# MVDT Proof in economics

1. Mar 14, 2012

### Bipolarity

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:

$S' > 0 > D'$
$S(x_{1}) = D(x_{1}) = Q_{1}$
$S(x_{2}) + \epsilon = D(x_{2}) + \epsilon = Q_{2}$

Then can I prove the following?
$x_{2} = x_{1}$
$Q_{2} > Q_{1}$

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.

BiP

2. Mar 14, 2012

### scurty

Intuitively I would think because S(x) is strictly increasing and D(x) is strictly decreasing, that if they intersect at one point ($Q_1$) then they cannot intersect at another point. So if you assume your assumptions, I would say, yes, $x_1 = x_2$ and because of that, $Q_2 > Q_1$, namely, $Q_2 = Q_1 + \epsilon$.

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

Someone else can chime in with a better proof idea if they want.

3. Mar 15, 2012

### Bipolarity

Yes my intuition was also similar, I just wanted to formalize it. Thanks though!

BiP