@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.
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 $cn$
@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
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.
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.
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).
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)...
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.
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)
The 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 ...
@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.
@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.
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.
@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.
@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.
@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.
@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.
@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.
@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?
@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
@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 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)
@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)