Chris M. Thomasson:

On 1/20/2025 2:48 PM, Moebius wrote:

Am 20.01.2025 um 23:40 schrieb Chris M. Thomasson:

[...] Fwiw, a fun part of GLSL is doing stuff like:

vec3 a = vec3(.25, 1, .75);
vec2 b = a.xz;
vec2 c = b + vec2(.75, .25);

c now equals (1, 1)

Nice.


In math:

a := (.25, 1, .75) ,

b := (a_1, a_3) ,

c := b + (.75, .25) .

Then c = (1, 1) .

:-P


Question. What if

vec4 a = vec4(.25, 1, .75, .999);

vec3 b = a.xz<?>;

I'd like to get b == (.25, .75, .999). :-P

That would be:

vec4 a = vec4(.25, 1, .75, .999);
vec3 b = a.xzw;

;^)

C'mon... lol

And then?

vec4's in GLSL are (x, y, z, w) or

vec4 a = vec4(1, 2, 3, 4);
a[1] = 3.f;

now, a.y = 3   :^)

xyzw ... uv ...

I usually use the "series"

      x, y, z, u, v, w

in math/physics.

GLSL has x, y, z, w or [0], [1], [2], [3]

A useful convention, I guess, when doing graphics in (3D) space; for
doing it in space-time I'd prefer t instead of w here. :-P

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Am 21.01.2025 um 01:24 schrieb Moebius:

Am 21.01.2025 um 00:50 schrieb Tom Bola:

Am 21.01.2025 00:25:01 Moebius schrieb:

Am 21.01.2025 um 00:08 schrieb Tom Bola:

Du hattest den Unsinn

"By the binominal formulas we have: (a, b)^2 = a^2 + 2ab + b^2"

geschrieben. Korrekt wäre:

"By the binominal formula we have: (a + b)^2 = a^2 + 2ab + b^2"

Das wäre zwar richtig, aber hier irrelevant. Though this would be
correct, it's immaterial here. (See below.)

https://www.math.kit.edu/iana2/~mandel/seite/schnupperkurs/media/
komplex.pdf

Gutes pdf.

Da steht:

"Der Division liegt die _dritte Binomische Formel_ (a + b)(a − b) =
a^2 - b^2 zugrunde."

Ja, mir ging es ganz allein um ...

Ja, ja. Verstehst Du auch, warum es da heißt: "Der Division liegt die _dritte Binomische Formel_ (a + b)(a − b) = a^2 - b^2 zugrunde"?

Kannst Du das mal etwas genauer ausführen/zeigen?

Falls nicht. Es geht hier darum:

z1/z2 = (z1 . z2*)/(z2 . z2*) = (z1 . z2*)/((x + iy) . (x - iy)), wenn
z2 = x + iy =/= 0 ist. Und HIER kommt nun -bei (x + iy) . (x -
iy)-tatsächlich die "dritte Binomische Formel" zum Einsatz.*) Demnach ist:

(x + iy) . (x - iy) = x^2 - (iy)^2 ,
also
(x + iy) . (x - iy) = x^2 - (-1)y^2 = x^2 + y^2 .

Damit haben wir:

z1/z2 =(z1 . z2*)/(x^2 + y^2) (mit z2 = x + iy =/= 0) .

Siehe dazu auch: https://nemoweb.net/jntp?EAsBh7E4-FgqHBbEPrgfaV9LEbI@jntp/Data.Media:1

bzw. (ausführlicher): https://en.wikipedia.org/wiki/Complex_number#Complex_conjugate,_absolute_value,_argument_and_division

Aber auch hier muss man natürlich wissen, wie zwei komplexe Zahlen zu multipliziere sind. (Außerdem muss man wissen, dass (u + iv)/r = u/r +
iv/r ist für r e IR\{0}.)

_______________________________________________________________________

*) Hinweis: (a + b)(a − b) = a^2 - b^2 unterscheidet sich schon
erheblich von (a + b)^2 = a^2 + 2ab + b^2. Außerdem macht "(a, b)^2 =
a^2 + 2ab + b^2" nach wie vor keinen Sinn. Das ist schon Mückenheim-Preis-verdächtig!

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Am 21.01.2025 um 11:27 schrieb Moebius:

Außerdem macht "(a, b)^2 = a^2 + 2ab + b^2" nach wie vor keinen Sinn.

Let's write the complex number (a, b) in the usual form a + ib. Then

(a + ib)^2 = a^2 + i2ab + (ib)^2

by the binomial formula (!). Hence

(a + ib)^2 = a^2 + i2ab + (-1)b^2 = (a^2 - b^2) + i2ab .

And hence

(a, b)^2 = (a^2 - b^2) + i2ab (!)

or rather

(a, b)^2 = (a^2 - b^2, 2ab)

would be correct.

( Knapp daneben ist auch vorbei. :-P )

.
.
.

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Le 20/01/2025 à 20:29, Moebius a écrit :

Am 20.01.2025 um 20:21 schrieb Richard Hachel:

Le 20/01/2025 à 19:33, Jim Burns a écrit :

((a + ib)(c + id))⃰ =
(ac + i(ad+bc) + i²bd)⃰ =

No. You can't.

Yes, we can. :-)

You don't know what i is but

We know that i is i, i e C and that i² = -1, idiot.

If it is 1 then 1² = 1.
If it is -1 then -1² = 1

Yeah, but since i ISN'T 1 or -1, this is immaterial. Go that, idiot?

Again, i is i, i e C and i² = -1.

<facepalm>

No.

i²=-1 ----> (i)(i)=-1

This has no solution in reality, because we cannot, for example, be here
and there, or have two different ages at the same time.

We will then consider that another factor will intervene, for example
time. How many students are there in Mrs. Martin's class, 25 or 9? Both at
the same time, but not at the same moment.
We then set z=16+9i where i is both 1 and -1.
In the morning, it is 1, in the evening (catch-up class for adults), it is
-1.

Now, a mathematical error will occur (in my opinion, but I could be
wrong).
When you have to choose, you will have to choose between morning and
evening, to know the number of students, and so we will put 1 for morning,
and option -1 for evening.
SIMPLY, once the option is chosen, you can no longer say that 1=-1 in the
same equation. You start with 1, you stay with 1; you start with -1, you
stay with -1.

Now, a mathematical error will occur (in my opinion, but I could be
wrong).
When you have to choose, you will have to choose between morning and
evening, to know the number of students, and so we will put 1 for morning,
and option -1 for evening.
SIMPLY, once the option is chosen, you can no longer say that 1 = -1 in
the same equation. You start with 1, you stay with 1; you start with -1,
you stay with -1.

And you no longer have the mathematical right to say that (i)(i) = (1)(-1)
= -1.

Your two options are now only (1)² = 1 and (-1)² = 1

We then arrive at a logical equation that can eliminate i², since it is
the same thing on both sides that we will choose.

z1.z2=aa'+bb'+i(ab'+a'b)

where aa'+bb' is the real part and ab'+a'b the imaginary part.

In this regard, let's take z1=1+i and z2=2-2i

What becomes of Z=z1+z2?

What becomes of Z=z1*z2 (in usual mode and in Hachel mode)?

z1.z2=aa'-bb'+i(ab'+a'b) (classic)

Or,

z1.z2=aa'+bb'+i(ab'+a'b) (Hachel)


R.H.

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Le 20/01/2025 à 23:28, Tom Bola a écrit :

Am 20.01.2025 23:20:38 Moebius schrieb:

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

This is not a definition.

Saying i is the imaginary unit that implies that i²=-1 is not a
definition in the proper sense of the term.

It is as if we said:
- Mr. Teacher, what is 3?

And the teacher replied:
- It is the cube root of 27.

As a definition, can't we do better.

I say this, I say nothing, I am a crank, let's not forget it.

R.H.

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Le 21/01/2025 à 11:37, Moebius a écrit :

Am 21.01.2025 um 11:27 schrieb Moebius:>

Let's write the complex number (a, b) in the usual form a + ib. Then

(a + ib)^2 = a^2 + i2ab + (ib)^2

Vous allez trop vite.

(a + ib)^2 = a²+2abi+(ib)(ib)

YOU say (ib)(ib), it's the same thing as i²b² with i²=-1.

With careful reflection, I realize that if we change the definition of i,
that is to say if we define it as a unit being at the same time its unit,
but also its opposite, its inverse, the inverse of its opposite and the opposite of its inverse, all at the same time, we realize that, not
knowing i, and by attributing to it the two values
in the same equation, we have i²=(i1)(i2)=(1)(-1)=-1.

But there are two solutions which are i1 and i2.

If I take i1, I must maintain i1, and the same for i2.

It comes that if the imaginary part remains correct, i(ab'+a'b), it is no longer true of the real part, because whether we take i1 or i2, each time i²=1, and we come across a position simplification of i, and not a
negative simplification.

We can then write:
Z=aa'+bb'+i(ab'+a'b) which seems to me to be the correct solution
(verified by statistics (example of the students of Plougastel).

The same for the division:
Z=z1/z2
We obtain another mathematical appearance, and
Z=[(aa'-bb')/(a'²-b'²)] + i [(ba'-ab')/(a'²-b'²)]

by the binomial formula (!). Hence

(a + ib)^2 = a^2 + i2ab + (-1)b^2 = (a^2 - b^2) + i2ab .

No.

And hence

(a, b)^2 = (a^2 - b^2) + i2ab (!)

No.

or rather

(a, b)^2 = (a^2 - b^2, 2ab)

would be correct.

Incorrect.

z1=2+2i
z2=1-i

Z=z1*z2

You say that Z=4, and me Z=0.

Le produit de deux complexes orthogonaux étant nul chez moi, égal à 4
chez vous.

R.H.

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Am 21.01.2025 um 13:39 schrieb Tom Bola:

Am 21.01.2025 11:27:25 Moebius schrieb:

"[...] we have: (a, b)^2 = a^2 + 2ab + b^2" (Tom Bola)

No, we don't.

[...] it is meant: Z_complex = (a, b) = a + bi

Yes. Then (a, b)^2 = (a + bi)^2 = (a^2 - b^2) + i2ab

You see: Knapp daneben ist auch vorbei. :-o

So und jetzt Schluss mit Dieser Farce, Du Möchtegern-Mückenheim.

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Am 21.01.2025 um 12:19 schrieb Richard Hachel:

Le 20/01/2025 à 20:29, Moebius a écrit :

Am 20.01.2025 um 20:21 schrieb Richard Hachel:

You don't know what i is but

We know that i is i, i e C and that i² = -1, idiot.

If it is  1 then  1² = 1.
If it is -1 then -1² = 1

Yeah, but since i ISN'T 1 or -1, this is immaterial. Got that, idiot?

Again, i is i, i e C and i² = -1.

<facepalm>

No.

Yes. EOD.

*plonk*

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On 1/20/2025 2:21 PM, Richard Hachel wrote:

Le 20/01/2025 à 19:33, Jim Burns a écrit :

On 1/20/2025 6:02 AM, Richard Hachel wrote:

Division of two complex numbers.

Now let's set Z=(a+ib)/(a'+ib')

((a + ib)(c + id))⃰  =
(ac + i(ad+bc) + i²bd)⃰  =

No. You can't.

If planar (complex) multiplication has
certain reasonable.looking properties,
it is a theorem that i² has certain properties.

If (w⋅z)* = w*⋅z*
then, for the north (imaginary) unit i
i² is on the east,west (real) axis, i²=(i²)*

If, for each non.0 z,
there is an inverse w = z⁻¹, z⋅w = 1
then i²<0

Those aren't all the reasonable.looking properties
which planar multiplication should have.

If, for each planar pair v,w
there is exactly one two.dimensional z = v⋅w
then your Schrödinger's cat never exists,
not even in a Schrödinger's.cat semi.existent way.

=(ac + i(ad+bc) + (ib*id)⃰
You don't know what i is
but at this moment,
you know that it is both
i=1 and i=-1.
If it is 1 then 1²=1.
If it is -1 then -1²=1
In any case, its square will be 1
because in your two solutions,
you will have to give your choice,
but not both at the same time.

Reasoning from incomplete information is
just another day in Mathematics.

How to say i=1 or i=-1 and not say which
is i ∈ {1,-1}

Planar "multiplication" with i ∈ {1,-1}
is not the multiplication we are looking for.

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Le 21/01/2025 à 20:39, Ross Finlayson a écrit :

On 01/20/2025 03:02 AM, Richard Hachel wrote:

Division of two complex numbers.

Now let's set Z=(a+ib)/(a'+ib')
with
z1=a+ib
and
z2=a'+ib'

What becomes of Z=A+iB?

R.H.

Like I said, division of complex numbers is under-defined.

It's kind of like dividing by zero, about "roots of zero".

544 / 5 000
If you want to know if a chosen mathematics is consistent, it is necessary
that Z=z1*z2 implies that z1=Z/z2 and that z2=Z/z1

We notice that the mathematics of mathematicians is consistent.

Mine too.

Who relies on the best principles? It would seem that it is me.

For me, the problems posed remain true if we solve them differently, for example with statistics (see the problem of the Plougastel college); for
me, the product of two orthogonal complexes is zero.

Mathematicians cannot do it.

R.H.

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Am 21.01.2025 um 22:12 schrieb Chris M. Thomasson:

On 1/21/2025 1:17 AM, Moebius wrote:

GLSL has x, y, z, w or [0], [1], [2], [3]

A useful convention, I guess, when doing graphics in (3D) space; for
doing it in space-time I'd prefer t instead of w here. :-P

Just pulling your legs. :-)

;^) well, shit happens. GLSL chose the symbol w for the 4d component of
a vec4. Fwiw, in some of my work I use the 4'th dimension for plotting
some of my fields.

Thought so!

For instance, my cover of the AMS calendar used the 4'th dimension. In other words the w component was non-zero. It really
casts an interesting "mutation" to the 3d components during iterations of field lines...

Again, thought so! :-)

A vec4 should suffice for your work, I guess.

Keep on the good work, Chris!

.
.
.

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Am 21.01.2025 um 22:49 schrieb Moebius:

Am 21.01.2025 um 22:12 schrieb Chris M. Thomasson:

On 1/21/2025 1:17 AM, Moebius wrote:

GLSL has x, y, z, w or [0], [1], [2], [3]

A useful convention, I guess, when doing graphics in (3D) space; for
doing it in space-time I'd prefer t instead of w here. :-P

Just pulling your legs. :-)

;^) well, shit happens. GLSL chose the symbol w for the 4d component
of a vec4. Fwiw, in some of my work I use the 4'th dimension for
plotting some of my fields.

Thought so!

For instance, my cover of the AMS calendar used the  4'th dimension.
In other words the w component was non-zero. It really casts an
interesting "mutation" to the 3d components during iterations of field
lines...

Again, thought so! :-)

A vec4 should suffice for your work, I guess.

Keep on the good work, Chris!

It's clearly art.

.
.
.

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