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Flezz Help with math plz

a dallop a d-
aisy

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Level 6 Camwhore

“Training Broad”

Plz hlp I hav a problem in math i dun understand plz hlp me here

let M(S) denote the monotone clbum generated by a set of sets S

i dun understand why ∀S S is a ring ∀A x∈M(S) ∃{A_n} (A_n⊂A_(n+1)∧S_n∈S∧(∪(S_n)=S)

plz explain this to me i dun understand why this is true, its a fact used for proving the uniqueness of extention of measure

Ziro

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Level 10 Emo Kid

“Gloomy Gus”

lol?

a dallop a d-
aisy

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Level 6 Camwhore

“Training Broad”

acutually i made a mistake plz forgiv me the actual statement i dun understand is (corections are bolded):

∀R R is a ring∀A A∈M(R )∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)

thar i think i wroat it in proper formal language plz tell me why this is true

a dallop a daisy edited this message on 03/07/2008 8:40PM

Warpvader

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Level 10 Troll

“Pain in the ASCII”

[This message has randomly exploded, causing the “divide by 0” effect, and thus creating a pime taradox.]

Warpvader edited this message on 03/10/2008 9:52PM

Opportunity

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Level 10 Troll

“Pain in the ASCII”

Do a barrelroll.

Harix

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2

[Totally Cool Guys -
from UGOP
]

Level 10 Camwhore

“Leave it to Cleavage”

The problem is that there are upside down “A”s in your statement…

dreamserpent

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12

Level 10 Camwhore

“Leave it to Cleavage”

the upside down A just means “for all” LOLOL NOOB

a dallop a d-
aisy

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Level 6 Camwhore

“Training Broad”

Warpvader Posted:

I would love to help you with your math problem, but I don’t understand Greek/Latin/or any other dead languages.

o its very simple

it means “for all rings R, for all A within M(R ), there exists a sequence of sets A_n which are elements of R such that their union is A.”

to see wat im talking about, here is an example in which it is implicitly stated:

http://www-maths.swan.ac.uk/staff/vl/c98.pdf go to page 18, theorem 4.2

Theorem 4.2 Let A be an algebra of sets, μ and ν two measures defined on the σ-algebra A_σ generated by A. Then μ ↾ A =  ν ↾ A implies μ = ν.

Proof. Choose S ∈ A_σ, then S=lim (n⇒∞) S_n, S_n ∈A, for A_σ=M(A ). Using continuity of measure, one has

μ(S) = lim (n⇒∞) μ(S_n) = lim (n⇒∞) ν(S_n) = ν(S).

a dallop a daisy edited this message on 03/07/2008 9:01PM

Miggle

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Level 17 Emo Kid

God won't kill me, he likes to see me suffer.

the answer is jesus

Skyreal

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Level 24 Camwhore

“The Lady is a Tramp”

a dallop a daisy Posted:

acutually i made a mistake plz forgiv me the actual statement i dun understand is (corections are bolded):

∀R R is a ring∀A A∈M(R )∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)

thar i think i wroat it in proper formal language plz tell me why this is true

(A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) has me pretty confused.

a dallop a d-
aisy

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Level 6 Camwhore

“Training Broad”

Skyreal Posted:

(A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) has me pretty confused.

it means the sequence of sets A_n is increasing and each A_n is in the ring R and the union of it is A.

PhineasPoe

Avatar: 12179 2010-01-24 16:27:57 -0500
8

[70 Character Story-
tellers
]

Level 35 Troll

You got a smudge there Phin... oh wait thats Trouts feces

Drop the clbum before it does any more damage.

VIVa-ChiCk

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aka FireFury

The game

Balloon

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28

[Grey Goose Mafiosi]

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Inflate my ovaries until they pop out of me and float away

The only thing I remember from math was not to use my teeth.

crackednut

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“Pain in the ASCII”

T∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) ∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)

Math Attack.

Ego -40^e-(ln(1+x^2))

Damage +50^e(x^(1/3)+x^(ln(1+mod(x^.5))))

oh.. where x is the number of times you fapped in the past 1.32 days

PWNED THIS THREAD! Log in to see images!

Leadpipe

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How about you stop trying to look smart on the internet there

Skyreal

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Level 24 Camwhore

“The Lady is a Tramp”

Leadpipe Posted:

How about you stop trying to look smart on the internet there

Eeee! I was found out! How?

Oh wait, were you referring to me?

Abyssinth

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[Sisterhood of the -
Quivering Rose
]

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“Leave it to Cleavage”

do your own algebra homework and don’t expect the internets to do it for you Log in to see images!

Also, “An alternative way to show that an open subset X of R is the disjoint union of open intervals is to consider the set S of all sets of disjoint open intervals with union contained in X, with an appropriate order, apply Zorn’s lemma to S to obtain a maximal one, and then show that the maximal one in fact covers all of X.” If that helps.

Abyssinth edited this message on 03/10/2008 8:25PM

Neurotic Nin-
ja

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I think this Log in to see images!

nanalatinoje-
sus gets you-
JUSTICE IN -
YOUR FORUMS

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[Full of SbumSS]

Level 26 Troll

I SHOULD POST MORE BUT I DON'T BECAUSE YOU'RE ALL REALLY LAME

Leadpipe Posted:

How about you stop trying to look smart on the internet there

LETS ALL QUOTE OUR ACT SCORES!!

i got the highest score on everything possible and am currently developing a unified theory of everything. i attend super genius school and am the top in my clbum.

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