Re: Equation complexe

On 2/25/2025 3:07 PM, Richard Hachel wrote:

Le 25/02/2025 à 18:21, Jim Burns a écrit :

On 2/25/2025 9:23 AM, Richard Hachel wrote:

x^4=-81
>
What is x?

>
x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

>
Oui, c'est ce que dis aussi l'Intelligence artificielle,
mais sans trop expliquer pourquoi.

(x-(+1+𝑖)⋅3/√2)⋅(x-(+1-𝑖)⋅3/√2)⋅(x-(-1+𝑖)⋅3/√2)⋅(x-(-1-𝑖)⋅3/√2) =
x⁴+81
In real and in complex numbers,
if a⋅b⋅c⋅d = 0
then one of a,b,c,d is 0
otherwise, a⋅b⋅c⋅d ≠ 0
If x⁴+81 = 0
and a⋅b⋅c⋅d = x⁴+81
then one of
a = x-(+1+𝑖)⋅3/√2 = 0
b = x-(+1-𝑖)⋅3/√2 = 0
c = x-(-1+𝑖)⋅3/√2 = 0
d = x-(-1-𝑖)⋅3/√2 = 0
is true.
Hence,
x ∈ {(+1+𝑖)⋅3/√2, (+1-𝑖)⋅3/√2, (-1+𝑖)⋅3/√2, (-1-𝑖)⋅3/√2}

Personally, I propose only one root,
but it is not in conformity with what is said about "complex numbers".

Therefore, despite appearances,
your question is not "What is x?"
and it is not "what is a complex number?"
Your question is
"How can it be possible
for one person to speak to another
and be understood?"
The possibility of understanding is
greatly facilitated where
that which a speaker means by a word and
that which a listener thinks they mean by it
are in conformity.
This is not a deep philosophical point,
This is similar to noting that
phone calls with non.operating phones
are completely unsatisfactory.
A point doesn't need to be deep to be true.

I remind you that
I do not admit the definition i²=-1,

Your question assumes a certain common background,
because that's how language works.
If your question was "How do I get to the post office?"
but you didn't admit the usual definitions of
'left', 'right', and so on,
odds are you don't get to the post office,
but that would have nothing to do with
the directions you were given.

which, in itself, is not false, but so narrow
that I do not understand its semantic interest.

I think you are asking why 𝐢²=-1
and why not = something else.
Most of the answer is that
we want a 2.dimensional field which extends
the 1.dimensional field of the real numbers.
That is to say, we want 2.dimensional
addition '+' and multiplication '⋅'
which satisfy the same laws which
our 1.dimensional '+' and '⋅' satisfy:
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'
We have what we want  if
we have a vector 𝐯 not on the real axis such that,
for this 2.dimensional multiplication '⋅'
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐯 = 𝐯
𝐯⋅𝟏 = 𝐯
𝐯⋅𝐯 = -α𝟏-2β𝐯
such that  α > β²
Pick 𝐯 ∈ ℝ×(ℝ\{0}), β, α > β²
Define
𝐯⋅𝐯 = -α𝟏-2β𝐯
(a𝟏+b𝐯)⋅(c𝟏+d𝐯) = ac𝟏+(ad+bd)𝐯+bd(𝐯⋅𝐯)
We have what we want,
a 2.dimensional field extending ℝ
However,
suppose 𝐯 ≠ ⟨0,1⟩ and 𝐯⋅𝐯 ≠ -𝟏
Then 𝐯 ≠ 𝐢
But 𝐢 still exists,//////////////////
determined by choices 𝐯, β, α
𝐢 = ±(𝐯+β𝟏)/(α-β²)¹ᐟ² (either works)
𝐯 = ±(α-β²)¹ᐟ²𝐢-β𝟏
Each  a𝟏+b𝐯 has a corresponding a′𝟏+b′𝐢
and vice versa.
𝟏⋅𝟏 = 𝟏
𝟏⋅𝐢 = 𝐢
𝐢⋅𝟏 = 𝐢
𝐢⋅𝐢 = -𝟏
(a𝟏+b𝐢)⋅(c𝟏+d𝐢) = (ac-bd)𝟏+(ad+bd)𝐢
in the usual way.

For me, the definition must be
extended to all powers of x such that i^x=-1.

For me, the definition of 'left' must be
extended to all directions.
Wish me luck!

The other four roots being incorrect
(in the proposed system).

The proposed system do not have
2.dimensional '+' and '⋅' satisfying:
⎛ associativity and commutativity for both,
⎜ identities 𝟎 𝟏, inverses -𝐱 𝐱⁻¹ except 𝟎⁻¹,
⎝ distributivity of '.' over '+'