/sci/ - Science & Math

>>12448449
In the UK, but I would have applied to the places I mentioned should they have had open positions back then. I've been to seminars with folk those places, really liked the Copenhagen Jespers and Markus Szymik as speakers, mathemagicians and nice people.

>>9677326
Blah blah blah. There were no mistakes on my behalf, and if there were I would simply have admitted them. This is an anonymous board. I have nothing to lose by saying I was wrong on something (assuming I was), but I wasn't so there is no reason for me to make such a ridiculous claim. Cry some more, and then get that job I told you to get.

>>9554419
Occasionally. I'm basically just watching the faggotry unfold. You are entertaining, but too much below me in the universal hierarchy of Thulean hyperdoctrines for me to actually interact with you.

>>9430304
I'll
>>9430306
check
>>9430314
these.
Thanks.

>>9264178
Nope, it's because then y wouldn't be an integer. If you have any claim of the form "for every ... there is some ...", all you need is one counter-example. In this case it would be any positive integer whose squareroot is not an integer, 2, 3, 5, 6, 7, and so on. The order of words is important, too. You can have "for every integer x, there is an integer y such that x-y=0", and you can have "there is an integer x such that, for every integer y, x-y=0". The first one is true, but the second one is false.

>>9133212
Oh... I'm sorry to hear that, anon,

>>9044249
Let your journey be pleasant, little one.

Mine isn't even ranked for math, but I still BTFO PhD students on this board.

>>9013812
You know seppuku, that number game? You can make a computer program such that it fills the grid, and checks if it is a valid solution to the problem (that is, there are no repetition of numbers in a row, column or a smaller square). Well, if you have a finite group (which you would now have) you can write down its multiplication table. Such tables follow the same logic: each element should be found exactly once in each row and each column. Basically, you would want to solve an [math]n\times n[/math]-game with no smaller squares.

>>8990162
[math][\pi ]=\frac{[\text{Perimeter}]}{[\text{diameter}]} = \frac{m}{m} = 1[/math]

>>8973434
(i) no prime ideal contains the whole ring, every prime ideal contains the trivial ideal.
(ii) [math]\mathfrak{p} \in V(\bigcup_\alpha I_\alpha)\Leftrightarrow \bigcup_\alpha I_\alpha \subset \mathfrak{p} \Leftrightarrow \forall\alpha : I_\alpha\subset\mathfrak{p} \Leftrightarrow \forall\alpha : \mathfrak{p}\in V(I_\alpha) \Leftrightarrow \mathfrak{p} \in \bigcap_\alpha V(I_\alpha)[/math]
(iii) suppose [math]\mathfrak{p}\in V(I_1\cap I_2)[/math] and [math]\mathfrak{p}\not\in V(I_1)[/math]. If now [math]\mathfrak{p}\not\in V(I_2)[/math], then there are elements [math]a\in I_1, b\in I_2[/math] such that [math]a\not\in\mathfrak{p}\not\ni b[/math], but [math]ab\in I_1\cap I_2\subset\mathfrak{p}[/math], which is a contradiction, and so [math]V(I_1\cap I_2)\subset V(I_1)\cup V(I_2)[/math]. On the other hand, if [math]\mathfrak{p}\in V(I_1)\cup V(I_2)[/math], we may assume [math]\mathfrak{p}\in V(I_1)\subset V(I_1\cap I_2)[/math], and so [math]V(I_1)\cup V(I_2)\subset V(I_1\cap I_2)[/math].

The proof took about 10 minutes. You actually missed a typo. [math]\bigcup_\alpha I[/math] should be [math]\bigcup_\alpha I_\alpha[/math].