@amWhy I don't even know what this is

what is that question

@Holo Look at OP's comments on this answer what the actual heck

@RushabhMehta I think he called John a bully at the end there but I couldn't follow his line of though...

@RushabhMehta Question is on its way to deletion.

Deleted!

Blatant PSQ with five up votes and four answers.

A PSQ from a high rep user.

@amWhy@Did@Holo@paulplusx@TheSimpliFire@user21820@XanderHenderson These posts need a few more closure votes: Κ1, Κ2, Κ3, Κ4, Κ5, Κ6

@AlexFrancisco All but 5 and 6 are closed

@Holo Thanks!

Also Κ7, Κ8, Κ9

For deletion: δ1, δ2, δ3, δ4, δ5,

δ6, δ7, δ8, δ9, δ10,

δ11, δ12, δ13, δ14, δ15

Caution: Goldbach's conjecture.

[ SmokeDetector | MS ] Blacklisted username, few unique characters in answer, no whitespace in answer (277): What is the inclusion-exclusion principle for 4 sets? by fuck on math.SE

@amWhy "Construct a function shaped like this: W" would in my opinion be much clearer. It took several users, including me, more than one attempt to get what is meant. As a rule I have difficulty to consider an answer to a question as good that takes me "ages" to understand while I perfectly well know how to answer the question and there is nothing in the answer that would not have been clear to me, if only I had understood that this is what is intended.

Beyond that, given the question I doubt that a sketch of the function would have been sufficient, and it is possible, even quite plausible, for somebody to even know graphically what the function should look like but to still not know how to actually write this down. In that sense even the version of the answer I gave above is arguably not an actual answer.

Still, it can possibly considered kind of a nice curiosity and an inside-joke.

I agree that the answer has a certain appeal, which is why it is popular, and I think it is well within reason to do something like this occasionally (and this was also the spirit in which it was given), but evaluate according to usual criteria this answer is not a good example for an answer.

@MartinSleziak yes. That's what I meant to write as final comment. See above. But that is orthogonal.

Personally, I liked that answer and I consider it quite nice. IIRC that question has several other answers which have sufficient details. (I am not advocating "joke" or "funny" answers. But if it is this good, I would definitely prefer it to stay.)

@MartinSleziak yes but that's orthogonal.

And it is important to be clear about that.

My flag was declined for this PSQ

@paulplusx One of them always chooses to leave open.

v1, v2, v3.

v4, d1, d2.

@AlexFrancisco All clean. Left with the above and those pinned.

I understand why someone would say that this question had no context. But I'm also curious as to what additional context one can put there. Yes, this is the type of thing one is expected to search online before posting, but at the same time I think that question is right in that gray area of what is also reasonable to post on MSE without putting hours of search first.

(Even though I agree that ideally the OP should have searched first, and come across one of the many related answers on MSE/MO which may or may not have been enough for them to deduce the answer on their own.)

@AsafKaragila I would expect more context (why are you interested in the problem) and effort (what do you already know about it), but I can't tell what the close/delete-voters had in mind. Arguably, it's such an easy question that the asker must have been asking out of curiosity rather than because it's their homework, but again we can't tell because not enough context was provided.

@user21820 Well. It's not that easy, at least if you're not very familiar with choiceless set theory. I think the formulation could have been used a little bit of sprucing up, but all in all it's a natural question to ask when you see the proof that the groups are isomorphic.

@AsafKaragila Hmm I guess that's fair if one does not know that choice can succeed at lower levels but fail arbitrarily high up.

@user21820 And even then, if there is a Hamel basis, then you still get isomorphism. And those can exist even if choice fails for the reals. For example in Cohen's model.

@AsafKaragila Mmm but that's not necessary to answer that particular question, right?

@user21820 My point is that in Cohen's model, there is a Dedekind-finite set of reals, but (R,+) and (C,+) are still isomorphic. So even if choice fails for the reals, the two might still be isomorphic.

@AsafKaragila Yes but that question only requires knowing that we can have choice for reals while anti-choice for some other set, making the answer "unprovable".

Anyway I don't disagree that it's interesting. =)

@user21820 The question was whether or not ZF proves there is an isomorphism. So you need to know that there is a specific anti-choice axiom that implies that the answer is "no".

@AsafKaragila Ah I missed that, oops.

@user21820 It's a single line! How could you have missed that? :D

I get that it got closed and deleted for lack of context, but that at least makes the actual question very easy to read. :P

@AsafKaragila I like questions that have two parts, the first part clearly and succinctly stating the question, and the second part sketching the context. =)

@user21820 I like questions that have three parts. Title, body, and answer.

Lol.

I prefer those three parts to be head, body and tale.

@user21820 Tale?

The tale of how the question came to be, in place of a waggy tail.

@user21820 I like my cats to have head, body, and tail. The tail is optional, but preferable. But head and body are pretty mandatory. But we're getting off topic. So I'll go back to figuring out some symmetric model with Sacks reals instead. Hopefully, the people who closed/deleted the question I linked will come forward to speak their mind.

@AsafKaragila: even without hours of research, for the question you linked, the OP could have started with something like "We know that the additive groups of the real numbers and of the complex numbers are isomorphic, because both the reals and complex numbers are vector spaces of the same dimension over the rationals. However, the construction of this isomprphism requires us to choose bases for R and C over Q, which cannot be done in ZF without the axiom of choice.

Therefore, I wonder whether (R,+) and (C,+) can be proved isomorphic in ZF without the axiom of choice? I don't know enough set theory to answer the question myself, but it seems like something that ought to be known. " In other words they could just come out and say what they already know, for the benefit of those who don't.

And if they don't know even the basics from that post, then they really do need to do some more research first.

@CarlMummert Thank you Carl. I agree with you. But let me also point out that none of this was explained to the OP. The only comment about the quality of the question came from user1729, after the question was initially closed (and user1729 was not one of those voting to close). Your comments here are excellent. I think it would be good for the OP to also see them, and maybe even react to them by editing their post.

(Just to clarify, it's not that I want this question to be undeleted because my answer is there. I just noticed that it was deleted because my answer was there. And as I said, I find this question to be somewhere on the gray area of context-less posts.)

Thanks everyone who solved my homework, I erased my question.

As the body shows, how can I fill a too contextless question on Math SE?

Can anyone please help come see what's up with this contextless question? thanks (:

@user21820 Had to chuckle at the titles of the questions you present!

@amWhy It happens when I get bored/tired of seeing such, and it's not hard to learn to use the same phrasing. =P

Let F be a field theory homework question. How to prove that you can solve F for me..

math.stackexchange.com/q/2942395/29335 Do my homework

Please remember that you can collect a lot of rep by aswering among the homework questions is the asker is generous with the tick...

@user21820 Well, if I smirk, or chuckle, that's a good thing, because it does liven up this room, and the crap we see day in and day out.

Maximize title+body subject to the condition that title=body..

Another copycat.

Three is a crowd.

Time for some refreshments: Good question on the HNQ!

Find the value of a great sum of questions like this 1.

Do this for me, declares the potato girl

@amWhy it got expanded now. There is a substantive attempt now. I thus reversed my vote.

@quid Thanks for letting me know.

Can anyone help me understand why my question was closed? I know it is currently in the review queue and might be reopened with what I added, but the original version of the question should have been detailed enough.

@DavidCoffron A good starting point for you to better understand what is expected in questions on this site, please read How to ask a good question. What I can say is that a question that looks like it was copied and pasted from an assigned exercise, with absolutely no context from the asker, which describes your initial question, is not a good question.

@amWhy I mean it wasn't. It came out of a discussion with a friend

Is it worth adding that context? I felt that removing the baggage of the use case would be more useful for answers that can apply to a broader range of topics

It still reads like a problem/exercise statement. That's not sufficient for asking a good question on MSE.

@amWhy I suppose I can mention the souce

source*

@DavidCoffron That would be helpful, and describing and linking to the chat discussion. (I meant the original question still read like a problem/exercise statement).

@amWhy Gotcha

@AlexFrancisco 3 in fact this time

The user of this post, which needs to be closed, has a string of helpless sounding homework questions, like the one I link here.

Since MH's discussion was deleted, I'll repost his comments, because we must have evidence aavailable:

Note, the Chatroom: "Discussion between amWhy and Michael Hardy" has since been deleted by site mods, or SE Community Managers; but these comments were preserved two days ago, the same day MH wrote them.

Seems like the newer room is gone too.

@MikeMiller Yes, the major reason I repeat the most egregious comments here. I could no longer access them from that room, due to its deletion. But the Mods, and SE staff, can surely verify these are permalinks from a room, which were permalinks from the now deleted discussion.

Hello, @WillHunting!

Hello @amWhy I hope you are well.

@WillHunting Indeed. I hope you too are well.

@amWhy I'd be willing to argue that this is not a duplicate of the linked question.

@amWhy Look at the answers on both questions, and tell me what you think.

@amWhy actually nvm, I'm a goof

ignore that

@RushabhMehta No problem!

This question is about proving $(x+y)^n \leq x^n + y^n$ for $n\in[0,1]$ using "induction", I believe it should be dup of this question about general "induction" over interval, but I can see why it is not necessarily duplicate, please vote as you see fit