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Math Help - inequality proof

  1. #1
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    inequality proof

    Using the triangle inequality show a and b are in \mathbb{R} with a \neq b, then \exists open intervals U centered at a and V centered at b, both with radius \epsilon=\frac{1}{2}|a-b|

    so far all I have is

    |a-b|=2\epsilon

    ||a|-|b|| \leq |a-b|

    now I'm not sure if what I'm doing is right:

    -\epsilon<||a|-|b|| < \epsilon \rightarrow ||a|-|b|| < 2\epsilon \rightarrow ||a|-|b|| < |a-b| ?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    Using the triangle inequality show a and b are in \mathbb{R} with a \neq b, then \exists open intervals U centered at a and V centered at b, both with radius \epsilon=\frac{1}{2}|a-b|
    You do realise that as this stands it is not the question that you think it is, don't you?

    Do you want to show that:

    For all a,b \in \mathbb{R},\ a \ne b, then the two disjoint open intervals U centered at a and V centered at b, both with radius \epsilon=\frac{1}{2}|a-b| are disjoint?

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post

    Do you want to show that:

    For all a,b \in \mathbb{R},\ a \ne b, then the two disjoint open intervals U centered at a and V centered at b, both with radius \epsilon=\frac{1}{2}|a-b| are disjoint?

    RonL
    I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof.

    This is what I found:

    U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)

    U \bigcap V = \emptyset

    suppose z is in  U \bigcap V

    |x-y| \leq |x-z|+|z-y| < \epsilon + \epsilon = 2\epsilon = |x-y|

    which is apparently a contradiction.

    If anyone could clarify this, it would help me out a lot.
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  4. #4
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    Quote Originally Posted by lllll View Post
    I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof.

    This is what I found:

    U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)

    U \bigcap V = \emptyset

    suppose z is in  U \bigcap V

    |x-y| \leq |x-z|+|z-y| < \epsilon + \epsilon = 2\epsilon = |x-y|

    which is apparently a contradiction.

    If anyone could clarify this, it would help me out a lot.
    The bit that is missing is:

    |x-y|=|x-z+z-y|=|(x-z)-(y-z)|

    Now by the triangle inequality:

    |x-y|=|(x-z)-(y-z)| \le |x-z|+|y-z|

    etc.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    The bit that is missing is:

    |x-y|=|x-z+z-y|=|(x-z)-(y-z)|

    Now by the triangle inequality:

    |x-y|=|(x-z)-(y-z)| \le |x-z|+|y-z|

    etc.

    RonL
    That solves a part of the proof, but where is the contradiction?

    It lies is the fact that z is assumed to be in U \cap V so lies in U and in V so |x-z|<\epsilon, and |y-z|<\epsilon, so:

    |x-y|<2 \epsilon

    But |x-y|=2 \epsilon, so we have:

    |x-y|<|x-y|

    which is a contradiction.

    RonL
    Last edited by CaptainBlack; October 5th 2008 at 12:19 AM.
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