Mathematics - 2017-08-29 (page 3 of 6)

Let (G,*) be a set with an associative binary operation ,
1) there exist an element e in G such that x*e = x for all x
2) for every x in G there exist y in G such that xy= e

show that G is a group

I mean what is left to prove? ( dont answer if there is something really )

Tim The Enchanter

3:18 PM

left identity and inverses

Kasmir Khaan

I just want to see if that Question has a point or not =p

Tim The Enchanter

it need not be commutative

Leaky Nun

@TimTheEnchanter it isn't required to be commutative

@TimTheEnchanter he said don't answer

Kasmir Khaan

exactly tim -.-

=P

Tim The Enchanter

he wanted to know what to prove right?

Leaky Nun

3:19 PM

really, Tim

Tim The Enchanter

i didnt say the answer

Leaky Nun

@TimTheEnchanter no, he didn't

Kasmir Khaan

Ehm from my point of view those assumptions are enough to make it a group

closure and associativty (given )

existence of identity

Tim The Enchanter

oh crap ive messed up havent I

sorry

Leaky Nun

@KasmirKhaan read your group axioms again.

Kasmir Khaan

3:20 PM

and invers for all elements in G

Leaky Nun

@TimTheEnchanter no you haven't

Kasmir Khaan

@LeakyNun Ermm I did , close associative, unique identity, and unique inverse

let me think for a while =p

Leaky Nun

@KasmirKhaan read the precise statement

Kasmir Khaan

Okay ill will again =p

@LeakyNun I noticed something but i think it is because I assumed too much things when I first read the question

Well if x*e = x for all x , should'nt that be enough to be sure that e is the identity ?

Leaky Nun

@KasmirKhaan no

Kasmir Khaan

3:23 PM

does the other order (e*x = x ) is nessasry ?

Leaky Nun

read the definition of the identity from your group axioms

@KasmirKhaan yes

Kasmir Khaan

hmm am pretty sure that my teacher today said that it is enough that x*e = x (can only be accomplshed by the identity )

hmm this is so odd =p

Will Hunting

Sometimes, some of the requirements given in axioms can be relaxed a little, that is some can be deduced from others. But I can't recall which axiom where now.

Kasmir Khaan

and if say ab =e , can we have ba = something else other than e ?

Leaky Nun

@KasmirKhaan that's a theorem.

that you need to prove.

Will Hunting

3:26 PM

In general group operation is not commutative though.

Leaky Nun

@KasmirKhaan then either your teacher is wrong or there's some missing context you forgot

Kasmir Khaan

Yes I get that part , i mean if it was ab =x and ba = y , then it is possible on non-abelian groups

@LeakyNun ill vote for the second thing =p

Leaky Nun

@KasmirKhaan it isn't possible.

and that's a theorem you need to prove.

you can't just assert that it's impossible

Kasmir Khaan

Hmm I need to get more regor =p

Secret

[Chemistry] The hessian is numerically singular. This is so weird because this time, everything is really on the stationary point of the multivariable function from the previous calculation

Kasmir Khaan

3:28 PM

Okay ill give it another try @LeakyNun thanks alot for help btw! :)

Leaky Nun

@KasmirKhaan good luck

Will Hunting

@Secret Sorry, what has hessian got to do with chemistry?

Secret

The hessian is what is needed to check whether your optimised molecule geometry (which gives function of energy in terms of the positions of all the atoms) is at a minima, maxima or a first order saddle point (which is often called a transition state)

It's entries also gives you the vibrational frequencies of the normal modes of the molecule, which is important in predicting its IR, UV etc. spectrum

Will Hunting

Very interesting stuff.

Good morning @Faust7

Secret

3

Q: What is the difference between an undulation point and other critical values?

A necessary but not sufficient condition for a point of inflection is that $$f''(x)=0$$ If the second derivative is 0 and the point is not a point of inflection, Wikipedia tells me that is called an undulation point, which apparently means a point on a curve where the curvature vanishes but ...

Hmm, since I have tell gaussian to not reorientate my molecule on calculation, which means the error form the integration grid will not come into play, it is possible that those local minima are undulation points...

but is this flat enough to be an undulation point...?

Maybe the flatness is from the other 60 something degrees of freedom...

Faust7

3:41 PM

Morning will

@BalarkaSen Nice rage today

Balarka Sen

chat was on drama mode today

Kasmir Khaan

Hey yall

Faust7

woh where did u come from?

Was like i casted a spell n u appeared lol

Kasmir Khaan

@ barlarka can you tell me if i did correct or not ? ( abstract algebra )

Will Hunting

I don't think there was much drama today.

Kasmir Khaan

3:47 PM

@BalarkaSen .

Secret

Begin Astatine calculation

Balarka Sen

@KasmirKhaan I need to get some work done; maybe someone else can help?

Kasmir Khaan

@BalarkaSen all right thanks anyway , and good luck =p

leaky nun was helping me but he is afk =p

Faust7

@BalarkaSen you sure your not a genie?

Balarka Sen

I was lurking and replied to your ping.

Faust7

3:49 PM

;P

Whatcha working on today?

Balarka Sen

stuff

Faust7

Alright, good luck with your stuff ;)

Balarka Sen

ktnxbai

Faust7

i gotta get to work too just thought id say hello to everyone.

can someone explain colum space and ker under a linear mapping whats preserved and not etc?

i have a decent understanding im just missing a couple pieces

preserved wrong word

Kasmir Khaan

how to make blank space on latex?

im trying to write a phrase it making it like long word

._.

Will Hunting

3:59 PM

Hey @Daminark, lol.

Ah it is midnight here.

Faust7

\space

@KasmirKhaan

Kasmir Khaan

thanks fraust7

I wish I could meet the other 6 frausts:D

Haha

=p

Faust, not Fraust

4:01 PM

@KasmirKhaan ~

Will Hunting

LOL

Kasmir Khaan

@LeakyNun Hello ! I think I got it now, let me show you my proof

Yes!

Leaky Nun

show me

Faust7

It was a name given to me by a german friend he always said i played go like i was possessed by the devil

Kasmir Khaan

x*e =x , we multiply by x inverse on the left and by x on the right

using associativity we get e*x =x

Ill try to put it on latex so i can send it here

Leaky Nun

4:04 PM

@KasmirKhaan you haven't proved that left inverse exists

and you also haven't proved that left identity exists

(you used left identity in the step (x'x)x = ex = x)

Kasmir Khaan

hmm I think I should post em in right order

xy= e , xyx=x , x'xyx=x'x , yx=e

Leaky Nun

xy=e => xyx=x is invalid

Kasmir Khaan

hmm

Leaky Nun

you really need to make sure that you use the axioms one at a time to prevent things like this from happening in your proof

aka step by step

Kasmir Khaan

I only multiplied by x on the left

isint that the definition of the equiality sign ?

I can do anything on both sides?

Tim The Enchanter

4:07 PM

@KasmirKhaan You used ex = x

you only have xe = x from axioms

Kasmir Khaan

aaaaaaaa

Mary Star

Hello!! Is the number 1,34 344 3444 34444... rational?

Leaky Nun

@KasmirKhaan no, you multiplied by x on the right

@MaryStar no it isn't: it neither terminates nor repeats

Kasmir Khaan

okay let me do it again

Leaky Nun

@KasmirKhaan good luck

Kasmir Khaan

4:09 PM

@LeakyNun Thanks :D

Mary Star

@LeakyNun Ah ok! If we have a number that does not terminate but repeats, it would be rational?

Leaky Nun

@MaryStar yes

Faust7

i was replying to the original question didnt see your reply

Mary Star

@LeakyNun Ah ok! Thank you!

Leaky Nun

@MaryStar no problem

Faust7

4:14 PM

Why are homomorphisms useful?

Leaky Nun

@Faust7 they let you see the group structure

Faust7

uh i know that uhh

why is that useful?

Leaky Nun

they let you know more about the group

(they also produce normal subgroups)

(as well as quotients)

Faust7

yeah i know both those things as well it just feels like

i dunno

Leaky Nun

everything in math is useless

Faust7

4:18 PM

i call bs

Leaky Nun

what kind of answer are you expecting?

Faust7

it giuves us insight into blank or shows us how blank works

Leaky Nun

@Faust7 you mean, I should relate it to other disciplines of mathematics?

Faust7

well that would be one way to make me happy lol

Leaky Nun

or else I'd just say it gives us insight into the structure of the group...

isomorphism would be more useful

Faust7

4:20 PM

i agree

i guess i just dont understand how insight into the group structure helps us understand the orginonal group

i know what it means concretely if its isomorphic

Kasmir Khaan

Okay x*x' =e @LeakyNun

now xx' *x =ex

Faust7

what is x'?

Kasmir Khaan

by associativity we get x=e*x

Faust7

a right inverse or left or both?

Kasmir Khaan

I used the right inverse on x' witch is x

Fraust7

I posted the question like 1 hour ago >< there are missing stuff from your point of view

Leaky Nun

4:24 PM

@KasmirKhaan no, you don't get that by associativity

Kasmir Khaan

ill post it again =p

Leaky Nun

don't skip any step.

Kasmir Khaan

we only have right inverse and right identity right?

Leaky Nun

@KasmirKhaan yes

Kasmir Khaan

I used only those

Leaky Nun

4:25 PM

@KasmirKhaan no you didn't

show me all your steps

Kasmir Khaan

okay one second

Balarka Sen

@AkivaW An elliptic isometry of $\Bbb H^2$ fixes two complex points, right?

Kasmir Khaan

@LeakyNun my argument was if x is in G , means that x' is also in G , so it also have the right inverse property

Leaky Nun

@KasmirKhaan yes

Kasmir Khaan

x*x' =e
(x)*x'*x = (x) - ( the new element is in parenteses )
from the point of view of x' , x is it own inverse , so i can do x' *x to get e

Leaky Nun

4:28 PM

you already skipped like 2 steps here

Kasmir Khaan

x*e = x ( right identity was given )

grrrrrrr

I did it that way really

Let me keep trying

at least now I get the question ><

Leaky Nun

just don't skip any steps

Kasmir Khaan

Okay ! :)

xx' =e
xx'x=ex
x(x'x)=ex
xe=ex
x=ex

@LeakyNun Now i think its all there :D

Leaky Nun

justify line 4

Kasmir Khaan

Oh because we are given that xe= x

right identity law

Leaky Nun

4:33 PM

I said line 4

not line 5

Faust7

lol

Kasmir Khaan

oh hmm, x'x is by the right inverse law =p @LeakyNun

Leaky Nun

@KasmirKhaan no that isn't what the law says

Faust7

That would only be true if x' is a right and left inverse

Kasmir Khaan

xy=e for all x in G

Leaky Nun

4:35 PM

@KasmirKhaan yes

here y = x'

Secret

3

Q: Applications of the concept of homomorphism

What are some interesting applications of the concept of homomorphism? Example: If there is a homorphism from a ring $R$ to a ring $r$ then a solution to a polynomial equation in $R$ gives rise to a solution in $r$. e.g. if $f:R \rightarrow r$ and $X^2+Y^2=0$ then $f(X^2+Y^2)=f(0), f(X^2)+f(Y^2)...

abstract-algebra

Kasmir Khaan

Yes that what I meant to say =p

Leaky Nun

@KasmirKhaan so you have xx'=e

not x'x =e

Kasmir Khaan

hmm for x' , x is its right inverse

Leaky Nun

@KasmirKhaan why?

Kasmir Khaan

4:36 PM

since x' is also in G i can use that law

Faust7

@Secret THANK YOU

Leaky Nun

@KasmirKhaan the law only said that the right inverse of x' exists

the law doesn't say that the right inverse is x

Kasmir Khaan

hmm true ><

Let me think :D

Akiva Weinberger

What are you trying to prove?

Leaky Nun

@AkivaWeinberger one sided identity and inverse => two sided

Akiva Weinberger

4:39 PM

One-sided identity as well?

Faust7

oh is the identity both sides?

Leaky Nun

@AkivaWeinberger yes

Akiva Weinberger

Which side

Leaky Nun

@AkivaWeinberger the same side

Kasmir Khaan

right side for both

Akiva Weinberger

4:39 PM

So we know xx'=e and xe=x

Faust7

lol

Akiva Weinberger

…OK

Leaky Nun

@AkivaWeinberger please delete

Will Hunting

LOL

Semiclassical

oy vey

Will Hunting

4:40 PM

Is this a test?

Faust7

so much anger :0

Leaky Nun

@WillHunting Kasmir wants to figure it out by himself

Akiva Weinberger

I didn't realize it would be that much of a hint

I don't even know how to do it

Will Hunting

@LeakyNun Actually, to figure out himself he should not even come to this chat, lol.

Leaky Nun

@WillHunting I'm helping him proofread

he tends to make logical errors

Will Hunting

4:42 PM

In my entire undergrad studies I never asked anyone a single problem's solution.

Leaky Nun

@WillHunting speak for yourself

Balarka Sen

@AkivaWeinberger I pinged you a question shortly above in the transcript.

Will Hunting

If I can't solve it, I just don't, lol.

Faust7

@WillHunting to figure out himself he should goto a psychologist not this chat we r liable to make em worse

Balarka Sen

I think I am right, but I am not sure how those isometries look like.

Leaky Nun

4:42 PM

@WillHunting that isn't the right attitude

Will Hunting

Yeah but this one-sided thing is very tricky indeed.

It takes a looong time to get it done.

Leaky Nun

@WillHunting I second you

Kasmir Khaan

I cant find a clever way to make sense of
x(x'x) =ex to xe =ex

but hmm I feel like am on the right path so far

Will Hunting

I think I will try to sleep from 6 AM to 2 PM from now.

Kasmir Khaan

Can I use that the inverse of inverse is the element itself or not allowed to ? @LeakyNun

Leaky Nun

4:45 PM

@KasmirKhaan not allowed

you need to prove every theorem you use.

Will Hunting

You can't use the group results when you haven't shown it is a group yet.

Kasmir Khaan

Hmm well if i prove it then i can use it right?

Leaky Nun

@KasmirKhaan of course.

Kasmir Khaan

:D

Faust7

@KasmirKhaan too much work

Leaky Nun

4:45 PM

have you read the proof of the theorem you mentioned?

Kasmir Khaan

Did not think about it like a theorem , just a fact

><

Leaky Nun

@KasmirKhaan this shows your lack of understanding towards groups

Kasmir Khaan

Yepp =p

But I only did 1 lecture and it was today

Faust7

what makes you belive that x is a right inverse of x' ?

Leaky Nun

do you follow a textbook or something?

Kasmir Khaan

4:47 PM

so am a bit fresh on the subject

Leaky Nun

@Faust7 intuition :P

Kasmir Khaan

Yes by dummit and floote

Akiva Weinberger

@BalarkaSen I don't remember which one's elliptic

Leaky Nun

@KasmirKhaan how long have you been reading it?

Faust7

it is by why assume it to be so?

Akiva Weinberger

4:47 PM

Ask Ted. Or Google

Faust7

why not prove it to be so

Semiclassical

this conversation makes my head hurt

Kasmir Khaan

We only did an intro lecture so we did not have to read anything @LeakyNun

Leaky Nun

@Faust7: is this a rhetorical question?

@Semiclassical then leave

@KasmirKhaan alright

then who told you that the inverse of the inverse is itself?

Faust7

@Semiclassical Akiva solved it already but leaky dont want us to give him the hint

Will Hunting

4:48 PM

@Semiclassical How is dealing with uni bureaucracy that you mentioned?

Kasmir Khaan

its from other areas of math

Balarka Sen

@AkivaWeinberger So, uh, it says if the isometry fixes a point in the interior of $\Bbb H^2$ then it's elliptic; it's easy to check from writing $g\cdot z = z$ for a $g \in \text{SL}_2(\Bbb Z)$ that that means the trace has absolute value less than $2$.

Kasmir Khaan

like (f^-1 )^-1 = f

Leaky Nun

@KasmirKhaan ...

Semiclassical

@WillHunting Waiting on an email from them right now

Kasmir Khaan

4:49 PM

@LeakyNun Sorry ill keep working serious =p

Balarka Sen

But this also means $g$ has to fix two complex points, because $g \cdot z = z$ is a quadratic equation on $z$ with negative determinant.

Leaky Nun

TIL using results from other areas of math is a thing @KasmirKhaan

Kasmir Khaan

TIL means?

Balarka Sen

Which leaves me a bit confused about how they look like. Apparently if it fixes $p$, it's "rotation about $p$", which shouldn't really fix other points unless I am dumb and don't understand hyperbolic rotation?

Semiclassical

right now the issue is that it'll take a while for the refund appeal to be processed, and in the meantime I need to register for classes. hopefully they'll give me permission to do while I wait for the situation to be resolved.

Leaky Nun

4:51 PM

@KasmirKhaan today I learnt

Semiclassical

otherwise, I'll need to pay the balance of my student account out of pocket and wait to be refunded

Faust7

im assuming raising that much coin as a student wouldnt be easy?

Will Hunting

@Semiclassical I see, but you will get your PhD eventually right?

Semiclassical

actually, I've got enough in savings at this point that I could take the hit (so long as I eventually do get the money back)

Kasmir Khaan

@LeakyNun Ill keep working on it , I did not mean it in that way, but the inverse of invers is the element itself is something intuiative , but ofc i need to prove it

Semiclassical

4:52 PM

@WillHunting right.

Leaky Nun

@KasmirKhaan alright

just don't assume anything you haven't proved/seen.

Will Hunting

@Semiclassical Good, good. Well, I hope your anxiety gets better, and then you can be a great mathematician. =D

Semiclassical

lol

Faust7

i thought u were doign physics?

Will Hunting

For myself, I would need a miracle...

Semiclassical

4:53 PM

I am.

Faust7

when did u switch?

nvm

Leaky Nun

@BalarkaSen sanity check: S^2 sans north pole and south pole isn't isomorphic to torus as their fundamental groups are not isomorphic

Secret

Today's calculation progress: 6 out of 40 batches will be completed in a few hours

Leaky Nun

namely, the latter is abelian

Balarka Sen

by sans I hope you mean minus

Akiva Weinberger

4:53 PM

@BalarkaSen If this is the Poincaré disk model maybe it fixes something inside the disk as well as something outside of it

Balarka Sen

S^2 - {north pole, south pole} is homotopy equivalent to the circle.

Semiclassical

sans means without

Leaky Nun

@BalarkaSen yes

Balarka Sen

Has the same $\pi_1$ as $\Bbb Z$.

Akiva Weinberger

Or, like, the half-plane model, maybe it fixes a conjugate pair

Semiclassical

4:54 PM

e.g. a sans serif font doesn't have a serif.

Leaky Nun

@BalarkaSen what?? :O

Kasmir Khaan

@LeakyNun Ill keep that in mind !, but last idea before I keep writing the proof, x' belongs to G , so its inverse is (x')' , can't I just use that as my y ? in xy=e for all x in G

Balarka Sen

@Semiclassical Oh

Leaky Nun

@KasmirKhaan yes you can

Secret

Hyperbolic discounting

In economics, hyperbolic discounting is a time-inconsistent model of discounting. It is one of the cornerstones of behavioral economics. The discounted utility approach states that Intertemporal choices are no different from other choices, except that some consequences are delayed and hence must be anticipated and discounted (i.e., reweighted to take into account the delay). Given two similar rewards, humans show a preference for one that arrives sooner rather than later. Humans are said to discount the value of the later reward, by a factor that increases with the length of the delay. This process...

Leaky Nun

4:54 PM

@AkivaWeinberger you see he finally figures it out

this is why I told you to delete

Will Hunting

@Semiclassical Arial is sans, Times is serif, Courier New is monospaced.

Semiclassical

a doubly punctured S^2 is homotopy equivalent to just a disk with a puncture

Kasmir Khaan

:D

Balarka Sen

@AkivaWeinberger Ohhh.