Re: Hello!

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.
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Let z=a+ib
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It is a number that has a real component and an imaginary component.
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I wonder if the terms "certain component" and "possible component" would not be as appropriate.
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What is i?
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It is an imaginary unit, such that i*i=-1.
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In our universe, this seems impossible, a square can never be negative.
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Except that we are in the imaginary.
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Let's assume that i is a number, or rather a unit, which is both its number and its opposite.
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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
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I remind you that we are in the imaginary. So why not.
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Let's set z=16+9i
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It then comes that at the same time, z=25 and z=7.
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It is a strange universe, but which can be useful for writing things in different ways.
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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.
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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?
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We can nevertheless give z a real part, which is the average of the two classes. a=16.
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And ib then becomes the fluctuation of the average.
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If we set i=1 then ib=+9; if we set i=-1 then ib=-9.
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"i" would therefore be this entity, this unit, equal to both 1 and -1, depending on how we look at it (Schrodinger's cat).
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But what happens if we square i?
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It is both 1 and -1?
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Can we write i²=(1)*(1)=1?
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No, because i would only be 1.
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Can we write i²=(-1)(-1)=1?
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No, because i is not only -1, it is both -1 and 1.
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We then have i²=(i)*(i)=(1)(-1)=(-1)(1)=-1.
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But here, we will notice something extraordinary, the additions and products of complex numbers can be determined.
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Z=z1+z2
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Z=(a+ib)+(a'+ib')
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and, Z=(a+a')+i(b+b')
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All this is very simple for the moment.
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But we are going to enter into a huge astonishment concerning the product of two complexes.
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How do mathematicians practice?
Z=z1*z2
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so, so far it's correct:
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Z=(a+ib)(a'+ib')
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So, and it's still correct for Dr. Hachel (that's me):
Z=aa'+i(ab'+a'b)+(ib)(ib')
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And there, for Dr. Hachel, mathematicians make a huge blunder by setting (ib)(ib')=i²bb'=-bb'
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Why?
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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.
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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)
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The real part of the product being aa'+bb' and not aa'-bb'
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With a remaining imaginary part where i is equal to both -1 and 1, which gives two results each time for Z.
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It seems that this is an astonishing blunder, due to the misunderstanding of the handling of complex and imaginary numbers.
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On the other hand, by going through statistics, statistics confirms HAchel's ideas, and the results usually proposed by mathematicians become totally false.
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I wish you a good reflection on this.
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Have a good day.
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R.H.

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