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7:00 AM
$b=\gcd(b,c)P$ and $c=\gcd(b,c)Q$
 
All he did was factor out gcd(b,c)
gcd(b,c) divides $b$ and $c$
 
OH
 
Factor out the greates common divisor from $b+c$
 
yea
 
and $\gcd(\frac{b}{\gcd(b,c)},\frac{c}{\gcd(b,c)})=1$
so P,Q are coprime
 
7:01 AM
now I understand - the OP's comment about symmetry mnakes sense now
sorry
 
I would like to see a proof without dirchlet though so you guys shouldn't give up.
 
I should probably go to bed, I pretty much wasted my whole day, I should have done some chemistry homework
I saw "the persuit of happiness"
The ending was fucking horrible
It was ok at the start, but there was never any climax
 
@user1 Yah. Sorry. I think I should get some slack for typing it up into chat though :).
 
@Bageer lol
 
I win.
2
 
7:03 AM
@Ethan ya its pretty much merh
@anon whatd you win?
@anon jerk
xD
 
who names primes?
 
@anon how do you define $p$ there? just some random prime?
 
any prime
 
alright I am sufficiently bored, good night all
 
showing something = 1 is the same as showing it is not divisible by any prime
 
7:07 AM
Good night
 
@anon I didn't understand but now I do xD
 
:)
 
@anon Neat solution.
 
thanks
 
@anon I wish I could double upvote
 
7:14 AM
@anon "nor does it divide n, by definition"?
What definition are you refering to
 
that was witty as all heck!
 
@Ethan the definition of n
 
@Ethan definition of $n$
 
You don't seem to state $p$ in the construction of $n$
 
we assume there exists a prime p s.t. p|a,b+nc
 
7:15 AM
ok
 
@Ethan if $p \mid a$ and $p \mid b$ then $p$ is not in the product of $n$ so $p \not\mid n$
 
@anon Got to put that trick into my toolbox. Reminds me of the trick of looking at functions $\{x\} \to A$ to show a monomorphism is an injective function, only look at the small important piece of information.
 
I'm so jealous at how much math you guys know
 
@anon Clever construction
nvm
good job
why am i still awake
 
to answer your comment
my intuition for these sorts of things is sort of like venn diagrams that act as puzzle pieces that can be shifted
 
7:25 AM
I need to study alot more
 
the key idea is that $p\mid u,p\nmid v\implies p\nmid u+v$
 
yes the clever cancelation
(a+b)-b
 
so the trick is to make $b+cn$ not divisible by any prime dividing $a$, or in other words for any prime divisor $p\mid a$ make $p$ divide exactly one of $b$ or $cn$
 
alright, good night, again
 
I think this is your third time going to bed today
 
7:28 AM
heh
 
@anon Excellent!
Let me congratulate you for this wonderful solution!
 
:)
 
lol @awllower bageer employed dirichlets theorem
Scroll up
 
Yes I saw.
 
lol
 
7:30 AM
@Bageer As well excellent!
Thanks for this marvellous solution. Why not post it so that more people know of it?
 
its over kill
 
@awllower Its a silly overkill solution.
 
I guess if your reading about modular forms though you might have read the proof
 
@awllower Its like proving some trivial number theory fact with Fermats-Wiles theorem
 
Is it better to use more elementry proofs in general?
I don't know these things
 
7:31 AM
I think over-kills are also worth an answer, though.
 
anon's method provides an explict construction though
and it is elementry
 
Indeed.
 
@Ethan I don't think so necessarily. I think a proof that says a lot about what is actually going is is good. Mine doesn't say anything.
 
I think awllower would have read a proof of dirichlets anyways with his profile picture
 
Sometimes elementary solutions are way too complicated still.
 
7:33 AM
or he tells me its an L function for an elliptic curve or somthing else I don't know lol
 
Like the proof(s) by Erdös.
 
@awllower some of them, I think alot of it is presentation also
 
Okay, I guess I will post it.
 
@Ethan Indeed I have read a proof of Dirichlet.
That is why I think highly of the answer by Bageer : It is a linkage between two subjects, or else I am over-thinking? Haha
 
@awllower like the binomial coeiffient thing in his proof of bertands postulate I like, at first I didn't though, also having to check by hand the first 15 or so cases is a pain
 
7:34 AM
Agreed.
 
I think he wrote that proof when he was 17 though so whatever
 
In fact, on my way back, I was thinking about maybe we could prove the statement by modular forms? Haha
 
alright good night for the fourth time
 
Over-over-kill...
OK
Good night=Bonne nuit=Guten Abend!
 
Maybe that fact was used to prove dirchlet, in which case my proof would be cirular. Another bad thing about using high tech to prove (maybe "prove") small, elementary things.
 
7:41 AM
No, it's not used in the proof of DT. But I agree it doesn't shine light on the machinery in play.
Although often using powerful theorems to prove elementary things can run into those problems, especially concerning gcds, prime factorization etc.
 
@Bageer The square is missing, haha.
 
Yah I took it out and reworded some parts
When I first posted it I was just being a smartass
 
The first answer is oft ugly, again a lesson from history.
:)
 
8:34 AM
@robjohn Regarding my question on the "edit tags" hover link, cf. Meta.SO #182332. (I found this out by actually clicking it for the first time.)
 
9:29 AM
Anyone here ?
 
A good morning to you, @Dominic.
 
@Lord_Farin good morning too
 
How are you doing today?
 
I recently thought about the following:
not much better ...
 
@DominicMichaelis 't Makes me sad to hear that.
But let me listen to you; perhaps indulgence in mathematics can help you feel better :).
 
9:33 AM
There is a big difference when we switch from finite things to countable things. Like summation doesn't commute nor is defined in general. When we switch from countable to uncountable we lose even more properties, like really uncountable sums never converge, the measure is not uncountable additive and much more
Does something like that happen when we switch from $\mathfrak{c}$ to mh lets say $\aleph_3$?
 
@DominicMichaelis Interesting question. An indication that (we don't know of) anything interesting happening directly beyond $\mathfrak c$ is that we distinguish finite, countable, uncountable; this last category isn't usually split up further.
 
I think I ask it on main
 
There are all sorts of special set-theoretic properties that $\omega = \aleph_0$ has, that are very rare (exemplified by the hereditarily finite sets modelling ZFC except infinity).
I think you'll get some nice answers to that. :)
 
9:48 AM
Is it a biglist question ?
 
I think so. There will likely be examples from a considerable collection of mathematical fields.
 
10:16 AM
huhu
 
Yo, wazzup?
 
@skullpatrol The rules of grammar and intelligent conversation were just brutally molested. :P
 
@Lord_Farin (:-D)->-<
 
So how are you doing @skull?
 
10:20 AM
@Lord_Farin Fine thanks, how are you?
 
@skullpatrol Quite fine, reviewing for my set theory exam of next week.
 
Gema -.-
 
this video is not available in your country
 
10:28 AM
@DominicMichaelis Try here
 
ah there it works
 
(:-D)->-<
 
11:17 AM
Bloody... how can I make this guy realize that he doesn't have to solve for a fixed constant?
 
11:33 AM
Greetings
 
Greetings
and salutations :D
 
@skullpatrol Hello! How is it going? :-)
 
@Chris'swisesister Fine thanks, how are you?
 
@skullpatrol trying to compute some beautiful (and hard) integrals.
@skullpatrol did you see this one? $$\int_1^2 \frac{x+1-x^2\log(x)}{x^3+x^2}\cos(\log(x+1)) \ dx$$
I don't work on it right now since I have something much harder to play with.
 
You do like challenging integrals.
 
11:48 AM
$$ \int_0^1\frac{\log(1-x)}{x}\,\frac{2s}{\log^2 x+(2\pi s)^2}\, dx, \space s>0$$
@skullpatrol This one is really cute too. Yeah, I like integrals, series and limits in general.
 
CUTE? gee you have some serious problems :D
 
@DominicMichaelis hehe. am I wrong? :D
Hi
 
You are special ;)
 
unique
 
11:52 AM
:D
 
(:D)->-<
 
Does anyone know what a chain complex is ?
 
In mathematics, chain complex and cochain complex are constructs originally used in the field of algebraic topology. They are algebraic means of representing the relationships between the cycles and boundaries in various dimensions of a topological space. More generally, homological algebra includes the study of chain complexes in the abstract, without any reference to an underlying space. In this case, chain complexes are studied axiomatically as algebraic structures. Applications of chain complexes usually define and apply their homology groups (cohomology groups for cochain complexes)...
 
Have you worked with them? I am thinking about why they are only taking countable many objects in the sequenze
 
11:58 AM
@DominicMichaelis no, I didn't work with them. Do you study them at school now?
 
I gonna have to hold a seminar about diagramm lemmas in category theory
 
12:15 PM
Anyone know the origin of the sigma in "sigma finite" in measure theory?
 
i thought sigma stands for countable ?
 
yeah, but why the choice of the letter sigma?
My guess would be some French word beginning with s.
 
@PrimeRibeyeDeal Of relevance: this
The $\sigma$ tends to signify countable union.
And the French word for union is "somme".
 
The german word for summer is Sommer :D
 
Ahh, that's probably it. Thanks Lord_Farin.
 
12:25 PM
argh I hate it when i don't know how to translate stuff
 
stuff is just Stoff
 
oh I am trying to think of a good word for "composable arrows"
 
I thought you knew that already. Just change all u's into o's. There's German for you.
 
@PrimeRibeyeDeal there are no u's there that is the problem :D
 
Cleary it's already it German then.
in German
 
12:31 PM
@Dominic Possibly "komposierbare Morphismen"?
 
yeah that would be my first choice, but it doesn't sound as natural as i like it :D
 
Hm, indeed, Dutch doesn't do a terrific job either. Let me think some more.
 
Well maybe i just skip composable at all. If they are not composable it would be obviously nonsense :D
 
Is "zusammenstelbar" a German word?
You could always go for translations of "appropriate" or "suitable" or similar vague terms.
 
yeah but zusammenstellbar does have the sound of a location
 
12:39 PM
Then just go for my last suggestion :).
Or you remark in the introduction that "whenever compositions are written, it is implicitly assumed they make sense".
 
well if this is the hardest part in my topic i am a lucky guy :D
oh that is great
 
I presume you have already taken the habit to suppress composition symbols?
 
In handwriting surely but I think for the pdf version I will add them, so that there aren't to many conventions I have to mention
 
12:54 PM
@Dominic I'm leaving, see you around. Good luck with your topic!
Bye all.
 
@Lord_Farin Bye
 
@DominicMichaelis Should be "komponierbare" oder "verknüpfbare".
 
 
2 hours later…
2:55 PM
$0 \to 2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$ is a chain complex but not exact is it ?
 
what is the map?
 
from $2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ everything is mapped to zero
 
then yes, it is a chain complex but not exact
(if you take the 0-map everywhere, anything will be a complex, but almost never exact)
 
Oh that was to easy ... I am still a bit uncertain what i am allowed to use as arrows
 
any map of whatever objects you have
 
3:02 PM
In category theory because of
when i go to the dual category i switch the arrows
and well if I try to do this one with that mapping I am still a bit confused
 
what do you mean?
 
When $f:a\to b$ is an arrow in a category $C$, then the dual statement in $C^\ast$ is $f:b\to a$
 
right. But the dual category of the category of modules over some ring is not in general itself such a category
(so it is no wonder it is a hard to grasp thing to work with)
 
Oh I just play with abelian groups, but seems like the dual even doesn't allow functions on its
 
right, the dual is a very different category
 
3:17 PM
Even more surprising that sometimes things are easier in dual categories
 
4:43 PM
@robjohn Yello.
 
@PeterTamaroff good morning
 
@robjohn I didn't finish last night's problem.
In fact, I had another problem unfinished, and I went to sleep, and suddenly I realized I might had done something wrong, and jumped off bed to "correct" it, but it the end 'twas OK. I ended up adding a proof of the third item.
@robjohn What's with the banjo man's face?
 
@PeterTamaroff you've never seen Deliverance, I assume.
 
@robjohn Nope.
 
@PeterTamaroff you're probably not familiar with the idiosyncrasies attributed to the southern Appalachian region of the US.
 
4:51 PM
@robjohn Heh... no.
 
@PeterTamaroff Then there is too much to go into here without potentially offending someone
@PeterTamaroff Let's just say, "bad genes"
 
@robjohn =) I knew that was coming!
Is the word "redneck" offensive?
 
@PeterTamaroff To some, it probably is
 
@robjohn @peter how are you guys?
 
4:55 PM
@Charlie I am. How are you?
 
@PeterTamaroff I'm fine.
 
Extremely tedious but elementary proof.
Notice, as a contrary, how excellent other two elegant answers are.
[Here](http://math.stackexchange.com/questions/407383/if-a-b-c-1-is-there-n-in-mathbb-z-such-that-a-bnc-1/407855#407855)
 
@awllower You need http first.
 
@PeterTamaroff Excuse me, what do you mean?
Are you referring to the poor formulation?
 
@awllower I don't know why your link fails to work.
 
4:59 PM
Oh!
0
A: If $(a,b,c)=1$, is there $n\in \mathbb Z$ such that $(a,b+nc)=1$?

awllowerThe answer by anon is elegant and short, with the specific choice of $n$. On the other hand, the answer by Bageer is more "elementary" in the sense that it could reveal the essence of the question, at least in my view. So let me explan why I say so. Firstly $(a,b+nc)=(a,g(P+nQ))=(a,P+nQ)$, where ...

?
 
Mh...
Well, disregard that link then.
I am typically unable to correct such errors...
 
@PeterTamaroff I remember drawing an egg with circles in Mathematica. I will find it.
 
@robjohn Hi. What is your suggestion on how to deal with skullpatrol? First he drove Asaf into leaving this chat and now he's driven Jonas away. As of now, his mathematical contributions in here as well as on main are negligible. I'd much rather have an interesting conversation with Jonas, Asaf or both than having this bot in here.
8
And given his output he sure seems more bot-like than human.
Who is he going to attack next?
 
@MattN. I am siding with you, mister.
 
5:08 PM
@PeterTamaroff Thank you, brother.
 
@MattN. I have also realized Baire's theorem for complete metric spaces shows perfect subsets of complete metric space are uncountable.
 
@PeterTamaroff Hey, that's interesting. I don't even know what a perfect set is!
: )
 
@MattN. A perfect set is a closed set all whose points are limit points.
 
Ah.
 
like the cantor set
 
5:10 PM
@DominicMichaelis We all do...
 
a non trivial example may
 
@PeterTamaroff You are studying a book about topology at the moment?
@DominicMichaelis Ah, thanks for the example.
 
@MattN. Nope, from Apostol and Rudin - their books called "Mathematical Analysis".
But yeah, they have a little of Point Set Topology.
@DominicMichaelis LOL; I read that as "I like the Cantor set", hence my comment.
@MattN. Yeah, the Cantor set is a perfect set of measure zero.
Brainfuck.
 
@Peter I like the Cantor set, but it was only meant as an example
 
@MattN. I think that Asaf's reasons for leaving have little to do with skullpatrol, however, I won't say none. I have not been here when Jonas has been interacting with him to know what is going on there.
 
5:12 PM
@PeterTamaroff Not at all. Topology is so interesting. To me, anyway.
 
@DominicMichaelis Yes, I noticed.
 
Hello all, wassup?
 
@MattN. I mean, it is headtwisting to imagine a set of measure zero that is uncountable.
 
Hey @JayeshBadwaik! I'm sorry it didn't work out with the mod election. In retrospect we could have known in advance because the voting was entirely based on rep.
 
(In fact it has cardinality $\mathfrak c$)
 
5:13 PM
@PeterTamaroff Ah, not so much once you get used to it : )
@robjohn I'm in touch with Jonas and he's left because of a conversation with sp.
 
@MattN. Heh. :-) You know I am secretly smiling that I could afford to go on a hiking trip without worrying about what will happen to the site. :P But thanks for your support thought. It was nice knowing that atleast some thought I was a good candidate.
 
And let me add: I have him on my ignore list since he attacked Asaf.
 
@MattN. What did he do?
 
@PeterTamaroff Said some shit, bro, I don't want to go into it but the chat is public so you can look it up yourself.
 
@MattN. he left because of a conversation? Did he try to contact a moderator or the anon about it? That sounds pretty serious.
 
5:16 PM
@robjohn It bloody is. But Jonas is not the sort of drama queen that would contact a mod about it. Jonas has got a life outside math.SE, I suppose, so that he just chose to do whatever was the least amount of effort.
@JayeshBadwaik Just do it again once you have 40k. Then everyone will suddenly think you're the most suitable candidate ever.
 
@MattN. Yup! :-)
 
Good : )
@robjohn But apart from insulting users the main reason why I find him a chewing gum on my sole is that I see in him no more than random noise. If you look at his posts you will find that it's either random images, mindless comments like "n hours later..." or a question about division by zero.
I have him on my ignore list because I think if I get a glimpse of a single post of his my IQ would drop by 5 points.
(nearly wrote 5 pints there : O Cheers!)
 
:-O
@MattN. I agree with you about it. I am not sure of the random noise effect. I would not mind it if it was not so much.
 
who is sp ?
who are you talking about ?
 
@DominicMichaelis Skullpatrol.
 
5:22 PM
Too many inane questions about stuff which there is no serious answer too.
 
Hey: it is working! :D
 
@JayeshBadwaik But why do we need a random noise generator? It's already quite difficult to keep up with legit messages.
 
all those removed stuff is annoying me
 
@JayeshBadwaik So receiving emails about penis enlargement does also not annoy you?
 
@MattN. It annoys me yes.
 
5:24 PM
I think it should be the word "either".
A minor issue.
 
@JayeshBadwaik I don't see the difference between a penis enlargement ad and an image posted by sp.
@awllower "does not annoy you either" would be ok, yes. But I think as I wrote it is fine, too.
 
@Bageer So how do you think of my answer?
Is it too tedious to read it?
 
@MattN. Ahh. He has angered many people before too.
 
@MattN. I see. Thanks for telling me.
 
@JayeshBadwaik He has indeed. But now that he's personally attacked two users I think he's crossed a line.
 
5:27 PM
@MattN. I do not know about Asaf (I was not here), but I agree with you about Jonas. Also, the annoying part is not offensive at all, it is kind of passive-aggressive which is even more difficult to get back at.
 
@JayeshBadwaik Link to discussion?
 
@JayeshBadwaik I couldn't agree more.
2
If I remember correctly he is a scientology member. Or believer, whatever you call them. Shouldn't he get banned, just for that?
 
How are you @dominic ?
 
@PeterTamaroff let me find.
 
@Charlie mh making pancakes
 
5:30 PM
@MattN. yes
@DominicMichaelis :9
 
@MattN. What? Scientology?
 
@Charlie Hi!
 
@JayeshBadwaik hi
 
@robjohn Another thing about the situation that bothers me is this: if someone writes "fuck" or "cunt" or similar words they might get flagged and possibly subsequently banned from chat. On the other hand, if you're a passive-agressive little cunt that personally attacks single users without using any swear words then... nothing happens. Something is messed up here, me thinks.
 
our freshmen got a very nice exercise in linear algebra
 
5:33 PM
@PeterTamaroff The clan that persuades people into giving them money in exchange for nothing and when you want to leave they blackmail you and drive you into suicide?
 
@DominicMichaelis what exercise?
 
Fact is, I really like Jonas in here. His funny. And he's decent. And he's clever. And now he's gone.
 
@PeterTamaroff I guess a lot of stuff before here.
 
@MattN. Did know about that...
 
Let me ask a question to sooth the atmosphere.
Does anyone know about some simple examples of non-Euclidean PIDs?
 
5:36 PM
I knew what scientology kinda was, but not more that than.
 
@PeterTamaroff Ah. If you want the details I think there is a documentary, perhaps on youtube. It's about a person who wanted to leave.
@awllower : D
 
@Charlie Let $V$ be a finite dimensional $\mathbb{K}$ vector space (over the reals or complex) and let $f$ be cyclic endomorphism such that there is a $v\in V$ with $$ V= \operatorname{span}(v,fv, f^2v, \ldots)$$ Prove that every eigenspace of $f$ is one dimensinoal
 
Interesting question I guess. :)
 
Afk, going to scavenge for food. Bbl
 
@DominicMichaelis aah
 
5:40 PM
@Charlie I think I found a really neat solution
 
@DominicMichaelis Do you mean neat?
 
@MattN. Bye Bye
 
@DominicMichaelis show me :)
 
@DominicMichaelis cool one.
 
@awllower yeas
 
5:41 PM
:D
 
@Charlie From Cayley hamilton we know that the characteristical polynomial annilates the matrix hence we know that $V=\operatorname{span} (v,\dots,f^{n-1} v)$
 
@DominicMichaelis How do you define cyclic endomorphism?
 
If any Eigenspace is 2 dimensional the minimal polynomial of will have degree strictly smaller than $n$, which would make the span linear dependent
@Peter a endo is cyclic when there is a $v$ such that $V=\operatorname{span} (v,f v, f^2 v, \ldots)$
 
@DominicMichaelis Oh, OK.
 
But if the span is linear dependent it won't be a basis. QED :D
 
5:47 PM
Good proof!
 
@DominicMichaelis :D
 
good luck i am not a tutor because I think most students will write bullshit there :D
 
@DominicMichaelis a really tough linear algebra problem: Let $V$ be the vector space of all polynomials over $R$ in the $n$ variables $x_1,\dots,x_n$ and let $p = \prod_{i < j}(x_i - x_j)$. Let $W$ be the subspace of $V$ spanned by all possible partial derivatives of $p$ with respect to all possible combinations of variables.
Prove that $W$ has dimension $n!$
(It is a tricky but doable exercise to show that it has dimension at least $n!$. I have no idea how to prove that the dimension actually equals that)
 
@DominicMichaelis oh...that's not nice...nevermind...
 
Hi @TobiasKildetoft
:)
 
5:52 PM
@Theorem hi
 
@Charlie they just can't handle polynomials :/
 
@Theorem hi theo
 
@TobiasKildetoft : today prof introduced puiseux series , but i didn't quite get it clearly . Do you have any online material or suggestions ?
@Charlie Hi
 
@DominicMichaelis then teach them
 
@Theorem I have no idea what that is
 
5:54 PM
@Tobias If we write it's hessian matrix there should be n! linear independent entries hsouldn't it ?
 
@amWhy Don't make an answer more hard than they should be! =)
 
@Charlie I am not a tutor :/
 
@DominicMichaelis possibly
@DominicMichaelis I am not familiar with Hessians
 
@TobiasKildetoft : we are doing little bit on algebraic curves on a plane , it comes when you want to parametrize a curve locally.
 
@DominicMichaelis ok
 
5:55 PM
its specially very useful to determine singularity etc . @TobiasKildetoft
 
@Theorem It is not really something I know anything about
@DominicMichaelis the only reason I know it holds is that the lecturer mentioned it followed as a special case of the $n!$ conjecture (which was proven not too many years ago)
 
@TobiasKildetoft : no prob :)
 
@tobias just for my intution i have a $n^n$ matrix in mind with a lot of symmetries
 
he had set it as a hard exercise because he thought there was a nice argument for it, but it turned out he didn't actually have one
 
My head is hitting the computer now.
I should sleep until I have again energy.
Later.
 
5:59 PM
well and by schwartz the order of differention doesn't change anything
 

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