Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality.

If we find as a root x'=2 and x"=4 we can include these two roots in a
single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say (+ib)
for one and (-ib) for the other, everything will go very well.

Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the same
sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

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Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality.

If we find as a root x'=2 and x"=4 we can include these two roots in a single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say (+ib) for one and (-ib) for the other, everything will go very well.

Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the same sign
in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has
strictly NO content.

BTW, the result is correct in C, by the way C is defined. And it makes a
lot of sense. And it can be *shown*.

Another formula could make sense too, in R(j) or R(epsilon) for instance.

But switching from a formula to another one depending of the sign of the imaginary part will lead to an inconsistent structure (i.e. il will not be
an associative algebra on R, so it will be useless). BTW there are only
three of them, see: https://www.youtube.com/watch?v=r5mccK8mNw8

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Le 03/03/2025 à 22:47, Python a écrit :

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Z=(aa')-(bb')+i(ab'+a'b)

Question: Why does this formula become incorrect for complexes of the same sign
in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has strictly NO
content.

Well.

Z=(3+i)(4+2i)

Z=12+4i+6i+2i²
Z=12+10i+2i²=10+10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(6+4)
Z= 10+10i ? ? ? ?

Cela n'est plus cohérent avec la logique mathématique alors que ça le restait avec les produits à signe inversés.

Z=(3-i)(4-2i)
Z=12-6i-4i+2i²
Z=10-10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(-6-4)
Z= 10-10i ? ? ? ?

Là encore, les résultats ne sont pas cohérents avec la logique des
produits mathématiques.

La formule correcte est ici (si les signes sont de même nature) :
Z=(aa')+(bb')+i(ab'+a'b)

Il y a donc un problème sur le calcul de la partie réelle.

Maintenant, une question se pose : pourquoi dans ces deux cas là,
i²bb'=1bb' ce qui n'est pas logique.

D'où vient le problème de signe?

Si par exemple, tu places tes deux nombres complexes sur un repère quadrillé, tu vas t'apercevoir que la
formule que j'ai donnée ne marche plus pour les complexes de signe
opposé, mais marche pour les complexes
de même signe (négatifs ou positifs).

Et inversement pour la formule des mathématiciens qui marche pour les complexes de signes opposés, mais pas pour ceux de même signe.

Qu'est ce qu'il se passe?

R.H.

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Le 03/03/2025 à 23:17, Richard Hachel a écrit :

Le 03/03/2025 à 22:47, Python a écrit :

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Z=(aa')-(bb')+i(ab'+a'b)

Question: Why does this formula become incorrect for complexes of the same sign
in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has strictly NO
content.

Well.

Z=(3+i)(4+2i)

Z=12+4i+6i+2i²
Z=12+10i+2i²=10+10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(6+4)
Z= 10+10i ? ? ? ?

Cela n'est plus cohérent avec la logique mathématique

En quoi ? Why ?

Z=(3-i)(4-2i)
Z=12-6i-4i+2i²
Z=10-10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(-6-4)
Z= 10-10i ? ? ? ?

Là encore, les résultats ne sont pas cohérents avec la logique des produits
mathématiques.

En quoi ? Why ?

Qu'est ce qu'il se passe?

You are confronted with the inconsistance of your "system". Complex, and
duals, and perplex numbers are going perfectly fine.

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XPost: fr.sci.maths

Le 03/03/2025 à 23:24, Python a écrit :

Le 03/03/2025 à 23:17, Richard Hachel a écrit :

Le 03/03/2025 à 22:47, Python a écrit :

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Z=(aa')-(bb')+i(ab'+a'b)

Question: Why does this formula become incorrect for complexes of the same sign
in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has strictly NO
content.

Well.

Z=(3+i)(4+2i)

Z=12+4i+6i+2i²
Z=12+10i+2i²=10+10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(6+4)
Z= 10+10i ? ? ? ?

Cela n'est plus cohérent avec la logique mathématique

En quoi ? Why ?

Z=(3-i)(4-2i)
Z=12-6i-4i+2i²
Z=10-10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(-6-4)
Z= 10-10i ? ? ? ?

Là encore, les résultats ne sont pas cohérents avec la logique des produits
mathématiques.

En quoi ? Why ?

Qu'est ce qu'il se passe?

You are confronted with the inconsistance of your "system". Complex, and duals,
and perplex numbers are going perfectly fine.

Il faut garder l'axe des ordonnées tel qu'il est.

C'est mathématiquement très simple, et facile à comprendre.

Tu prends un repère cartésien, Oxy, et tu définis tes ordonnées y sur
l'axe ascendant, et tu ne touches plus à RIEN.

Ton axe i'Oi tu le confonds avec ton axe x'Ox mais en le plaçant en sens inverse.

Ainsi, c'est toujours très simple, ton point A(4,0) est le même que ton point A(-4i,0), ton point B(-2,0) est le même que ton point B(2i,0) mais écrit différemment.

Tout point Z=a+ib se trouve forcément sur x'Ox. Par exemple le point C(4+9i,O) se trouve tout simplement en C(-5,0) dans le repère cartésien.

Niveau classe quatrième pour des enfants éveillés.

On fait les additions de complexes très facilement sur x'Ox selon la
méthode habituelle.

Maintenant... Maintenant, plus difficile, nous allons multiplier les
nombres complexes.

Il est évident que nous ne pouvons pas le faire sur un simple plan cartésien, car y, c'est y, et x est déjà alloué pour les i.

Il faut alors utiliser un nouvel axe (axe z'Oz) en profondeur, sur
lequel, comme pour x'Ox, je vais placer mes complexes, et de la même
façon inversée.

Je vais alors multiplier mes complexes sur ce nouveau plan, et je vais
obtenir des "surfaces".

On remarque alors que Z=(aa')-(bb')+i(ab'+a'b), ça marche très bien
pour les complexes de signe opposés,
mais que pour les complexes de même signe, c'est Z=(aa')+(bb')-i(ab'+a'b)
qui marche.

Je pense qu'il y a un problème avec les signes du côté des
mathématiciens (qui ne considèrent pas que i
est l'antithèse de 1, et que son axe se trouve en contre-sens), et pas perpendiculairement du moins dans les problème de géométrie analytique.


Par contre, les retraits ou les ajouts de i se font effectivement
toujours perpendiculairement à y.

Et c'est peut-être là qu'intervient le repère d'Argand-Gauss
traditionnel.

R.H.

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XPost: fr.sci.maths

Le 03/03/2025 à 23:54, Richard Hachel a écrit :

Le 03/03/2025 à 23:24, Python a écrit :

Le 03/03/2025 à 23:17, Richard Hachel a écrit :

Le 03/03/2025 à 22:47, Python a écrit :

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Z=(aa')-(bb')+i(ab'+a'b)

Question: Why does this formula become incorrect for complexes of the same sign
in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has strictly NO
content.

Well.

Z=(3+i)(4+2i)

Z=12+4i+6i+2i²
Z=12+10i+2i²=10+10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(6+4)
Z= 10+10i ? ? ? ?

Cela n'est plus cohérent avec la logique mathématique

En quoi ? Why ?

Z=(3-i)(4-2i)
Z=12-6i-4i+2i²
Z=10-10i ? ? ?

Z=(aa')-(bb')+i(ab'+a'b)
Z=(12)-(2)+i(-6-4)
Z= 10-10i ? ? ? ?

Là encore, les résultats ne sont pas cohérents avec la logique des produits
mathématiques.

En quoi ? Why ?

Tu n'as pas répondu à ma question : en quoi « les résultats ne sont
pas cohérents avec la logique des produits mathématique ».

[snip délire]

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XPost: fr.sci.maths

Le 03/03/2025 à 23:52, "Chris M. Thomasson" a écrit :

On 3/3/2025 2:24 PM, Python wrote:

Le 03/03/2025 à 23:17, Richard Hachel a écrit :

Le 03/03/2025 à 22:47, Python a écrit :

Le 03/03/2025 à 22:37, Richard Hachel a écrit :

Z=(aa')-(bb')+i(ab'+a'b)

Question: Why does this formula become incorrect for complexes of
the same sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

You are "saying". Only saying. No argument for your claim. So it has
strictly NO content.

 Well.

 Z=(3+i)(4+2i)

 Z=12+4i+6i+2i²
 Z=12+10i+2i²=10+10i  ? ? ?

 Z=(aa')-(bb')+i(ab'+a'b)
 Z=(12)-(2)+i(6+4)
 Z= 10+10i           ? ? ? ?
 Cela n'est plus cohérent avec la logique mathématique

En quoi ? Why ?

 Z=(3-i)(4-2i)
 Z=12-6i-4i+2i²
 Z=10-10i        ? ? ?

 Z=(aa')-(bb')+i(ab'+a'b)
 Z=(12)-(2)+i(-6-4)
 Z= 10-10i           ? ? ? ?
 Là encore, les résultats ne sont pas cohérents avec la logique des
produits mathématiques.

En quoi ? Why ?

 Qu'est ce qu'il se passe?

You are confronted with the inconsistance of your "system". Complex, and
duals, and perplex numbers are going perfectly fine.

One of my friends, David Makin, is/was working on so-called multiplex,
or multex numbers. It's been a while since I talked to him. Btw, he programmed the game Crystal Dragons way back on the Amiga:

https://youtu.be/zxfirQwl-Jw

I need to try to find his old writings on it.

I spent hours on Amiga.
I even made a lot of plug-ins on some games... (X-It, Pacman Deluxe, Dune II).... I still have some...

R.H.

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Le 03/03/2025 à 23:38, "Chris M. Thomasson" a écrit :

On 3/3/2025 1:37 PM, Richard Hachel wrote:

Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality.

If we find as a root x'=2 and x"=4 we can include these two roots in a
single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say
(+ib) for one and (-ib) for the other, everything will go very well.

Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the
same sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

Where would you plot say, 1+.5i on the plane? I would say at 2-ary point
(1, .5), right where x = 1 and y = .5. Say draw a little filled circle
at said coordinates in the 2-ary plane where:

(+y)
^
|
|
|
(-x)<---0--->(+x)
|
|
|
v
(-y)

<http://nemoweb.net/jntp?myUYryqxSTftKUa3owdcVRxGpRA@jntp/Data.Media:1>

R.H.

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Le 04/03/2025 à 00:16, "Chris M. Thomasson" a écrit :

On 3/3/2025 3:10 PM, Richard Hachel wrote:

Le 03/03/2025 à 23:38, "Chris M. Thomasson" a écrit :

On 3/3/2025 1:37 PM, Richard Hachel wrote:

Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality.

If we find as a root x'=2 and x"=4 we can include these two roots in
a single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say
(+ib) for one and (-ib) for the other, everything will go very well.

Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the
same sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

Where would you plot say, 1+.5i on the plane? I would say at 2-ary
point (1, .5), right where x = 1 and y = .5. Say draw a little filled
circle at said coordinates in the 2-ary plane where:

       (+y)
         ^
         |
         |
         |
(-x)<---0--->(+x)
         |
         |
         |
         v
       (-y)

<http://nemoweb.net/jntp?myUYryqxSTftKUa3owdcVRxGpRA@jntp/Data.Media:1>

R.H.

So, you try to embed it all in the real line? So, why even have your y
axis showing in your graphic? Show me a single point, color it yellow,
on your real line that shows the plotted point 1+.5i.

Don't try to make sense of Hachel's asinine products. There is none.

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Le 04/03/2025 à 00:16, "Chris M. Thomasson" a écrit :

On 3/3/2025 3:10 PM, Richard Hachel wrote:

Le 03/03/2025 à 23:38, "Chris M. Thomasson" a écrit :

On 3/3/2025 1:37 PM, Richard Hachel wrote:

Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality.

If we find as a root x'=2 and x"=4 we can include these two roots in
a single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say
(+ib) for one and (-ib) for the other, everything will go very well.

Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the
same sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

Where would you plot say, 1+.5i on the plane? I would say at 2-ary
point (1, .5), right where x = 1 and y = .5. Say draw a little filled
circle at said coordinates in the 2-ary plane where:

       (+y)
         ^
         |
         |
         |
(-x)<---0--->(+x)
         |
         |
         |
         v
       (-y)

<http://nemoweb.net/jntp?myUYryqxSTftKUa3owdcVRxGpRA@jntp/Data.Media:1>

R.H.

So, you try to embed it all in the real line? So, why even have your y
axis showing in your graphic? Show me a single point, color it yellow,
on your real line that shows the plotted point 1+.5i.

Point 1+5i is A(-4,0).

R.H.

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Le 04/03/2025 à 00:32, Python a écrit :

Le 04/03/2025 à 00:16, "Chris M. Thomasson" a écrit :

On 3/3/2025 3:10 PM, Richard Hachel wrote:

Le 03/03/2025 à 23:38, "Chris M. Thomasson" a écrit :

On 3/3/2025 1:37 PM, Richard Hachel wrote:

Complex numbers and products of different complex signs.

What is a complex number?

It is initially an imaginary number which is a duality.

The two real roots of a quadratic curve, for example, are a duality. >>>>>
If we find as a root x'=2 and x"=4 we can include these two roots in >>>>> a single expression: Z=3(+/-)i.

Z is this dual number which will split into x'=3+i and x"=3-i.

As in Hachel i^x=-1 whatever x, we have: x'=2 and x"=4.

Be careful with the signs (i=-1). If we add i, we subtract 1.

If we subtract 5i, we add 5.

But let's go further.
A small problem arises in the products of complexes.

Certainly, if we take complexes of inverse spacings, that is to say
(+ib) for one and (-ib) for the other, everything will go very well. >>>>>
Let's set z1=3-i and z2=4+2i.

We have z1*z2=12+6i-4i-2i²=14+2i

Let's set the inverse by permuting the signs of b:

z1=3+i and z2=4-2i.

We have z1*z2=12-6i+4i-2i²=14-2i

We notice that each time, we did:
Z=(aa')-(bb)+i(ab'+a'b)
and that it works.

Question: Why does this formula become incorrect for complexes of the >>>>> same sign in b?

Example Z=(3+i)(4+2i) or Z=(3-i)(4-2i)

The formula given by mathematicians is incorrect.
I am not saying that it does not give a result.
I am saying that it is incorrect.

Where would you plot say, 1+.5i on the plane? I would say at 2-ary
point (1, .5), right where x = 1 and y = .5. Say draw a little filled
circle at said coordinates in the 2-ary plane where:

       (+y)
         ^
         |
         |
         |
(-x)<---0--->(+x)
         |
         |
         |
         v
       (-y)

<http://nemoweb.net/jntp?myUYryqxSTftKUa3owdcVRxGpRA@jntp/Data.Media:1>

R.H.

So, you try to embed it all in the real line? So, why even have your y
axis showing in your graphic? Show me a single point, color it yellow,
on your real line that shows the plotted point 1+.5i.

Don't try to make sense of Hachel's asinine products. There is none.

Merci de bien vouloir laisser les gens disposer d'eux mêmes.

R.H.

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Le 06/03/2025 à 21:52, "Chris M. Thomasson" a écrit :

On 3/3/2025 4:23 PM, Chris M. Thomasson wrote:

On 3/3/2025 3:58 PM, Richard Hachel wrote:

Le 04/03/2025 à 00:16, "Chris M. Thomasson" a écrit :

[...]

Where would you plot say, 1+.5i on the plane? I would say at 2-ary >>>>>> point (1, .5), right where x = 1 and y = .5. Say draw a little
filled circle at said coordinates in the 2-ary plane where:

       (+y)
         ^
         |
         |
         |
(-x)<---0--->(+x)
         |
         |
         |
         v
       (-y)

<http://nemoweb.net/jntp?myUYryqxSTftKUa3owdcVRxGpRA@jntp/Data.Media:1> >>>>>
R.H.

So, you try to embed it all in the real line? So, why even have your
y axis showing in your graphic? Show me a single point, color it
yellow, on your real line that shows the plotted point 1+.5i.

Point 1+5i is A(-4,0).
R.H.

I said 1+.5i. Where is that?

Any answer?

I don't understand your answer.
What is 1+.5i. ?

R.H.

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Le 06/03/2025 à 23:31, "Chris M. Thomasson" a écrit :

You have used graph paper and a ruler before, right?

Yes, a hundred years ago...
Otherwise, I don't consider complex numbers in the same way as
mathematicians. I use them in a much, much simpler way, which is a problem
for them because they are so tangled up in their difficulties that they
don't understand anything at all when I try to teach them.
The current scientific problem is very particular, in the sense that
everything has become so complicated that we don't even understand simple things anymore.

R.H.

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