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12:14 AM
@Pedro Someone is trying to convince me Euler's formula is beautiful.
@AlexanderGruber are you here?
 
@Mike Is it the usual, "it involves all of these special numbers"?
 
howdy @Karl @Mike @seaturtles @Pedro
 
Hi @Ted.
 
Anything interesting happening here?
 
@Karl His argument is that it depends on how you define things, and then defined $e^x$ by its power series, and $\pi$ as the ratio of a unit circle's circumference to its diameter.
because choosing specific definitions gonna convince me. :)
@Ted Never.
 
12:20 AM
Well, in that case ... I'll be back later.
 
Bye. :D
enjoy your dinner.
 
@Mike That kinda makes it uglier, imo.
 
@Karl I think it makes it less surprising, because the power series for sine and cos jump right out.
 
@Mike in and out, what's up?
 
@AlexanderGruber I'm going to give your cfrac bounty a shot this weekend.
 
12:24 AM
@Mike cool, what are you going to write about?
 
I actually know of an interesting algorithm in combinatorial commutative algebra that you might like
eh, I'm most going to point out theorems in various parts of math that use cfracs.
they show up in bizarre places.
 
I'm planning to make a post about Schor's algorithm at some point over there, I did a class in quantum computing where we had to implement it in Mathematica and I still have some neat visuals somewhere. (i'm not eligible for the bounty though of course)
@Mike cool what is it?
 
I need to remember, but IIRC one can make a lattice ordered by set inclusion called the "intersection algebra" of a ring, and an algorithm counting the minimum number of generators depends on the continued fraction expansion of a number associated to the algebra
of a rational number associated to the algebra, I should say. which is cool. most contfrac stuf you see is about infinite continued fractions
 
12:39 AM
@Mike Yeah! that's pretty neat
 
@AlexanderGruber Hola! How is Florida?
 
@JayeshBadwaik HOT.
it's april and i have three air conditioners on in my apartment
 
@AlexanderGruber WOW....
It's warm in here too... at night cool winds make it lovely though...
How is it going otherwise?
 
@JayeshBadwaik here we have alligators at night. :)
it's not bad. today was my last class of commutative algebra, which is
very good
 
Hahaha, hope one doesn't show up in your bathroom! :-P
I've got measure theory exam in 30 hours.... I don't think I'll be able to sleep till then... or atleast till 6 hours before it...
 
12:47 AM
@JayeshBadwaik take a yardstick with you. it'll be good luck.
 
@AlexanderGruber Yeah.. tried that last exam, my prof was more interested in non-measurable sets then... :-(
 
@JayeshBadwaik Which college are you in?
 
as a discrete mathematician, i refuse to acknowledge the existence of non-measurable sets.
 
@Sawarnik Tata Institute of Fundamental Research, Bangalore.
 
Ok :)
 
12:50 AM
@AlexanderGruber Good for you.... sigh me has to deal with infinities all around...
@Sawarnik howz it going with you? just got up? or going to sleep?
 
@JayeshBadwaik i highly recommend it. we don't need to worry about choice either.
 
@JayeshBadwaik Just missed my school bus .. now enjoying!
 
@AlexanderGruber Half the proofs I have to write tomorrow depend on choice.... I also have an algebra exam a fortnight later... I'll complain about different shit then... :-P
@Sawarnik cool....
@AlexanderGruber the PSA guideline appears to be new.. what changed?
 
@JayeshBadwaik that PSA is just something i put up this week because people kept starring a ton of messages in the row for now reason
like this
 
ohh, I see.....
 
12:55 AM
the location pinning one was just a fun idea i had a few weeks ago, I guess people like it because it keeps getting re-pinned/starred
I don't know how long it'll keep up there
 
That's a cool one, I just pinned my approximate location...
 
@JayeshBadwaik Why are you up so early?
 
@Sawarnik I haven't slept yet... I have a class at 11... and an exam on 25th at 10 am...
So, I'm planning on staying awake till today evening and then going to sleep around 2 in the night tomorrow...
 
@JayeshBadwaik Yes, I see you have corrected!
 
@Sawarnik Nopes, east of Bangalore.
@Sawarnik You are in Patna it seems...
 
1:07 AM
@JayeshBadwaik Yes!
 
I thought you were Parth... :-/
 
Wat!
No, no, he is my enemy :D
@JayeshBadwaik Why did you think so? I guess because of my age and avatar?
 
@Sawarnik I had my reasons, anyway, I cleared my confusion now...
@Sawarnik among others yes....
 
1:59 AM
@KarlKronenfeld
 
@PedroTamaroff sup
 
@KarlKronenfeld I am using Nakayama's lemma to show that if $A$ is local and $M,N$ are finitely generated $A$-modules then $M\otimes_A N=0$ implies $M=0$ or $N=0$.
If $\mathfrak m$ is our maximal ideal, and $k:=A/\mathfrak m$, we have by associativity that, denoting $M_k=k\otimes_A M\simeq M/\mathfrak mM$
$M\otimes_A N=0\implies M_k\otimes_A N_k=0$.
Now, I want to obtain that $M_k\otimes_k N_k=0$.
@KarlKronenfeld
 
@PedroTamaroff I'd guess that it's true in general that $M_k\otimes_A N_k=M_k\otimes_kN_k$
 
@KarlKronenfeld Yes, I buy that.
 
Show that $M_k\otimes_AN_k$ satisfies the universal property of the other one.
 
2:06 AM
I mean, clearly multiplying by elements of $A$ in $M_k$ or $N_k$ obviates elements of $\mathfrak m M$.
 
what is up with this question?
 
@AlexanderGruber Oh?
What is up?
 
@PedroTamaroff viewed: 23672 times
 
holy shit
 
@AlexanderGruber Yeah, that's ridiculous.
 
2:09 AM
and it's attracting so many weird troll posts
 
@AlexanderGruber Sorry, I wrote a script that creates new accounts just to view that question... and write trollish answers. :P
 
i just protected it so that should stave those off but i wonder why they're even coming, that is such a mundane question. :p
@KarlKronenfeld i wouldn't even be surprised if that's what was happening.
 
@KarlKronenfeld Please let this be true.
 
can I see a screenshot of the deleted answers?
 
Haaaaaaaaaaaaaaaaalllo
 
2:13 AM
"hi you can eat pinnaples or try eating your green gooey boogers!"
"L'O'L
L
'O'
L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L"
 
@pedro really?
 
@Mike Yes.
 
weird.
 
Anyone have advice on how to show a function is smooth?
 
unfortunately we left out the best one... heh
 
2:16 AM
@Anthony Sit on it.
Does it hurt?
 
@Anthony put it through a power sander, if it wasn't before, it will be.
 
Then it is not smooth.
 
hoho
 
@PedroTamaroff That's the same analogy my prof. uses. :)
 
is it equal to its smoothification?
 
2:17 AM
@AlexanderGruber LULZ.
 
I've got $e^{-1/x}$ for $x>0$ and $0$ elsewhere.
 
le classic
you merely need to check it is smooth at $0$.
 
@Anthony so you want to show the $n$th derivative exists
for starters, what's the $n$th derivative outside, at $x\ne 0$?
 
So that was along the lines of what I was asking-I should find a formula for the nth derivative?
 
No.
You should show $f^{(n)}$ exists and is continuous for every $n$.
 
2:19 AM
Indeed... Is there a better way to show existence then to explicitly give them a formula?
Not that I know there is one.
 
you could find the $n$th derivative if you want though, i would just recommend grouping together whatever doesn't depend on $x$ as a constant
 
@Anthony Here.
Spoilers ahead.
 
i'm probably taking the hard way, though. as a constructivist i tend to do that.
 
@AlexanderGruber Alex?
Constructivist.
Breathing accelerates.
 
@PedroTamaroff that's where the finitism comes from my friend
:)
 
2:22 AM
@AlexanderGruber You said you liked finite groups. Never thought you were into finitism.
I was being fun though.
 
i sort of am, i acknowledge continuums i just avoid them if i can help it
 
@PedroTamaroff, should I just say the rest of the function is smooth-just because? I mean 1 is obviously the problem point, but shouldn't I say something about the rest?
 
same for non-constructive proofs, they work logically, i just like constructive better
 
@Anthony No, it is smooth because $e^x$ and $-1/x$ are.
 
I don't think I've learned a composition rule
 
2:25 AM
Yes, you have.
 
Nor have I shown that e and 1/x are smooh.
 
$D(f\circ g)=(Df\circ g)\cdot Dg$.
 
hohoho
<3
 
Composition and product of smooth is smooth.
Inversion is smooth.
Product and composition of continuous is continuous.
So you have all there.
Done diddly doo.
 
@PedroTamaroff whatchyou know bout semisimplicity?
 
2:26 AM
I still have shown that e and inversion are
But
<3
Gracias.
 
semisimple rings are simple, but not as simple as simple rings
 
@AlexanderGruber Of rings?
 
@PedroTamaroff yeah, should be close to where you are in CA right?
 
Nah, not that much.
But Jacobson has a great section on them.
I scanned over it.
Will read it thoroughly sometime.
 
i only have two more homeworks until i can burn my commutative algebra book and never speak of it again.
 
2:28 AM
why do you dislike CA so much @AlexanderGruber?
 
@FernandoMartin because i'm taking it from an awful professor.
 
Ahh, makes sense
 
@AlexanderGruber Damn bad professors ruining math for everyone.
@FernandoMartin AYO.
 
someday i will pick it up again once my gag reflex has become less irritated
 
@AlexanderGruber much graphic
 
2:31 AM
i think this summer i'll be focusing on analysis actually.
there are some analytic subjects that I want to study that require reviewing a bunch of fundamental material
next year there'll be a diffgeo course here, and i'm going to fit in some type of probability if I can.
 
@FernandoMartin Dude.
 
maybe stochastic something.
 
sup @Pedro
 
$$\frac{M}{\mathfrak a M}\otimes_A \frac{N}{\mathfrak a N}\simeq \frac{M}{\mathfrak a M}\otimes_{A/\mathfrak a} \frac{N}{\mathfrak a N}$$
 
Jesus Christ Pedro
 
2:33 AM
What.
 
Uhh, ok?
 
Agree?
 
I guess, I should try to write the UP
 
@FernandoMartin It is pretty obvious informally.
 
I don't see why tbh
I had overlooked that detail when I solved that since I didn't write subindices for the tensors
shame on me
 
2:46 AM
@FernandoMartin Maybe I am wrong then. Essentially, multiplying by elements of $A$ or of $A/\mathfrak a$ is the same in $M/\mathfrak aM$ and $N/\mathfrak aN$.
 
yeah but you can't make scalars jump from one side to the other
 
Where?
 
it's kinda like what happens with $\Bbb C\otimes_{\Bbb C}{\Bbb C}$ vs $\Bbb C\otimes_{\Bbb R}{\Bbb C}$
 
No, I don't think so.
 
maybe it works in this case
 
2:49 AM
My point is that if we multiply by something in $\mathfrak a$, those modules obviate it.
For $am\in \mathfrak aM$ if $a\in\mathfrak a$.
So we might as well just multiply by things lying over $\mathfrak a$.
 
right
it makes sense
that actually proves that the obvious map is well defined
 
 
1 hour later…
4:19 AM
hello?
 
yo, wazzup?
 
doing homework :/
like I sort of get what's going on but something isn't clear
 
4:43 AM
0
Q: Prove that if $y,z \in Q$ then $y^z \in A$

usukidollQuestion : Prove that if $y,z \in Q$ then $y^z \in A$ My attempt: Definition 2.7.8 states that a number s is an algebraic number when there exists some $p \in Z[x]$ such that $p(s) =0$. Let us denote the set $ A = [x \in C: x $ is algebraic] The set of all polynomials in x with coefficients fr...

 
Is this using induction?
 
I don't think so
this is rational numbers and algebraic numbers
If I was using induction I need the basis step, P(k) and P(k+1)
 
.
 
sup @Mike
 
yeah youre right. Idk why I said that. It's late.
 
4:53 AM
@FernandoMartin Getting rid of the '21 new messages' thing
 
there's this tiny thing that's throwing me off
I was doing fine until I saw that $y^z \in A$
 
lel
you're popular
 
huh?
 
5:15 AM
@_@
@usukidoll what's the question in gist?
 
gist?!
 
the gist
^_^
 
math.stackexchange.com/questions/766935/… trying to provide the answer in a comment hold up ^^
 
ok, lemme know if you want help
 
I'm trying to latex a huge line xD
am I on the right track @enjoysmath
 
5:22 AM
idk
 
<_<
 
if you mean is math the right path to be on and to that I say yes
>_<
 
yeah but what about the proof part... well the recent one scroll down
 
you'll have to let me know specifically what's tripping you up
 
$y^z \in A$
I have to find a polynomial root of $\frac{a\b}^\frac{c}{d}$
nguh I donn't wanna edit
I mention the root in here
http://math.stackexchange.com/questions/766935/prove-that-if-y-z-in-q-then-yz-in-a/766949?noredirect=1#comment1593438_766949
 
5:25 AM
you have that 'start chatJax' link?
 
yessum
 
5:52 AM
pokes @enjoysmath
 
ur hot :|
 
help me D:
 
@_@
 
5:55 AM
@usukidoll First thing I don't understand is : "Thus each element p∈Z[x] has the form

p(x)=anxn+an−1xn−1+...+a1x+a0

where n is a non-negative integer, ai∈Z for each i∈[0,1,2,...n] and an≠0 except when n=0
"
 
what the hell? that's part of the def i the book
 
polynomials by def can have $a_i \neq 0$
Do you mean for $x \in A$ its associated poly is such that $a_i \neq 0$ for $0 \lt i \leq n$?
also you should use \Bbb{Q} for $\Bbb{Q}$
 
no
whatever I typed from the book is correct
anywayyyyyyyyyy
sigh view this.. I've latex an example

http://openstudy.com/study#/updates/5358994fe4b0f8d836a765a9
 
Question: are you trying to say $y^z \in A$ when both are rational or algebraic???
@usukidoll
 
yessssssssssssssss by that prop man
 
6:00 AM
I can't read that from your post. You mean when $y, z \in A$, $y^z \in A$ or no?
I think you mean when they're rational but not sure
just a question
 
oh gawd
 
what?
 
can you read the question?
what does it say?
Question : Prove that if $y,z \in Q$ then $y^z \in A$
 
I got you
 
y and z is flipping rational numbers... I gotta prove that $y^z \in A$ which means that y to the z power is algebraic
by the prop...all rational numbers are algebraic
makes sense because y and z belong to the set of rational numbers
 
6:03 AM
Okay, first show that if $z \in \Bbb{Z}$ then $z^y\in A$ when $y \in \Bbb{Q}$ and you're half-way there
 
if it was irrational, I can't prove this now can I because it break the prop
wth is that?
z is an integer y is a rational ... your point?
 
Show that $\sqrt[b]{z} \in A$ for all $b \in \Bbb{N}, z \in \Bbb{N}$ and you're almost there, do you see that?
 
:(
I like the other way
can we do it after we had the z and y in Q?!!?!?!
 
@usukidoll : $y = \frac{a}{b}$ where the $a,b$ are integers. So if you prove that integers to a certain power are algebraic then you already know that $1/b^{n}$ is algebraic since it's equal to $(b^n)^{-1}$
No, you start with smaller problems then build the bigger problem. It makes sense mostly that way
I'll write up an answer
 
noooooo I want to find a polynomial with the root $(\frac{a}{b})^{\frac{c}{d}}$ in $Z[x]$
 
6:07 AM
..
 
@usukidoll since $x^d-\left(\frac{a}{b}\right)^c$ has that as a root, multiply thorugh by $b^c$ to get the polynomial you want
$b^cx^d - a^c$
 
there ^ that's what I want. so I multiply all of that by $b^c$
but I feel like I'm skipping steps.. there's this identical problem that multiplies, rearrange the coefficents, and then muliply the last equation by the denominator
it's something like ... $p(x) =c_nx^n+c_{n-1}x^{n-1}+...+c_1x+c_0$
such that p(y) = 0 and there exist integers a and b such that $b \neq 0$ and $ y=\frac{a}{b}$
 
Hello Nando
 
6:15 AM
Thus...
$c_ny^n+c_{n-1}y^{n-1}+...+c_1y+c_0=0$
 
@FernandoMartin its u
 
now I'm hungry
thanks for nothing @Mike
 
which we multiply by $y^n$ to obtain
$\frac{a^n}{b^n}y^nc_n+\frac{a^n}{b^n}c_{n-1}y^{n-1}+...+\frac{a^n}{b^n}c_1y+\frac{a^n}{b^n}c_0=0$
@Mike
then rearrange the terms and multiply by $b^n$
 
@usukidoll look at my post, much simpler than dealing with polys
 
6:19 AM
@Mike Sup.
 
lol
three mikes in a minute
 
@Mike Mik
e's a bad name. You should get a new one.
 
I'll be Gaddis.
 
broken code wait gotta fix
 
@Mike Lame. You should be, Michaela.
 
6:22 AM
$c_n(\frac{ay}{b})^n+c_{n_1}\frac{a}{b}$
 
I'll consider it.
 
@Mike And then, go to Coachella.
 
$(\frac{ay}{b})^{n-1}+...$
 
No.
 
$ +c_1 \frac{a^{n-1}}{b^{n-1})$ $(\frac{ay}{b} + \frac{a^n}{b^n}c_0=0$
ugh long code went explode
 
6:24 AM
@Mike They don't sleep anymore on the beach.
 
I don't Do 'chella.
 
@AlexYoucis ;____;
 
@mike do you see the polynomial in parts ...? should I do that for the y and the z so that $y^z \in A$?
 
@FernandoMartin I have never understood what that emoticon meant.
 
a crying face
 
6:25 AM
that means crying
 
@AlexYoucis Here's my favorite emoticon
.-.
 
@FernandoMartin What happened to :'( Did it go the way of the dinosaur?
@Mike That's terrible, and I hate you for showing it to me.
@FernandoMartin Also, why the crying face? Because that song's sad?
 
eerrrr guys what about my problem? ^^
 
I think he was crying because he saw you chatting, @AlexYoucis
 
@Mike Your half-hearted burns are like sweet rays of sun on a perfect summer's day.
 
6:27 AM
Because they don't do that anymore - things changed
 
@FernandoMartin True, true.
 
I don't understand wy someone going to 'chella would sleep.
 
cap
i have a question. If $f(x) = \tau(n*p^x) - \tau(n)\tau(p^x)$, where $(n,p)=1$ and $\tau$ is ramanujan's function. I am trying to show $f$ is always $0$, and I was advised to show $f(x)$ is a linear combination of $f(x-1)$ and $f(x-2)$ . How does this suffice.
 
@Mike I don't know if you are serious. But, in case not. I references a GY!BE song, and then Fernando commiserated with me on a more global level, and then you misunderstood. That's the world as I see it.
 
All of this is true
 
6:32 AM
@Mike A confession eh? Cuff 'im boys.
 
takes out hand cuffs
 
6:45 AM
Are you going to come peacefully @Mike?
 
 
1 hour later…
8:05 AM
@usukidoll any luck yet?
 
come here and get this done for meh >:D
I also have another one... I might have some clue or not X>X
Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Hint: consider decimal expansions of $a$ and $b$

Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has an infinitely repeating sequence of digits.

Set $I$ is irrational numbers

$I =[x \in R: x \notin Q]$

Set $Q$ is rational numbers

$Q = [ \frac{a}{b}:a,b, \in Z$ and $b \neq 0]$
oh geez if there is an irrational number d... the decimal expansion doesn't terminate D:
If the division process terminates, then we are done. Otherwise, since each

digit of the quotient determines in turn its duccessor, and since there are

at most $b-1$ possible remainders when dividing by $b$ (by the division

algorithim), some digit of the remainder must show up again, forcing a

sequence of digits to repeat forever. The length of the repeating cycle is

at most $b-1$
Since $d$ is irrational, the decimal expansion will terminate. Therefore,

for $r=0$, we have $a_1,a_2,...a_k$ where each $a_i \in

[0,1,2,3,4,5,6,7,8,9]$ and $a_k \neq 0$. Then

$ r = \frac{a_110^{k-1}+a_210^{k-2}+...+a_k}{10^k}$
 
8:25 AM
..
@usukidoll Are you a girl or a boy?
 
no
but I'm trying to get this together x.x like I sort of know how to approach it but not sure if it's going to work points up
oh no I meant to put since $c$ is rational, the decimal expansion will terminate.. since $d$ is irrational the expansion won't terminate
girl
 
Ok :)
 
Greetings.
 
I'm stuck D:
 
Today I just received a short movie that was meant to impress me profoundly, but it didn't. It's related to numbers.
See this entirely (without skipping) and tell me if it works for you - youtube.com/watch?v=5Qg24O864lI
For me it didn't work at all.
 
8:34 AM
rawr
 
@Chris'ssis I would like to take a Black chainsaw to that video :D
 
@Chris'ssis Nonsense. And how the hell is it related to numbers?
 
@Sawarnik It's probably related to the way mind usually works.
 
It's not related to anything, imo.
A hammer is a common tool
 
@Chris'ssis I didn't see any connection with the calculations and the final question. Many people prefer red and hammer is a common tool could have been said in 1 line.
 
8:46 AM
and red is a common color
 
@685-252 Indeed.
@Sawarnik Yeah.
 
@Chris'ssis I was not impressed at all.
 
@685-252 Neither do I.
 
@685-252 Neither will any sensible person.
 
In fact, I want a refund on the 1:28 of my attention back :D
damn internet.
8 mins ago, by 685-252
@Chris'ssis I would like to take a Black chainsaw to that video :D
 
8:52 AM
@685-252 :-))))
 
1
Q: Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$.

usukidollLet $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Hint: consider decimal expansions of $a$ and $b$ Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has ...

 
Anyone know how to access the following paper?
"An extension of integration by parts", American Mathematical Monthly, vol. 67 , 1960 , p. 372
 

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