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Calculus II - Real numbers proofing

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that |a-b|<= |y-a|+|x-y|+|x-b|, for all x,y in ℝ


    2. Relevant equations



    3. The attempt at a solution
    |a-b| <= |y-a+x-y+x-b| (correct? Not sure about this one...is it not part of the triangle rule?)
    |a-b| <= |2x-b-1|
    |a-b+2x-2x| <= |2x-b-1|
    |a-2x| + |2x-b| <= |2x-b-1| really have clue what I am doing here, no idea if I am on the right track...doubt it is right. Please help. Thanks!
     
  2. jcsd
  3. Jan 15, 2013 #2
    No, that's not correct at all.

    Start from the expression |a-b| and use the triangle inequality (I assume you've proved this already)

    Note that |a-b|*is the distance between points a and b. Therefore |a-b| ≤ |a-x| + |x-b|
     
  4. Jan 15, 2013 #3
    Oh sorry you are not supposed to start on the right hand side. But would it start out like this then:

    |a-b+x-x+y-y| <= |y-a| + |x-y| + |x-b|

    |a-b+x-x+y-y| <= |a-y| + |y-x| + |x-b|
    |a-y+y-x+x-b|<= |a-y| + |y-x| + |x-b|

    According to the triangle inequality thing this is true?
     
  5. Jan 16, 2013 #4
    Well, |a-b| is the left side but you can start on whichever side you prefer.

    Clearly it is true as you are asked to prove it. However you need to also give the proof.
     
  6. Jan 16, 2013 #5
    Wait so is the proof I shown above correct? That's what I meant.
     
  7. Jan 16, 2013 #6
    It certainly looks like you're on the right track, but you are still missing the actual proof. Why is LHS≤RHS? You should be able to take a form of triangle inequality (which I again assume has been proven in your lecture notes or somewhere) and use that explicitly to show that this holds.
     
  8. Jan 16, 2013 #7
    So I have to state that since |a+b| <= |a| + |b| therefore this is true?
     
  9. Jan 16, 2013 #8

    jbunniii

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    Your first line is already incorrect, as can be verified by choosing a = 2, b = 0, y = 0, x = 1. In that case the LHS is |a - b| = |2 - 0| = 2, and the RHS is |y - a + x - y + x - b| = |0 - 2 + 1 - 0 + 1 - 0| = 0.

    I suggest starting as follows: [itex]|a - b| = |a - y + y - x + x - b|[/itex]. Now think about how to apply the triangle inequality to the right hand side.
     
  10. Jan 16, 2013 #9
    Didnt I already say that? Please read the full post before posting please
     
  11. Jan 16, 2013 #10

    jbunniii

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    OK, it wasn't clear what you were doing in your second post. So you have:
    [tex]|a - b| = |a - y + y - x + x - b|[/tex]
    What if I suggestively add parentheses as follows:
    [tex]|a - y + y - x + x - b| = |(a - y) + (y - x + x - b)|[/tex]
    Now apply the triangle inequality to that.
     
  12. Jan 16, 2013 #11
    what do you mean by applying the trinagle inequality to that? what does the parentheses do? Sorry never seen parenthesis in these questions before.
     
  13. Jan 16, 2013 #12

    jbunniii

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    The parentheses were a hint to suggest the following: the triangle inequality says [itex]|u + v| \leq |u| + |v|[/itex]. So let [itex]u = a - y[/itex] and [itex]v = y - x + x - b[/itex].
     
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