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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: New equation
Date: Tue, 25 Feb 2025 15:07:16 -0800
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On 2/25/2025 3:05 PM, Chris M. Thomasson wrote:
> On 2/25/2025 3:03 PM, Chris M. Thomasson wrote:
>> On 2/25/2025 1:58 PM, Richard Hachel wrote:
>>> Le 25/02/2025 à 22:36, "Chris M. Thomasson" a écrit :
>>>> On 2/25/2025 6:20 AM, Richard Hachel wrote:
>>>>> Le 25/02/2025 à 09:21, Barry Schwarz a écrit :
>>>>>> On Mon, 24 Feb 25 21:11:40 +0000, Richard Hachel 
>>>>>> <r.hachel@tiscali.fr>
>>>>>> wrote:
>>>>>>
>>>>>>> Le 24/02/2025 à 21:23, Barry Schwarz a écrit :
>>>>>>>> On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel 
>>>>>>>> <r.hachel@tiscali.fr>
>>>>>>>
>>>>>>>> A quartic always has four roots.
>>>>>>>
>>>>>>> Here, I would still put a small caveat.
>>>>>>> The fact of saying that an equation of degree n has n roots is 
>>>>>>> perhaps not entirely correct.
>>>>>>> I ask myself the question.
>>>>>>> If for example we write f(x)=x^3+3x-4, it is indeed an equation 
>>>>>>> of degree 3.
>>>>>>> But how many roots, and what are they?
>>>>>>> I asked this question to mathematicians, and to artificial 
>>>>>>> intelligence, and I was given three roots, but they are 
>>>>>>> incorrect, because those who answer do not seem to understand the 
>>>>>>> real concept of imaginary numbers.
>>>>>>> There is in fact only one root.
>>>>>>> A very strange root composed of a real root and a complex root. 
>>>>>>> Both placed on the same point A(1,0) and A(-i,0).
>>>>>>>
>>>>>>> R.H. 
>>>>>>
>>>>>> A cubic has three roots.
>>>>>
>>>>> This is what is generally said, but is it always true?
>>>>>
>>>>>> The roots of your equation are 1, (-1+i*sqrt(15))/2, and
>>>>>> (-1-i*sqrt(15))/2.
>>>>>
>>>>> That one of the roots is 1, and that it can be represented on a 
>>>>> Cartesian coordinate system, is obvious. I then set the point A(1,0).
>>>>> I then look for the other two roots of the equation, but I realize 
>>>>> that I can't find any others, even complex ones, and that the two 
>>>>> complex roots given are fanciful.
>>>>> I then start from the principle that the complex roots are the real 
>>>>> roots of the mirror curve, and that the real roots are the complex 
>>>>> roots of this other curve, and I find a complex root which is x'=-1.
>>>>>
>>>>> I therefore obtain the point A(-i,0) which is exactly the same as 
>>>>> the point A(1,0) knowing that i=-1 and -i=+1.
>>>>
>>>> Point (1, 0) = 1+0i
>>>> Point (-1, 0) = -1+0i
>>>> Point (0, 1) = 0+1i
>>>> Point (0, -1) = 0-1i
>>>>
>>>>
>>>>
>>>>>
>>>>> It seems that this curve is its own mirror.
>>>>>
>>>>> R.H.
>>>
>>> No, no, no, no, no...
>>> I see that you did not understand what I am saying about complex 
>>> numbers, and how I would use them in a Cartesian coordinate system.
>>> I use them longitudinally, on the x'Ox axis, but in the opposite 
>>> direction.
>>> The complex roots are therefore on the x'Ox axis like the real roots 
>>> and are found where the curve g(x) mirror of f(x) passes.
>>>
>>>
>>> Point (1, 0) = Point (-i,0)
>>> Point (-1, 0) = Point (i,0)
>>> Point (0, 1) = Point (0,1)
>>> Point (0, -1) = Point (0,-1)
>>>
>>> Point (5,3) = Point (-5i,3)
>>> Point (-2,-4) = Point (2i,-4)
>>>
>>> Imaginary number i is purely ON the x'Ox axe, never elsewhere in 
>>> cartesian reference points.
>>> Then there are Argand's representations, where the components of the 
>>> complex are perpendicularly dissociated.
>>> But that's something else.
>>
>> No. The x axis is the real, the y axis is the imaginary. Why do you 
>> seem to insist on flipping the two?
>>
> 
> What are you trying to do here? Mess up complex numbers?

Keep in mind that i = Point (0, 1)