CURED - 2020-04-07

2:29 AM

HNQ question. The body of the post contains no question (it seems that the entire question is in the title). I would try to edit the title into something reasonable, but I am going to bed right now.

rschwieb

2:23 PM

PSQ math.stackexchange.com/questions/3614057/…

amWhy

@rschwieb Closed and open for deletion.

Xander Henderson

3:32 PM

This answer to an unclear question is almost certainly wrong (though it is hard to say, because the question is deeply flawed).

cv-pls general-topology Needs Details Or Clarity What does $\mathbb{Q}^{*}$ mean?‭ - Juan Cruz Carrau‭ 2020-04-06 23:24:06Z

amWhy

@XanderHenderson I addressed as best I could, both messages above.

Xander Henderson

@amWhy Thanks.

amWhy

3:49 PM

@XanderHenderson NP :-)

SmokeDetector

[ SmokeDetector | MS ] Blacklisted user (71): For which values of $x$ is $x = x!$ by Mr Majeed on math.SE

user21820

5:44 PM

@XanderHenderson Agreed.

D1, D2, D3.

D4, D5, D6.

D7, D8, D9.

D10, D11, D12.

D13, D14, D15.

C1, C2, C3.

C4, C5, C6.

C7, C8, C9.

rschwieb

6:08 PM

Would it be unreasonable to mark this a NAA: https://math.stackexchange.com/a/3614418/29335

I mean, a solution should not be something that is l-i-t-e-r-a-l-l-y applicable to any proposition of the form "Show A implies B". At least, not in this case. Proving the contrapositive offers nothing more than proving the original statement by, say, contradiction, so the suggestion seems empty, to me. A comment, at best. It is not really even a *hint*.

amWhy

@rschwieb No it is not unreasonable to mark as NAA, for the reasons you give. I plan to do so too. It is now deletable.

rschwieb

@amwhy I'm not sure when the downvote arrived for the problem, but was that you? I just noticed the user attempted their own solution in the solutions area, and I"m inclined to take that as a token of context.

... although I bet this is a duplicate anyway...

amWhy

@rschwieb No, I just downvoted the answer, which makes deletion much easier.

rschwieb

@amWhy OK, never mind me then

amWhy

@rschwieb But I saw the OP attempted their own solution; just not clear why they deleted it. Also upvoted your comment to the OP of the question.

rschwieb

6:35 PM

VLQ, nearly PSQ math.stackexchange.com/q/3613985/29335

amWhy

7:14 PM

Pretty blantant PSQ, times two (original title uninformative, written in all caps).

Xander Henderson

@rschwieb This has been deleted.

@rschwieb Sigh...

@amWhy Did what I could.

amWhy

@XanderHenderson tankya... errr, Thank you!

Xander Henderson

@user21820 I don't really have a problem with C5. It is uninteresting, but I think it meets the bar for context (it is sourced, and the asker has pinpointed where their issue is). Maybe if there were a "too local" close reason?

In any event, it will be automatically deleted in a couple of days if no other action is taken.

amWhy

7:31 PM

Blatant PSQ math.stackexchange.com/q/3614568/9003

Xander Henderson

@amWhy It is frightening to me that someone who is learning about convolutions does not understand the meaning of the verb "to solve".

"Evaluate" or "find" or "determine" or even "expand" or "simplify", maybe. But solve? No.

amWhy

@XanderHenderson Right!?

Xander Henderson

I mean, this is an error that my precalc students make all the time (at least, for the first couple of weeks of class). But they probably can't even spell "convulution"... "canvolution"... "canine"... THAT WORD, let alone define it.

amWhy

@XanderHenderson ...let alone "convulsion"!

Xander Henderson

Heh.

José Carlos Santos

8:26 PM

@rschwieb It is closed now, and ready to be deleted.

amWhy

8:45 PM

→ 6 messages moved to Trash

→ 1 message moved to Trash

Bill Dubuque

@amWhy Thanks. I am open to chatting elsewhere to attempt to clear up any misunderstandings.

amWhy

9:13 PM

@BillDubuque You are free to create a chatroom if you'd like to clear up matters, in which case, if you invite me, I'll enter it.

Xander Henderson

9:44 PM

This has to be a duplicate many times over...

I am out of votes, but I posted a possible dupe target as a comment.

Bill Dubuque

@amWhy Will do since that will likely be productive (but probably not right away since I am also volunteering in the local covid-19 fight, besides here, so spare time is shorter these days)

amWhy

@BillDubuque No problem. Thanks!

@Xander, thanks for catching the "anonymous user" suggesting a problematic edit on a post of mine.

Xander Henderson

10:01 PM

@amWhy it came through the review queue.

amWhy

@XanderHenderson Well glad you caught it after it had an accept and a reject!

Xander Henderson

Yikes!

An accept?!

amWhy

@XanderHenderson Yes, unfortunately.

@XanderHenderson See math.stackexchange.com/review/suggested-edits/1353819

Xander Henderson

Oi...

amWhy

10:20 PM

@XanderHenderson Such suggested edits on my answers have been rather frequent lately, usually caught by a reviewer or me (because I get messaged about suggested edits on my answers, as do all answerers).

@XanderHenderson Yes, I selected the most relevant of your links and summoned my dupe-hammer power on all question tagged discrete-math to close it as a dupe.

Xander Henderson

10:40 PM

@amWhy Thanks.

amWhy

11:12 PM

@XanderHenderson Sorry, @Xander, but I felt compelled to undelete the question I deleted with a dupe hammer. "not reflexive" is not equivalent to "irreflexive" (the term used by the OP.) A relation that is not reflexive needs only one $x$ in the set S for which $(x, x)\notin R$. A relation is R irreflexive if and only if there is no $x\in S$ such that $(x, x) \in R$.

Xander Henderson

@amWhy shrugs I guess I found the question to be unclear :\

That being said, I have no great attachment to the question, one way or the other.

amWhy

11:33 PM

@XanderHenderson Look over this: math.stackexchange.com/questions/3614746/…. The language of properties of relations can sometimes be confusing. Just as not-symetric is not the same property as anti-symmetric, not-reflexive is not the same as irreflexive. I had to double check definitions, which led me to reopen and answer the question.

@XanderHenderson You're right, in that OP did not thoroughly carry out why $(a, b)$ could not belong to the relation, because by the symmetric property, we'd need $(b, a) \in R$ but that would require, by transitivity, that $(a, a) \in R$. But if $(a, a) \in R$, then the relation would not be irreflexive....

But, per my most recent comment, the best reason for closure (if closure is desired) is "needs details and clarity"...

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$Mathematics$ CURED