@amWhy @Did @LeakyNun: Seems like some people don't like questions that don't have a question mark in them... My recent post got closed so quickly, though I had implicitly made a reference request. Anyway I've edited it. Do you think it's okay now?
@quid: If you have any feedback on my recent self-answered question ^ please let me know. =)
@Joffan I support the way amWhy suggests (which some people like Martin Sleziak have done very well), which is to post a new question of your own that starts off with the problem and what you have tried. Since it is your question, it has nothing to do with the original question, and as long as it is a good quality question that's great. I personally do not agree with editing the original question to improve it if the asker has not shown any effort.
@amWhy Michael R added the tags again :) – There is the opinion that "tagging a question according to the techniques used in the answers" is useful: math.meta.stackexchange.com/q/16681, math.meta.stackexchange.com/q/2612. But even then: Why [polynomials] and [tangent-line-method]?
Here math.stackexchange.com/posts/2464089/revisions is a question were [roots] was added and later changed to [radicals] (after a now deleted discussion between me and Michael R). I still don't see the value of adding [radicals] to all questions where a square root is used somewhere in an answer ...
@quid Thank you for your response! The answer is actually an improved version of the one I used in the discussion in the Logic room last weekend, and I felt it was a waste not to post it on the main site.
@SimplyBeautifulArt So what do you think? I don't understand what's wrong with those users who upvote Rob's comment after I have already edited the post to be strictly a Q&A.
@SimplyBeautifulArt Yes when I post answers I either aim to just answer the question and provide points that I feel other answers are missing, or I aim to write as complete and general an answer as possible, and would improve it over time as I get better myself.
Given a prime $p > 2$ and positive integers $x,y \le \frac{p − 1}{2}$, prove that if $x(p − x)y(p − y)$ is a perfect square, then $x = y$.
This question was previously asked without context , and so was closed, but I did start towards a solution.
The primary insight is that the result depen...
@user21820 Hm, we were naive (or at least I was) to not check the profile of the vandalizing user Smokey caught. math.stackexchange.com/users/491981
I've also found it quite nice that the review queue has a little dot next to it indicating how bad the queue is at the moment (in terms of numbers). Hopefully it'll provide more incentive for users to check in on it.
@amWhy: Note that the original version of my question did not have a question mark, so Rob is 'justified' to say that it did not have a literal question, which is why I edited my question. Whether it is more suitable for a blog or not is a subjective question, but the SE system encourages what I'm doing, so...
Of course, I am slightly irritated that 6 users upvoted Rob's comment, 2 of which did so after my edit.
If people don't like logic they should just stay away from posts about logic. Anyway I too need to go off from Math SE for some while. Have a good day @amWhy!
@user21820 Yeah, that's what I was also irritated with, the upvotes to Rob's comment. I don't know about you, but I do find only a narrow interest in logic on this site, and very few of the more elementary logic questions, even those that attempt a deduction, translation, or simply want us to confirm their work, or point out errors. Elementary questions in other topics (derivatives, integration, elementary number theory), get many answers, many upvotes. :-?
@Joffan Very nice question! Thanks for posting! +1
@Did is it necessary to add comments to your answers when they get downvoted (see math.stackexchange.com/a/2483396/66096 for example) ? I'm not the downvoter btw
@GabrielRomon comments are never necessary, they are always an option. I believe your question is more of "is it proper etiquette to add comments..."
I'm not quite sure if you could leave more than a "Hm, well I wonder why I got downvoted" if you were downvoted and didn't have any other comments to go off of.
@SimplyBeautifulArt I wish the mentions to stay because they correspond to four revenge downvotes given in short succession and because I have come to the conclusion that making such events as visible as possible was the best way. For example the OP of another of the downvoted answers reacted by commenting: "btw i did not downvoted the post.. but maybe someone saw a mistake"... to which I replied firmly, hence the mention at the beginning of my answer proved useful to dispel at least this OP's doubts.
@SimplyBeautifulArt This maybe how you would handle this, but I wish to proceed otherwise. To make some noise about these unpleasant actions, one could call it. The mention I had added are a simple way to express simultaneously the fact that the answer is correct and the opinion that the DV should be ashamed of their behaviour.
@SimplyBeautifulArt Hmm, I AM interested to learn how you think about the matter and I took duly note of your remarks... but, as you can see, I have some (valid, I think) reasons to proceed otherwise.
Yeah, I've nothing to say as to how you'll proceed with your business. But over time I think I've become rather well at handling DV's that are very likely targeted. OTOH, I've never known my targeter personally.
At least handling them emotionally/internally and publicly.
@GabrielRomon Yeah I know. But my aim is not to chastise the culprit (and not even to nullify the -2 points, gee -2 points, that is, 0.001%...) but to try to raise the awareness of the community about revenge downvotes which, at present, seem to me to be considered as a viable option (or, at least, not a revolting one) by an awfully large proportion of said community.
@Did you have a point. Btw, do you recommend a book with a rigorous exposition of Markov chains ? (this is way off-topic, so feel free to answer or not)
@SimplyBeautifulArt I think @Did makes a good point; when one flags, one risks it is handled wrongly, or is dependent on who handles the flag; commenting in the mod chat can often prove useless. For you, I think you'd get some results. I think Did gets the point across to the folks who need to hear it, the most immediately and effectively.
@SimplyBeautifulArt One always here's, back from a flag, or in the mod chat: "Wait for the script to run; hopefully it'll reverse them. Then, if it doesn't, you can always flag us then, or comment in the mod chat."
@amWhy Well for me, the relief of telling the mods first hand (and semi-publicly (okay, hardly anyone visits the mod chat, but...)) is enough for a problem I've only dealt with less than a handful of times in my entire existence on this site.
I can understand if Did has had more and its a problem for him though. Being relatively new leads to less grudges by default.
@SimplyBeautifulArt You had a couple of instances in which you had 6 or more in a shore succession of time, and you've gone to the mod chat a number of times even when you've only received one or two.
I think you might want to check what you state is the answer. I believe the answer is 1, but I did it very quickly, and perhaps I'm wrong here. — amWhy7 hours ago
@amWhy I believe the limit is indeed 10.
Hm
@amWhy Actually, I kinda want to take my vote back. We were kinda fast on that one, and the OP has not been seen for 12 hours... I want to give them time between their question being [closed] and [deleted].
Of course, to be fair, the comment was placed 12 hours ago, so they were told...
Let $a_k = 1$ if $k = n!$ for some $n > 1$, $a_1 = 2$, and $a_k = 0$ otherwise, so the sum is $\sum_{k=0}^\infty a_kz^k$. Then $\sqrt[k]{a_k} = 0$ if $k$ is not a factorial, and $\sqrt[k]{a_k} = 1$ if $k > 1$ is a factorial. So $\limsup \sqrt[k]{a_k} = 1$. — Daniel Fischer ♦Sep 11 '13 at 23:33
@SimplyBeautifulArt, @user21820, @Did, and anyone reading this: This post is an image from a text or powerpoint etc, followed by a PsQ. math.stackexchange.com/questions/2488408/…. Asker got an answer and, yet, had the nerve to ask the answerer for more details. Votes to close are welcome.
Given a prime $p > 2$ and positive integers $x,y \le \frac{p − 1}{2}$, prove that if $x(p − x)y(p − y)$ is a perfect square, then $x = y$.
This question was previously asked without context , and so was closed, but I did start towards a solution.
The primary insight is that the result depen...
