Hello!

[Richard Hachel]

Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning to
read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.

Let z=a+ib

It is a number that has a real component and an imaginary component.

I wonder if the terms "certain component" and "possible component" would
not be as appropriate.

What is i?

It is an imaginary unit, such that i*i=-1.

In our universe, this seems impossible, a square can never be negative.

Except that we are in the imaginary.

Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.

Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9

I remind you that we are in the imaginary. So why not.

Let's set z=16+9i

It then comes that at the same time, z=25 and z=7.

It is a strange universe, but which can be useful for writing things in
different ways.

Explanations: We ask Mrs. Martin how many students she has in her class,
and she is very bored to answer because she does not know if
Schrödinger's cat is dead.

It has two classes, and depending on whether we imagine the morning class
or the evening class (catch-up classes for adults), the answer will not be
the same. There is no absolute answer. What is z?

We can nevertheless give z a real part, which is the average of the two
classes. a=16.

And ib then becomes the fluctuation of the average.

If we set i=1 then ib=+9; if we set i=-1 then ib=-9.

"i" would therefore be this entity, this unit, equal to both 1 and -1,
depending on how we look at it (Schrodinger's cat).

But what happens if we square i?

It is both 1 and -1?

Can we write i²=(1)*(1)=1?

No, because i would only be 1.

Can we write i²=(-1)(-1)=1?

No, because i is not only -1, it is both -1 and 1.

We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.

But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.

Z=z1+z2

Z=(a+ib)+(a'+ib')

and, Z=(a+a')+i(b+b')

All this is very simple for the moment.

But we are going to enter into a huge astonishment concerning the product
of two complexes.

How do mathematicians practice?
Z=z1*z2

so, so far it's correct:

Z=(a+ib)(a'+ib')

So, and it's still correct for Dr. Hachel (that's me):
Z=aa'+i(ab'+a'b)+(ib)(ib')

And there, for Dr. Hachel, mathematicians make a huge blunder by setting
(ib)(ib')=i²bb'=-bb'

Why?

Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.

We must therefore write, for the product of two complexes:
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)

The real part of the product being aa'+bb' and not aa'-bb'

With a remaining imaginary part where i is equal to both -1 and 1, which
gives two results each time for Z.

It seems that this is an astonishing blunder, due to the misunderstanding
of the handling of complex and imaginary numbers.

On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians become
totally false.

I wish you a good reflection on this.

Have a good day.

R.H.

[sobriquet]

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning to
read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component" would
not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things in
different ways.
Explanations: We ask Mrs. Martin how many students she has in her class,
and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the two
classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and -1,
depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by setting
(ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1, which
gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of conjugation
would no longer be a homomorphism with respect to the field of complex
numbers under multiplication.

So conj(z1*z2) would not be equal to conj(z1)*conj(z2).

https://www.desmos.com/calculator/kqzgbliix1

[Richard Hachel]

Post by sobriquet

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning to
read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component" would
not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things in
different ways.
Explanations: We ask Mrs. Martin how many students she has in her class,
and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the two
classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and -1,
depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by setting
(ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1, which
gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of conjugation
would no longer be a homomorphism with respect to the field of complex
numbers under multiplication.
So conj(z1*z2) would not be equal to conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

Thank you for your answer.

But nevertheless, I continue to certify that there is an extremely fine
mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.

Because as long as we do not know what i is worth, which can be BOTH equal
to 1 or -1 in this imaginary mathematics, we must pose i²=-1.

But once we pose i=1, it is no longer possible to say i²=-1; and in the
same way, when we pose i=-1, it is no longer possible to say 1²=-1.

It is necessary, at this instant where we have defined i (whether it is 1
or -1 but defined at this instant, it is necessary to set Z=z1*z2 such
that:
Z=(a+ib)(a'+ib')=aa'+bb'+i(ab'+a'b) to have the correct result, otherwise
the real part becomes very incorrect.

You tell me: yes, but it does not work with the conjugate.

Of course it does.

If it does not work, it is because you make a sign error, and the computer
does the same because it is not formatted on the right concept giving the
right real part.

Mathematical proof that Z(conj)=z1(conj)*z2(conj)

We set:
z1=16+9i
z2= 14+3i

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

Z=251+174i.

Let z1(conj)=16-9i and z2(conj)=14-3i

Z(conj)=aa'+bb'+i(ab'+a'b)
Z(conj)=(16)(14)+(-3)(-9)+i[(16)(-3)+(14)(-9)]
Z(conj)=251-174i

R.H.

[Jim Burns]

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
I was recently thinking,
because of a poster named Python,
about what complex numbers were,
wondering if teaching them was so important and useful,
especially in kindergarten
where children are only learning to read.
What is a complex number?
Let z=a+ib
It is a number that has
a real component and an imaginary component.
I wonder if the terms
"certain component" and "possible component"
would not be as appropriate.

The terms 'real' and 'imaginary' are hand.me.downs
from the fabled.past of mathematics.
They might not always be a perfect fit.

If I decided to update our vocabulary,
for the sake of the children, perhaps,
I would select terms suggesting two directions.
For example, 'east' and 'north'.

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
What is i?
It is an imaginary unit, such that i*i=-1.

It is a unit north.
Multiplication by i is a left turn.
Multiplication by i twice reverses direction.

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
Let's assume that i is a number, or rather a unit,
which is both its number and its opposite.

Let's not assume that a unit north is
both a unit east and a unit west.

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
Thus,
if we set z=9i
we see that z is both,
as in this story of Schrödinger's cat,
z=9 and z=-9
I remind you that we are in the imaginary.
So why not.

Because z = z but 9 ≠ -9

⎛ Schrödinger's cat asks questions other than
⎜ how far north.south.east.west.
⎜
⎜ Experiments and Bell's theorem seem to say that,
⎜ before we look,
⎜ the cat doesn't have
⎜ either a past in which it died
⎜ or a past in which it survived.
⎜
⎜ But then we look, and _when we look_
⎜ we see a cat with one of those pasts.
⎜
⎜⎛
⎜⎜ The Moving Finger writes; and, having writ,
⎜⎜ Moves on: nor all thy Piety nor Wit
⎜⎜ Shall lure it back to cancel half a Line,
⎜⎜ Nor all thy Tears wash out a Word of it.
⎜⎝
⎜ -- Omar Khayyam, trans. Edward FitzGerald
⎜
⎜ What if, for just a moment,
⎜ we can hold the door to the cat shut, and,
⎜ for just that moment, the finger pauses,
⎜ and history isn't written -- yet.
⎜
⎜ But always our hand slips, the door opens,
⎜ the finger writes and moves on.
⎜ Holding that door shut
⎝ is very challenging, technically.

Post by Richard Hachel

Post by sobriquet
If we define complex multiplication
in the way you suggest
instead of the conventional way,
that would mean that
the operation of conjugation would no longer be
a homomorphism with respect to
the field of complex numbers under multiplication.

Conventionally,
when you stand on your head (conjugation),
all this makes (conjugated) sense.

Post by Richard Hachel

Post by sobriquet
So conj(z1*z2) would not be equal to
conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

Because as long as we do not know what i is worth,
which can be BOTH equal to 1 or -1
in this imaginary mathematics, we must pose i²=-1.

For ⅈ=1 or ⅈ=-1, ⅈ²=1 or ⅈ²=1

Post by Richard Hachel
yes, but it does not work with the conjugate.

I vaguely recall, from a course in complex analysis
I took approximately When Dinosaurs Ruled The Earth,
that, by requiring the complex field axioms
to be satisfied, one can generate ⅈ²=-1 algebraically.

[Richard Hachel]

Post by Jim Burns
For ⅈ=1 or ⅈ=-1, ⅈ²=1 or ⅈ²=1

I answered this in the previous post.

I said that mathematicians confuse, when squaring i, the following two
objects.

i²=|i²|
and i²= |i|²

I repeat, when we do not know z, which can have two values
​​simultaneously (as in my example of the Plougastel high school),
we must give i an imaginary value, and i is neither 1 nor -1, but both at
the same time.

And its square becomes i²=(-1)(1)=-1

But as soon as we know it (we can photograph it when it is -1 or when it
is 1), we must no longer consider it as a duplicate.

If it is -1, we must follow the idea through to the end:
(a+ib)(a'+ib')=aa'-(ab'+a'b)+bb'
Same for i=1
(a+ib)(a'+ib')=aa'+(ab'+a'b)+bb'

We then realize that for the real part, we always have,
A=aa'+bb' whatever the value given to i.

On the other hand, and here, let's reintroduce i to make only one of the
two equations:

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

where i can resume its bipolarity without any problem, while aa'+bb' is a
correct real part for the equation.

Do you understand these things?

R.H.

[Richard Hachel]

Post by sobriquet

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning to
read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component" would
not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things in
different ways.
Explanations: We ask Mrs. Martin how many students she has in her class,
and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the two
classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and -1,
depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by setting
(ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1, which
gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of conjugation
would no longer be a homomorphism with respect to the field of complex
numbers under multiplication.
So conj(z1*z2) would not be equal to conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

Thank you for your answer.

But nevertheless, I continue to certify that there is an extremely fine
mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.

Because as long as we do not know what i is worth, which can be BOTH equal
to 1 or -1 in this imaginary mathematics, we must pose i²=-1.

But once we pose i=1, it is no longer possible to say i²=-1; and in the
same way, when we pose i=-1, it is no longer possible to say 1²=-1.

It is necessary, at this instant where we have defined i (whether it is 1
or -1 but defined at this instant, it is necessary to set Z=z1*z2 such
that:
Z=(a+ib)(a'+ib')=aa'+bb'+i(ab'+a'b) to have the correct result, otherwise
the real part becomes very incorrect.

You tell me: yes, but it does not work with the conjugate.

Of course it does.

If it does not work, it is because you make a sign error, and the computer
does the same because it is not formatted on the right concept giving the
right real part.

Mathematical proof that Z(conj)=z1(conj)*z2(conj)

We set:
z1=16+9i
z2= 14+3i

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

Z=251+174i.

Let z1(conj)=16-9i and z2(conj)=14-3i

Z(conj)=aa'+bb'+i(ab'+a'b)
Z(conj)=(16)(14)+(-3)(-9)+i[(16)(-3)+(14)(-9)]
Z(conj)=251-174i

R.H.

[Moebius]

Post by Richard Hachel
But nevertheless, I continue to certify that there is an extremely fine
mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.

Nope. i² = -1 was not invented/introduced by physicists, but by
mathematicians.

Post by Richard Hachel
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we [...]

i is neither 1 nor -1.

Hint: If it were 1 or -1 we would get i² = 1 (in bot cases).

.
.
.

[Moebius]

Post by Moebius

Post by Richard Hachel
But nevertheless, I continue to certify that there is an extremely
fine mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.

Nope. i² = -1 was not invented/introduced by physicists, but by
mathematicians.

Post by Richard Hachel
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we [...]

i is neither 1 nor -1.
Hint: If it were 1 or -1 we would get i² = 1 (in bot cases).

Of course, such a number does not exist in the real number field. But
there is such a number in the complex number field:

"In mathematics, a complex number is an element of a number system that
extends the real numbers with a specific element denoted i, called the
imaginary unit and satisfying the equation i² = -1; ..."

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

Post by Moebius
.
.
.

[Richard Hachel]

Post by Moebius

Post by Richard Hachel
But nevertheless, I continue to certify that there is an extremely fine
mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.

Nope. i² = -1 was not invented/introduced by physicists, but by
mathematicians.

Yes, I made a slip of the tongue.

:))

Post by Moebius

Post by Richard Hachel
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we [...]

i is neither 1 nor -1.

I consider (I could be wrong) that this is a case where we do not know
what i is worth, and we hesitate between two values.
In short, that for the moment, i is "imaginary", and can take two values.
It is neither 1 nor -1, but both at the same time potentially, and its
square is (1)(-1)=1.
Which means that z is also a number and that it can take two values
​​at the same time.
Example z=16+9i gives z=25 AND 7 at the same time.
See the example of the Plougastel college where we do not know which class
we are talking about because the class is occupied by both 25 students in
the morning and 7, who come to the evening class.

Post by Moebius
Hint: If it were 1 or -1 we would get i² = 1 (in bot cases).

It' that I said.

R.H.

[Moebius]

Post by Richard Hachel

Post by Moebius

Post by Richard Hachel
But nevertheless, I continue to certify that there is an extremely
fine mathematical error [...]
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we [...]

i is neither 1 nor -1.

I consider (I could be wrong) that this is a case where we do not know
what i is worth, and we hesitate between two values.

No, we don't.

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

Then
i := (0, 1) .

Hence i =/= 1 and i =/= -1.

Simple as that.

See:
https://www.quora.com/Are-complex-numbers-the-same-as-ordered-pairs-of-real-numbers-or-is-there-more-to-them-than-that-Is-there-anything-that-we-can-do-with-complex-numbers-that-its-impossible-to-do-with-real-numbers

and: https://www.math.uh.edu/~etgen/ComplexNos.pdf

.
.
.

[Richard Hachel]

Post by Moebius
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

i=1=-1

C'est complétement différent.

Post by Moebius
Simple as that.

Vous dites que i n'est ni 1, ni -1.

Je dis qu'il est les deux.

C'est complétement différent.

Et je gagne, parce que si je vous dis, mais alors, pour vous, il est
quoi? Comment définissez-vous cette "autre chose"?

Vous ne savez pas répondre de façon positive.

R.H.

[Python]

Post by Richard Hachel

Post by Moebius
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.
i=1=-1
C'est complétement différent.

Post by Moebius
Simple as that.

Vous dites que i n'est ni 1, ni -1.
Je dis qu'il est les deux.
C'est complétement différent.

So you're talk talking about Complex Numbers.

Post by Richard Hachel
Et je gagne, parce que si je vous dis, mais alors, pour vous, il est quoi?
Comment définissez-vous cette "autre chose"?
Vous ne savez pas répondre de façon positive.

Of course we can. There are even several ways to define "positively" what
i is, what C is.

[Moebius]

Post by Python

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

 No.

Yes.

Post by Python

 i = 1 = -1

Nonsens. Hint: 1 =/= -1.

Post by Python

 C'est complétement différent.

Post by Moebius
Simple as that.

 Vous dites que i n'est ni 1, ni -1.
 Je dis qu'il est les deux.
 C'est complétement différent.

So you're talk talking about Complex Numbers.

"So you're NOT talking about Complex Numbers." - right?

Post by Python

Et je gagne, parce que si je vous dis, mais alors, pour vous, il est
quoi? Comment définissez-vous cette "autre chose"?
Vous ne savez pas répondre de façon positive.

Of course we can. There are even several ways to define "positively"
what i is, what C is.

Indeed.

.
.
.

[Python]

Post by Richard Hachel

Post by Moebius
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.
i=1=-1
C'est complétement différent.

Post by Moebius
Simple as that.

Vous dites que i n'est ni 1, ni -1.
Je dis qu'il est les deux.
C'est complétement différent.

So you're NOT talking about Complex Numbers.

Post by Richard Hachel
Et je gagne, parce que si je vous dis, mais alors, pour vous, il est quoi?
Comment définissez-vous cette "autre chose"?
Vous ne savez pas répondre de façon positive.

Of course we can. There are even several ways to define "positively" what
i is, what C is.

[Moebius]

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

Yes.

i = 1 = -1

Hear, hear, hence 1 = -1 now?

Fuck off, idiot!

[Moebius]

Post by Moebius

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

Yes.

Hint: Then (in this context) i * i = (0, 1) * (0, 1) = (-1, 0), the
latter is "identified" with -1, hence we may write:

i^2 = -1 .

That's exactly what we want.

Post by Moebius

i = 1 = -1

Hear, hear, hence 1 = -1 now?
Fuck off, idiot!

[Moebius]

Post by Moebius

Post by Moebius

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

Yes.

Hint: Then (in this context) i * i = (0, 1) * (0, 1) = (-1, 0), the
                 i^2 = -1 .
That's exactly what we want.

Hint: No "extremely fine mathematical error" at all.

Post by Moebius

Post by Moebius

i = 1 = -1

Hear, hear, hence 1 = -1 now?
Fuck off, idiot!

[Python]

Post by Moebius

Post by Moebius

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

Yes.

Hint: Then (in this context) i * i = (0, 1) * (0, 1) = (-1, 0), the
                 i^2 = -1 .
That's exactly what we want.

Hint: No "extremely fine mathematical error" at all.

Richard "Hachel" Lengrand suffers of a lot of mental diseases. One of them
is pathological hubris. When he fails to understand something (and he
failed to understand complex numbers when in high schools, then he
studied... medicine) instead of thinking and studying the subject he
systematically decides that everyone is wrong and he pretend to reinvent
the subjet.

He's doing this for Special Relativity for decades and ended up with an
atrocious bunch of nonsense and contradiction.

He cannot realize that there cannot be an "error" in a definition, such as
the ones for complex numbers.

Either it is consistent, compatible with general algebraic structures and
useful or not.

Complex numbers are consistent, they form a Field extending R, allowing to
factorize all polynomials better than in R[X]. They allow to utterly
simplify a lot of geometric problems and deduce trigonometric identity.

They are not the only interesting structures on R^2 btw, dual numbers
defined by a+b*epsilon where epsilon^2 = 0 (more rigorously this is the
ring R[X]/X^2, while epsilon is the equivalence class of the polynomial X,
quite a similar approach to the algebraic construction of C as
R[X]/(X^1+1))

Dual numbers provide a very elegant introduction of calculus without
having to deal with limites (at least for polynomials).

Hachel/Lengrand's "ideas" may have had some sense, but they are 100%
unrelated to complex numbers. Moreover the idea of something having two
distinct values -1 and 1 is, of course, totally absurd and contradictory.
Given his deranged mental states Richard does not bother.

[Moebius]

[Mr. NN] suffers of a lot of mental diseases. One of them is pathological hubris. When he fails to understand something [...] instead of thinking and studying the subject he systematically decides that everyone is wrong and he pretend to [understand] the subject [better than everyone else].

Talking about Mückenheim? :-)

[Chris M. Thomasson]

Post by Moebius

[Mr. NN] suffers of a lot of mental diseases. One of them is
pathological hubris. When he fails to understand something [...]
instead of thinking and studying the subject he systematically decides
that everyone is wrong and he pretend to [understand] the subject
[better than everyone else].

Talking about Mückenheim? :-)

Oh man! Puke!

[Richard Hachel]

the idea of something having two distinct values -1 and 1 is, of course, totally
absurd and contradictory. Given his deranged mental states Richard does not
bother.

Ce n'est pas plus absurde que de dire que le chat de Schödinger est à
la fois mort et vivant.

De plus, nous sommes en univers imaginaire, c'est à dire "au-delà du
réel".

Je trouve tes griefs un peu déplacés quand ils ne sont pas carrément
mensongers ou diffamatoires.

Je rappelle le problème du collège de Plougastel?

Combien y a-t-il d'élèves dans la classe de madame Martin?

Madame Martin ne peut pas répondre, car cela dépend de l'heure de la
visite de l'inspecteur d'académie.

La réponse proposée est donc z, qui peut prendre deux valeurs, selon
qu'on imagine que l'inspecteur vient
le matin, ou à l'heure du cours de rattrapage pour adulte, le soir,
après le travail.

z1=16+9i

Idem pour Mlle Watson, élémentaire, mon cher Python, z2=14+3i

Reste la notion d'addition, qui est celle de monsieur le proviseur, qui
lui supervise les quatre classes.

Combien a-t-il d'élèves?

Z=z1+z2

Z=(a+a')+i(b+b')

Passons au produit de deux complexes (en utilisant la bonne formule).

On imagine qu'on va envoyer, si l'inspecteur d'académie se pointe, sans
qu'on sache le jour, ni l'heure (Jésus-Christ copyrights) un couple
formé d'un garçon de madame Martin et d'une fille de Melle Watson.

Z=z1*z2

Docteur Hachel, immense génie dans ce monde de crétins dit:
Z=(a+ib)*(a'+ib') et pose quelque chose d'hallucinant devant le monde
entier médusé:
Z=aa'+bb'+i(ab'+a'b) en affirmant qu'il faut poser ici i²=1 et non
i²=-1 car i qu'il soit identifié
comme i=1 ou i=-1 ne pourra plus avoir qu'un carré positif.

Allons plus loin.

Le produit de deux complexes étant connus, on fait l'opération inverse.

Si z1*z2=Z on doit forcément avoir Z/z1=z2 et Z/z2=z1

Comment calcule-t-on z2=Z/z1 ou z1=Z/z2 ?

Posons par exemple :

Z=A+iB

z1=a+ib

z2=Z/z1

z2= ?



R.H.

[Python]

the idea of something having two distinct values -1 and 1 is, of course, totally
absurd and contradictory. Given his deranged mental states Richard does not bother.

Ce n'est pas plus absurde que de dire que le chat de Schödinger est à la fois
mort et vivant.
De plus, nous sommes en univers imaginaire, c'est à dire "au-delà du réel".
Je trouve tes griefs un peu déplacés quand ils ne sont pas carrément
mensongers ou diffamatoires.
Je rappelle le problème du collège de Plougastel?
Combien y a-t-il d'élèves dans la classe de madame Martin?
Madame Martin ne peut pas répondre, car cela dépend de l'heure de la visite
de l'inspecteur d'académie.
La réponse proposée est donc z, qui peut prendre deux valeurs, selon qu'on
imagine que l'inspecteur vient
le matin, ou à l'heure du cours de rattrapage pour adulte, le soir, après le
travail.
z1=16+9i
Idem pour Mlle Watson, élémentaire, mon cher Python, z2=14+3i
Reste la notion d'addition, qui est celle de monsieur le proviseur, qui lui
supervise les quatre classes.
Combien a-t-il d'élèves?
Z=z1+z2
Z=(a+a')+i(b+b')
Passons au produit de deux complexes (en utilisant la bonne formule).
On imagine qu'on va envoyer, si l'inspecteur d'académie se pointe, sans qu'on
sache le jour, ni l'heure (Jésus-Christ copyrights) un couple formé d'un garçon
de madame Martin et d'une fille de Melle Watson.
Z=z1*z2
Z=(a+ib)*(a'+ib') et pose quelque chose d'hallucinant devant le monde entier
Z=aa'+bb'+i(ab'+a'b) en affirmant qu'il faut poser ici i²=1 et non i²=-1 car
i qu'il soit identifié
comme i=1 ou i=-1 ne pourra plus avoir qu'un carré positif.
Allons plus loin.
Le produit de deux complexes étant connus, on fait l'opération inverse.
Si z1*z2=Z on doit forcément avoir Z/z1=z2 et Z/z2=z1
Comment calcule-t-on z2=Z/z1 ou z1=Z/z2 ?
Z=A+iB
z1=a+ib
z2=Z/z1
z2= ?
R.H.

Abruti ! Tu es pathétiquement complètement à côté de la plaque, pour
ne pas changer. Si tu réponds en français fais le sur fr.sci.maths (et
non pas en cuistre sur sci.*) où je te répondrai volontiers en
français.

Idiot! You are pathetically off track, as usual. If you answer in French,
show some respects to people on Usenet do so on fr.sci.maths not here. I
could answer you there.

[Richard Hachel]

Post by Moebius

Post by Moebius
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
         i := (0, 1) .
Hence i =/= 1 and i =/= -1.

No.

Yes.

i = 1 = -1

Hear, hear, hence 1 = -1 now?
Fuck off, idiot!

Ce n'est pas ce que je dis, O saint crétin.

J'ai dit que dans le monde des nombres complexes (qui ne sont pas si
complexes que ça si l'on comprend ce que l'on est en train de faire) nous
parlions de nombres i et z, imaginaires.

i, selon l'immense Hachel, triple pris Nobel dans vos gueules quand même,
est l'unité imaginaire, qui est, à la fois, égal à -1 et 1.

z est un nombre imaginaire qui contient une partie ferme, compacte,
réelle, et une partie bicéphale imagianaire qui le rend, lui aussi,
double. Ainsi, par exemple, z vaut 25 et 7 tout à la fois si l'on donne
à z l'identité suivante : z=16+9i.

Tu comprends, O crétin?

R.H.

[sobriquet]

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning
to read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component"
would not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things
in different ways.
Explanations: We ask Mrs. Martin how many students she has in her
class, and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the
two classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and
-1, depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by
setting (ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1,
which gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of
conjugation would no longer be a homomorphism with respect to the
field of complex numbers under multiplication.
So conj(z1*z2) would not be equal to conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

Thank you for your answer.
But nevertheless, I continue to certify that there is an extremely fine
mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we must pose i²=-1.
But once we pose i=1, it is no longer possible to say i²=-1; and in the
same way, when we pose i=-1, it is no longer possible to say 1²=-1.
It is necessary, at this instant where we have defined i (whether it is
1 or -1 but defined at this instant, it is necessary to set Z=z1*z2 such
Z=(a+ib)(a'+ib')=aa'+bb'+i(ab'+a'b) to have the correct result,
otherwise the real part becomes very incorrect.
You tell me: yes, but it does not work with the conjugate.
Of course it does.
If it does not work, it is because you make a sign error, and the
computer does the same because it is not formatted on the right concept
giving the right real part.
Mathematical proof that Z(conj)=z1(conj)*z2(conj)
z1=16+9i
z2= 14+3i
Z (equation correct)=aa'+bb'+i(ab'+a'b)
Z=251+174i.
Let z1(conj)=16-9i and z2(conj)=14-3i
Z(conj)=aa'+bb'+i(ab'+a'b)
Z(conj)=(16)(14)+(-3)(-9)+i[(16)(-3)+(14)(-9)]
Z(conj)=251-174i
R.H.

I see I had an error in the desmos demonstration where it specified your
alternative way of defining complex multiplication, and it seems that
you're right with respect to conjugation remaining a homomorphism under
your alternative definition.
But taking the modulus would no longer be a homomorphism under your
alternative definition, while it would be under the conventional definition.

https://www.desmos.com/calculator/kijg1kvt75

[sobriquet]

Post by sobriquet

Post by Richard Hachel

Post by sobriquet

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about
what complex numbers were, wondering if teaching them was so
important and useful, especially in kindergarten where children are
only learning to read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component"
would not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things
in different ways.
Explanations: We ask Mrs. Martin how many students she has in her
class, and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the
two classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and
-1, depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by
setting (ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1,
which gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of
conjugation would no longer be a homomorphism with respect to the
field of complex numbers under multiplication.
So conj(z1*z2) would not be equal to conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

Thank you for your answer.
But nevertheless, I continue to certify that there is an extremely
fine mathematical error, at the moment when physicists pose
i²=-1 to quickly simplify what seems a convenient operation.
Because as long as we do not know what i is worth, which can be BOTH
equal to 1 or -1 in this imaginary mathematics, we must pose i²=-1.
But once we pose i=1, it is no longer possible to say i²=-1; and in
the same way, when we pose i=-1, it is no longer possible to say 1²=-1.
It is necessary, at this instant where we have defined i (whether it
is 1 or -1 but defined at this instant, it is necessary to set Z=z1*z2
Z=(a+ib)(a'+ib')=aa'+bb'+i(ab'+a'b) to have the correct result,
otherwise the real part becomes very incorrect.
You tell me: yes, but it does not work with the conjugate.
Of course it does.
If it does not work, it is because you make a sign error, and the
computer does the same because it is not formatted on the right
concept giving the right real part.
Mathematical proof that Z(conj)=z1(conj)*z2(conj)
z1=16+9i
z2= 14+3i
Z (equation correct)=aa'+bb'+i(ab'+a'b)
Z=251+174i.
Let z1(conj)=16-9i and z2(conj)=14-3i
Z(conj)=aa'+bb'+i(ab'+a'b)
Z(conj)=(16)(14)+(-3)(-9)+i[(16)(-3)+(14)(-9)]
Z(conj)=251-174i
R.H.

I see I had an error in the desmos demonstration where it specified your
alternative way of defining complex multiplication, and it seems that
you're right with respect to conjugation remaining a homomorphism under
your alternative definition.
But taking the modulus would no longer be a homomorphism under your
alternative definition, while it would be under the conventional definition.
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration, so you can drag the blue and yellow
point to pick any combination of two complex numbers and the resulting
product of the two numbers will be indicated in cyan (the conventional
definition) and green (your alternative definition). I think it's
helpful to pick points within the unit circle, since that will result in
a product that is also within the unit circle (or slightly outside of
the unit circle in case of the alternative way to define a product).

https://www.desmos.com/calculator/mptrtggp7x

It's a desirable property of complex numbers that they multiply in a
nice way in polar form, where the polar form of the product of two
complex numbers will have a modulus that is the product of the modulus
of the inputs and have an argument that is the sum of the arguments of
the inputs.

[Richard Hachel]

Post by sobriquet

Post by sobriquet
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration

Basée sur une équation fausse.

On tourne en rond.

R.H.

[guido wugi]

Post by Richard Hachel

Post by sobriquet

Post by sobriquet
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration

Basée sur une équation fausse.
On tourne en rond.

Not "on", but "only" you ;)

The definition of "i" is simple: it's a solution of the equation i^2=-1.

So it's clear from the start that it's not a member of the real number
line, and thus
neither a doubting case, or Schrödinger cat limbo state case, between
[+1 and -1],
neither a case where it should be alternatively +1 and -1, to have
i*i=-1 work...

It's simply some "other number" outside the real number line, one
creating its own number line.
What's there more to brood on?

[Moebius]

Post by guido wugi

Post by Richard Hachel

Post by sobriquet

Post by sobriquet
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration

Basée sur une équation fausse.
On tourne en rond.

Not "on", but "only" you ;)
The definition of "i" is simple: it's a solution of the equation i^2 = -1.

If there is such a solution. :-P

Actually, there are 2 such solutions in C. So which one is i?

Hence I'd perefer tp claim that i is a number such that i^2 = -1. (-i is
another one.)

Post by guido wugi
So it's clear from the start that it's not a member of the real number line [...]

Right.

Post by guido wugi
It's simply some "other number" outside the real number line, one
creating its own number line.

I guess you mean ix (with x e IR). Right.

Post by guido wugi
What's there more to brood on?

[Moebius]

Post by guido wugi

Post by Richard Hachel

Post by sobriquet

Post by sobriquet
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration

Basée sur une équation fausse.
On tourne en rond.

Not "on", but "only" you ;)
The definition of "i" is simple: it's a solution of the equation [x^2 = -1].

If there is such a solution. :-P

Actually, there are 2 such solutions in C. So which one is i?

Hence I'd perefer tp claim that i is a number such that i^2 = -1. (-i is
another one.)

Post by guido wugi
So it's clear from the start that it's not a member of the real number line [...]

Right.

Post by guido wugi
It's simply some "other number" outside the real number line, one
creating its own number line.

I guess you mean ix (with x e IR). Right.

Post by guido wugi
What's there more to brood on?

[Richard Hachel]

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

Nothing prevents me from thinking that i is a special unit, existing in an
imaginary world, and which is neither 1 nor -1.

If it were -1 or 1, its square would be 1.

This is also what it becomes once the equation is posed, when it is
presented under the visible face of the moon, we immediately have i²=1
whatever the observed value (-1 or 1).

As long as it is not observed, that we do not know what it is, we can only
imagine two solutions, and in the imagination, it is worth both 1 and -1.

At this precise moment of imagination, i having both values, we have
(i)(i)=(-1)(1)=-1=i².

The immense error of mathematicians is then to consider that i remains
forever unknown. This is false. We can then give TWO solutions to z (which
is an imaginary double number) z=a+ib

But if we give the first solution, with i=-1, we get
as a product (a+ib)(a'+ib')=aa+bb+i(ab'+a'b). And exactly the same thing
for i=1. Because each time, i being known, we can simplify like this.
We then just have to reintroduce the uncertainty, in the equation, which
now contains only one ito have the two solutions.

But in each solution, the real part remains the same, and it is aa+bb',
and certainly not aa'-bb".

R.H.

[sobriquet]

Post by Richard Hachel

Post by sobriquet

Post by sobriquet
https://www.desmos.com/calculator/kijg1kvt75

It's an interactive demonstration

Basée sur une équation fausse.
On tourne en rond.
R.H.

The latest desmos links should have the correct equation. If not, could
you please be specific and point out which equation in the list of
expressions on the left side is incorrect?.

One striking feature of conventional products is that the product of
complex numbers on the unit circle ends up on the unit circle and your
alternative way to define multiplication doesn't have that feature:

https://www.desmos.com/calculator/g4nxgeszqc

[Richard Hachel]

Post by sobriquet
The latest desmos links should have the correct equation. If not, could
you please be specific and point out which equation in the list of
expressions on the left side is incorrect?.
One striking feature of conventional products is that the product of
complex numbers on the unit circle ends up on the unit circle and your

I gave yesterday the formula for the addition of complex numbers as I
gave, a long time ago already, the correct formula for the general
addition of relativistic speeds.
I put it here again as an example.
Addition formula:
(a+ib)+(a'-ib')=(a+a')+i(b+b')
Multiplication formula:
(a+ib)(a'+ib')=aa'+(ib')a+(ib)a'+(ib)(ib')

Here, during the simplification, we must be careful not to make a colossal
blunder. It is the product (ib)(ib') that will produce this blunder,
because the mathematician will not take care that (ib)(ib') is NOT
|i²|bb BUT |i|²bb'.

Here Hachel (it's me) then poses:
(a+ib)(a'+ib')=aa'+i(ab'+a'b)+bb'
and no longer:
(a+ib)(a'+ib')=aa'+i(ab'+a'b)-bb' which is a false result in its real part
aa'+bb'.

R.H.

[Ross Finlayson]

Post by sobriquet

Post by Richard Hachel
Hello friends of mathematics.
I was recently thinking, because of a poster named Python, about what
complex numbers were, wondering if teaching them was so important and
useful, especially in kindergarten where children are only learning to
read.
What is a complex number? Many have difficulty answering, especially
girls, whose minds are often more practical than abstract.
Let z=a+ib
It is a number that has a real component and an imaginary component.
I wonder if the terms "certain component" and "possible component"
would not be as appropriate.
What is i?
It is an imaginary unit, such that i*i=-1.
In our universe, this seems impossible, a square can never be negative.
Except that we are in the imaginary.
Let's assume that i is a number, or rather a unit, which is both its
number and its opposite.
Thus, if we set z=9i we see that z is both, as in this story of
Schrödinger's cat, z=9 and z=-9
I remind you that we are in the imaginary. So why not.
Let's set z=16+9i
It then comes that at the same time, z=25 and z=7.
It is a strange universe, but which can be useful for writing things
in different ways.
Explanations: We ask Mrs. Martin how many students she has in her
class, and she is very bored to answer because she does not know if
Schrödinger's cat is dead.
It has two classes, and depending on whether we imagine the morning
class or the evening class (catch-up classes for adults), the answer
will not be the same. There is no absolute answer. What is z?
We can nevertheless give z a real part, which is the average of the
two classes. a=16.
And ib then becomes the fluctuation of the average.
If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
"i" would therefore be this entity, this unit, equal to both 1 and -1,
depending on how we look at it (Schrodinger's cat).
But what happens if we square i?
It is both 1 and -1?
Can we write i²=(1)*(1)=1?
No, because i would only be 1.
Can we write i²=(-1)(-1)=1?
No, because i is not only -1, it is both -1 and 1.
We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
But here, we will notice something extraordinary, the additions and
products of complex numbers can be determined.
Z=z1+z2
Z=(a+ib)+(a'+ib')
and, Z=(a+a')+i(b+b')
All this is very simple for the moment.
But we are going to enter into a huge astonishment concerning the
product of two complexes.
How do mathematicians practice?
Z=z1*z2
Z=(a+ib)(a'+ib')
Z=aa'+i(ab'+a'b)+(ib)(ib')
And there, for Dr. Hachel, mathematicians make a huge blunder by
setting (ib)(ib')=i²bb'=-bb'
Why?
Because at this point in the calculation, we impose that i will
indefinitely remain
both positive and negative, and the correct formula
Z=aa'+i(ab'+a'b)+(ib)(ib') will become incorrect written in the form
Z=aa'+i(ab'+a'b)+(i²bb') and a sign error will appear.
Z=aa'+bb'+i(ab'+a'b) and not aa'-bb'+i(ab'+a'b)
The real part of the product being aa'+bb' and not aa'-bb'
With a remaining imaginary part where i is equal to both -1 and 1,
which gives two results each time for Z.
It seems that this is an astonishing blunder, due to the
misunderstanding of the handling of complex and imaginary numbers.
On the other hand, by going through statistics, statistics confirms
HAchel's ideas, and the results usually proposed by mathematicians
become totally false.
I wish you a good reflection on this.
Have a good day.
R.H.

If we define complex multiplication in the way you suggest instead of
the conventional way, that would mean that the operation of conjugation
would no longer be a homomorphism with respect to the field of complex
numbers under multiplication.
So conj(z1*z2) would not be equal to conj(z1)*conj(z2).
https://www.desmos.com/calculator/kqzgbliix1

The definition of division of complex numbers is contrived.

There's a way to make objects like complex-complex numbers
or left-complex and right-complex numbers,
like a + bi and ai + b, with regards to that
the definition of division of complex numbers
is contrived.

There's a consideration of this in my podcasts
under the "Descriptive Differential Dynamics"
about an inner "original analysis" and not so much
about the De Moivre-Euler-Gauss complex analysis.

[Jim Burns]

Post by Richard Hachel
I was recently thinking,
because of a poster named Python,
about what complex numbers were,
wondering if teaching them was so important and useful,
especially in kindergarten
where children are only learning to read.
What is a complex number?
What is i?
It is an imaginary unit, such that i*i=-1.

⟨ℝ,+,⋅⟩ is a field, which means:
'+','⋅': ℝ×ℝ → ℝ
are associative and commutative, and
have identities 0,1 and inverses -x,x⁻¹,
except there's no 0⁻¹,
and '⋅' distributes over '+'.

⟨ℝ²,⨢,∘⟩ is a field, which means:
'⨢','∘': ℝ²×ℝ² → ℝ²
are associative and commutative, and
have identities 𝟎,𝒆₁ = [1 0] and inverses -𝐱,𝐱⁻¹,
except there's no 𝟎⁻¹,
and '∘' distributes over '⨢'.

⟨ℝ²,⨢,∘⟩ is defined to be
an extension of ⟨ℝ,+,⋅⟩, which means:
'⨢' and '∘' agree with '+' and '⋅' on ℝ×{0}
and, on the whole ℝ², is a field.

'∘' is bilinear, and
𝒆₁ is left.unit and right.unit.
⎛ (c⋅𝐫)∘𝐬 = c⋅(𝐫∘𝐬)
⎜ (𝐫⨢𝐭)∘𝐬 = 𝐫∘𝐬 ⨢ 𝐭∘𝐬
⎜ 𝐫∘(c⋅𝐬) = c⋅(𝐫∘𝐬)
⎜ 𝐫∘(𝐬⨢𝐭) = 𝐫∘𝐬 ⨢ 𝐫∘𝐭
⎜ 𝒆₁∘𝐬 = 𝐬
⎝ 𝐫∘𝒆₁ = 𝐫

That's enough information to determine a lot of,
but not all of the definition of '∘'.

Let 𝒆₂ = [0 1]
and 𝒆₂∘𝒆₂ = [-μ₁ -μ₂].
(a𝒆₁⨢b𝒆₂)∘(c𝒆₁⨢d𝒆₂) = (ac-μ₁bd)𝒆₁⨢(ad+bc-μ₂bd)𝒆₂

μ₁ > μ₂²/4
iff
a point 𝒊 exists such that 𝒊∘𝒊 = -𝒆₁
𝒊 = ±[μ₂/2 1]/(μ₁-μ₂²/4)¹ᐟ²
and
(a𝒆₁⨢b𝒊)∘(c𝒆₁⨢d𝒊) = (ac-bd)𝒆₁⨢(ad+bc)𝒊
and
the constants μ₁,μ₂ disappear from view,
disappear into 𝒊, in effect,
and
for each 𝐫 ≠ 𝟎, 𝐫⁻¹ exists, 𝐫∘𝐫⁻¹ = 𝒆₁
and
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.

There are different values possible for 𝒆₂∘𝒆₂ = [-μ₁ -μ₂],
but, as long as μ₁ > μ₂²/4,
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.

For two plane.multiplications ∘′ and ∘″
we can map 𝒊′ ⟷ 𝒊″
and then
(a𝒆₁⨢b𝒊′)∘′(c𝒆₁⨢d𝒊′) = (ac-bd)𝒆₁⨢(ad+bc)𝒊′
(a𝒆₁⨢b𝒊″)∘″(c𝒆₁⨢d𝒊″) = (ac-bd)𝒆₁⨢(ad+bc)𝒊″

And the two ℝ.extending plane.multiplications
are isomorphic.
Therefore, there is
only one extension of ℝ to ℝ², up to isomorphism,
and, for that extension, 𝒊² = -1

⎛ What I got wrong initially was that 𝒊 ≠ 𝒆₂,
⎜ at least, not necessarily equal.
⎜ That it's not doesn't matter, though.
⎜ All the different '∘' with their different 𝒊
⎝ map to each other very neatly.

[Richard Hachel]

Post by Jim Burns

Post by Richard Hachel
I was recently thinking,
because of a poster named Python,
about what complex numbers were,
wondering if teaching them was so important and useful,
especially in kindergarten
where children are only learning to read.
What is a complex number?
What is i?
It is an imaginary unit, such that i*i=-1.

'+','⋅': ℝ×ℝ → ℝ
are associative and commutative, and
have identities 0,1 and inverses -x,x⁻¹,
except there's no 0⁻¹,
and '⋅' distributes over '+'.
'⨢','∘': ℝ²×ℝ² → ℝ²
are associative and commutative, and
have identities 𝟎,𝒆₁ = [1 0] and inverses -𝐱,𝐱⁻¹,
except there's no 𝟎⁻¹,
and '∘' distributes over '⨢'.
⟨ℝ²,⨢,∘⟩ is defined to be
'⨢' and '∘' agree with '+' and '⋅' on ℝ×{0}
and, on the whole ℝ², is a field.
'∘' is bilinear, and
𝒆₁ is left.unit and right.unit.
⎛ (c⋅𝐫)∘𝐬 = c⋅(𝐫∘𝐬)
⎜ (𝐫⨢𝐭)∘𝐬 = 𝐫∘𝐬 ⨢ 𝐭∘𝐬
⎜ 𝐫∘(c⋅𝐬) = c⋅(𝐫∘𝐬)
⎜ 𝐫∘(𝐬⨢𝐭) = 𝐫∘𝐬 ⨢ 𝐫∘𝐭
⎜ 𝒆₁∘𝐬 = 𝐬
⎝ 𝐫∘𝒆₁ = 𝐫
That's enough information to determine a lot of,
but not all of the definition of '∘'.
Let 𝒆₂ = [0 1]
and 𝒆₂∘𝒆₂ = [-μ₁ -μ₂].
(a𝒆₁⨢b𝒆₂)∘(c𝒆₁⨢d𝒆₂) =
(ac-μ₁bd)𝒆₁⨢(ad+bc-μ₂bd)𝒆₂
μ₁ > μ₂²/4
iff
a point 𝒊 exists such that 𝒊∘𝒊 = -𝒆₁
𝒊 = ±[μ₂/2 1]/(μ₁-μ₂²/4)¹ᐟ²
and
(a𝒆₁⨢b𝒊)∘(c𝒆₁⨢d𝒊) = (ac-bd)𝒆₁⨢(ad+bc)𝒊
and
the constants μ₁,μ₂ disappear from view,
disappear into 𝒊, in effect,
and
for each 𝐫 ≠ 𝟎, 𝐫⁻¹ exists, 𝐫∘𝐫⁻¹ = 𝒆₁
and
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
There are different values possible for 𝒆₂∘𝒆₂ = [-μ₁ -μ₂],
but, as long as μ₁ > μ₂²/4,
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
For two plane.multiplications ∘′ and ∘″
we can map 𝒊′ ⟷ 𝒊″
and then
(a𝒆₁⨢b𝒊′)∘′(c𝒆₁⨢d𝒊′) =
(ac-bd)𝒆₁⨢(ad+bc)𝒊′
(a𝒆₁⨢b𝒊″)∘″(c𝒆₁⨢d𝒊″) =
(ac-bd)𝒆₁⨢(ad+bc)𝒊″
And the two ℝ.extending plane.multiplications
are isomorphic.
Therefore, there is
only one extension of ℝ to ℝ², up to isomorphism,
and, for that extension, 𝒊² = -1
⎛ What I got wrong initially was that 𝒊 ≠ 𝒆₂,
⎜ at least, not necessarily equal.
⎜ That it's not doesn't matter, though.
⎜ All the different '∘' with their different 𝒊
⎝ map to each other very neatly.

Thank you for all these details.

R.H.

[guido wugi]

Post by Jim Burns

Post by Richard Hachel
I was recently thinking,
because of a poster named Python,
about what complex numbers were,
wondering if teaching them was so important and useful,
especially in kindergarten
where children are only learning to read.
What is a complex number?
What is i?
It is an imaginary unit, such that i*i=-1.

'+','⋅': ℝ×ℝ → ℝ
are associative and commutative, and
have identities 0,1 and inverses -x,x⁻¹,
except there's no 0⁻¹,
and '⋅' distributes over '+'.
'⨢','∘': ℝ²×ℝ² → ℝ²
are associative and commutative, and
have identities 𝟎,𝒆₁ = [1 0] and inverses -𝐱,𝐱⁻¹,
except there's no 𝟎⁻¹,
and '∘' distributes over '⨢'.
⟨ℝ²,⨢,∘⟩ is defined to be
'⨢' and '∘' agree with '+' and '⋅' on ℝ×{0}
and, on the whole ℝ², is a field.
'∘' is bilinear, and
𝒆₁ is left.unit and right.unit.
⎛ (c⋅𝐫)∘𝐬 = c⋅(𝐫∘𝐬)
⎜ (𝐫⨢𝐭)∘𝐬 = 𝐫∘𝐬 ⨢ 𝐭∘𝐬
⎜ 𝐫∘(c⋅𝐬) = c⋅(𝐫∘𝐬)
⎜ 𝐫∘(𝐬⨢𝐭) = 𝐫∘𝐬 ⨢ 𝐫∘𝐭
⎜ 𝒆₁∘𝐬 = 𝐬
⎝ 𝐫∘𝒆₁ = 𝐫
That's enough information to determine a lot of,
but not all of the definition of '∘'.
Let 𝒆₂ = [0 1]
and 𝒆₂∘𝒆₂ = [-μ₁ -μ₂].
(a𝒆₁⨢b𝒆₂)∘(c𝒆₁⨢d𝒆₂) = (ac-μ₁bd)𝒆₁⨢(ad+bc-μ₂bd)𝒆₂
μ₁ > μ₂²/4
 iff
a point 𝒊 exists such that 𝒊∘𝒊 = -𝒆₁
𝒊 = ±[μ₂/2 1]/(μ₁-μ₂²/4)¹ᐟ²
and
(a𝒆₁⨢b𝒊)∘(c𝒆₁⨢d𝒊) = (ac-bd)𝒆₁⨢(ad+bc)𝒊
and
the constants μ₁,μ₂ disappear from view,
disappear into 𝒊, in effect,
and
for each 𝐫 ≠ 𝟎, 𝐫⁻¹ exists, 𝐫∘𝐫⁻¹ = 𝒆₁
and
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
There are different values possible for 𝒆₂∘𝒆₂ = [-μ₁ -μ₂],
but, as long as μ₁ > μ₂²/4,
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
For two plane.multiplications ∘′ and ∘″
we can map 𝒊′ ⟷ 𝒊″
and then
(a𝒆₁⨢b𝒊′)∘′(c𝒆₁⨢d𝒊′) = (ac-bd)𝒆₁⨢(ad+bc)𝒊′
(a𝒆₁⨢b𝒊″)∘″(c𝒆₁⨢d𝒊″) = (ac-bd)𝒆₁⨢(ad+bc)𝒊″
And the two ℝ.extending plane.multiplications
are isomorphic.
Therefore, there is
only one extension of ℝ to ℝ², up to isomorphism,
and, for that extension, 𝒊² = -1
⎛ What I got wrong initially was that 𝒊 ≠ 𝒆₂,
⎜ at least, not necessarily equal.
⎜ That it's not doesn't matter, though.
⎜ All the different '∘' with their different 𝒊
⎝ map to each other very neatly.

Not sure I "got" it all. I once did, I guess, a similar thinkthing about
not necessarily fields, but multidimensional numbers alright, as
n-vectors and as "autovariant" nxn matrix families:
https://www.wugi.be/hypereal.htm

[Jim Burns]

Post by guido wugi

Post by Jim Burns
There are different values possible for 𝒆₂∘𝒆₂ = [-μ₁ -μ₂],
but, as long as μ₁ > μ₂²/4,
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
For two plane.multiplications ∘′ and ∘″
we can map 𝒊′ ⟷ 𝒊″

𝒊′ = ±[μ′₂/2 1]/(μ′₁-μ′₂²/4)¹ᐟ²

𝒊″ = ±[μ″₂/2 1]/(μ″₁-μ″₂²/4)¹ᐟ²

Post by guido wugi

Post by Jim Burns
and then
(a𝒆₁⨢b𝒊′)∘′(c𝒆₁⨢d𝒊′) = (ac-bd)𝒆₁⨢(ad+bc)𝒊′
(a𝒆₁⨢b𝒊″)∘″(c𝒆₁⨢d𝒊″) = (ac-bd)𝒆₁⨢(ad+bc)𝒊″
And the two ℝ.extending plane.multiplications
are isomorphic.
Therefore, there is
only one extension of ℝ to ℝ², up to isomorphism,
and, for that extension, 𝒊² = -1
⎛ What I got wrong initially was that 𝒊 ≠ 𝒆₂,
⎜ at least, not necessarily equal.
⎜ That it's not doesn't matter, though.
⎜ All the different '∘' with their different 𝒊
⎝ map to each other very neatly.

Not sure I "got" it all.

My explanations evolve.
To a higher, purer state, I hope,
but to a different state, at least.

Richard Hachel's question "What is i?"
is a good one.
Others will ask it, others have asked it.
I think I might have, a million or so years ago.

When I spot another excuse to try,
I'll likely try explaining again.
It'll likely be different. Again.
Maybe even better.

Thank you for your attention.

Post by guido wugi
I once did, I guess, a similar thinkthing about
not necessarily fields, but
multidimensional numbers alright,
as n-vectors and
https://www.wugi.be/hypereal.htm

That deserves more than a glance.

I'm a great fan of vector spaces.
So many theorems, so broadly applicable.
Fourier transforms are rotations in function space!

You:
/ I called these the "Autovariance conditions",
| assuring that the matrix family embraces
| any product of its members.
| Without these conditions,
| a random product would “leave”
| the n-dim matrix family into
\ the n x n matrix space!

Perhaps you are talking about
an n.dimensional subspace closed under
the usual matrix multiplication?

Yes,
it reminds me of how I describe complex numbers.
For me, "mediating" 2x2 matrices are inserted
to define the not.the.usual product.
[a b]∘[c d] :=
[ [a b]ᵀ𝑴₁[c d] [a b]ᵀ𝑴₂[c d] ]

[ 𝑴₁ 𝑴₂ ] has eight degrees of freedom, enough,
we find, to impose field conditions on '∘'

I wonder what we can get if
something like that is done with nxn matrices.

[guido wugi]

Post by Jim Burns

Post by guido wugi

Post by Jim Burns
There are different values possible for 𝒆₂∘𝒆₂ = [-μ₁ -μ₂],
but, as long as μ₁ > μ₂²/4,
⟨ℝ²,⨢,∘⟩ is a field extending ⟨ℝ,+,⋅⟩.
For two plane.multiplications ∘′ and ∘″
we can map 𝒊′ ⟷ 𝒊″

𝒊′ = ±[μ′₂/2 1]/(μ′₁-μ′₂²/4)¹ᐟ²
𝒊″ = ±[μ″₂/2 1]/(μ″₁-μ″₂²/4)¹ᐟ²

Post by guido wugi

Post by Jim Burns
and then
(a𝒆₁⨢b𝒊′)∘′(c𝒆₁⨢d𝒊′) = (ac-bd)𝒆₁⨢(ad+bc)𝒊′
(a𝒆₁⨢b𝒊″)∘″(c𝒆₁⨢d𝒊″) = (ac-bd)𝒆₁⨢(ad+bc)𝒊″
And the two ℝ.extending plane.multiplications
are isomorphic.
Therefore, there is
only one extension of ℝ to ℝ², up to isomorphism,
and, for that extension, 𝒊² = -1
⎛ What I got wrong initially was that 𝒊 ≠ 𝒆₂,
⎜ at least, not necessarily equal.
⎜ That it's not doesn't matter, though.
⎜ All the different '∘' with their different 𝒊
⎝ map to each other very neatly.

Not sure I "got" it all.

My explanations evolve.
To a higher, purer state, I hope,
but to a different state, at least.
Richard Hachel's question "What is i?"
is a good one.
Others will ask it, others have asked it.
I think I might have, a million or so years ago.
When I spot another excuse to try,
I'll likely try explaining again.
It'll likely be different. Again.
Maybe even better.
Thank you for your attention.

Post by guido wugi
I once did, I guess, a similar thinkthing about not necessarily
fields, but
multidimensional numbers alright,
as n-vectors and
https://www.wugi.be/hypereal.htm

That deserves more than a glance.
I'm a great fan of vector spaces.
So many theorems, so broadly applicable.
Fourier transforms are rotations in function space!
/ I called these the "Autovariance conditions",
| assuring that the matrix family embraces
| any product of its members.
| Without these conditions,
| a random product would “leave”
| the n-dim matrix family into
\ the n x n matrix space!
Perhaps you are talking about
an n.dimensional subspace closed under
the usual matrix multiplication?

Yes, that would be the proper description I think.

Post by Jim Burns
Yes,
it reminds me of how I describe complex numbers.
For me, "mediating" 2x2 matrices are inserted
to define the not.the.usual product.
[a b]∘[c d] :=
[ [a b]ᵀ𝑴₁[c d] [a b]ᵀ𝑴₂[c d] ]
[ 𝑴₁ 𝑴₂ ] has eight degrees of freedom, enough,
we find, to impose field conditions on '∘'
I wonder what we can get if
something like that is done with nxn matrices.

There you've lost me :)
Still, my interest now (and since then:) is really the "true" geometry
of complex functions w=f(z).
Here's my recent little "paper" on that topic, with all relevant links
in the references:
https://www.wugi.be/mijndocs/compl-func-visu.compleet.pdf

Greetingz,