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Math Help - Proving a metric

  1. #1
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    Proving a metric

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

    <br />
\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<br />

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

    <br />
\frac{|x-z|}{1+|x-z|}<br />
= \frac{|x-y+y-z|}{1+|x-y+y-z|}<br />
\le \frac{|x-y|+|y-z|}{1+|x-y+y-z|}<br />

    <br />
= \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} = \ldots<br />

    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:

    <br />
\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<br />

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

    <br />
\frac{|x-z|}{1+|x-z|}<br />
= \frac{|x-y+y-z|}{1+|x-y+y-z|}<br />
\le \frac{|x-y|+|y-z|}{1+|x-y+y-z|}<br />

    <br />
= \frac{|x-y|}{1+|x-y+y-z|} + \frac{|y-z|}{1+|x-y+y-z|} = \ldots<br />

    Any nudge in the right direction is appreciated. Thanks.
    <br />
  \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
    <br />
  \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 |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 \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:
    \begin{gathered}<br />
  0 \leqslant a \leqslant b \Rightarrow \quad a + ab \leqslant b + ab \hfill \\<br />
   \Rightarrow \quad a(1 + b) \leqslant b(1 + a) \Rightarrow \quad \frac{a}<br />
{{1 + a}} \leqslant \frac{b}<br />
{{1 + b}} \hfill \\ <br />
\end{gathered}

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

    \frac{{\left| {x - z} \right|}}<br />
{{1 + \left| {x - z} \right|}} \leqslant \frac{{\left| {x - y} \right| + \left| {y - z} \right|}}<br />
{{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {x - y} \right|}}<br />
{{1 + \left| {x - y} \right|}} + \frac{{\left| {y - z} \right|}}<br />
{{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 \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 d(x,y)=\frac{|x-y|}{1+|x-y|} then to show that the triangle quality hold here you must show that \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 \mathbb{R}, automatically thought you must mess with the |x-y|'s.
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