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seen Jul 18 at 15:00

"Points don't matter"... and yet because of these points DIRECTLY most of the misunderstandings and fights among participants happen in this site.

I believe the above would be greatly avoided if people were required to justify, at least shortly, her/his downvote, making it clear it is not personal or petty but more or less well reasoned. If the site would also require from people to write down shortly their personal background in mathematics and this would appear in peoples' profiles then we'd avoid many, many misunderstandings in answers.

I can't agree anymore with the site's outlines of working and neither with the site's administration "making of justice" and direct insinuations of misuse, without possibility of defending oneself and without being able to retort absurd accusations and other nonsenses.

I hereby shall stop my participation in this site at once. This is sad to me because I enjoyed participating here and I myself learned a lot, both from other participants' answers but also from some askers who happened to ask very interesting questions and/or in some very interesting fashion.

Perhaps in the future I shall take part once in a while in this or that thread, but I wouldn't wait for this to happen without getting a chair and a fat book to read.


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awarded  Refiner
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30
awarded  Explainer
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awarded  Nice Answer
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24
awarded  Autobiographer
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awarded  Good Answer
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awarded  induction
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awarded  Nice Answer
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awarded  Nice Answer
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awarded  Necromancer
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awarded  Good Answer
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awarded  Curious
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awarded  Nice Answer
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awarded  Revival
Jun
18
revised FC groups with infinite derived subgroup which are not constructed by direct product of finite groups.
edited body
Jun
18
comment Any Help would thanks : Product of continuous and Riemann integrable function
Yes @user....but remember that first you have to prove $\;f\;$ is Riemann integrable (it is almost trivial).
Jun
18
comment Why must a field with a cyclic group of units be finite?
I think I got it now, @JyrkiLahtonen: "since otherwise" means "if $\;char\neq 2\;$ then $\;-1\neq 1\;$ and ...etc. ". Thanks.
Jun
18
comment Integral domain (rings and fields)
Perhaps the easiest and shortest way: we have that $\;\Bbb Z[i]\le\Bbb C\;$ , and since the last one is a field we're done.
Jun
17
comment Prove the cartesian product $K \times L \subset \mathbb R^{m+n}$ is also a compact set
"bounded", not limited.
Jun
17
revised Prove the cartesian product $K \times L \subset \mathbb R^{m+n}$ is also a compact set
added 50 characters in body; edited title
Jun
17
comment Dimension and basis of a vectorial subspace of the polynomial vector space
No, it's not "the polynomial" but what is called the evaluation homomorphism (or linear transformation): $$\phi(p(x)):=p(a)$$. A linear functional is only a linear tranformation which image is inte hdefinition field.