Re: Equation complexe

On 2/26/2025 1:43 PM, Richard Hachel wrote:

Le 26/02/2025 à 18:57, Jim Burns a écrit :

>

>
But the error will quickly be found
if we use the unit i
as it must be used in this case.

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 satisfied by
our 1.dimensional '+' and '⋅':
⎛ 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,
and it's determined by choices 𝐯, β, α
𝐢 = ±(𝐯+β𝟏)/(α-β²)¹ᐟ² (either sign 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.

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

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