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MVDT Proof in economics

  1. Mar 14, 2012 #1
    I'm trying to prove something small in economics using MVDT but can't seem to work it out:

    So consider: S(x) and D(x) such that:

    [itex] S' > 0 > D' [/itex]
    [itex]S(x_{1}) = D(x_{1}) = Q_{1} [/itex]
    [itex]S(x_{2}) + \epsilon = D(x_{2}) + \epsilon = Q_{2} [/itex]

    Then can I prove the following?
    [itex] x_{2} = x_{1} [/itex]
    [itex] Q_{2} > Q_{1} [/itex]

    I've tried to define a function that is the difference between S and D and then apply the mean value theorem, but it seems like I can't get where I want.

    By the way there may not be an answer since I made the problem up. But instinct tells me this is doable, I just can't figure out how.

    Thanks for your help.

  2. jcsd
  3. Mar 14, 2012 #2
    Intuitively I would think because S(x) is strictly increasing and D(x) is strictly decreasing, that if they intersect at one point ([itex]Q_1[/itex]) then they cannot intersect at another point. So if you assume your assumptions, I would say, yes, [itex]x_1 = x_2[/itex] and because of that, [itex]Q_2 > Q_1[/itex], namely, [itex]Q_2 = Q_1 + \epsilon[/itex].

    Here would be my pseudo proof: Show that [itex]x_1 = x_2[/itex] by contradiction (by assuming intersection at two points). Then show [itex]Q_2 = Q_1 + \epsilon[/itex] which implies [itex]Q_2 > Q_1[/itex].

    Someone else can chime in with a better proof idea if they want.
  4. Mar 15, 2012 #3
    Yes my intuition was also similar, I just wanted to formalize it. Thanks though!

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