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Triangle ineuality, in need for help.

  1. Jan 12, 2006 #1
    i need help in proving this essential ineuality which i don't know how to prove (quite trivial isn't it):
    ||x|-|y||<=|x-y|

    i know that my first trick is this:
    |x|=|(x-y)+y|<=|(x-y)|+|y|
    |x|-|y|<=|x-y|
    and iv'e been told to do so also with |y| and i get to this:
    |y|-|x|<=|y-x|
    and then adding both inequalities give us with this:
    |x-y|>=-|y-x|

    here i'm pretty much puzzled, so can you help, or add some insight or hints.

    btw, i'm in a hurry so if i have angried anyone by posting here, sorry!
     
  2. jcsd
  3. Jan 12, 2006 #2
    You were almost there! You just needed this little step:
    |y|-|x| <= |y-x|
    => |x|-|y| >= -|y-x| = -|x-y|

    So you have -|x-y|<=|x|-|y|<=|x-y|.
     
  4. Jan 12, 2006 #3

    mathman

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    Another way of seeing it is that ||x|-|y||=|x|-|y| or =|y|-|x|, depending on which is bigger , |x| or |y|.
     
  5. Jan 12, 2006 #4

    Galileo

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    You don't have to add the inequalities. You've shown that for any x and y that |x-y| is greater-equal to both |y|-|x| and |x|-|y|. So isn't it obvious it's also greater-equal to the absolute value of |y|-|x|?
     
  6. Jan 13, 2006 #5
    apparently no, you need to prove up untill you can say q.e.d or מ.ש.ל
     
  7. Jan 13, 2006 #6
    i think you also jumped too quickly here, don't you need to divise the options when |y-x|= y-x when y-x>=0 and |y-x|=x-y when y-x<0 and then you can see that -|y-x|=-|y-x|.

    ok, i get it.
    thank you.
    a great site we have here, please don't change physicsforums. (-:
     
  8. Jan 13, 2006 #7

    arildno

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    Remember that ||x|-|y||=max(|x|-|y|,|y|-|x|)
    Reconsider Galileo's post.
     
  9. Jan 13, 2006 #8
    well, arlidno i think your notation is with regard to vector calclulus which iv'e seen your notation from, but my question is about vector calclulus and this is why i don't think it's related here, but you are the expert here not me.

    cheers.
     
  10. Jan 13, 2006 #9

    Galileo

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    So are x and y vectors or numbers? (Note: It doesn't matter!)

    Ok, if you have some number a that is larger than b and larger than -b, you don't think it's obvious it's also larger than |b|?

    Same thing here: You have some number a=|x-y| larger than b=|x|-|y| and larger than -b=|y|-|x| (or equal ofcourse). So a>=|b| or |x-y|>=||x|-|y||

    Also, reread arildno's post. Clearly |a| is either a or -a, whichever is positive and thus the mximum max(a,-a) of a and -a.
     
  11. Jan 13, 2006 #10
    ok i understand your reasoning, but then again this question if we would be on topic was a help on pure mathematical question, and the real question is:
    will the checker approve this obviously logical assertion, this i don't know.

    but it looks that your way is the same as of devious and mathman has proposed, i myself not yet convinced it's any different, just other notation.

    but i'm willing to be disproven.
     
  12. Jan 13, 2006 #11
    Or you can just square both sides:

    [tex]||x|-|y||^2=(|x|-|y|)^2=x^2-2|x||y|+y^2[/tex]
    [tex]|x-y|^2=(x-y)^2=x^2-2x\cdot y+y^2[/tex]

    where of course, multiplication is the scalar dot product in Euclidean space, or more generally the inner product.
     
  13. Jan 18, 2006 #12
    cant you just square both sides?
     
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