A field is a spacial distribution of a quantity. Numbers can be represented as coordinates within a field, but that in itself is not a proof of their orthogonality.
I'm talking about a field in the context of number theory. A field that satisfies field axioms. Stop being stupid on purpose.
Show me some references to back up your "rotational operator" definition of i. Otherwise you have lost this game.
What is there to explain? jω is a point on complex plane (0,ω). It has a real part of 0 and imaginary part ω. You can think ω was rotated 90 degrees if you want and get to the same conclusion.How does it explain jω or ωj algebraically?
Both Eric Gibbs and I have shown you references which you ignore.
I have also explained that you need rotation to get orthogonality.
What is there to explain? jω is a point on complex plane (0,ω). It has a real part of 0 and imaginary part ω. You can think ω was rotated 90 degrees if you want and get to the same conclusion.
None of the references prove that i is an operator. Treating it as an operator works, but none of the references show why i is an operator.
Yes you have, but you are wrong. You don't need rotation to get orthogonality. You can just define orthogonality.
That question is just some ******** you keep ranting about.That is why you cannot explain the algebraic significance of "j number of items each ω in length".
The references explain the orthogonality of number operated on by j.
You only get orthogonality only if you consider j as a rotational operator.
Things cannot just be defined because you want them to be in existence.
No? Then how about this: I define objects p and q so that p⊥qOrthogonality is obtained by rotation, not by definition.
When defining j as a operator, the results are still correct, but j is not an algebraic constant, so there is no conflict.
Googling for j operator will also bring up lots of links
That question is just some ******** you keep ranting about.
".. algebraic significance of 'j number of items each ω in length'."
what a ******** question. stop embarrassing yourself.
No they don't. The references do not even include the word "orthogonality".
That is not true.. you need to show some proof on that one!
I agree, i (imaginary unit) cannot be defined as an operator just because you like it that way.
What is the conflict you refer to?
And do not talk some nonsense about orthogonality and rotations. You need to show some proof to keep that argument alive.
As I said: I let you call j as an operator (in the context of electrical engineering). But don't tell me that: "The symbol 'i' does not mean √-1. 'i' is a mathematical operator, not a finite value." You know it is ********.
As I said before, that is too simple to prove. Rotate some quantity by 90° to get a orthogonal direction. Can't get more basic than that.
... That is how orthogonality is defined.
Show me the proof then. Saying so does not make it so. As you said: "Things cannot just be defined because you want them to be in existence."
Let me get this right. What you are saying is:
If I have objects p and q so that p||q, then p⊥jq always holds (for all objects {p,q | p||q})? Here j is an operator defined as you say it is.
What does {p,q | p||q} mean? Perhaps it would be clearer if you gave a verbal description of what you mean.
I told you, I cannot prove it because it is too basic. Every mathematician knows that the simplest things are the hardest to prove. However, what I said is consistent with its mathematical concepts.
I did not have a signature before.. but that is an instant classic!
What does {p,q | p||q} mean? Perhaps it would be clearer if you gave a verbal description of what you mean.
The set of p and q such that p and q are parallel.
It tells that we are only concerned about objects p and q that are parallel (with each other).
So, If objects p and q are parallel, then p⊥jq always holds, if j is an operator defined as you say it is? (p⊥jq means that objects p and jq are orthogonal)
I would agree with the above. Continue.
Ok, so if p and q are 2-dimensional real vectors ( p,q ∈ ℝ² ), say:
p = [1 2]
q = [2 4]
What in your opinion is jq?
1) If j is an operator that rotates q by 90 degrees, then jq = [-4 2] (and p is orthogonal to qj).
2) If jq = [2j 4j], then it follows that p is not orthogonal to qj.
dot product of p and jq: dot(p, qj) = 2j+8j ≠ 0
jq ≠ [2j 4j]
Show me a book, or a reference, or anything, where it is said that a vector [2 4] multiplied by j equals [-4 2], and does not equal [2j 4j].
The rotation operator concept just does not work.
First of all, to be pedantic, complex numbers are not truly vectors, but the quantities they represent can be described by orthogonal vectors.
Now, the number in your notation was [2 4]. It must be assumed that it means a real component of 2 and a orthogonal component of 4. In my rotational notation, that would be [2 j(4)]. All I did was rotate both components of the number, which is a legitimate operation. So the real part 2 ===> j(2) and the orthogonal part j(4) ===> -4 . The final rotational sum is [-4 j(2)] or [-4 2] by your notation.
And because complex numbers are not vectors, the multiplication is done componentwise: [2 4]j = [2j 4j]
Now you are just making stuff up. I defined the vectors to be purely real ( p,q ∈ ℝ² ). Show me any example, a web page or a book, that shows the kind of calculations that you just did.. just plain wrong. A vector multiplied by imaginary unit i, or j operator, or a scalar, is done componentwise.
I would also like to know how you would calculate a three dimensional vector multiplied by j: [1 2 3] j = ?
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