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.
Op 18/01/2025 om 11:34 schreef 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.
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
Op 18/01/2025 om 11:34 schreef 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.
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
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 [...]
Am 19.01.2025 um 08:51 schrieb 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.
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).
.
.
.
Am 19.01.2025 um 08:51 schrieb 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.
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).
Le 19/01/2025 à 02:53, sobriquet a écrit :
Op 18/01/2025 om 11:34 schreef 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.
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.
Le 19/01/2025 à 12:04, Moebius a écrit :
Am 19.01.2025 um 08:51 schrieb 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.
Op 19/01/2025 om 08:51 schreef Richard Hachel:
Le 19/01/2025 à 02:53, sobriquet a écrit :
Op 18/01/2025 om 11:34 schreef 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.
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.
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
https://www.desmos.com/calculator/kijg1kvt75
It's an interactive demonstration
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.
Simple as that.
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.
No.
i=1=-1
C'est complétement différent.
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.
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.
No.
i=1=-1
C'est complétement différent.
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.
Am 19.01.2025 um 16:30 schrieb Richard Hachel:
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
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!
Le 19/01/2025 à 16:30, Richard Hachel a écrit :
Le 19/01/2025 à 15:28, Moebius a écrit :
Complex numbers can be defined as (ordered) pairs of real numbers.
Then [usually]:
i := (0, 1) .
Hence i =/= 1 and i =/= -1.
No.
i = 1 = -1
C'est complétement différent.
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.
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.
Am 19.01.2025 um 16:39 schrieb Moebius:
Am 19.01.2025 um 16:30 schrieb Richard Hachel:
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
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.
i = 1 = -1
Hear, hear, hence 1 = -1 now?
Fuck off, idiot!
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
Complex numbers can be defined as (ordered) pairs of real numbers.
Then
i := (0, 1) .
Hence i =/= 1 and i =/= -1.
No.
i = 1 = -1
Am 19.01.2025 um 16:50 schrieb Moebius:
Am 19.01.2025 um 16:39 schrieb Moebius:
Am 19.01.2025 um 16:30 schrieb Richard Hachel:
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
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.
Hint: No "extremely fine mathematical error" at all.
On the other hand, this one is an "extremely severe mathematical error":
Op 19/01/2025 om 16:32 schreef Richard Hachel:
Le 19/01/2025 à 15:18, sobriquet a écrit :
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 [...]
It's simply some "other number" outside the real number line, one
creating its own number line.
What's there more to brood on?
Le 19/01/2025 à 15:18, sobriquet a écrit :
https://www.desmos.com/calculator/kijg1kvt75
It's an interactive demonstration
Basée sur une équation fausse.
On tourne en rond.
Le 19/01/2025 à 15:18, sobriquet a écrit :
https://www.desmos.com/calculator/kijg1kvt75
It's an interactive demonstration
Basée sur une équation fausse.
On tourne en rond.
R.H.
Op 19/01/2025 om 16:32 schreef Richard Hachel:
Le 19/01/2025 à 15:18, sobriquet a écrit :
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].
So it's clear from the start that it's not a member of the real number line [...]
It's simply some "other number" outside the real number line, one
creating its own number line.
What's there more to brood on?
[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].
Le 19/01/2025 à 02:53, sobriquet a écrit :
Op 18/01/2025 om 11:34 schreef 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.
What is i?
It is an imaginary unit, such that i*i=-1.
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.
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
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.
You tell me:
yes, but it does not work with the conjugate.
Am 19.01.2025 um 16:30 schrieb Richard Hachel:
Le 19/01/2025 à 15:28, Moebius a écrit :
Am 19.01.2025 um 12:37 schrieb Richard Hachel:
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!
cOn 1/19/2025 2:41 AM, Richard Hachel wrote:
Le 19/01/2025 à 02:53, sobriquet a écrit :
For ⅈ=1 or ⅈ=-1, ⅈ²=1 or ⅈ²=1
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:
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?
Le 19/01/2025 à 17:04, Moebius a écrit :
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.
Le 19/01/2025 à 17:21, Python a écrit :
Le 19/01/2025 à 17:04, Moebius a écrit :
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.
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.
On 1/18/2025 5:34 AM, Richard Hachel wrote:
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.
On 1/18/2025 5:34 AM, Richard Hachel wrote:
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.
Op 30/01/2025 om 22:55 schreef 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″
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
On 1/31/2025 6:31 AM, guido wugi wrote:
Op 30/01/2025 om 22:55 schreef 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
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.
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
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.
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