@Pedro: To see the order, think orbit/stabilizer ?

@PedroTamaroff it looks like an extension of klein $4$ by $C_2$

You just show that each side is a subset of the other side @usukidoll

You can actually find the vertices of a square, too, @Pedro. :P

it's symmetric in two ways, horizontal and vertical, with the stitchings

Oh, no, @Alex, it's not

each side a subset?

Vertical, yes, but not horizontal.

It's a nonobvious composition of rotations or reflections.

I showed $(C_4\times C_4)/\langle (2,2)\rangle\simeq C_2\times C_4$

@TedShifrin right that's not what i'm trying to say

it's hard without showing the tennis ball i mean

OK, @Alex ... Indeed, I was about to type the identical words :P

look at the middle of those... stitch arcs

But I think it might help Pedro to mark four vertices that form a square. :)

I gave those stitches a symmetry group. Stitches love symmetry groups.

there are two pairs of them that are identical (that's $C_2\times C_2$) and those pairs could be interchanged but the interchange would be nonabelian (so extend by $C_2$)

@Pedro ...

@Mike I thought you'd laugh.

good problem

$x\in B\iff x\notin B'\iff x\in (B')'$ @usukidoll

@Ted What causes problems if your exact sequences aren't 'natural'? Hatcher proves this for the ones we studied but not why we care.

Scratch that.

Oh I see where this is going...

OK, I"m heading out for now. You all have fun!

@TedShifrin cya

@usukidoll Essentially that one line is the proof, lol.

@JasperLoy have you seen balarka sen

So X belongs to B, X doesn't belong to B, but if we negate the statement again, then the the (B')' belongs to B

@Ethan Not in person.

I must've interpreted it wrong

@TedShifrin What is the homomorphism one uses to show (p-1)|SL_n(F_p)|=|GL_n(F_p)|?

he study too fast? or is he legit lol

@Ethan He seems to be legit, lol. Don't worry about him, you are my Ramanujan!

haha I always feel stupid when I come on here

ahh there is so much math out there anyway who knows

nvm

@PedroTamaroff how bout $\operatorname{det}:\operatorname{GL}_n(\mathbb{F}_p)\rightarrow \mathbb{F}_p^\times$?

@JasperLoy what about that proof above ^^^^ this time I used definitions...

@AlexanderGruber Yiss.

yay :D

we could have one or the other... in this one because of that or

@AlexanderGruber So GL_n(F)/SL_n(F) is iso to the multiplicative group of F for any F.

@usukidoll Yes, of course "all" here is wrt the universal set.

:D

ok so for a proof to happen it needs definitions... no wonder my (B')' = B was strange.... no def... no back up... no proof!

In general to show that $A=B$ show that $x\in A\iff x\in B$ @usukidoll

@PedroTamaroff yeah. in particular, this i think is interesting: in a coset of $SL_n$ in $GL_n$ every matrix has the same determinant.

there are elements in A and there are elements in B

For the universal set and the empty set one of the directions is trivial.

but I gotta be careful...there's another def that says that $A=B$ are empty

if there are no element

s

Trivially, the empty set is a subset of every set and every set is a subset of the universal set.

which implies that x doesn't belong in A or x doesn't belong in B

Essentially the laws for sets follow from the laws for implications.

so what do I do after this line? ??? of $(A \cup B)'$

@usukidoll You got that wrong, that final or should be an and, lol.

:S but I'm starting from the left

unless I should start from the right

DAMN

of course FACEPALM

Yes, that's what Morgan is about.

because I'm doing a negator.. so it shouldn't be or it's an AND LOL!

now I see the ending

LOL

$(A \cup B)' = [x: x \notin A \land \notin B]$

Anyone watching the superbowl?

my family is

I wish though I got this bs to finish... there was a typo on the homework so one of the problems I did was unecessary ah crap spelling

I got most of it though

they seem to be straight forward

96 users in 41 rooms, keeping the same ratio almost down to the very last number (2.34 this time). I guess the listening guy is falling asleep.

trick to da proofs...read read read read read read over and over and over and over and think outside the box

that's how I got the easy proofs ^^

Just think of the definitions and try to apply them.

yeah I think the typo revision homework is easier than the original

I still have to finish proving the associative law and do h which is a killer because the answer to g is long

Anyhow, I am a bit stuck with some question and if anyone here could help would appreciat it greatly (in set Theory). I managed to get everything down to what was needed, yet I lack one thing - showing an order-keeping function from one group unto another. Said ordered groups are: <Z,~> where a~b iff a>=0 and a<b, or b<a<0, or b<0<=a. The second group is N, ordered so first the even number come, then the odd numbers come.

Any ideas?

@usukidoll It's good to practise writing out the complete answers to get your foundation right.

Proved that both are strictly and well ordered

math.stackexchange.com/questions/661296/… stuck D:

@usukidoll Try to draw Venn diagrams to help you write the proof.

I figured out why my previous proof for (B')' = B didn't work.. I didn't use the universal set complement def

XD

Have you ever seen someone's photo and feel as if you have known him or her forever and fall in love immediately?

I downloaded miktex and got a virus alert. Surely, that's a false positive?

@PaulEpstein Where did you download it from?

And what is the virus alert?

@Jasper miktex.org Rising Anti-Virus said something like Trojan Worm destroying.

@Jasper I haven't had the experience before but I hear it happens

Has anyone had any experience with cryptolocker? I'd like to avoid it. Also, there's a theory going around that people who work for antivirus companies deliberately spread viruses to create a market for their own products. Any truth behind that one, or urban legend? Seems like it would be a tough conspiracy to keep secret.

@PaulEpstein Well, that is really weird.

@Jasper What is weird?

@PaulEpstein The virus alert, lol.

@Jasper I hope it was a false positive. False positives are not unusual? Why is it so weird?

@PaulEpstein Well, I no longer use antivirus programs now but I would not expect any alert with miktex.

@Jasper Re the photo question. If you have no prior connection to the person pictured. For example, if it's a photo in a magazine article, then emotional attachment would seem unusual.

Hey @paul if you are using Windows, I just recommend Microsoft Security Essentials.

However, if there is some reason for a connection, for example a new penfriend, then it would be very normal to get excited or have some other emotional response.

@PaulEpstein Yes, but it happens to me, I guess I fall in love too easily.

@Jasper So you see a photo in a magazine article, then "fall in love". Then what happens?

Do you try and contact the person?

@PaulEpstein Well, it wasn't exactly a magazine, it was online, and in the end I chatted with the person in a chat room. It's been years but I still can't forget her, lol.

@Jasper, Well, that sounds completely normal.

@Jasper Why would you want to forget her, anyway?

Anyway I prefer TeX Live, it is cross platform too.

And I prefer TeXworks.

@Jasper Microsoft Security Essentials is probably incarcerated.

@Jasper The product I have installed is called TexWorks

@PaulEpstein Ah, should be the same editor.

And the installation was to read files by who? A) Elton John, B) Terry Tao, C) John Grisham, D) Arianna Huffington E) Katy Perry.

What do you think?

does anybody know how to show that, if $K$ and $L$ are simplicial complexes with $\operatorname{dim}(K)=m$, any continuous map $h:|K|\rightarrow |L|$ is homotopic to a map taking $K$ into the $m$-skeleton of $L$?

I should have said by whom?

@PaulEpstein Probably B, lol.

@Jasper you are correct!!!

That was a bad riddle, lol.

@Jasper Was it bad because it was too easy?

@PaulEpstein Well, because I would not expect it to be any of the others.

I woul have a hard time picking between B and E.

@Jasper I wanted to bring joy to the world by letting others guess correctly.

@PaulEpstein Haha, now you sound like a nutcase, like me, lol.

How cute

so I have a If $A \cap B = \emptyset$, then $A \subseteq B$... hmmmm the elements in A and B are empty XD

@mike What editor do you use?

I liked that Katy Perry film -- it was called a part of me or something like that, but I can't remember the exact title. I liked the bit where she has to choose between being an algebraic topologist or a pop star.

so it's telling me that if the elements of A and the elements of B are empty, then $A \subseteq B$ which is that A is a subset of B... I guess this statement can be true since both of the sets are nothing...

@jasper how is it going?

@Charlie Hmm, nothing new, I hope I get better this year.

@Mike Are you a major expert in elliptic curves?

@usukidoll No!

@jasper you must make efforts too :)

Wait what is the question again @usukidoll

@Paul I am not a major expert in anything.

@Charlie Yes I know.

If $A \cap B = \emptyset$ then $A \subseteq B'$ @JasperLoy

@Mike Have you studied in elliptic curves?

@usukidoll And you are asked to prove that?

yeah

like I know that A is a subset of B

@usukidoll Erm, if there is no typo, that is wrong.

but the set intersection of A and B is that the elements belong in A and B but they are empty sets?!

I didn't typo

oh wait damn it

I did fffffff

If A and B are disjoint, A need not be a subset of B.

Ah, typo!!!

I'm sorry I can't READ super small ' or exponentials

so holy crap this changes everything

@Paul I know a little.

Yes, holy shit.

let's pull the definition of $A \subseteq B$ which is $(\forall x) [x \in A \rightarrow x \in B$

but I have a B' which is a negatior

$A \subseteq B'$ which is $(\forall x) [x \in A \rightarrow x \notin B$

Aha.

so x belongs to A and what.... it doesn't belong to B

What have you been doing @jasper?

KITTY! hugs

@Mike Do you have a basic understanding of why Fermat's theorem is true?

@Charlie I have been sorting out my thoughts as usual. It is difficult to explain why it takes so long to heal.

$ A \cap B = \emptyset $ so that means that there are no elements in A and B because it's empty

@usukidoll Wanna see my proof?

for?

Nvm, you continue with this/

@Paul That is a far harder thing to have than just knowledge of elliptic curves. You need to know about modular forms and modular curves as well.

maybe you should stop thinking @jasper

I know a little bit about elliptic curves but not that stuff.