Results 1 to 3 of 3

Math Help - Euclidean metric is a metric

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    Euclidean metric is a metric

    I am trying to prove that the euclidean metric is a metric. I was able to prove the first three properties easily but the fourth is giving me trouble.

    I have to prove that

    \displaystyle\sqrt{\sum_{i=1}^{n}(x_i-z_i)^2}\leq \sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}+\sqrt{\sum_{i=1}^{n}(y_i-z_i)^2}

    Now in the outlined proof they ask you to first prove the CBS inequality and obtain the form

    \displaystyle\sum_{i=1}^{n}x_iy_i \leq \sqrt{\left(\sum_{i=1}^{n}x_i^2\right)\left(\sum_{  i=1}^{n}y_i^2\right)}

    And I've done this.

    What I don't get is the jump from this to the proof of the fourth condition.

    Could someone give me a hint (I'd actually prefer a hint rather then the whole solution if possible or a guide I want to get as much of this on my own as possible.)

    Also, do I need to use \sqrt{a+b}\leq \sqrt{a}+\sqrt{b} and if so how should I approach proving that?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    d^2(x,z)=\displaystyle\sum_{i=1}^n{(x_i-z_i)^2} =\displaystyle\sum_{i=1}^n(x_i-y_i+y_i-z_i)^2=

    \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)

    Now, use Cauchy Schwartz inequality:

    \displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)\leq \left|\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)\right|\leq

    \sqrt{\displaystyle\sum_{i=1}^n{(x_i-y_i)^2}}\sqrt{\displaystyle\sum_{i=1}^n{(y_i-z_i)^2}}

    You'll obtain:

    d^2(x,z)\leq \left(d(x,y)+d(y,z)\right)^2


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    d^2(x,z)=\displaystyle\sum_{i=1}^n{(x_i-z_i)^2} =\displaystyle\sum_{i=1}^n(x_i-y_i+y_i-z_i)^2

    Makes everything crystal clear.

    So from

    \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)

    We get

    \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\sqrt{\displaystyle\sum_{i=1}^n{(x_i-y_i)^2}}\sqrt{\displaystyle\sum_{i=1}^n{(y_i-z_i)^2}}

    Which becomes

    \displaystyle\sum_{i=1}^n{(x_i-z_i)^2\leq\left(\sqrt{\displaystyle\sum_{i=1}^n(x_  i-y_i)^2}+\sqrt{\displaystyle\sum_{i=1}^n(y_i-z_i)^2}\right)^2

    \sqrt{\displaystyle\sum_{i=1}^n{(x_i-z_i)^2}}\leq\sqrt{\displaystyle\sum_{i=1}^n(x_i-y_i)^2}+\sqrt{\displaystyle\sum_{i=1}^n(y_i-z_i)^2}

    Thank you so much for the help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: September 17th 2011, 04:44 PM
  2. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2010, 03:04 PM
  3. Show that the Euclidean plane is a metric space?
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 27th 2010, 05:42 AM
  4. Sets > Metric Space > Euclidean Space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 25th 2010, 11:17 PM
  5. standard metric and discrete metric
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 24th 2009, 08:25 AM

/mathhelpforum @mathhelpforum