@BillDubuque Then we'd have to disagree on what is the right motivation. In my opinion, it is because the recursive definition of x^(n+1) = x^n·x for natural n demands x^0 = 1. Also, I am sure that if you had the chance to question the author of that post, you would find that he/she didn't really understand the basics. I can tell.
@user21820 But your motivation is simply a concrete special case (S = N) of the motivation that I gave (i.e. adjoining a neutral element to a semigroup S, or enlarging to the group of differences). I don't think it is wise to downvote and delete answers based on guesses about what the authors understands. Rather, that should be based only on what is written in the answer.
@BillDubuque I have seen too many students misled by such erroneous pedagogy. I (and any other mathematician) can easily 'derive' lots of interesting questions from most crank posts, as long as they are not too nonsensical, but that doesn't mean their posts should be preserved. We will just have to disagree on this post.
In case it isn't clear, what you can see in a post is not what a student reader sees. I judge a post on Math SE based on its estimated effect on its intended audience, not on its effect on me or experts in the relevant areas.
@user21820 As I mentioned, it is most certainly not misleading. This is how algebraists think about these matters (peruse any book on universal algebra to learn more about such universal properties of algebraic constructions e.g. Bergman's online book)
@BillDubuque You're just supporting my point; as I said, I judge a post on Math SE based on its estimated effect on its intended audience (here it is student readers including the asker), not on its effect on me or experts in the relevant areas (which you have indeed named).
@user21820 Not true. You wrote it is "pedagogically nonsense because x^0 is not defined as x^n/x^n, nor is it right to invoke the exponents 'laws' in order to justify the value of x^0". But - as I explained - not only is it not nonsense, it is very natural view from an algebraic standpoint. Apparently you think that this view has some bad effect on the "intended audience" but I have no clue what you mean by that since you have not explained it. Please strive to be more precise.
@user21820 Let's consider your objection in additive form, e.g. adjoining a neutral element 0 (and/or additive inverses) to the additive semigroup (N,+) of naturals. Your claim translated there is that it is pedagically nonsense because "0 is not defined as n-n nor is it right to invoke laws of (N,+) to justify the value of 0". Do you really support that claim?
I find your claims to be far more vague and misleading than anything written in the answer that you critiqued.
What is written in that answer is based upon solid mathematical (and pedagogical) foundations. I cannot say the same for your claims since they are so imprecise that they cannot even be understood at this point.
Please be much more careful about soliciting down/deletion votes. These should not be based merely on your guess that "if you had the chance to question the author of that post, you would find that he/she didn't really understand the basics. I can tell"](chat.stackexchange.com/transcript/message/52675363#52675363). Rather, they should be based on mathematics. I can no longer be confident in your recommendations if you continue to do things like that.
$X_1,X_2,\ldots,X_n$ are independently and identically $N(0,\sigma^2)$ distributed. I want to prove that the distribution of $$W=\frac{\bar{X}}{\sqrt{\sum_{i=1}^{n}{X_i^2}}}$$ is symmetric about the origin.
I have shown that $$\frac{n(n-1)W^2}{1-nW^2}=T^2,\quad\text{where}\quad T\sim t(n-1)$$
T...
and here is the answer that I find wrong math.stackexchange.com/q/3448311 now, as a new user of this chat room, what is the difference between the "send" and the "upload" button?
... that is, there is NO "wrong answer" flag, only "not an answer" (which is not quite the same).
@Mirko Gamification is evil, and your imaginary internet points are not actually worth anything.
That being said, if you think an answer is wrong, there isn't a lot you can do:
(1) downvote (at a cost of 1 XP to yourself)
(2) flag it as "Not an Answer" (which should send it to the review queue)
(3) flag it as low quality (which should also send it to the review queue)
and / or (4) leave a comment requesting a correction or otherwise pointing out the error.
user12692
11:16 PM
@XanderHenderson When the imaginary internet points are greater than 25K, maybe yes. When the imaginary internet points are less than 10K, one cannot vote to delete a question.
@Jack If the concern is one's ability to vote to delete, then 1 point is a drop in the bucket compared to 10k (to vote to delete a question) or 20k (to vote to delete an answer).
@XanderHenderson I already did (2) and multiple (4). I do not see a "low quality" or "wrong answer" flag. I do have 10K but do not see a button for "vote to delete an answer" (that last thing addressed to @Jack).
@XanderHenderson Thanks, I guess I just couldn't let it go without making some fuss about it first. It didn't help that my first comment to that answer was addressed but I believe inadequately, by the answerer (perhaps some misunderstanding, but I think my comment was clear, and the answerer has plenty rep, so I expected they would easily understand my criticism). Thank you for reminding about the word "politely", I hope I have not crossed some boundaries that I shouldn't ...
$X_1,X_2,\ldots,X_n$ are independently and identically $N(0,\sigma^2)$ distributed. I want to prove that the distribution of $$W=\frac{\bar{X}}{\sqrt{\sum_{i=1}^{n}{X_i^2}}}$$ is symmetric about the origin.
I have shown that $$\frac{n(n-1)W^2}{1-nW^2}=T^2,\quad\text{where}\quad T\sim t(n-1)$$
T...
