@robjohn I mean that I always forget this kind of formulae so I need to calculate them from the Sarrus rule, wish there was some fast visual way...there must be...
alternatively: view the $3\times 3$ matrix as the points of the affine plane over $\mathbb F_3$: then the terms in the determinant correspond to affine lines which are neither horizontal nor vertical
and there is some way to pick up the sign from the geometry, but I don't remember it
@tb Good LaTeX advice. The sentence does not sound right though. "to prevent the period from being moved to" is the minimally adjusted correct sentence I believe.
@hhh Draw a triangle with vertices i, j, k with arrows i->j, j->k, k->i. Multiplying two of these units is done as follows: if they are the same, 0; if they go with the flow of the triangle, the third in the line; if the go opposite the flow, the third one as well but with a minus sign
@tb that's how I do cross products. I just remember that if it is an even permutation of $i,j,k$ that the sign of $i=j\times k$ and $i\times j=k$ is right.
If it's any help, I point my right thumb in the direction of the first vector, and my right forefinger in the direction of the second vector. My right middle finger can then point in the direction of the cross product.
user19161
12:17 AM
@AsafKaragila Because he did not want to be suspended.
@tb Either way, I think it's pretty effin' cool that the complement of the Cantor set is a union of countably many disjoint intervals but has uncountably many boundary points.
@tb Yes not bad way to preprocess ideas first in Chat and then try to build quality threads that way -- making threads many times more deeper (and killing stupid qs) :)
\footnote{$(A+B)\cdot (A+B) = A\cdot A + A\cdot B + B\cdot A + B\cdot B$. The cross terms are $A\cdot B$ and $B\cdot A$. This has nothing to do with the cross product.}
@hhh: When editing the OP's comments into their answer for them, you might want to mention it's the *OP*s comments, not yours, in the suggested edit description box. Generally I (and others I think) would reject an edit that put someone's personal comments into someone else's answer, so you risked getting your edit rejected. Also, I still don't see how category theory is relevant to pictorial mnemonics. :)
@anon Yes that is good point, I wish I could somehow clarify that it was his comments. At the time of writing, I thought it was obvious.
@anon Wish I knew that :P (I just know that it can be done if really wanted...but sure perhaps a bit over-engineering, well hopefully some gets more inspired by my comments :P
@anon: now I tried to apologize the op due to my mischief, now better? Here. (hopefully the term "mischief" does not infer intentional because it was accidental choice of vocabulary)
@hhh: You've misunderstood me; there was zero issue with what you did, in fact it was nice of you to help out the answerer and it was totally clear to those reading the answer and its comments what was going on. However, the people who review suggested edits (e.g. me) don't get to see this context, so I didn't know they were the OP's own comments until I investigated your suggested edit. I only wanted to forewarn you for your own sake that you risked getting in rejected because a reviewer...
...might not have bothered to check what was going on like I did.
@anon What is the correct terminology here to use? Look I asked the question, can I still use the term "op's comments" (not referring to myself) when I want to make answers to my questions better/clearer? Perhaps in the future, I will use the term "answerer's comments" -- is that safe?
I would use "user" rather than "op," because the latter generally refers to the question-asker (even though technically it is still applicable to the user who submitted an answer).
I don't know what an LSAT or a GRE is; and certainly couldn't tell you (until just now) what each of them is scored out of. Can't we say "100%" or something sensible?
i have to go afk for a bit. if anyone can help me here math.stackexchange.com/questions/124108/… i would be much obliged. even an upvote to give it more attention would be appreciated
Usually, if I want to know the answer to something, I want it immediately. If it doesn't already exist on Stack Exchange or Stack Overflow, I go and look somewhere else.
Which is kind of selfish of me, and unfair to future users who might want an answer to the same question. I make up for it by answering more than my fair share of questions.
@David : If i click on your photo icon in the chat it shows that your primary account is on StackOverflow, but it showed the other day as math.SE and even you said that you never posted anything on stackoverflow
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?
(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)
Thanks!
@DavidWallace I know the 6 answers of mine that the deleted user upvoted. I see no trace except there is one comment by user13838, who no longer exists.
Yes, I think the basic algorithm is wrong. If someone's reputation points are supposed to reflect reputation, then the votes (up or down) of former users should count just as much as the votes of current users.
Presumably then, when a person quits the site, all their questions and answers disappear. If the purpose of SE/SO is to compile a database of questions and answers, then this is the completely wrong thing to do. Imagine if somebody set fire to everything Isaac Newton ever wrote, on the day that he died. Or Thomas Jefferson. Or anyone else who's made a great contribution to the world.
@WillHunting solve the problem means solve your particular problem or it solves your problem but not fully? In the latter case, I see no issue, in the other case, I think you may have to think if you should really wait!
Nothing to apologise for; if the formula's right, then it's right. If this is the case, then I'm glad you pointed it out. I have removed my comment from the question.
Yes, I've just managed to prove that the formula is right.
Thank you @Kan for pointing this out. You've saved me from making a complete fool of myself.
Some other dictionaries list that sense, too. For example, Chambers (the 11th edition) has this: "intransitive verb To make a conscious attempt to provoke controversy or disagreement on the Internet", Merriam-Webster's Advanced Learner's Dictionary says “a person who tries to cause problems on an Internet message board by posting messages that cause other people to argue, become angry, etc” for troll, Cambridge Advanced Learner's Dictionary says “to leave an insulting message on a part of the Internet for no reason”, and NOAD has "send (an e-mail message or posting on the Internet) intended…
Could someone explain why I am missing one term there? (Notice the question mark) Related to polar coordinates and partially things we have done with @robjohn.
@tb Beats me. I think we just talked about some theorem that says that if a set is $\Pi^1_1$ (lightface, of course) then it can be used as a selector kind of set or whatnot.
@N3buchadnezzar I am not sure if I can get anything on the top of my head. But, yes, I can if I got to my house. Look up some of the books I solved when I was young! : )
@AsafKaragila Did you look at that call for Duplicate, I feel guilty for having to retype the same argument, I'd like if you please went through and told me if they are the duplicates of each other or not!
Regarding the permutation questions, the older question you linked seems much more general than the recent question, and at too high a level for the op (probably).
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?
(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)
Thanks!
