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Proof the indices

  1. Apr 2, 2010 #1
    Let {x1, x2, x3, x4, x5} be distinct real numbers. Prove there are indices a, b with 0< xa-xb<1+xaxb.

    Seriously I have no idea how to even start...

    I tried subbing random numbers in... but nope...

    Can anyone give a hint?

    Hey wait, the sets do not need to be ordered right? Can I do this?

    Direct proof:

    {x1, x2, x3, x4, x5} = {-1, -2, -3, -4, -5}

    proven.
     
    Last edited: Apr 2, 2010
  2. jcsd
  3. Apr 3, 2010 #2
    Try to use the fact that we can put an order on the set, say x1<x2<..<x5.
     
    Last edited: Apr 3, 2010
  4. Apr 3, 2010 #3
    Then I dun see any combination that can prove it...
     
  5. Apr 3, 2010 #4
    You might want to start with xi-xj>0, for some i,j. You can make such a choice. I would be inclined to consider three cases separately: when the set consists of all positive real nr., all negative and a combination of both.
     
  6. Apr 3, 2010 #5
    Erm... how is tat possible...

    Example for xi - xj, if they are an ordered set...

    For a positive case... a smaller number - a bigger number I will get negative... which is smaller than 0

    For a negative case... a smaller number - a bigger number I will get negative... which is smaller than 0

    if I mix them up... which I can't...because if u say I should do the sets in order the set would look sumting like... {-2, -1, 0, 1, 2} <- juz an example, its not the right anwer. and since i and j have to be consective elements in the set... halfway thru' the set they would unmixed... and besides, I still get a negative number...
     
  7. Apr 3, 2010 #6
    The question is asking only two show that such indices exist, not to show that such a thing is true for all of them. So, given any two elements from your set, say x_i, x_j, since they are distinct, then precisely one of the following should hold: xi<xj or xj<xi.

    None of them holds a special status, so we can choose either one, say the secon holds, then certainly xi-xj>0.
     
  8. Apr 3, 2010 #7
    Okay... but I still don't see how it can be done if the set is ordered... like I alreadi explained above.
     
  9. Apr 3, 2010 #8
    I am working under the assumption that R has the simple order (the natural order).
     
  10. Apr 3, 2010 #9
    Yes I understand ur saying if u mean natural order as ...-1,-2,0,1,2,3,4...

    But I dun see how the logic can be proven true using this order?

    Do u have a set of number in mind tat can work? Coz' I can't see ani.
     
  11. Apr 3, 2010 #10
    {1,2,3,4,5}={x1,x2,x3,x4,x5}

    0<x3-x1=2<1+3*1=1+x3*x1

    However, the point here is to show that this works for any such set.
     
  12. Apr 3, 2010 #11
    hold on a sec!!

    when it says indices, a, b.

    I am pretty sure it doesn't meant any 2 elements in the set.

    But 2 consecutive elements in the set?

    Fine, I mistyped abit, they actually stated i, j instead of a, b. if i, j actualli means anything...
     
  13. Apr 3, 2010 #12

    Mark44

    Staff: Mentor

    No, they don't have to be consecutive. Here is a revision of the problem description you posted at the beginning of this thread.

    I interpret this to mean that the inequality doesn't have to hold for all choices of i and j, but it has to hold for at least one choice of i and j. It doesn't say anything about i and j being consecutive values.
     
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