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Set of Ordered pairs question

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    (!) Determine the set of ordered pairs (x,y) of nonzero real numbers such that x/y + y/x >= 2.


    2. Relevant equations

    x/y + y/x >= 2

    3. The attempt at a solution
    Relatively new to set notation and proving so, I merely am seeking reassurances that what I am doing is correct.

    x/y + y/x >= 2
    *Made everything a common denominator*
    x^2/xy + y^2/xy >=2xy/xy

    (x^2+y^2-2xy)/(xy) >= 0
    (x-y)^2/(xy)>=0
    (x-y)^2 >= 0
    x-y >= 0
    x >= y

    Set would be written therefore as:
    [tex]\left\{\left(x,y\right) \in\Re: x \geq y\right\}[/tex]

    But x and y cannot equal zero (not sure how to depict that in set notation).

    Sincerely,

    NastyAccident
     
  2. jcsd
  3. Aug 26, 2010 #2

    Mark44

    Staff: Mentor

    The inequality above is true for all real x and y.
     
  4. Aug 26, 2010 #3
    I think the answer is actually all positive real numbers. X need not be greater than Y. To see this set t = x/y then the equation becomes
    [tex] t+ \frac{1}{t}\geq2[/tex]
    [tex] t^{2} -2t+1\geq0[/tex]
    [tex] (t-1)^{2}\geq0[/tex]
    Drawing the parabola you can see that x and y simply have to be positive and non zero and that is sufficient. We have to disregard negative t's for obvious reasons.

    To show you an example consider x=0. 5 and y=1. This satisfies the inequality but x is less than y.
     
    Last edited: Aug 26, 2010
  5. Aug 26, 2010 #4

    Mark44

    Staff: Mentor

    Starting from x/y + y/x >= 2, when you multiply both sides by xy, you need two cases: one for which xy > 0, and another for which xy < 0. You get two different inequalities in these two cases.
     
  6. Aug 26, 2010 #5
    I just realized that x and y can be negative but they have to negative together so that t is not negative. So the actual answer is all real numbers such that x/y is positive.
     
  7. Aug 27, 2010 #6
    So, cases:

    1.) xy <= 0

    2.) xy >= 0

    x/y + y/x >= 2
    x^2/xy + y^2/xy >=2xy/xy
    (x^2+y^2-2xy)/(xy) >= 0
    (x-y)^2/(xy)>=0
    (x-y)^2 >= 0

    The first case contains all real numbers in the first and third quadrants so results like (1,1), (-3,-1), et cetera work. The second case contains all real numbers in the second and fourth quadrants so results like (-2,1) and (3,-4) work.

    However, wouldn't you have to evaluate the original inequality and determine that (0,0) would not work due to both x and y being found on the denominator?

    If this is not the case would the resulting set simply be the union between the two case? Or my original guess: [tex]
    \left\{\left(x,y\right) \in\Re: x \geq y\right\}
    [/tex]

    Sincerely,

    NastyAccident
     
  8. Aug 27, 2010 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You have multiplied the first inequality with xy to get the new one in the second line. It is true if xy>0, but fails when xy<0. Multiplying an inequality with a negative number will reverse it. So it is (x-y)^2 <= 0 if one of x and y is negative, the other positive.
    (x-y)^2 is always positive so this second inequality cannot be true.

    The second case does not work. Yes, you should check in the original inequality x/y + y/x >= 2 if it is true for (-2,1) and (3,-4) ?

    ehild
     
  9. Aug 27, 2010 #8
    Okay, so I think I follow but I want to make sure that my train of thought is correct.
    We have one case:
    (x-y)^2/(xy)>=0
    with two subcases:
    xy<0
    xy>0

    There is no >= in the sub cases since zero no matter what cannot be in the denominator.

    When you introduce a negative x or negative y, the first sub case fails since that turns (x-y)^2/(xy) from being greater than or equal to zero to being less than or equal to zero causing the inequality to no longer be true.

    In the second subcase, xy>0 holds true when both x & y are positive and when both x&y are negative since a negative times a negative is a positive.

    Hence, the set contains values of x & y when they are both positive and the set contains values of x & y when they are both negative.

    So, if that is the correct conclusion then the proper set notation when therefore be:
    [tex]\left\{(x,y) \in\Re: -\infty < x < 0 and -\infty < y < 0 or 0 < x < \infty and 0 < y < \infty \right\}[/tex]

    Sincerely,

    NastyAccident
     
  10. Aug 27, 2010 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct now.


    ehild
     
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