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Thread: Proving a metric

  1. #1
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    Proving a metric

    To prove that a function is a metric, I must show that:

    $\displaystyle
    \frac{|x-z|}{1+|x-z|} \le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} \quad \forall x,y,z \in \Re
    $

    However, I cannot quite get there. Here's what I come up with:

    $\displaystyle
    \frac{|x-z|}{1+|x-z|}
    = \frac{|x-y+y-z|}{1+|x-y+y-z|}
    \le \frac{|x-y|+|y-z|}{1+|x-y+y-z|}
    $

    $\displaystyle
    = \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} = \ldots
    $

    Any nudge in the right direction is appreciated. Thanks.
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by drumist View Post
    To prove that a function is a metric, I must show that:

    $\displaystyle
    \frac{|x-z|}{1+|x-z|} \le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} \quad \forall x,y,z \in \Re
    $

    However, I cannot quite get there. Here's what I come up with:

    $\displaystyle
    \frac{|x-z|}{1+|x-z|}
    = \frac{|x-y+y-z|}{1+|x-y+y-z|}
    \le \frac{|x-y|+|y-z|}{1+|x-y+y-z|}
    $

    $\displaystyle
    = \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} = \ldots
    $

    Any nudge in the right direction is appreciated. Thanks.
    $\displaystyle
    \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} \leq \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} $.
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  3. #3
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    Quote Originally Posted by Sampras View Post
    $\displaystyle
    \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} \leq \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} $.
    Sorry but I just don't see how that is true. How can you claim that $\displaystyle |x-y+y-z| \ge |x-y|$?
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  4. #4
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    So, I just did some crunching, and it turns out that $\displaystyle \frac{|x-y|+|y-z|}{1+|x-y+y-z|} \le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}$ is NOT a true statement, so clearly I'm on the wrong track from the beginning. (Use x=1, y=2, z=0 for a counterexample.)
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  5. #5
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    Lemma:
    $\displaystyle \begin{gathered}
    0 \leqslant a \leqslant b \Rightarrow \quad a + ab \leqslant b + ab \hfill \\
    \Rightarrow \quad a(1 + b) \leqslant b(1 + a) \Rightarrow \quad \frac{a}
    {{1 + a}} \leqslant \frac{b}
    {{1 + b}} \hfill \\
    \end{gathered} $

    Using that lemma, we know that $\displaystyle |x-z|\le |x-y|+|y-z|$ so

    $\displaystyle \frac{{\left| {x - z} \right|}}
    {{1 + \left| {x - z} \right|}} \leqslant \frac{{\left| {x - y} \right| + \left| {y - z} \right|}}
    {{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {x - y} \right|}}
    {{1 + \left| {x - y} \right|}} + \frac{{\left| {y - z} \right|}}
    {{1 + \left| {y - z} \right|}}$
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by drumist View Post
    So, I just did some crunching, and it turns out that $\displaystyle \frac{|x-y|+|y-z|}{1+|x-y+y-z|} \le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}$ is NOT a true statement, so clearly I'm on the wrong track from the beginning. (Use x=1, y=2, z=0 for a counterexample.)
    You're problem here was the confusion of what the actual metric is. Let $\displaystyle d(x,y)=\frac{|x-y|}{1+|x-y|}$ then to show that the triangle quality hold here you must show that $\displaystyle \frac{|x-y|}{1+|x-y|}=d(x,y)\leqslant d(x,z)+d(y,z)=\frac{|x-z|}{1+|x-z|}+\frac{|y-z|}{1+|y-z|}$. You, probably used to using the triangle inequality for the usual metric on $\displaystyle \mathbb{R}$, automatically thought you must mess with the $\displaystyle |x-y|$'s.
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