1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Triangle Inequality for a Metric

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove the triangle inequality for the following metric [itex]d[/itex]

    [itex] d\big((x_1, x_2), (y_1, y_2)\big) = \begin{cases}
    |x_2| + |y_2| + |x_1 - y_1| & \text{if } x_1 \neq y_1 \\
    |x_2 - y_2| & \text{if } x_1 = y_1
    \end{cases}, [/itex]

    where [itex]x_1, x_2, y_1, y_2 \in \mathbb{R}.[/itex]

    2. Relevant equations

    We may assume the triangle inequality for real numbers. That is, [itex]|x + y| \leq |x| + |y|[/itex] for [itex]x, y \in \mathbb{R}[/itex]
    [itex]d[/itex]

    3. The attempt at a solution

    We wish to show that

    [itex]d\big( (x_1, x_2), (z_1, z_2) \big) \leq d\big( (x_1, x_2), (y_1, y_2) \big) + d\big( (y_1, y_2), (z_1, z_2) \big)[/itex]

    We may write
    [itex]d\big( (x_1, x_2), (z_1, z_2) \big) \leq |x_2| + |z_2| + |x_1 - z_1|[/itex]
    [itex]d\big( (x_1, x_2), (y_1, y_2) \big) \leq |x_2| + |y_2| + |x_1 - y_1|[/itex]
    [itex]d\big( (y_1, y_2), (z_1, z_2) \big) \leq |y_2| + |z_2| + |y_1 - z_1|[/itex],
    although I am not sure if this is helpful.

    There are essentially two cases to consider: (i) [itex]x_1 = y_1[/itex] and [itex]y_1 = z_1[/itex], and (ii) [itex]x_1 \neq y_1[/itex] or [itex]y_1 \neq z_1[/itex]

    I can do the first case, since [itex]|x_2 - z_2| = |x_2 - y_2 + y_2 - z_2| \leq |x_2 - y_2| + |y_2 - z_2|[/itex], which completes this case.

    I have problems with the second case though. I feel like there is some trick that I am missing. Or maybe I shouldn't be handling cases at all? Any help would be appreciated!
     
  2. jcsd
  3. Feb 2, 2012 #2
    For your case (ii) there are 3 cases
    1. x1≠y1≠z1, where
    d1+d2=|x1-y1|+|y1-z1|+|x2|+|y2|+|y2|+|z2|
    ≥|x1-y1+y1-z1|+|x2|+0+0+|z2|
    =d
    2. x1=y1≠z1, where
    d1+d2=|x2-y2|+|y1-z1|+|y2|+|z2|
    =|x2-y2|+|y2|+|x1-z1|+|z2|
    ≥|x2-y2+y2|+|x1-z1|+|z2|
    =|x2|+|x1-z1|+|z2|
    =d
    3. x1≠y1=z1, same as 2
     
  4. Feb 2, 2012 #3
    That makes much more sense. Thank you so much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Triangle Inequality for a Metric
  1. Triangle inequality (Replies: 10)

  2. Triangle inequality (Replies: 2)

  3. Triangle Inequality (Replies: 2)

Loading...