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Thread: 1 = -1

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    1 = -1

    1 = -1

    Proof:

    -1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1

    QED!

    Prove me wrong!

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    Elder Contributor barbos's Avatar
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    Quote Originally Posted by SLD View Post
    1 = -1

    Proof:

    -1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1

    QED!

    Prove me wrong!
    SQR(1) != 1

    SQR(1)={1,-1}

    Also, i != SQR(-1), but SQR(-1)={i,-i}

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    As in algebra texts is
    sqrt(1) =+-1.
    sqrt(1) = 1
    SQRT(-1) = i, A complex number [0 + i]. i^2 = -1. i^3 = 1...

    sqrt(4) = 2
    sqrt(-4) = sqrt(-1*4) = [0 + i2] or just i2
    Last edited by steve_bank; 12-08-2019 at 03:37 AM.

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    Quote Originally Posted by SLD View Post
    1 = -1

    Proof:

    -1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1

    QED!

    Prove me wrong!

    1=\=-1.

    I just proved you wrong.



    So, there is something wrong with your computation, and the only candidate is the SQR function, not defined for negative numbers (though if you want to try to define such a function ).


  5. Top | #5
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    If your calculator has a complex number mode it will take sqrt(-1).

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    Quote Originally Posted by steve_bank View Post
    If your calculator has a complex number mode it will take sqrt(-1).
    You can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.

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    Quote Originally Posted by Angra Mainyu View Post
    Quote Originally Posted by steve_bank View Post
    If your calculator has a complex number mode it will take sqrt(-1).
    You can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.
    All numbers are complex. It is the top category. Reals are complex numbers with zero imaginary. My HP calculator does complex numbers as does most software packages. It is just trigonometry.
    I
    In software you can code the sqr function nay way you like. If you limit t to solely reals sqr(-1) is undefined.

    The OP argument is counter to any high school algebra text. The quadratic equation. If the discriminant is negative the solution is complex, if not the solution is real.

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    Veteran Member Wiploc's Avatar
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    Quote Originally Posted by SLD View Post

    Prove me wrong!
    Division by zero is undefined.

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    Quote Originally Posted by steve_bank View Post
    Quote Originally Posted by Angra Mainyu View Post
    Quote Originally Posted by steve_bank View Post
    If your calculator has a complex number mode it will take sqrt(-1).
    You can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.
    All numbers are complex. It is the top category. Reals are complex numbers with zero imaginary. My HP calculator does complex numbers as does most software packages. It is just trigonometry.
    I
    In software you can code the sqr function nay way you like. If you limit t to solely reals sqr(-1) is undefined.

    The OP argument is counter to any high school algebra text. The quadratic equation. If the discriminant is negative the solution is complex, if not the solution is real.
    Again, you can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the sqrt function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.

  10. Top | #10
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    sqr(-4 * 9) = sqr(-1) * sqr(4) * sqr(9) = [0 + i6]

    [0 + i6]^2= [0 + i6] * [0 + i6] = i^2 * 6^2 = -1*36 = -4 * 9

    sqr(-1) = i, i^2 = sqr(-1) A definition.

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