9:00 PM
(you are free to be wrong) haha

3 mins ago, by Mariano Suárez-Alvarez
the fact that equality is transiytive is at the basis of everything we do
2 mins ago, by John Junior
I thought you said "uniqueness" was?

@MarianoSuárez-Alvarez What is the semantic difference?
You did not say what you mean with "is" or "to be".
In mathematics it means many things.

@JonasTeuwen Can I ask you something?

You can always ask.
And I can always not answer. And I can always be wrong and always disagree or whatever.
And always drink beer, but that is not possible 8-(.

“Beer Is God’s Way Of Saying He Loves Us” – Benjamin Franklin

9:06 PM
@JonasTeuwen Hehehe OK. Remember the problem that went "Suppose that $f(x+y)=f(x)+f(y)$ and $f$ is continuous at $0$. Show that $f$ is continuous for all $x$."
I think I have a good solution.

@JohnSenior :D.
@PeterTamaroff You prove it is linear?
Then done.

@JohnSenior and wants us to be happy!

yep

@JohnSenior I thought it was because he figured he was too sadistic giving all the diseases and such and this is like to make it up a bit. To make it bearable.

@JonasTeuwen that's a nice idea too ... I guess ...

9:07 PM
@JonasTeuwen PROOF First note that $f(0)=f(0)+f(0)$ so that $f(0)=0$. Then we have that $$\lim_{x\to 0}f(x)=0$$. Now, this is equivalent to $$\lim_{x\to 0}(f(x)+f(p))=f(p)$$ for any fixed $p$. Then we obtain $$\lim_{x\to 0}f(x+p)=f(p)$$, but that is equivalent to $$\lim_{x\to p}f(x)=f(p)$$ $\blacktriangle$

@JonasTeuwen Stop telling God what to think.

@JohnJunior Stop telling me what to do.
@PeterTamaroff Why is that limit $=0$?
For the limit only the values in the neighborhood matter.

@JonasTeuwen $f$ is continuous at $0$, JT

And you do not know those yet. That is what you are trying to prove?
Oops.
Hmm, it feels a bit fishy. Ask John.
I am a bit... whatever.

@JonasTeuwen Fishy?

9:10 PM
Looks okay actually.
Yes, because it is easier than my idea if you have to prove everything.
But that could well be my retardedness.

@JonasTeuwen I have already proven that $$\lim_{x\to0}f(x+p)=l\iff \lim_{x\to p}f(x)=l$$

Under what conditions?
Well, actually basically by definition right?

@JonasTeuwen That is just a general statement for any $f$
@JonasTeuwen Yuppy.

@PeterTamaroff I think it is OK - and rather neat, actually

Yes, I think it is okay, but I don't trust my judgement now.
Right, it is neat :-).
My way would be to prove something more heavy and take the special case.
You just skip that. A characteristic of the great! 8-).

9:13 PM
@JohnSenior I wanted to ask Mariano, actually.

I once (many years ago) spent hours trying to find a discontinuous solution of that functional equation ...

@JohnSenior But you directly see it is linear!

@JohnSenior Spivak says there are examples.

@JohnSenior And you know for linear...

But that they are very complicated.

9:13 PM
@PeterTamaroff What??
Examples for what?

@JonasTeuwen Of discontinuous $f$ with $f(x+y)=f(x)+f(y)$

Without continuity at one point right?
Because that fucks it up.

@JonasTeuwen Yep. That's probably why it is so hard.

Hmm.
Right, take domain $(0, 1)$. Now let us mess with $n$-ary expansions of $x$.

I think Hamel bases came into the discontinuous case - but my memory might well be totally wrong here

9:15 PM
SO now $f(x) = 0$ except when...
Some shit about expansions and stuff.
Anyway. Beer.
2

@JonasTeuwen LAWL

@JohnSenior The existence of the Hamel base, isn't that equivalent to AC?
The existence of discontinuous functions also can be shown using AC right?
So... in any way. There is some truth in the statement!

@JonasTeuwen I think so - and I think that might be why Halmos (or whoever) said that examples were complicated

@JohnSenior Nah - that way he looks smarter when he constructs one.
It's all about the status...

@JonasTeuwen :)

9:22 PM
How's the beer?

Good question! Let me rinse the glass.

@JonasTeuwen What is a Hamel base?

@PeterTamaroff Just a base for a finite dimensional vector space.

@JonasTeuwen Oh, OK.

@JohnJunior I don't know yet, but if there is a heaven... this is how it must smell like!
@PeterTamaroff Actually for all vector spaces.

9:25 PM
maybe get a discontinuous solution by something like this:
define $f$ to be some random real for each $b_\alpha$ in the Hamel basis, and extend to all of $\mathbb{R}$ by the functional equation
I think the Hamel basis we are talking about here is a base for $\mathbb{R}$ as a vector space over $\mathbb{Q}$

This guy mick is unbelievable.

That must be a great beer!

@JonasTeuwen Are you going to chill the glass?

Why would I?

Some do.

9:27 PM
@JohnJunior Yes, that will work.
@JohnJunior Why would I do that... then it would have to heat up again.

@EdGorcenski oh yes! (but he is not unique in that respect)

Funny how we return to uniqueness ;-D

General life lesson: when one alone encounters problems everywhere, look to the only consistent factor: themselves.
4

@JohnJunior I did that on purpose :) (following our earlier discussion)

@JohnSenior Smooth.

9:31 PM
@JohnJunior not really - I am actually remarkably hairy :)

@JohnSenior I think "infinity" is the farthest thing from "unique."

@JohnJunior Is this legit? $$\sum\limits_{k = n + 1}^\infty {{a_k}} = \sum\limits_{k = 1}^\infty {{a_{n + k}}}$$
For a convergent $a_n$

@JohnSenior My beard does not want to be a good beard :-(.
@PeterTamaroff Why would you think it is not ligit?
Do you know what a series means?

@JonasTeuwen Yes, I do.

Then you have proven this already.

9:38 PM
@JonasTeuwen Very well then.

@JonasTeuwen It takes patience and old-age to grow a really good beard - I am still working on it

@JohnSenior Damn! Should I use Pokon to help it?

Pokon?

9:42 PM
$$\sin^2(|(\ln 2 + \pi x)^2|) + \cos^2(|(\ln 2 + \pi x)^2|) = 1$$ is undecidable!
@JohnSenior Page does not load.
Give it to students with the task "find $x$".
Only the ones who draw an arrow to $x$ and write "here it is!" get points.

@JonasTeuwen ok - not worth looking at actually :) (cyclops)

Pseudocyclops!

9:59 PM
General life lesson: when one alone encounters problems everywhere, look for a place where you can get a beer.
3

@robjohn good grief ...

When can I rain all sorts of havoc here and be immune from retribution?

@robjohn exactly - worrying, isn't it ?

10:03 PM
@robjohn Depends what you count as retribution.

@JohnSenior I don't think there is much worry that he will get to 20K, though.

@robjohn So the answer is "20K".

@robjohn not in my lifetime :)

@JonasTeuwen well, there is no reputation that prevents you from being suspended or removed.

@robjohn Yes, but you know the thing that will happen...
People with a higher reputation will have more "friends" 8-). Or people that like that person because of a certain skill.
So that might compensate for harsh responses.
As the way it is more difficult to arrest the ambassador from a foreign country then an illegal immigrant of that country.

10:07 PM
@JonasTeuwen being an SE employee might be a similar situation, but even then, I think that if they stirred too many pots, they'd be gone as well.

Or called too many kettles black ;-)
but that's politics.
>8(

Who wants chowder?

@MarianoSuárez-Alvarez: we are on the same wavelength :-)

aha - I am 20% trustworthy :)

@JonasTeuwen @robjohn That reminds me of "There's a rope around your neck."
Brings me shivers everytime

10:12 PM
Is there?

that doesn't sound very good - I wouldn't buy a car from someone who was 20% trustworthy :P

trusted user 65%
protect questions 87%

@robjohn That's the politically correct answer.
Who would ban Jeff? 8-).

@JohnSenior You mean 20% thrust worthy ;-D

Incidentally - the number plate of the car I got today is MK 62 ...

10:14 PM
@JohnSenior And...?

M akoto K ato

@JohnSenior "Is it the thrusted badge perhaps that you need ? Did i overread that ? If so im sorry."

@JohnJunior Oh, he's sorry, alright...

@JohnJunior don't be sorry - we all talk rubbish here :)

@JohnSenior It haunts you.

10:16 PM
I have to go. Time to get my wife from the airport. BBL

later

@robjohn later
@PeterTamaroff :)

So upvoted comments don't gain reputation?
(I am trying to figure out where 10 rep came from today.)

@peoplepower That is from a $(+1)$ in an answer, or $(+2)$ in a question.

@PeterTamaroff You're right, somebody upvoted an answer I was planning to delete.

10:28 PM
@peoplepower there is probably a badge for that :)
I seem to have 36 badges - mostly for things I don't understand the relevance of

@JohnSenior Beer.

@JonasTeuwen I want a whisky badge :)

8-).

how about a badge for electrocuted whisky loving wannabe-mathematicians ?
(I would have one of those)

Lightning struck, whisky loving cranks (100% luck)

10:35 PM
Good night brothers and sisters (and @peoplepower). Bye!

Uh oh.
They know that I am a bird.

@JonasTeuwen Bye Jonas - have a good evening

@JohnSenior It is past 0AM!

@JonasTeuwen ah yes- an hour difference - I forget - still evening here :)
must sleep -night all

11:32 PM
Hello folks.

Hello Kannappan
you're up early today?

@FortuonPaendrag Yep.

Any particular reason?

I woke up pretty early today. Awaiting my mom's arrival.

Aha. What classes are you in this semester? :)

11:34 PM
(She is coming by an early morning train...)
@FortuonPaendrag We have an Analysis course -- Green, Stoke, Implicit and Inverse; Algebra -- Rings, Modules; Physics -- Electrodynamics; Computer (Science, perhaps) -- Programming Basics, Data Structures.

Nice and well-rounded. :)

There's also a Statistics Course.

Even more well-rounded.

Well, what are your courses?

algebraic number theory, a second course. representation theory, probability and a Problem solving seminar
and a division requirement for which I'm taking Islam.

11:41 PM
@FortuonPaendrag Representations <3
So, what text are you guys using?

Sorry about the delay
I was getting food

I see.

Do you know if it is a good book?

Well, \me thinks, it is a good book.
Anyway, I cannot say because, I have not read the book fully or thoroughly.
Perhaps @Mariano might have something to add here?

I trust your word :) What are you planning for next year?
You are a third year now, right?

11:50 PM
Nah! I am in my third semester, which in the Indian System translates to the first half of the second academic year in college.

Oh wow. For some reason, I always thought you were my year.
But eitherway, I'm glad that you are back on M.SE :)

Well, I'll come to this chat room...
I'll be around, perhaps later, OK...

yeah, for sure :) Anyway, I need to be off. So, I'll talk to you later