Math Problems

[Najdorf]

Since schools out for the term, all my references are pretty much gone for a month or so. Been seein some math topics in the past, so thinking I'll post any questions over some math I'm doin on the break here; course, if anyone else has and Q&A themselves feel free :smokin:

Enough talk, got my first Q on the scan below, be needin that proofin. Headsup on the thanks!

http://i141.photobucket.com/albums/r59/Najdorf/Math%20Problems/9712212-1dd.jpg?t=1166612863

[Ketsuban]

What level math are you studying? That problem looks pretty advanced, perhaps even graduate level. I'd help you but I haven't quite got all my group theory down yet :/

[defcon]

you need grundle power. honey bbq flavored grundle power.

[Najdorf]

I'm sure its grad lvl, but its just for indie studies Im juicin. And where the fucks the grundle, local Lowes said they were out of stock!? McDs said they only hand honey bbq cunt sauce nothin worthy of sir ronins damn time.

[8S8]

Its a trick question.

[Rufus]

It's been a while since I've done any abstract algebra. You could try http://www.physicsforums.com/ which is likely to give you better results.

[Night]

I stopped reading after the first sentence lol. After graduating, I pretty much don't know anything anymore w00t!

[HoneyBBQGrundle]

you need grundle power. honey bbq flavored grundle power.


I did just take abstract algebra last year, but it's kind of a busy time of the year so I can't really start learning new or semi new definitions and proving shit today. Maybe over the next few days I can check it out, but I think the OP is best equipped to prove it still, since it's so fresh on his mind.

Maybe warlock or rsigley, etc can help. In the meantime try planetmath.com or some other more specific board

[Adam Warlock]

I did just take abstract algebra last year, but it's kind of a busy time of the year so I can't really start learning new or semi new definitions and proving shit today. Maybe over the next few days I can check it out, but I think the OP is best equipped to prove it still, since it's so fresh on his mind.

Maybe warlock or rsigley, etc can help. In the meantime try planetmath.com or some other more specific board

I did abstract/advanced Calc. I'm not really familiar with some of these definitions.

[thurst]

i'm 80% positive that question is from your take home final

[Bluehaze]

I'm not too sure if this is correct, so if someone sees anything wrong with it, please don't hesitate to let me know lol ><

But regardless, I think the proof should go something like this:

Suppose that X does not cross Y. We can replace one or both of X and Y by its complement if needed, and we can assume that X ∩ Y projects to a finite subset of E\G. The fact that Y is non-trivial implies that E\Y is an infinite subset of E\G, so there is a point z in E\Y which is not in the image of X ∩ Y. Now we need to use some choice of generators for G and consider the corresponding Cayley graph T of G. The vertices of T are identified with G and the action of G on itself on the left extends to an action on T. We consider z and the image of X ∩ Y in the quotient graph E\T. As X ∩ Y has finite image, there is a number d such that each point of its image can be joined to z by a path of length at most d. As the projection of T to E\T is a covering map, it follows that each point of X ∩ Y can be joined to some point lying above z by a path of length at most d. As any point above z lies in X*, it follows that each point of X ∩ Y can be joined to some point of X* by a path of length at most d. Hence, each point of X ∩ Y lies at most distance d from deltaX. Thus, the image of X ∩ Y in A\T lies within the d–neighborhood of the compact set delta(A\X), and so must itself be finite. If this is all correct, then it would seem to follow that Y does not cross X, a symmetry result. m^5

I hope this helps and good luck with your math this break ^_^