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Thread: Verify that it defines a distance

  1. #1
    Member thaopanda's Avatar
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    Verify that it defines a distance

    Verify that d : $\displaystyle R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\displaystyle \frac{|X-Y|}{1+|X-Y|}$ for $\displaystyle X,Y \in R^3$ defines a distance on $\displaystyle R^3$

    So, I haven't actually learned this yet in class... so I don't really understand what this is saying...
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  2. #2
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    Quote Originally Posted by thaopanda View Post
    Verify that d : $\displaystyle R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\displaystyle \frac{|X-Y|}{1+|X-Y|}$ for $\displaystyle X,Y \in R^3$ defines a distance on $\displaystyle R^3$
    So, I haven't actually learned this yet in class... so I don't really understand what this is saying...
    You must show three properties.
    $\displaystyle \begin{gathered}
    d(X,Y) = 0\text{ if and only if }X=Y \hfill \\
    d(X,Y) = d(Y,X) \hfill \\
    d(X,Y) \leqslant d(X,Z) + d(Z,Y) \hfill \\
    \end{gathered} $
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    Verify that d : $\displaystyle R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\displaystyle \frac{|X-Y|}{1+|X-Y|}$ for $\displaystyle X,Y \in R^3$ defines a distance on $\displaystyle R^3$

    So, I haven't actually learned this yet in class... so I don't really understand what this is saying...
    Quote Originally Posted by Plato View Post
    You must show three properties.
    $\displaystyle \begin{gathered}
    1.~d(X,Y) = 0\text{ if and only if }X=Y \hfill \\
    2.~d(X,Y) = d(Y,X) \hfill \\
    3.~d(X,Y) \leqslant d(X,Z) + d(Z,Y) \hfill \\
    \end{gathered} $
    The first two are trivial. The third may be a bit tricky to prove if you get hung up on the absolute value signs. All you need to know is that $\displaystyle |X-Y|$ is a metric. It would be easiest to label them: $\displaystyle |X-Y|=a$, $\displaystyle |Y-Z|=b$, $\displaystyle |X-Z|=c$. You know from the triangle inequality that $\displaystyle c\leq a+b$.

    So now here's what you have to prove:

    $\displaystyle \frac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}\impli es\frac{a}{1+a}+\frac{b}{1+b}-\frac{c}{1+c}\geq0$ given that $\displaystyle c\leq a+b\implies a+b-c\geq0$.

    Spoiler:
    Make common denominators and combine the fractions: $\displaystyle \frac{a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)}{(1+a)(1+b)(1+c)}$.

    You only have to show that the numerator is positive, as the denominator is clearly positive (because $\displaystyle a,b,c>0$).

    Multiply out the numerator: $\displaystyle a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)$ $\displaystyle =a+ab+ac+abc+b+ab+bc+abc-c-ac-bc-abc$ $\displaystyle =a+b-c+2ab+abc$

    Now we use the fact that $\displaystyle a+b-c\geq0$ to say that $\displaystyle a+b-c+2ab+abc\geq 2ab+abc$.

    Clearly $\displaystyle 2ab+abc\geq0$, because $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ are all positive. So we're done.
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  4. #4
    Member thaopanda's Avatar
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    that was a lot easier than I thought it was, thanks! I think it was all the R's and the wording that kinda threw me off
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  5. #5
    Member thaopanda's Avatar
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    actually, I'm still kinda lost on the third part... I don't really understand how to prove the triangle inequality
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