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Math Help - inequality problem

  1. #1
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    inequality problem

    The real number c is positive and the real numbers a and b are such that ab > 0 and a > b. How can I disprove the case a - c < b + c for all such a, b, and c? (The case \frac{c}{a} < \frac{c}{b} works). When I tried it, I can't seem to get numbers that disprove it??
    Last edited by mr fantastic; March 23rd 2010 at 03:24 AM. Reason: Restored deleted question.
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  2. #2
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    Quote Originally Posted by donnagirl View Post
    The real number c is positive and the real numbers a and b are such that ab > 0 and a > b. How can I disprove the case a - c < b + c for all such a, b, and c? (The case \frac{c}{a} < \frac{c}{b} works). When I tried it, I can't seem to get numbers that disprove it??
    How hard did you try? How about a= 100, b= 1, c= 1?
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  3. #3
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    Hello donnagirl
    Quote Originally Posted by donnagirl View Post
    The real number c is positive and the real numbers a and b are such that ab > 0 and a > b. How can I disprove the case a - c < b + c for all such a, b, and c? (The case \frac{c}{a} < \frac{c}{b} works). When I tried it, I can't seem to get numbers that disprove it??
    I'm not quite sure I understand the problem here, because it seems very easy to find a counter-example. In order that a-c>b+c, we must ensure that a-b > 2c. So, for instance:
    a=10, b = 1, c = 3
    gives a-c=7 and b+c = 4. So a-c> b+c.

    Grandad
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  4. #4
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    Thanks again Grandad!
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