But yeah, you're right.

@BrianM.Scott Where is it ?

@BrianM.Scott The core of the idea is clear tough: if $G$ has nonempty intersection at nbhs of $0$, we can translate it and make it intersect in any other real number.

@JasperLoy thanks

@user43418 Here.

@BrianM.Scott Post it if you want ?

@PeterTamaroff Yes $-$ but it seems to be easier to work backwards from a gap around $x$ to a gap around $0$ than forwards from no gap around $0$ to no gap around any $x$!

@user43418 Nah $-$ it’s identical to the answer that’s already been posted.

@BrianM.Scott Ok Thank you

@BrianM.Scott Yeah =)

@JasperLoy I couldn’t: I wrote it here before the question was posted to the main site, and essentially the same answer was already up there by the time I discovered that the question had been posted.

@user43418 You’re welcome.

Hello, @Brian ! Good to see you here!

@amWhy Peter dragged me in, kicking and screaming. :-)

@BrianM.Scott KICKING AND SCREAMING!?

@BrianM.Scott That's Peter for you!!! ;-)

@BrianM.Scott What does your middle M stand for?

@PeterTamaroff Maynard.

I've got a very short, sweet middle name: Jo

@BrianM.Scott Do people call you Brian May?

Hehehehe

@PeterTamaroff Most people have no idea what the M stands for.

@BrianM.Scott Could you refresh my memory re: the domain of your email address?

@amWhy I gave you <[email protected]>, I think.

Yes, I think that was it...I think you mentioned you still have access to your campus account, as Emeritus...

You can delete it, if you'd like, I've got it down! (again)

@amWhy It’s the second-most useful privilege; the most useful is access to the online library, including journal subscriptions.

@BrianM.Scott Don't I know it!!! What a bonus ;-)

ahoy!

Hi, @anon !!

ello

or...yellow?

or L'O

@jasper ;)

@anon I like your answers, especially the expository-ish ones! I first noticed back when you were explaining the basics of permutations to an OP who was having trouble grasping the different notations. You're explanation could serve as a "reference" entry.

thanks

Of course, you "stole" an accept from me, by doing so, but it was certainly a better answer for the OP than mine, at the time.

*your @amWhy

@anon I like your answers, especially the expository-ish ones! I first noticed back when you were explaining the basics of permutations to an OP who was having trouble grasping the different notations. Your explanation could serve as a "reference" entry.

:/

@Charlie I tried correcting...but it was too late and reposted as a new post!

@anon OP's should feel no loyalty to an earlier upvote if a subsequent post is far more helpful and informative. ;-)

Suppose i have a system of equations At=b representing two planes in the euclidian space R^3 where A is 2x3 and t is (x y z). My same system has a 1-dimension NullSpace, that means the intersection of the planes shifted at the origin is a line.If the only possible free variables are x and y for example, does it mean the line is in the x-y plane ? If the only possible free variables are x and z for example, does it mean the line is in the x-z plane?

@anon what happened?

reposting due to a simple typo

Indeed

@user43418 No, only I say that!!!

@Charlie <innocent> I thought that it was pretty nearly a unanimous refrain. </innocent>

@BrianM.Scott :-O

@Charlie hi

@pourjour Soufian!!! Hi, I'm fine and you?

@BrianM.Scott Are you there?

@Charlie pretty good thanks

step 1: what is the set of positive integers m for which (m,5)=1? step 2: restrict to odd m.

@pourjour excellent!

@PeterTamaroff I am now.

@anon m are those that cannot be divised by 5 am I right?

alternatively, (2n+1,5)=(2n-4,5)=(n-2,5), so whenever n is 2 more than a number coprime to 5

@BrianM.Scott OK. We assume $B(x,\epsilon)\cap G$ is empty. This means that there are no numbers inside $(g,h)$ for $g=\sup\{g':g'<x-\epsilon\}$ and $h=\inf\{g:g>x+\epsilon\}$.

But before, I said that there were some $g,h\in G$ that worked.

I think it needs to be justified.

@anon can't understand this "so whenever n is 2 more than a number coprime to 5"

@PeterTamaroff All that has to be justified is that $G\cap(\leftarrow,x)\ne\varnothing\ne G\cap(x,\to)$.

@BrianM.Scott OK, but we said that such $g,h$ were such that no $\ell \in G$ satisfied $g<\ell <h$; or did I say it and wasn't corrected?

@PeterTamaroff See my first comment in the one where I described a sentence as awful.

@BrianM.Scott =)

@BrianM.Scott What I'm wondering then is: how do we prove $G\cap (h,g)=\varnothing$?

@PeterTamaroff What is there to prove? Every element of $G$ is either $\le h$ or $\ge g$ by the definition of $h$ and $g$.

@BrianM.Scott Sorry, how did we define $g,h$?

@anon I think the set of n is $S=\{ n : n=5p+k : p\in \mathbb{N}$ and$ k\in\{0,1,3,4\}\}$

@PeterTamaroff You did it about half a dozen comments back.

@BrianM.Scott the infs and sups?

@anon does p must be a prime?

@PeterTamaroff Yes.

@pourjour $p\in\Bbb N$ is what you wrote, and what you wrote is correct

@BrianM.Scott Oh, sorry. I thought that didn't work. I think I lost confidence.

@anon and k=0 or 1 or3 or 4

??

yes

@PeterTamaroff Oops! My fault: no, those don’t have to belong to $G$, so a little more work is needed.

@BrianM.Scott That's why.

@anon thanks

@anon is there any other way without noticing this $(n-2,5)=1$

?

Maybe if I drink more beer a miracle will happen! A leprechaun will give me the solution!

@pourjour yes, 2n+1 must be among 1,2,3,4 mod 5, so solve 2n+1=k mod 5 for n for each of k=1,2,3,4. (note the multiplicative inverse of 2 mod 5 is 3)

@anon that's what already did

you're idea was great too

@PeterTamaroff Take sequences of group elements $g_n$ and $h_n$ converging to the sup and inf. Look at the group elements $h_n-g_n$. They’re positive, so they have an infimum $y$. Show that $y>0$ and that $G\cap(0,y)=\varnothing$.

@BrianM.Scott You win the mathwebs Brian! You win!

@PeterTamaroff :-)!!

@anon BTW, do you have any experience with complex numbers?

yes

@pourjour They are good guys, I swear.

hahahahaha^

@PeterTamaroff are you drunk?

@pourjour No. It takes a lot of alcohol to get me drunk.

I barely had two glasses of beer.

@PeterTamaroff hhh

I don't need alcohol to get drunk

@pourjour "hhh"?

@PeterTamaroff laugh

@PeterTamaroff nevermind

yeah, don't you laugh with your mouth closed like a normal person?

@Charlie That is a laughter?

@PeterTamaroff yes

@anon When I'm on the internet, I just exhale fast through my nose.

Rare cases make me LOL

@PeterTamaroff like kkkk rsrsrsrsr

@PeterTamaroff i laugh all the time

the easiest thing is to make me laugh

I found this hilarious for example.

@anon 8750393 I'm trying to find fixed point of this transformation $g$in a complex plan :as $g:z\rightarrow(1+i)\overline z -1+3i$

@pourjour So you have $(1+i)\bar z-1+3i=z$.

Solve that.

It will be easier if you put $z=a+bi$ and solve for $a,b$

@PeterTamaroff yeah but I couldn't

because of the bar over z

$\bar{z}=a-bi$

I will retry and show you the result

@anon seems that you are always in a bad mood, because of your gravatar

other way around

oh, you mean seems to others

@anon just like me every morning

exactly

so I think $z=1-i$

is that correct

no

@anon so z=?

ahh $z=-1+i$

am I wrong again?

@PeterTamaroff that's a good one!

@pourjour firstly, expand (1+i)(a-bi)-1+3i=z to be (a+b-1)+(a-b+3)i=a+bi, so a+b-1=a and a-b+3=b, so b=1 and a=-1, i.e. z=-1+i (correct)

thanks :D

@Charlie It's hilarious.

@PeterTamaroff so much!!!!

now I'm struggling to prove that $z'=2i\overline z -5 +5i$ is an homotecy

@JasperLoy :D

@anon any ideas?

@pourjour what's a homotecy?

@anon I mean homothecy

homothety

either or

@JasperLoy aah

@JasperLoy you change the mewaning of things when you blink

@JasperLoy I see

@JasperLoy I know

@JasperLoy }:)

ok people I've got to sleep

@JasperLoy did you like that cute bears i used to send you?

@pourjour good night, sleep tight!

@Charlie good night! :D

@pourjour :D

@JasperLoy ow

@JasperLoy :DDD

@JasperLoy goodnight

@JasperLoy sleep tight! :)

@JasperLoy HAPPY EASTER!!!!

are you Russian charlie

@Ethan No.

oh, but you have seen swan lake

@Ethan of course :)

@κρανίοπεριπολία hello

Good night everyone!

@JasperLoy 哥哥

@Karl'sstudents what kind of math do you do

@Ethan differential geometry :D

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$\Huge\text{(}$removed$\Huge\text{)}$

@Karl'sstudents Hi honey.

@κρανίοπεριπολία hey bee :-)

@Karl'sstudents :-D

bzzz...

@κρανίοπεριπολία Please up vote it for me. :-)

@Karl'sstudents Do you mean star it?

0.999...hours later...

@κρανίοπεριπολία Yes. :D

@JasperLoy I see. Thanks for pointing it out.

BTW, has anyone read the book sensual quadratic forms, by Conway?

I think of it as the most elementary approach to this subject!

is swaping the rows of a matrix then doing RREF the same as doing the RREF of a matrix then swaping its rows ?

@awllower yeah it's one of my favorite books

first time I read it I wrote programs to test equality of forms and things like that

I also understood the value of Zoltarevs reciprocity proof from that book

I'd seen it before but not realized its importance

@caveman hi

@caveman how are you?

:-D

@caveman Does the book contain thi topic?

I have not seen it?

lets suppose i have a system Ax=b with x being ( x y z w ).Suppose i also know that the intersection of all graphs is 2-dimensional, that is, the nullspace has 2 free variables and 2 lead variables.Suppose i know that z and w could be made lead variables and i want a RREF matrix such that the pivots have z and w as corresponding variables.What should i do ?

@nerdy Post the question on main.

He yes, how about you use the main site for that? It's the purpose.

Not that you can't ask mathematical questions in chat; it's just that you should just bloody wait like anybody else that wants an answer from strangers.

Has anybody read Stephen Smale's biography?

@nerdy +1 :-D

what are those zeros of

the Riemann zeta function

almost 8 lol

yes almost 8

saa

Yo, does the idea of a pivot in linear algebra has only to do with coeficient 1

or it laos has to do with leftmost element ?

does a pivot need only to be 1, or does it need also to be the leftmost element and 1 ?

wtf ?

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(removed) hours later...

hi

is there anyone please

@Karl'sstudents do you find him annoying?

graveyard....

hello

the radius of convergence of z^2/(e^z+1) is \pi?

hi ,can i ask a question ?

askaway

what is a differential manifold in dimension 1

?

@κρανίοπεριπολία No. He is a handsome guy.

There is a new joke in LaTeX:

\let\ea\expandafter

\ea\sports\to\the\game

@Karl'sstudents what;s that?

@Ilya a joke.

I mean, what;s the point?

shall I complile it with Math Jax?

@Ilya: It is better to ask chat.stackexchange.com/rooms/41/tex-latex-and-friends. :-)

Hmm.

What kind of distribution do you get if you uniformly sample a finite interval in projective space?

E.g. you generate pairs of random variables, of which one lies in an interval $X=[a,b]$ and the other in an interval $Y=[c,d]$ and you divide the two $\frac{X}{Y}$

I see. That stuff is called ratio distrubitons.

@Karl'sstudents why would you want me to star that?

@κρανίοπεριπολία just 4 fun

@Karl'sstudents But he can be very sensitive sometimes...

@κρανίοπεριπολία does kranioperipolia have any meaning?

oh whoops

wrong one

@κρανίοπεριπολία :-)

fixed

@kram1032 skullpatrol

heh. Does that explain the HUGE Skullpatrol in the favorites on the right?

yip

nice

hi

hi

lo

the black square is back

Is $\cos(\frac{\pi}{7})$ irrational ? and if no is $\cos^3(\frac{\pi}{7})$ irrational ?

huhu

this one is a nice question even though the op really shouldn't get reputation for that

I suppose I should pick a username.

oh what a waste i gave such a nice bijection and he only wants one for natural numbers

here

@user67848 hastings will be nice... :-)

my function is a bijection from $\mathbb{Q} \to (100,200) \cap \mathbb{Q}$ ;)

mh i guess it is how the op meant it although i unterstood it a bit different

but my answer is neater cause of it sends rationals to rationals and irrationals to irrationals :)

i gonna fix that ;)

a banananda :D

i didn't upvote the other answer, he posted imho a worse answer then mine 10 minutes after me

yeah yours is good

but arctan is a stupid function ^^

Let $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. For each $x \in \mathbb{R}$, define a function $g_x: \mathbb{R} \rightarrow \mathbb{R}$ by $g_x(y)=f(x,y)$. Suppose that for each $x$, there is a unique y such that $g_x'(y)=0$; let $c(x)$ be this $y$.

How do I show that c is differentiable ?

i hate when someone don't know an approximation and i need to prove it -.-

This reminds me to much on numerics -.-

As we know, an invertible matrix could be rewritten into the product of several elementary matrices.

As we know.

Can we discover an analogous one for 1-1 continuously differentiable mapping?

regards to where you are mapping

For example, $f\colon\mathbb R^n\to\mathbb R^n$.

you know what lie algebras are ?

sry lie groups

I hope if there's some result locally at, say, $0$.

No.

Is that necessary?

i guess that is what you are searching for ;)

I want to summarize theorems in multi-variable calculus, say, reverse/implicit function theorem, rank theorem, etc.

Happy easter everyone!!!!

@κρανίοπεριπολία Hi :)

Goodbye!

@dominic hi

hi :)

@DominicMichaelis how are you?

tired :D

@DominicMichaelis oh

@JasperLoy thanks! It's a swan :)

@JasperLoy I think so...

@JasperLoy it's a "black swan"

@JasperLoy lhf?

low hanging fruit

easy questions

@JasperLoy precisely

i do have such a lot answers with 8 upvotes ...

@JasperLoy Glad you like it. I am not all that creative. :)

@JasperLoy mathoverflow has a link in its faq, I think.

And I have no idea what led me to MO.

hi

I need a diagram to solve differential equation

can someone please help me

I think it's simple

you should be able to solve simple tasks ;)

Gee that preimages of closed sets under continuous functions are closed is not liked on this site

here i bet when i would use am > gm i would have 9000 upvotes now

but i like the preimage definition

nah I need just clear manner