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Thread: inequality proof

  1. #1
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    inequality proof

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

    so far all I have is

    $\displaystyle |a-b|=2\epsilon$

    $\displaystyle ||a|-|b|| \leq |a-b|$

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

    $\displaystyle -\epsilon<||a|-|b|| < \epsilon \rightarrow ||a|-|b|| < 2\epsilon \rightarrow ||a|-|b|| < |a-b|$ ?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Using the triangle inequality show $\displaystyle a$ and $\displaystyle b$ are in $\displaystyle \mathbb{R}$ with $\displaystyle a \neq b$, then $\displaystyle \exists$ open intervals $\displaystyle U$ centered at $\displaystyle a$ and $\displaystyle V$ centered at b, both with radius $\displaystyle \epsilon=\frac{1}{2}|a-b|$
    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 $\displaystyle a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $\displaystyle U$ centered at $\displaystyle a$ and $\displaystyle V$ centered at $\displaystyle b$, both with radius $\displaystyle \epsilon=\frac{1}{2}|a-b|$ 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 $\displaystyle a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $\displaystyle U$ centered at $\displaystyle a$ and $\displaystyle V$ centered at $\displaystyle b$, both with radius $\displaystyle \epsilon=\frac{1}{2}|a-b|$ 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:

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

    $\displaystyle U \bigcap V = \emptyset$

    suppose $\displaystyle z$ is in $\displaystyle U \bigcap V$

    $\displaystyle |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:

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

    $\displaystyle U \bigcap V = \emptyset$

    suppose $\displaystyle z$ is in $\displaystyle U \bigcap V$

    $\displaystyle |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:

    $\displaystyle |x-y|=|x-z+z-y|=|(x-z)-(y-z)| $

    Now by the triangle inequality:

    $\displaystyle |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:

    $\displaystyle |x-y|=|x-z+z-y|=|(x-z)-(y-z)| $

    Now by the triangle inequality:

    $\displaystyle |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 $\displaystyle z$ is assumed to be in $\displaystyle U \cap V$ so lies in $\displaystyle U$ and in $\displaystyle V$ so $\displaystyle |x-z|<\epsilon,$ and $\displaystyle |y-z|<\epsilon$, so:

    $\displaystyle |x-y|<2 \epsilon $

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

    $\displaystyle |x-y|<|x-y|$

    which is a contradiction.

    RonL
    Last edited by CaptainBlack; Oct 4th 2008 at 11:19 PM.
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