• Roots of =?UTF-8?Q?f=28x=29=3Dy=3Dsqrt=28x=29+=31?=

    From Richard Hachel@21:1/5 to All on Tue Feb 11 13:34:51 2025
    x'=x"=i

    <http://nemoweb.net/jntp?dlMs7UrkzQgTSiLDMc9gkEuqYSk@jntp/Data.Media:1>

    R.H.

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  • From Python@21:1/5 to All on Tue Feb 11 15:11:44 2025
    Le 11/02/2025 à 14:34, Richard "Hachel" Lengrand a écrit :
    x'=x"=i

    <http://nemoweb.net/jntp?dlMs7UrkzQgTSiLDMc9gkEuqYSk@jntp/Data.Media:1>

    R.H.

    So i = -1. No need for a new symbol then :-)

    And still not a root of sqrt(x) + 1 (not even defined at x=-1)

    You are loosing your marbles Richard.

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  • From Richard Hachel@21:1/5 to All on Tue Feb 11 18:49:36 2025
    Le 11/02/2025 à 16:11, Python a écrit :
    Le 11/02/2025 à 14:34, Richard "Hachel" Lengrand a écrit :
    x'=x"=i

    <http://nemoweb.net/jntp?dlMs7UrkzQgTSiLDMc9gkEuqYSk@jntp/Data.Media:1>

    R.H.

    So i = -1. No need for a new symbol then :-)

    We need a new imaginary symbol.

    You cannot say the root of the curve f(x)=sqrt(x)+1 is x'=-1 because it is
    not true in the real world. It is only true for the mirror curve.

    We must therefore say that it is indeed the root, but that it is
    imaginary, and for that, you must use the notation i.

    Note that i'Oi is on the x'Ox axis, but is declined in the opposite
    direction, according to what I have already said.

    The point A in real coordinates, for example, is noted A(8,0).

    The same point in imaginary coordinates is noted A(-8i,0) to show that it
    is a point designated is in imaginary notation.


    And still not a root of sqrt(x) + 1 (not even defined at x=-1)

    You are loosing your marbles Richard.

    The point A(-1,0) is the real root of the imaginary curve, and in notation A(i,0) it is the imaginary root of the real curve.

    Wait, let me say it again.

    The point A(-1,0) is the real root of the imaginary curve, and in notation A(i,0) it is the imaginary root of the real curve.

    R.H.

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