Bahamut

Well, if you're not staying for the overnight, you don't need to bring the Wii & stuff - that was for the overnight.

are you over the age limit to sign up for these forums? He likes Kalmah, the power metal band, so I'm going to have to go with yes. http://www.metal-archives.com/band.php?id=702

are you over the age limit to sign up for these forums? He likes Kalmah, the power metal band, so I'm going to have to go with yes.

Oh actually, I do have a suggestion!

Bump! Updated the first post. Also, with such a fun group as OCRers, there should be plenty of things to do at night ...and no, not alcohol related.

mememe

I have no idea how the unlockables thing works. I *did* do most of the quests though, including beating Nest of Evil...

Actually, Athair has a Wii too . I think he wants to stay over too, but I forget. He'd actually offer his place but he said it's too small (he lives in White Plains). I'll edit the first page with the list of people interested in crashing overnight at zircon's after I come back from this FINAL BOSS.

Eh, if you own a PS3, you probably own a PS2 anyway.

I like how your only posts are posts related to meetups . In any case, I was ahead of you .

Well, looks like we'll be seeing Atma though so shall I put you down?

Two words: Thousand Blade Hmm? Clarify. I actually figured out what you meant sometime later when looking at the special moves. I use Greatest Heroes so all the other moves are kinda meh to me.

Wow Dracula is incredibly hard as Richter...

I vote for brunch at Pizza Brew though!

I'll bring a sleeping bag to stay over too! ...But I'd feel a bit sheepish walking around NYC carrying a big bag, one of the items being a sleeping bag. Is it possible if I could drop it off earlier sometime?

Oh I have Advance Wars DS and Metroid Prime: Hunter too.

I have Tetris & Mario Kart too. Got Castlevania for some multiplayer action too.

Well, looking at J&R's online music catalog, it seems to show a somewhat weaker selection than Virgin - Virgin just seems to have some rare stuff that is hard to find here in the US (i. e. J&R only carries Rhapsody's newest CD). So, how many people going to the meetup have DSs btw? Bring them if you have them.

Just finished Intro to Abstract Algebra I. My professor freaking sucked. Fortunately his tests were ridiculously easy, so I got an A without learning a single thing. Let's see if I can figure this out... I'll let C=(A intersect . To show G/C is abelian, I must show that ab=ba for a,b in G/C. a=gC and b=hC for g,h in G. ab=gChC=gCh as C is normal. gCh=(gAh intersect gBh)=(hAg intersect hBg)=hCg=hCgC=ba. Done! Eh. Does that work? I've never had to write a solid proof for Abstract Algebra before, so I'm not really confident in it. Alright, so first you have to show that A intersect B is normal in G. Let x be in A intersect B. When you conjugate x by any element g in G, then x will be in A by normality of A and in B by normality of B, so x is in the intersection of A and B. Next, let x(A intersect and y(A intersect be in G/(A intersect (coset representation). Then xy(A intersect = yx(A intersect B since xyA = yxA and xyB = yxB, and the intersection of xyA and xyB is xy(A intersect and similarly yxA intersect yxB = yx(A intersect . Your argument fails because gC * hC = (gh)C by definition of coset multiplication in a quotient group. Another problem! Suppose U is a finite multiplicative subgroup of a finite field F. Prove that U must be cyclic.

See, minor mistakes in writing what I have in mind like that is what costs me huge on tests .

Well, think of a simple example, like the integers Z. Z, 2Z, 3Z, etc. are all subgroups of Z, and in those cases, they are isomorphic to Z sub n, the cyclic group under addition modulo n. Also, there exists other types of infinite groups, such as the infinite alternating group. Mathematicians treat infinity differently though - there are different levels of infinity. Set size is denoted by the cardinality of a set, and two sets have the same cardinality if there exists a bijection between them. The partial order that is assigned to cardinality is the less than or equal to symbol, in which a set A has smaller cardinality than a set B if there exists an injection from A to B. When talking about infinite sets though, you can have more interesting things happen. For example, take the set of natural numbers, and take the subset of all even natural numbers. Surely there exists a bijection between the two (notably mapping every natural number n to 2n), so they have the same cardinality, even though the set of all even natural numbers is surely a subset of the natural numbers. So, the cardinality of the natural numbers is the smallest cardinality that a set can have. Also noteworthy is that the real numbers have a strictly larger cardinality than the natural numbers, since the interval (0, 1) has a bijection into the real numbers, and the cardinality of (0, 1) is not equal to the cardinality of the natural numbers since there does not exist a bijection between the two (the famous Cantor diagonalization argument is used to prove this). Also, I figured out that question I just posed, it's quite easy.

Bahamut draws 1 vs Setz in a rofl game.

Alright, so here's a math question for math people. Suppose A, B are normal subgroups of a group G. Suppose G/A and G/B are abelian. Prove that G/(A intersect is abelian.

Bahamut wins game #2 vs Dama. Also, 1-1 vs. John.

Two words: Thousand Blade Hmm? Clarify. Also: