Re: Hello!

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.