Chess meet?

[chris_venturini]

Challenge me and the rest of HAI in the ultimate match of wits. Come, unless your a pussy.
BYOCB (bring your own chess board)
So whos in?
we'll plan the details later, i just want to see if theres any interest in this kind of thing.

[-=[Juztin]=-]

I'd be down for some Chess, but I'd have to stop by walmart for a cheap board seeing how I threw my last one away years ago

[chris_venturini]

just so you can lose again? like in starcraft.

(the shit talking has officially begun)

[SleeperZ]

I might be interested. Where are you getting together?

[tsitim]

count me in just depends when and where ;)

[David]

tim, why do you wink after every post? do you have an eye problem

[tsitim]

it's a birth defect ;)

[celica man]

tim, why do you wink after every post? do you have an eye problem
:rofl:

[V8SpankR]

Chess??? Boy,am I glad race season's almost here. :D

[wild.irish]

i might join, i like playing chess, but i'm not really good at it.

[Bedlam]

People still play that game? Damn..

-Bedlam

[tsitim]

;)

[celica man]

;)
+1

http://nutria.msn.no/gallery/emoticons/wink-.jpg

[celica man]

+2

http://www.denverracing.com/~aracheon/smilies/jerkoff1.gif ;)

[tsitim]

+1

http://nutria.msn.no/gallery/emoticons/wink-.jpg
hell yea I'm going to start using that http://nutria.msn.no/gallery/emoticons/wink-.jpg

[-=[Juztin]=-]

Chris I will whoop your ass in Chess, there aint no water to protect yo azz. Be sure to bring a barf bag cause your gonna hurl w/ all the humiliation you'll rake up trying to play chess.

[stu]

I"m not very good at chess, I guess that means I'll come.

[DrJones]

this better not be a joke because it sounds really fun.... anyone here play Go? I can bring my board.

[Storm Girl]

OH great....look what you started!

The winking reminds me of that commercial with the guy and the twitching eye....."Don't touch me"

[SleeperZ]

OH great....look what you started!

The winking reminds me of that commercial with the guy and the twitching eye....."Don't touch me"

Or Dr. Strangelove...

[foxxof]

im down for chess. i hardly have time for a game of tic tac toe lately since im buying a house but id love to whoop some of yer asses! ;)

damnit i think im catching the wink virus

[STIBungy]

Are we doing drunken chess matches too?

[stu]

This is like a Lan for POOR PEOPLE!!! :rofl: I'll be there! *wink*

[David]

Are we doing drunken chess matches too?
yea, sign me up for that league ;)

[chris_venturini]

this better not be a joke No joke, im dead serious. I havent played in a while, but before computers chess was one of my favorite pastimes. Yes, i was in chessclub. :suckit:

[Weston]

My first move will be E2 to SALSA IN YOUR EYE! bahahahaha! :rofl:

[DrJones]

No joke, im dead serious. I havent played in a while, but before computers chess was one of my favorite pastimes. Yes, i was in chessclub. :suckit:

Well well well... looks I have you beat yet again. Not only was I in both the chess and math clubs... but I also founded them and was president.

:fu:

[chris_venturini]

your still upset because you know im going to solve the computational complexity theory before you do. Then with the one million dollar prize i can purchase 4 million little JJJJuan burritos.

[-=[Juztin]=-]

which makes you a.... wanker

[-=[Juztin]=-]

damn u chris and your quantum posting, you phucked up my joke! hahaa, I could just see us playing chess drunk...

[DrJones]

your still upset because you know im going to solve the computational complexity theory before you do. Then with the one million dollar prize i can purchase 4 million little JJJJuan burritos.


How are you going to prove P = NP (or P != NP ) when you can't even show how the set of rational numbers can be represented using a subset of the positive integers... :rolleyes:

[chris_venturini]

im not even going to respond to that last statement because, for lack of better words, you know i showed you whats up biatch!

these are all big words for someone who didnt even show up to the LAN party after days of shit talking. *pussy*

[chris_venturini]

you can't even show how the set of rational numbers can be represented using a subset of the positive integers... :rolleyes:I already answered that question
From the point of view of pure set theory, the most basic question about a set is: How many elements does it have? It is a fundamental observation that we can define the statement “sets A and B have the same number of elements” without knowing anything about numbers.
Definition. Sets A and B have the same cardinality if there is a one-to-one function f with domain A and range B. We denote this by |A| = |B|. Definition. The cardinality of A is less than or equal to the cardinality of B (notation: |A| |B|) if there is a one-to-one mapping of A into B.
Notice that |A| |B| means that |A| = |C| for some subset C of B. We also write |A| < |B| to mean that |A| |B| and not |A| = |B|, i.e., that there is a one-to-one mapping of A onto a subset of B, but there is no one-to-one mapping of A onto B.
Lemma. If |A| |B| and |A| = |C|, then |C| |B|. If |A| |B| and |B| = |C|, then |A| |C|. |A| |A|. If |A| |B| and |B| |C|, then |A||C|.

i havent dealt much with cardinality, I understand that the cardinality of the rationals is the same as that of the integers and that is the introduction to diagonalization, but realy i never went too far into it

[DrJones]

I already answered that question
From the point of view of pure set theory, the most basic question about a set is: How many elements does it have? It is a fundamental observation that we can define the statement “sets A and B have the same number of elements” without knowing anything about numbers.
Definition. Sets A and B have the same cardinality if there is a one-to-one function f with domain A and range B. We denote this by |A| = |B|. Definition. The cardinality of A is less than or equal to the cardinality of B (notation: |A| |B|) if there is a one-to-one mapping of A into B.
Notice that |A| |B| means that |A| = |C| for some subset C of B. We also write |A| < |B| to mean that |A| |B| and not |A| = |B|, i.e., that there is a one-to-one mapping of A onto a subset of B, but there is no one-to-one mapping of A onto B.
Lemma. If |A| |B| and |A| = |C|, then |C| |B|. If |A| |B| and |B| = |C|, then |A| |C|. |A| |A|. If |A| |B| and |B| |C|, then |A||C|.

i havent dealt much with cardinality, I understand that the cardinality of the rationals is the same as that of the integers and that is the introduction to diagonalization, but realy i never went too far into it


I don't think that works at all. What is your explanation of the |A| = |C| for some subset C part? Are you saying that there exists a subset C for which |A| = |C|, or that |A| = |C| for any subset C? Is A the set of rationals and B is the set of positive integers? You haven't really done much there... you can't prove that any set has the same cardinality as any other set. In this case you need to look at the specifics of those sets.

[stu]

If you fags keep this up, you'll be the only two at your damn chess meet. :mad:

[Weston]

I'm picturing us getting drunk and throwing pieces at each other. :rofl: