A proof by contradiction is where you assume two things, both the hypothesis and the negation of the conclusion. Then, you prove any false statement by sound logic.
So, if for some reason I decided I was trying to direct proof this (even though tit's not the easiest way, clearly) I would start by re-writing A disjoin, not-empty subsets into something and trying to make it appear to prove the latter?
Although, righ toff the bat that looks like an enormous taks since I wouldn't even know how to break that down symbolically :)
So, the compliment of B would be A, obviously. And A-A = empty set This is because B cannot share elements with A, which means anything in its complement must have been in A.. or is there holes in that logic? "
I tried.. somewhat to explain something along those lines above :)
@VaughanHilts The "direct" method you're talking about is called a conditional proof. Assume $\psi$. Derive $\phi$. Then we are able to conclude $\psi \rightarrow \phi$ is true.
@VaughanHilts look: x.is in A, but not in B^c, then x is in B, but x.is in A, but ee are saying x is in B! Oh noes, contradiction, because they are ....
@VaughanHilts Yes. It a biconditoinal so you should prove both directions. Using the cases of $x = 2k$ and $x=2k+1$ show the relationship from left to right. Then do the same thing from right to left.
@VaughanHilts So it turns out ($3x + 1$ is even) is a false statement when $x = 2k$. What does this say about the statement If ($3x +1$ is even), then ($5x-2$ is odd)?
@VaughanHilts It's suggestive enough for me, but I don't know how strict you need to be
whatever you have to do to show that the right side is false (just trying to think more abstractly so that the general idea is clear, not just for this particular example)
If $f : X \to Y$ is a linear map then for any convex set $H \subset X$, $f_H: H \to f(H)$, inverse images of extreme subsets of $f(H)$ under $f_H$ are extreme subsets of $H$. I just thought you guys should know :)
@KevinDriscoll (my previous answer was wrong.) But the minimal path would be the same path taken by a negatively charged particle at A with a positively charged particle at B
@KarlKronenfeld That's what I'm trying to figure out - some constraint on the Lagrangian that forces the particle onto a path across the blue stuff colinear with the center of the circle.
@KevinDriscoll If I put two oppositely charged particles somewhere in space near each other, one of them fixed, will the unfixed one not move towards the fixed one by the shortest path possible?
@Bitrex I am still not sure how the 'blue stuff' will affect this though. Its certainly true for 2 guys in free space..... but put some other mass distribution between them and I don't know
@Bitrex actually I do know. The unfixed particle will accelerate straight towards the center of mass of the system, if it starts 'outside' the other mass dictributions
@KevinDriscoll The blue stuff is just a barrier that constrains the motion somehow. But even with it there the particle will still follow the shortest path, within the constraint set by the barrier
I just don't know how to formulate the constraint that it must stay on the same theta while it's in the barrier.
@KevinDriscoll I don't think the particle has to "know" anything. I think that by the principle of least action, the particle will take the shortest path from A to B, given whatever constraints I may put on its motion
@KevinDriscoll You were right. In the "gravity" interpretation, the particle will follow the shortest path to the outer ring. But then once it's constrained by the blue stuff, I see that all bets are off whether the total path length will be minimal.
@Pedro I still haven't started watching it. Maybe some day.... But I make the same joke, but instead I wonder if the instructions will be in "Canadian"
(I'm sure Jesse didn't intend it as a joke, being a bit of a dunce)
