Clearly because I'm putting two equations together! The only difference between them is a single letter on the LHS and a single sign on the RHS.

@MarianoSuárezAlvarez are the current mods listed somewhere by site?

\textcolor{red}{+}\hskip-1em\textcolor{blue}{\lower 3pt\hbox{$-$}}, or something like that

@robjohn math.stackexchange.com/about

@DylanMoreland Thanks :-) I had never seen that page before.

the skip and the raise/lower have to be adjusted for the font size; ideally, you should set the + in a box, measure it, and unskip exactly that...

I was actually looking for MSE usage but I'll store that away for document usage if need be.

you can use exactly the same in MSE

$\newcommand\cpm{\mathbin{\color{red}{\raise 2.5pt\hbox{$+$}}\hskip-1.75ex\color{blue}{\lower 2pt\hbox{$-$}}}} 1\cpm2$

ha

well, at some point mathjax will be able to deal with it correctly :)

On mse it's \color{Red} and I don't know if mathjax works with \cpm or \mathbin

\textcolor and \color do different things

not sure what those are anyway

oh

like \itshape and \textit, for example

there you go

Good night!

read is in fact my favorite color

8-).

you have to put { and } around \text{} !

$\color{Red}{\text{ding! ding! ding!}}$

Good night! $\color{red}{\text{Ding! Ding! Ding!}}$

@MarianoSuárezAlvarez needs more to be resizable :-)

$\Huge\cpm$

was that the "ha" previously?

$\cpm$?

@MarianoSuárezAlvarez that's what I just tried :-)

Well, if you really want that, you need to measure the first box to unskip exactly that

The Invasion of the $\color{red}{\text{Dingalings}}$?

@JonasTeuwen what the??

apparently \cpm now works in chat, but it gives messed up code on MSE previewing. lol...

I defined the command above

you need to define it wherever you want to use it

@anon That command will be usable until Mariano's \newcommand gets pushed off the scrollback

Interesting. I didn't know newcommand in chat would affect the mainsite.

it doesn't

@anon It shouldn't

Oh, then there's another explanation. I had previously entered the newcommand in the preview. Even though I then deleted it, it looks like it's still in effect.

if mathjax allowed changes in one tab/window to affect others, it would need to be fixed

soon

@anon the preview to an answer on the mainsite?

(and someone ought to report it to them!)

yes

(and probably to the browser makers, too)

@anon interesting. MathJax seems to remember the \newcommand in the preview until the page is refreshed/reloaded

I think I remember having some problem with \newcommand and that may have been the cause.

lol, that does sound like it would create problems

This election just gets better.

hmm?

no way - that's... so strange

@DylanMoreland A whole year? I wonder what happened.

@Holowitz: you got suspended?

yea

a year is the longest suspension I've ever heard of

@mixedmath a 100 point bounty with 1 point to use. I don't know if that is suspension-worthy. Ah, the one point is because of suspension. Never mind :-)

ah - that makes more sense. I'd be surprised if that got past the system

robjohn, are you considering running to be a mod?

@mixedmath perhaps. I've been reading up on the idea. It's useful whether for voting or actually knowing what to expect if running.

...!

whoa

for what? do you have any idea what would merit that?

@JasperLoy You mean the one point is not because of suspension?

@JasperLoy Yes. I could have deleted the comment. Instead, I just corrected myself. I get tired of looking at the chat and seeing lots of deleted comments.

3 years for sockpuppeting? :-|

@ZhenLin ah, I see. It is not a nice thing to do.

I believe Holowitz had just returned from a previous suspension.

Only to get suspended again.

@JasperLoy I agree; it is.

@JasperLoy Sorry for the multi-pings :-)

@JasperLoy Nah, my brain and my fingers were working apart and I had to correct when my brain finished the thought that my fingers had typed :-)

It's time to take Lilly for a walk. See y'all later.

@JasperLoy Indeed.

@ZhenLin remember that suspensions are sort of exponentially increasing each time due to SE policy.

@tb is there a progression schedule? :-)

Mmm, exponential backoff. Fun.

yes, I suppose I just don't imagine people get suspended very often

@robjohn Moderators can do it manually but the defaults are something like 2 days 7 days a month, a few months and then we're talking years.

I hadn't realized he'd been suspended before

@mixedmath I have replied to your answer.

@ymar: okay, I'll take a look

@JasperLoy No, I never was... It came up when a high-profile user was suspended for 30 days after a minor kerfuffle, which looked like a pretty harsh penalty for what actually happened.

Of course we do.

@tb That's why it was 30 days. I thought that was kind of odd. That explains things.

yes indeed

@robjohn I like the Lilly & Smoky picture.

@BrianMScott I do, too. They are quite cute together.

@JasperLoy If you look at the pictures, the names are there.

off to the park. bbl

@t.b.: Are you thinking of running for a mod position?

@mixedmath can you help me with something?

if I'm able to - what's up?

@mixedmath I want to prove that any automorphism of $\Bbb{R}$ must be the identity map

Now this is in a homework assignment

the problem is that my lecturer is telling me that my proof is wrong

He says I have to first show that such an automorphism must be continuous

What is your proof?

That sounds right.

What did you do instead?

@mixedmath I'm still awfully busy and I think I'd rather not commit myself to invest a lot of time in what I believe to be the less pleasant aspects of the site. But I haven't quite made up my mind, yet. Are you thinking about running again?

Ok

we know that $\phi$ has to fix the rationals

Yep.

Now I can also prove that for any positive real number $x$, such an automorphism $\phi$ must be such that $\phi(x) > 0$

similarly for negative real numbers

So suppose now we take a real number $x$

How?

Just for the sake of nailing everything down.

If $x$ is positive, set $x = y^2$ for some positive $y$

@DylanMoreland Then $\phi(x) = \phi(y^2) = \phi(y)^2 \geq 0$

Looks good.

For negative real numbers just use the same argument; if $x$ is a negative number, then $-x$ is positive and then bla

Now

Suppose given a real number $x$ we have that $\phi(x) \neq x$

Then by trichotomy either $\phi(x) > x$ or $\phi(x) < x$

I will show the first case

if $\phi(x) > x$

Then by the density of $\Bbb{Q}$ in $\Bbb{R}$ we have that there is a rational number $m/n$ such that

$x < m/n < \phi(x)$

Split this up into two parts

the first part we have $m/n - x > 0$

the second part $\phi(x) - m/n > 0$

Ah, very nice.

But then $\phi(x) - m/n = \phi(x) - \phi(m/n) = \phi(x - m/n) >0$

But then this is a contradiction because $x - m/n < 0$ so that $\phi(x - m/n)<0$

@DylanMoreland But then my lecturer is saying I first need to prove that an automorphism of $\Bbb{R}$ has to be continuous......

it seems to me that as long as you showed that an automorphism preserves ordering, then this proof works just fine

@t.b.: I was considering it, too. I think I will, but I'm going to sleep on it and see what I think in the morning.

@mixedmath I'm quite confused about this now, my proof seems just fine!!

@mixedmath It seems like all he needs is that $z > 0$ implies $\phi(z) > 0$. Which he's shown, unless I'm confused.

If two grad students are saying my proof is right, that makes me feel a lot better since I'm only second year undergrad

perhaps your lecturer is a big stickler on details. I could imagine a harsh grader demanding that you show that preserving nonnegativity implies that general order is preserved

Why? It's not needed for the argument.

@mixedmath Why do I need that ?

I mean you can use it immediately:

These are field automorphisms, right? Continuity popped into my head because I remembered this old thing but that's not what you're doing here.

yeah field automorphisms

@mixedmath If $x > y$ then $x - y > 0$ so by what I showed above, $\phi(x - y)>0$ and hence $\phi(x) > \phi(y)$

I'm just trying to play devil's advocate for a moment, as that is the only thing I can even think to poke at in your proof

it seemed fine to me

I'm wondering if the instructor was also unconsciously thinking of homogeneous but non-linear functions from $\Bbb R$ to $\Bbb R$.

@mixedmath Perhaps he's thinking of $\phi$ automorphism implies $\phi$ continuous which implies since a continuous function $\phi$ is constant on $\Bbb{Q}$ it is constant on $\Bbb{R}$

The proof is fine. Of course some analytic facts enter, such as the existence of the square root, but then it's a purely algebraic thing exploiting the interaction of the field structure and the order structure.

For complex numbers you have no chance without continuity, that might have been a different reason for the insistence on continuity.

@tb Yeah

If two grad students

and a pro is telling me that it's fine

I'm sticking with it

Two pros, actually.

hahahahahahaha!!!!

@BrianMScott Ok suppose I want to just follows what the lecturer is telling me and prove that an automorphism $\phi$ of $\Bbb{R}$ is continuous

it suffices to show that the inverse of $\phi$ is an open map

Yes.

It's too bad that there are only, like, two ordered fields.

It suffices to show that $\phi^{-1}$ takes a basis element $(a,b)$ of the usual topology in $\Bbb{R}$ to an open set

for the sake of convenience write $g$ for $\phi^{-1}$

@DylanMoreland what do you mean?

@BrianMScott

I knew that would happen.

And don't forget the ordered Field of surreal numbers!

@BenjaminLim ?

@BrianMScott take any $z \in U$

then $z = g(y)$ for some $y \in (a,b)$

Hence by the order preserving property

$g(a) < g(y) < g(b)$

Okay, assuming that you've proved that $\varphi$ is order-preserving.

Which I already have done above

by showing that for any positive real number $x$, $\phi(x) > 0$

then if $x > y$, just apply that to $x - y > 0$

Indeed.

But brian

you see if $y \in (a,b)$

then I can find rationals c,d such that

Okay.

Hence $g(a) < c < g(y) < d < g(b)$

Now by assumption $g( (a,b) )$ is some set $U$

I want to show that $U$ is open

Aren't you missing a couple of $g$'s there?

@BrianMScott $g$ is constant on the rationals

Oh, never mind: you know that $g$ fixes $\Bbb Q$ pointwise.

but then $c,d$ are in $U$

and $(c,d)$ is an open interval containing $g(y) = z$ that is completely contained in $U$

It follows that since $z$ was arbitrary that $U$ is open

@BrianMScott Isn't that the proof?

Hang on.

Since $c,d$ were in $(a,b)$

$g(c) = c$ and $g(d)=d$ are in $U$

How do you know that $(c,d)\subseteq U$?

hmmmm

yes

that is correct

I am tacitly assuming that $g$ takes connected sets to connected sets

Which is actually basically what you need to show.

Outline: An order-perserving map $\varphi:\Bbb R\to\Bbb R$ is continuous except where it has jumps. If it's onto, it has no jumps.

ok

hmmm

@BrianMScott If $\phi$ is order preserving, I think its inverse must be too no?

That's right.

But then ok

I will now prove that any automorphism of $\Bbb{R}$

must take connected sets to connected sets

Let $E$ be connected

and look at $U = \phi(E)$

Now take $x,y \in U$ and $z$ such that $x < z < y$

we want to prove that $z \in U$

Now since $\phi$ is surjective we can write $x = \phi(x'), z = \phi(z')$ and $y = \phi(y')$

Then $\phi(x') < \phi(z') < \phi(y')$

For some $x',y',z'$.

Now the inverse of $\phi$ is order preserving

yes

applying the inverse we get that $x' < z' < y' $

Yes.

But $x'$ and $y'$ live in $E$ that is connected implying that $z' \in E$

Hence $z = \phi(z') $ is in $U$ proving that $U$ is connected

Exactly.

@BrianMScott So I have just proven that $\phi$ is continuous!

The argument would need a bit of rearranging, but you have all of the necessary pieces, yes.

;p

Well let's go over it quickly

@mixedmath ?

I wanna show that the inverse of $\phi$ which I call $g$ is an open map

Let $g( (a,b) ) = U$

Now take any $z \in U$

Since $g$ is surjective we can write any $z \in U$ as $g(y)$ for some $y \in (a,b)$

Now there are rationals $c,d$ such that $a < c < y < d < b$

then $g(c) < g(y) < g(d)$

But then the interval $(c,d)$ is connected

Why work so hard? Why not extend your connectedness argument just a little to conclude that a monotone, onto map takes open intervals to open intervals?

Ah ok

I wanted to join in his excitement (I don't know emoticons, but ; and p are right next to each other - it's just so convenient)

@mixedmath My lecturer was just giving me an unnecessary headache, he told me there was a little trick in proving continuity, for me that just followed from definitions!

@BrianMScott Ok

let $(a,b)$ be an open interval

Is it true that the image of this under $g$ is $(g(a),g(b))$?

Yes, and it's not hard to prove.

@BrianMScott Take $y \in (g(a), g(b))$

Then $y$ is such that $g(a) < y < g(b)$

apply the inverse that is order preserving

$a < g^{-1}(y) < b$

Hence $g^{-1}(y) \in (a,b)$

Hence $y \in g((a,b))$

Note that you need to know here that $y$ is in the domain of $g^{-1}$, but that's clear, because $g^{-1}=\varphi$, whose domain is $\Bbb R$.

yes

So now for the reverse

Which is carbon copy almost exactly the same

@BrianMScott But one thing I am trying to understand is

nowhere have I used the fact that $g$ takes a connected set to a connected set

Or is that implicit from $g$ being order preserving?

I think it is

@BenjaminLim Yes. In fact, if you explicitly prove at the beginning that $\varphi$ is onto, then $\varphi$ and $g$ really can be treated exactly the same.

@BrianMScott We are talking about automorphisms

so everything is onto

now that was easy enough!

I don't know why my lecturer was confusing me!

@BenjaminLim You don't actually care about connectedness per se: what you care about is that $\varphi$ takes intervals to intervals. Of course this is equivalent to connectedness, but that's kind of beside the point.

yes

So the proof that $\phi$ our original automorphism being continuous is really easy!

So, what happens if you don't assume $\varphi$ to be an automorphism? Let's only assume that $\varphi: \mathbb{R} \to \mathbb{R}$ is a field homomorphism. Is it onto?

@tb well

Hm. But how does a field homomorphism have to be real linear?

@tb Not necessarily, that's why I deleted my comment :D

@tb But if say you have an extension $E/F$ such that $\phi$ is the identity on $F$

then it must be surjective

Hey @tb @BrianMScott I have to run now

will you guys be around soon?

say in maybe 30 minutes?

I really have to go, someone is shouting at me!

I'm late for class by 15 minutes too!!

@BenjaminLim Probably, but no promises.

thanks @BrianMScott ,@tb @mixedmath @DylanMoreland

@BenjaminLim cheater! :)

what cheater????

See you later. No promises from me either...

is there an example of a ring with two maximal ideals with trivial intersection?

@EricGregor What about the continuous functions on a two point space?

What about the product of two copies of $\Bbb Z/2\Bbb Z$?

i thought there was such an example, thanks @tb. the reason I ask is that Lang says that to show that Artinian rings have finitely many maximal ideals one should consider the set of finite intersections of maximal ideals and pick a minimum one

but this strategy doesn't make sense to me, because then if there's a trivial intersection that will be the minimal element and you seem to learn nothing from this approach

i think the idea is that you can use Zorn's lemma to conclude that the minimal element is of the form $I=\mathfrak{p}_1\cap\cdots\cap \mathfrak{p}_n$, and then for any maximal $\mathfrak{m}$ you have $\mathfrak{m}\supset \mathfrak{m}\cap I$. I want to conclude then that this implies that one of the $\mathfrak{p}_i=\mathfrak{m}$, but i don't know if that's possible in the case where the intersection is trivial

@tb, do you understand what i'm saying?

I don't see why you need Zorn's lemma. Doesn't the Artinian property give you that?

but still, my question remains

yeah, i just realized that

Hey @EricGregor. That approach through symmetric polynomials turned out to be pretty complicated :)

@anon, wow

thanks for following up on that. just have time to glance at it now, but i agree with your wish for something more elegant!

@tb do you see my concern? please tell me to shoo if you're busy

Hey

I'm back

@EricGregor My commutative algebra is very shaky, so I'm not eagerly following. But isn't $\mathfrak{p}_1 \cap \dots \cap \mathfrak{p}_n \supset \mathfrak{p}_1 \cdots \mathfrak{p}_n$, so you could use that $\mathfrak{m}$ is prime?

@tb I am confused about something

ok, that is what i was missing. i think that suffices. thanks @tb

If I have a continuous function $f$ such that $f(x) = x$ for every rational number $x$

then does it follow that $f(x) = x$ for every real number

functions can be anything you like

does this follow from the fact that if two functions are equal on a dense subset then they are equal on the whole spce?

i can say that f is zero everywhere else

do you mean a continuous $f$?

yes

@BenjaminLim yes, of course. (if the functions are both continuous)

if $f$ is continuous everywhere than yeah

@tb $f(x)$ is continuous everywhere

$x$ is continuous everywhere

if you want to find f(irrational) take the limit of $f(rational)$ as rationals approach your irrational

yeah ok

got it

Can I solve $\int_0^\infty \frac{1}{x^5+a^5}dx$ by factoring the bottom and using Cauchy Integral Formula?

anyone got a nick for a puzzle can try this -> math.stackexchange.com/questions/139220/… :)

@Byte what do you mean by "the lowest number on the list for each pair"?

@robjohn I can't be sure, but I think each ordered pair represents a range — so (3, 5) is composed of the third through fifth elements of the first list.

And then the "lowest number" is simply the minimum value in that sublist.

@BrianMScott I am looking at my proof again that $g$ is an open map; in particular that $g((a,b)) = (g(a),g(b))$, nowhere have I used that $g$ is surjective......is that a problem?

Oh no wait I used it in the existence of an inverse function

Ok

Hi @kahen

Welcome here.

Hi @robjohn. How do you do?

@KannappanSampath I have a field theory assignment

It is quite confusing!

@BenjaminLim How many problems are you assigned?

4

but then two of them are on galois theory

and we just started learning galois theory today

Oh, I understand, it is going to be a bit difficult.

yeah

and our lecturer likes to ask us to name

But, you'd get familiar if you work your way out!

examples of messed up things

possibly yeah

@BenjaminLim But, in algebra, you should have a prototype of everything, I think. So, it is probably the right thing to do. :)

yeah

Hi Alex.

but galois theory is beautiful

@KannappanSampath Hi there! I was napping, sorry.

@robjohn Np. :)

Now are you?

@KannappanSampath pretty good. A little groggy right now :-)

@robjohn Oh, OK. So, you probably just woke up?

Yes, I was napping.

I might get a bit of tea :-)

And, if you're following the mod story, we have not had nice people stepping in. Why don't you throw your hats in?

@robjohn Nice. I'll be heading out for lunch now, I guess.

@KannappanSampath I mentioned earlier that I was looking into it.

Moral: Read the transcript. :)

(I did not read the transcript, sorry about that.)

@KannappanSampath Oh, I was not implying that you should know what I said; I was just saying what I had said, so that you would know. :-)

:) Thanks for letting me know.

Are you having lunch with someone?

@Jonas hi there.

@robjohn Hi :-).

@robjohn No, but looking forward to meeting Srivatsan soon. : )

@KannappanSampath Is he showing up where you are soon?

or vice versa...

He said, he would be in Bangalore soon; but have not heard from his as yet.

@KannappanSampath well say hi to him from me when you see him :-)

@robjohn Sure, will do. :)

At 6.6k now, feeling better that rep points have increased with significantly less participation.

i am trying to kep my rep points to a minimum, but i still get some from time to time

@DavidWheeler Any reason?

Wow, moderator elections.

> because my mom would be so proud of me. :-)

Uuh, baby.

because i just want to know things, i am unconcerned with fame. the things i learn are their own reward.