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Math Help - Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

  1. #1
    Senior Member x3bnm's Avatar
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    Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    I'm reading a differential geometry book named "Elementary geometry of differentiable curves: An undergraduate introduction" written by Gibson.


    I'm on page 2. On that page, 4 properties of scalar product(dot product) of vectors are given.

    One of them is:

    \mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z  = \mathbf 0


    I know the proof of this and the proof shows that the statement is true for \geq

    I accept this.


    But if \mathbf z = \mathbf 0 isn't it definite that \mathbf z \cdot \mathbf z has to be equal to 0 and not greater than 0?

    Can anyone find a \mathbf z such that when \mathbf z = \mathbf 0 then \mathbf z \cdot \mathbf z is greater than 0?


    Why did the author said that \mathbf z \cdot \mathbf z \geq 0? Why did he mention greater and equal, and not just equal to 0?
    Last edited by x3bnm; June 25th 2012 at 01:31 PM.
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Quote Originally Posted by x3bnm View Post
    One of them is:
    \mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z  = \mathbf 0
    That is simply saying z\cdot z=0\text{ if and only if }z=0
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    Senior Member x3bnm's Avatar
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Plato said:
    That is simply saying z\cdot z=0\text{ if and only if }z=0
    So the > sign has no role here. I've to overlook it.

    Roger that. Thanks Plato.
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    z . z >= 0 for any z, and the z . z = 0 when z=0. Is this clearer?
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    it's just a concise of saying:

    if z ≠ 0, z.z > 0

    if z = 0, z.z = 0.

    from these 2 statements, we can conclude that if z.z = 0 (that is, z.z is not greater than 0), then z must itself be 0 (for if it were non-zero, then z.z > 0).

    in general, all we can say is that z.z ≥ 0, for an arbitrary vector z, because of the fact that z might be 0.

    it's like saying: "a square (of a real number) is always positive"....except for 0. Zero-objects are funny like that, they often introduce "exceptions" which make simple rules more complicated to say.
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Quote Originally Posted by x3bnm View Post
    So the > sign has no role here. I've to overlook it.
    It has role. That is a fairly standard notation.
    Here is another example:
    a^2+b^2\ge 2ab,~\text{ equality holds if and only if }a=b=1~.
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Quote Originally Posted by x3bnm View Post
    I'm reading a differential geometry book named "Elementary geometry of differentiable curves: An undergraduate introduction" written by Gibson.


    I'm on page 2. On that page, 4 properties of scalar product(dot product) of vectors are given.

    One of them is:

    \mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z  = \mathbf 0


    I know the proof of this and the proof shows that the statement is true for \geq

    I accept this.


    But if \mathbf z = \mathbf 0 isn't it definite that \mathbf z \cdot \mathbf z has to be equal to 0 and not greater than 0?

    Can anyone find a \mathbf z such that when \mathbf z = \mathbf 0 then \mathbf z \cdot \mathbf z is greater than 0?


    Why did the author said that \mathbf z \cdot \mathbf z \geq 0? Why did he mention greater and equal, and not just equal to 0?
    You are misreading the statement. He is NOT talking about only z= 0. The statement says that z\cdot z\ge 0 for all z and, additionally, z\cdot z= 0 if and only if z= 0.
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    Senior Member x3bnm's Avatar
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Thanks all for answering my question. So I understand that zero is an exceptional case that creates exceptional situations.
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    Senior Member x3bnm's Avatar
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    Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

    Quote Originally Posted by HallsofIvy View Post
    You are misreading the statement. He is NOT talking about only z= 0. The statement says that z\cdot z\ge 0 for all z and, additionally, z\cdot z= 0 if and only if z= 0.
    You are right. In that sense the statement has to use \geq 0 and not just = 0
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