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Homework Help: Prove that if m is the average of x1,x2, ,xk, then

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    http://img440.imageshack.us/img440/9242/captureyeq.jpg [Broken]

    Uploaded with ImageShack.us

    2. Relevant equations

    Given in problem statement.

    3. The attempt at a solution

    Basically I've come up with the contrapositive, but I'm unsure of how to proceed with the proof. Can anyone help?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 1, 2011 #2
    So x_1, x_2, ..., x_k are all less than m. You're trying to show that m cannot be (x_1 + x_2 +...+ x_k)/k. What can you say about the sum x_1 + x_2 +... + x_k?
  4. Nov 1, 2011 #3

    (x_1 + x_2 +... + x_k) is greater than (x_1 + x_2 +...+ x_k)/k ?

    thanks for the help btw :)
  5. Nov 1, 2011 #4
    Here are some more hints:

    Yes, (x_1 + x_2 +... + x_k) is greater than (x_1 + x_2 +...+ x_k)/k. But I think that's true in general, since basically whenever you divide something by a positive integer greater than 1 you make it smaller. But we want to somehow use the fact that x_1, x_2,...,x_k are all less than m. Is there anything you can say about the sum that involves m? Remember that we're trying to show that m cannot be the sum divided by k. Think it over for a while and see if you can come up with something. Also if you really get stuck, maybe play around with some examples, like k = 5 or something and then make something up for x_1 to x_k...I dunno. Anyway, good luck!
  6. Nov 1, 2011 #5
    Those are exactly my thoughts. It makes sense intuitively, I just don't know how to put it down in "math speak" haha.

    If anyone can help that'd be sweet...
  7. Nov 1, 2011 #6

    OK, one more hint until I'm off to do other things:
    Each of the x's are less than m. And there are k of them. So when you add all the x's together, what can you say? I mean, you want to show that m cannot be the sum of the x's divided by k. That's the same as: m times k is not the sum of the x's.
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