Re: New equation

On 3/1/2025 9:11 PM, Ross Finlayson wrote:

On 03/01/2025 03:27 PM, Jim Burns wrote:

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

On 03/01/2025 11:11 AM, Jim Burns wrote:

On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

The complex field does not fail at being a field.

>

Oh, I read a definition of complex numbers
and point out that division is non-unique.

>
(c+d𝑖)/(a+b𝑖) = (c+d𝑖)⋅(a+b𝑖)⁻¹
>
x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
is unique.
>
(x+y𝑖) = (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)

[...]

Not the other way around though

>
What do you mean by that?

I give up trying to get you to explain your self.

Oh, I read a definition of complex numbers
and point out that division is non-unique.

>
Non.zero complex division is uniquely.valued.
>
(c+d𝑖)/(a+b𝑖) = x+y𝑖

[...]

Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?

>
You started with "non-zero complex division
is uniquely valued", that's called "see rule 1".

I started by stating what I was about to prove,
and then I proved it.
⎛ You have math degree. Right?
⎝ Did you do any proofs while earning it?
----
Do you have a problem with any part of the following>
(c+d𝑖)/(a+b𝑖) = x+y𝑖
c+d𝑖 = (a+b𝑖)⋅(x+y𝑖)
c+d𝑖 = (ax-by)+(bx+ay)𝑖
ax-by = c
bx+ay = d
a²x-aby = ac
b²x+aby = bd
(a²+b²)x = ac+bd
x = (ac+bd)/(a²+b²)
abx-b²y = bc
abx+a²y = ad
(a²+b²)y = ad-bc
y = (ad-bc)/(a²+b²)
(c+d𝑖)/(a+b𝑖) = x+y𝑖 =
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?

Find various different formalisms of complex numbers
from all through the libraries and notice it's
given various ways.

Find a formalism of complex numbers _disagreeing with_
⎛
⎜ a+b𝑖 = c+d𝑖  ⇔  a=c ∧ b=d
⎜
⎜ (a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖
⎜
⎜ (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖
⎝
and show it to me.