8:00 PM
@Jorge same as deleting a downvoted answer: nothing.

but when you delete a downvoted answer you get rep back

I don't think so

speaking as someone who read their homological algebra book

“There was once a professor who taught at this school who was…really something else. I mean this guy would show up to his class and meetings completely wasted, it wasn’t a good thing. However, he had an amazing talent for multiplying matrices in his head. I’m talking like 6x6 matrices. I got to wondering ‘wow, how could somebody do something like that,’ and then I remembered that when you are intoxicated, your left eye can move right to left, and your right eye can move up and down.”

8:05 PM
@user153330 it did go down :(
@user153330 I'm rolling on the floor laughing

@JorgeFernández i hope you're not intoxicated, i don't want to feel intimidated
lol spivak's differential geometry n°5 cover :

“Overall you all did pretty good. The average was 62%.”
— Calculus professor
Why is this supposed to be funny?

@Huy Because normally averages in the U.S. are much higher. Anyway, don't read that blog. It's virtually never funny.

@Huy it's not.

8:17 PM
Aha. Good. I was scared I lacked humour.

@Huy me too

@Huy Lichtenberg Prinzip: Wenn ein Kopf und ein Buch zusammenstossen und es klingt hohl, ist es nicht immer das Buch.

@MikeMiller: Why are averages in the US much higher usually? They don't adjust the scale afterwards, do they?
@DanielFischer: I've never heard of that before.

8:35 PM

@Huy We're afraid to fail students. Average grade here nowadays is probably an 80 = B.
= 3.0/4.0

@MikeMiller: How many students usually fail, in the lower semesters, relatively spoken?

Few.

1%?

I suspect in the class I just taught, at most 10%. Probably far fewer.

8:38 PM
Why are they afraid to fail students? Less revenue?

in my school 80% fail at least one class in the first semester

@JorgeFernández: At mine, in the first year, you can only pass or fail all classes, not single ones. Usually, 50-60% fail all classes.

50% fail all of their classes?

@JorgeFernández: Usually, at least. But you can only fail either all or pass all.

oh! that's really weird, does this happen in the other semesters?

8:41 PM
@JorgeFernández: In the 3rd and 4th semester too. After that, classes will be counted individually.

so if you fail a single class they fail you automatically in everything else? or how does it work?

@JorgeFernández: It's supposed to make it a bit harder to pass.
@JorgeFernández: No, you will get a weighted average.

Oh, so you get a weighted average and that's what you get for all the classes?

8:43 PM
@evinda You can try. (Or, I can try to answer.)

@JorgeFernández: No. For each exam, you get a mark from 1 to 6, 6 being the best and 1 the worst. For example, after the first year, you must take exams in analysis 1&2, linear algebra 1&2, physics 1, physics 2, introduction to CS and numerical mathematics. Your mark for analysis 1&2 and for linear algebra 1&2 have weight 2, all other courses have weight 1. If the weighted average is less than 4.0, you fail and must repeat all exams (at the earliest half a year later).

@ArthurFischer I am looking at this proposition: The natural number $n'$ is the immediate successor of $n$, i.e. there is no natural number $m$ such that $n<m \lor m<n'$/ At the proof we assume that there is a $m$ such that $n<m \wedge m<n'$. Then $n \subset m$ and $m \subset n \cup \{n\}$.
Then there are take two cases: $n \in m$ and $n \notin m$.

Why do we distinguish the case $n \notin m$ although having supposed that $n<m$ ?

does anyone know what that image is?
no
I think it has to do with jewelry or clothes

@evinda I think it is mainly because you are likely seeing it in an introductory course/text, and they are trying to be "rigourous". You're right that $n < m$ is equivalent to $n \subsetneq m$ which is equivalent to $n \in m$. (This also means that one of the cases is exceedingly simple.)

So do we just have to check the first case, @ArthurFischer ?

8:57 PM
@JorgeFernández you're welcome

@user153330 thanks, tineye?

@evinda If you include the sort of reasoning I outlined above (and which you undoubtedly noticed), then yes: there is only one case.

Nice!!! Thank you very much!!! :) @ArthurFischer
Something else, if I am allowed to ask... Are you, @ArthurFischer & @DanielFischer relatives? :D

@evinda one is from germany, the other is from austria

9:02 PM
@evinda Ha! (See the last bit of my meta answer here.)
@evinda No problem.

@JorgeFernández "Modified bear" logo for Kid A by artists Stanley Donwood and Tchock (Thom Yorke)

A ok @ArthurFischer

Where's Gerd Fischer?

@DanielFischer Congratulations!!!!
@Huy :D

@user153330 I'm not really from Austria. I just live there (here?). Ich komme aus Kanada. (Though my paternal grandfather's father came to Canada from Germany, I've been told.)

9:04 PM
@evinda: Do you know Gerd Fischer? :D

@evinda Thanks. What for?

@ArthurFischer that explains it : P

@Huy I have heard of him but he isn't really on MS, is he?

I don't think he is.

@DanielFischer for the 100k !! and being the first user on SE to have 100k on two diffrnt accounts other than metas !
2

9:05 PM
@DanielFischer Arthur sent me a link where it says that you are the top user!!! :)

@user153330 Somebody had to be first.

@DanielFischer and you were that one and only !

and only

@DanielFischer What have you studied?

@evinda Mathematics

9:14 PM
@DanielFischer And did you also do a master's degree?

@evinda There was no master's degree in the olden times.

Aha! @DanielFischer

Can someone help me at Galois theory??

but his theory lives on...

9:25 PM
Maybe? @MaryStar
It's been a while

hello

@skullpatrol hi buddy

@HipsterMathematician Happy New Year

@skullpatrol Happy new year!!!

9:36 PM
@KajHansen We have the Galois group $G=\{\tau_{ij}, i=1,2,3 , j=1,2\}$, where $\tau_{ij}(\sqrt[3]{2})=\omega^{i-1}\sqrt[3]{2}, \tau_{ij}(\omega)=\omega^i$.

The subgroups $H$ of $G$ are:

#H=1: $H=\{id\}$
#H=6: $H=G$
#H=2,3

The order of the elements of $H$ is:

$ord \tau_{11}=1 \\ ord \tau_{21}=3 \\ ord \tau_{31}=3 \\ ord \tau_{12}=2 \\ ord \tau_{22}=2 \\ ord \tau_{32}=2$

$<\tau_{21}>=\{1, \tau_{21}, \tau_{21}^2\}$

Since $\tau_{31}=\tau_{21}^2, \tau_{31}^2=\tau_{21}$ we have that $<\tau_{21}>=<\tau_{21}>$.

@skullpatrol how are you?

Fine thanks @HipsterMathematician How are you doing?

@skullpatrol fine i think. ilost my voice though it's annoying

@JasperLoy Ah Yes, I have found some interesting examples!

@HipsterMathematician You sound fine to me :D

9:39 PM
@skullpatrol :DDDDDDD

@MaryStar, I'm guessing $\omega$ is a cube root of unity?

@KajHansen Yes!

I think I'm missing something. Where is the $j$ coming in with $\tau_{ij}$?
You have only $i$ in those exponents it looks like.

@skullpatrol what's new around?

@HipsterMathematician the mod elections are done

9:42 PM
@skullpatrol I realized

@KajHansen It should be as followed:

$\tau_{ij}(\sqrt[3]{2})=\omega^{i-1}\sqrt[3]{2}, \tau_{ij}(\omega)=\omega^j$

Oh ok

@skullpatrol seems everyone left

@HipsterMathematician yep, its pretty quite around here...

Let's see...so that Galois group is going to be isomorphic to $\langle \tau_{21}, \tau_{12} \rangle$ if I'm not mistaken.

9:45 PM
@skullpatrol quite odd

Which I believe is itself $\cong S_3$.

@HipsterMathematician these things go in cycles

Because we have a rotation of the "triangle" of roots in the complex plane, and a reflection as well @MaryStar

@skullpatrol hmmm... maybe skullex

The subgroup structure of $S_3$ is easily understood. You have either $\{id\}$, the whole group, and the only proper nontrivial subgroups are cyclic due to Lagrange's theorem.

9:47 PM
@HipsterMathematician skullex?

That information will help you match up the $\langle \tau_{ij} \rangle$ with one another.

@skullpatrol yup

At any rate, I'd highly recommend thinking about that Galois group as acting on the "triangle" of roots of $x^3 - 2$ in the complex plane and see what each element does to that triangle @MaryStar

Why is the Galois group isomorphic to $\langle \tau_{21}, \tau_{12} \rangle$?? @KajHansen Is it because $<\tau_{21}>=<\tau_{21}>$ ?? And $<\tau_{12}>=<\tau_{22}>=<\tau_{32}>$ ??

@HipsterMathematician what does "skullex" mean?

9:49 PM
Each one will be either a reflection about the real axis, a $120^\circ$ rotation, or a combination of those two.

@skullpatrol it's skull+ex

@MaryStar, essentially I'm getting that information from the fact that the symmetry group of a triangle is generated by a reflection together with a rotation.

@HipsterMathematician what does "ex" mean here?

@user153330 I never got in contact with Paul Nahin. I think he would love my questions though. :-)

@skullpatrol nothings skull

9:51 PM
(I had a very, very geometric introduction to group theory, so you'll have to forgive me). Or rather, blame @TedShifrin

hi @quid

hi @skullpatrol!

@robjohn do you plan to publish anything like a book on integrals, series and limits?

@KajHansen Ok... I will think about it... :-)

I have also an other question...

We are looking for an element $a$ such that $E^{<\tau_{21}>}=\mathbb{Q}(a)$, where $<\tau_{21}>=\{id, \tau_{21}, \tau_[31}\}$. Why does it stand that $a=\omega$ ??

$E=\mathbb{Q}(\sqrt[3]{2}, \omega)$, the splitting field of $x^3-2 \in \mathbb{Q}[x]$

@KajHansen Ooh, good. I can blame Ted ;-)

10:00 PM
Let's see. This is an exercise in Galois correspondence. So you're looking for an intermediate field between $\mathbb{Q}$ and $E$ that is fixed by $\langle \tau_{21} \rangle$. Notice that $\tau_{21}$ is fixing $\omega$.
So we know that this intermediate field at least contains $\mathbb{Q}(\omega)$ as a subfield.
But it has to be $\mathbb{Q}(\omega)$ and nothing more! Otherwise, we have the tower of fields $\mathbb{Q} \subset E^{\langle \tau_{21} \rangle} \subset E$, and in this situation we have $[E^{\langle \tau{21} \rangle}:\mathbb{Q}] > 3$. Can you see the problem?

See you later pal @HipsterMathematician

10:16 PM
@robjohn It would be nice one day to talk about our books here :-)
Maybe to post problems here from our books. Well, I already did it in advance! :D

@Chris'ssis Our books? Did robjohn write one?

@JasperLoy Maybe.
@robjohn I only hope you won't publish it when I publish it mine. You might ruin the success of my book. :-)

@robjohn What book are you publishing?

@JasperLoy I am not publishing anything...

@robjohn But Chris's Sis is saying otherwise...

10:27 PM
34 mins ago, by Chris's sis
@robjohn do you plan to publish anything like a book on integrals, series and limits?

@JasperLoy I was talking about a possibility ...

28 mins ago, by robjohn

@robjohn Thank you for being skullpatrol.

@JasperLoy just clarifying the situation

@JasperLoy How are you today? Happier? :-) These days you seemed depressed.

10:31 PM
hi @Jasper, @skull, @chris'ssis, @robjohn

@Chris'ssis No. But I plan to solve 99 per cent of my mental problems this year, start studying next year and solve the remaining 1 per cent next year. My condition is very, very bad now...

@TedShifrin Hello Ted

@Kaj: Don't blame me: I didn't teach your course. :) You can blame me for others.

@TedShifrin Hello, I just had breakfast.

I should keep better track of our time difference, @Jasper

10:32 PM
@TedShifrin I am GMT+8.

um, so it's 8:32 AM there?

No, 6.32 am.

oh, GMT is 7 hours ahead of me? wow

@JasperLoy Sorry to hear that. Is there anything you can do to improve that?

oh, oops, I made a sign error

10:34 PM
@TedShifrin Hey, Ted... we blame you

@Chris'ssis If you want to help me, you can pray for me. I need time to sort out my thoughts. I know I have been doing it for the past 7 years...

@TedShifrin, looks like Dr. Clark's going to be covering a decent bit in topology. Definitely a LOT faster pace than real.

oh, for sure, @Kaj ... I sure hope so. And I hope he assigns lots of homework.
of course you do, @robjohn ... This is why I shall abdicate in entirety!

@TedShifrin Don't do that!

@JasperLoy Ok, I can do that. You'll also need to pray for yourself.

10:35 PM
@TedShifrin That means you must be on the eastern side.

yes, @Jasper :P

If you count all the parts, we have 40ish problems due on Monday. It's all fairly easy set theory so far in the homework, but he said the length is indicative of future problem sets.

LOL, sounds like the bad old days with me, @Kaj :P

@KajHansen That's a lot of problems.

well, often 5- or 6-part problems aren't that much work ...

10:37 PM
@TedShifrin A mathematician never resigns, retires, abdicates ... :-)

Have you heard of the Moore method of teaching?

I don't think this week is all that bad. I'm just taking it one problem at a time, and I haven't gotten stuck yet. I am bracing myself though.

My advisee who didn't belong there (and wasn't there on my advice) is gone, @Kaj, so it should be a good class.
wanna bet, @Chris'ssis? :D

@TedShifrin Yeah! :D

yes @Jasper, quite well ... but it's just not my personality to do that. @Kaj will tell you I make students do a lot of answering and struggling, but I provide proofs and examples in lecture.

10:38 PM
@JasperLoy I am CUT-8

@TedShifrin I think the Moore method can be good but it is too time consuming to be practical.
@robjohn I am a mango.

It should be $22:40$ CUT

Interesting story for you, @Jasper: When I taught point-set topology (out of your least favorite book) about 20 years ago, I had some very talented undergraduates and a few graduate students from math education.
I would throw them questions and ask them to tell me how to do the proofs ... they'd say ideas, come to a consensus, and then eventually I'd write up the final proof on the board. The math ed grad students asked me just to give straight lectures. They were in favor of student discovery as an abstract thing for their high school students to engage in, but didn't like it for a university class :P
Oh, it was the undergraduates who came up with the ideas that turned into the proofs, of course, not the graduate students.

@TedShifrin Aha! I never understood why there is a need for math ed to be a graduate course. It's just teaching math, not a big deal to me.

These were people who get doctorate degrees, do research in math education, and then go train generations of teachers.

10:45 PM
I have a biased feeling that there are many masters programs that are actually very simple to do compared to a bachelors math program, and don't deserve to be called a masters.

Bachelors programs in math are very varied, at least in this country. Many of our students who get bachelors degrees know very little. Our best students know a ton.

The bachelors programs in my country suck too.

@TedShifrin How little?

@Chris'ssis Very little, considering you call Fubini 'high school'.

appallingly little, @Chris'ssis. I have a student in my differential geometry class who came to me with questions yesterday. He thought that $(1/f)' = 1/f'$ (disguised) and he didn't know the length of $v/|v|$. He hopes to graduate this semester. :(
LOL @Jasper ... I've berated @chris'ssis a lot. Perhaps she is recanting.

10:48 PM
@TedShifrin Even a mango like me knows those two things.

Yes, @Jasper, some mangoes are quite well-educated :)

@TedShifrin Interesting.

Depressing, @Chris'ssis.
This is one of the reasons I'm retiring. I have colleagues who've given such students passing grades from calculus through linear algebra through algebra and other upper-level courses.

@TedShifrin Why would one do mathematics if that one is not interested in knowing some more? Maybe they can get a good job after graduating.

Many are "interested." They just don't have much competence.

10:50 PM
@Chris'ssis Well, sometimes, interest and ability are different things.

Too many people go to college?

Jinx.

sorry, @Jasper :)
Yes, our average majors, provided they have computer skills, can get decent jobs.

@TedShifrin, at least I hear that your geometry course doesn't have 35 students this time around.

LOL, no, @Kaj, just 28 now.

10:52 PM
@TedShifrin I have an interesting theory about the human beings in general: I think inside every person lies a superman that can be brought to life by very much work and passion. For instance, I was not born a talented person as regards mathematics, but I forged my talent by very much work and passion.

@Kaj: You want to tutor that kid I described above? Tutoring $\ne$ doing his homework for him.

That's assuming I'd be able to do his homework! I didn't have the easiest time in all three of your courses @TedShifrin

Well, @Chris'ssis, I am not so optimistic. Work ethic makes a big difference, and passion is a necessity to do graduate work, but I believe there's a lot of mediocrity in our world.

@TedShifrin Mental health is also a necessity.

But in all seriousness, I'm not sure I have a firm enough grasp on differential geometry, especially moving forward, to charge people for my services. I did decide to extend myself into the Math 4000 crowd this semester though :D

10:54 PM
This guy, assuming he doesn't drop, has the drive to work hard, actually, @Kaj. But he's got to solidify his background, and I'm worried that some of the stuff in the course will just be too tough for him.

@ArthurFischer Could I also ask you something else?

Absolutely, @Jasper. Math is hard enough with a somewhat healthy mind.

@TedShifrin Sometimes, I think that solving my mental problems is harder than solving mathematical problems.

My caveat stands, @Kaj. Do not do their homework for them. That's not what a tutor should do.
@Jasper: Absolutely so.

When $p$ is an odd prime and $a=Re \left ( e^{\frac{2 \pi i}{p}} \right)$ then $[\mathbb{Q}(a) : \mathbb{Q}]=\frac{p-1}{2}$.

Let $\theta = \frac{2 \pi}{p}$.

If $\sin{\theta}$ is a constructable number show that $p=2^k+1$, $k \in \mathbb{N}$.

So that $\sin{\theta}$ is a constructable number, the degree of $Irr(\sin{\theta}, \mathbb{Q})$ should be a power of $2$.

Isn't $\sin{\theta}$ the imaginary part of $e^{\frac{2 \pi }{p}}$ ?? How can we use $[\mathbb{Q}(a):\mathbb{Q}(a)]$ ??

10:57 PM
I'm still a bit hesitant @Ted. I'd love to, but with my course load this semester, I feel more comfortable sticking to course material that requires minimal preparation on my part (precalc, calc 1-2, 4000)

No, @Kaj, my comment was with regard to 4000.

@TedShifrin Have you ever had the surprise to see one of your students that didn't get good marks in your class that managed to obtain some important achievements in his mathematical activity?

@Kaj: I presume you're helping @MaryStar here.
@Chris'ssis: Perhaps @Kaj would count :P

Hi @quid !!! I want to show that $m \leq n \leftrightarrow m' \leq n'$.
That's what I have tried so far:
$m \leq n \rightarrow m \in n \lor m=n$
$m \in n \rightarrow m \subset n \wedge \{m\} \subset n \rightarrow m \cup \{m\} \subset n \rightarrow m' \subset n \rightarrow m' \subset n \cup \{n\} \rightarrow m' \subset n' \rightarrow m' \leq n'$

oh, well, important achievements, not yet :P

10:58 PM
@TedShifrin :D

Oh, definitely. I admit that there is a fine line between too much and too little as far as hint-giving goes, and I've tried to put in a very conscious effort on my part to determine what's acceptable. @Ted

@quid Do I have to say also something for $m=n$ ?

I think there's an art to it, and I'm not an artist yet. But I aspire to be.

heya @teadawg ... not frozen solid yet?

Hi @evinda let me check.

10:59 PM
I hope you've learned something about that from me,@Kaj :P
Some students, quite seriously, @Kaj, think that if they're paying money, they can expect you to do their homework for them. Refuse.

Throw them into the refuse.

LOL @JasperLoy

@TedShifrin I had to go outside today. I forgot to put on gloves and a beanie. My hands are still cold

@Chris'ssis: I have had some students who've gone on to be highly successful and outshine me completely. But, no, they were excellent students.

@KajHansen Has your face changed much since the pic was taken?

11:01 PM
Poor @teadawg. Do you need hugs?

I'm giving this some thought @MaryStar. One sec

@KajHansen Ok!

@JasperLoy, I took that in August. So it's been about 5 months. My hair is a good bit longer now because I've been lazy about getting it cut.

Nah, what I need is a thick blanket

@TedShifrin From the lessons you posted on youtube you don't seem that kind of tough professor that scares the students. Well, we never know. I wonder if I'd successfully pass your classes. :D

11:02 PM
@Chris'ssis, but you haven't experienced the "silent look of disappointment" when he hands your test back. It's terrifying and motivating at the same time.

@KajHansen :-)))
@KajHansen I think the nicest challenge for a student is to try to be as good as his professor, or even better. :-)

@KajHansen I did not see the edit until I refreshed, weird.

@MaryStar, one quick note first because it's rather subtle: $\alpha$ being a degree power-of-two extension over $\mathbb{Q}$ doesn't mean $\alpha$ is constructible. That's a necessary, but not a sufficient, condition.
But yes, that's a good starting point.

+1 @Kaj
Some students get a "that was really nice" when I hand back the tests :P

@TedShifrin Have you ever had students that wanted to be as good as you are, or even to try to surpass you? I mean students that worked very hard for that.

11:09 PM
@MaryStar, you're certainly right. $\sin(\theta) = \operatorname{Im}(e^{2i\pi/p})$. Now let's see...

@Chris'ssis: I told you above that I have had some students who have far, far surpassed me.
One of my students in my class this term emailed me with a question a few hours ago, and I emailed him back, "This is why you're a born mathematician." He was curious how it could be that the boundary of some bounded open set in $\Bbb R^n$ would not have volume $0$ (or measure $0$).

And notice that $e^{2i\pi/p}$ is a root of unity, so you can apply the Pythagorean theorem to get the real part in terms of the imaginary part rather easily @MaryStar

I am trying to think of a username for my gmail address. I used jasperloy in the past and deleted the account, so now I cannot get it back. Any ideas?

one of my current students, whose name is ABC DEF, made his email address ABCCABDEFFED ... agh !
(@Jasper: To make it worse, his first name has 6 letters and his second has 5.)

11:16 PM
One of the hard things I cannot figure out is why I took 7 years to deal with my mental problems and still have not succeeded. Any ideas?
@TedShifrin I see. I need to stop the bad habit of deleting things one day. It's on my to-do list.

Well, @Jasper, you reject my naive notion that it is biochemical and therefore needs biochemical adjustment.

@TedShifrin I did not say it was naive. That is one possibility.

I think you're almost done at that point, but I'm getting hung up on small details. Basically you'll find that the imaginary part is a member of a quadratic extension of $\mathbb{Q}(a)$, so that bodes well @MaryStar

no, I said it was naive :P
@Kaj: I haven't looked at the question, so you'd better do it :P

I can barely see the screen. I'm out.

11:18 PM
@evinda yes for m=n it is rather simpler.
If m=n then m u {m} = n u {n} thus m' = n' and consequently m' < = n'

It is comforting to me to know that Nash spent a decade in and out of hospitals. It helps me understand that these things can take very long to fix.

@TedShifrin, I'm almost sure this is wrong, but suppose you have $a$ is constructible, and $\beta$ is a member of a quadratic extension of $\mathbb{Q}[a]$. Does it follow that $\beta$ is constructible?

Sure, @Kaj.
Remember that whenever $\gamma$ is constructible, so is $\sqrt{|\gamma|}$.

@quid I see.. So do I have to distinguish the case $m \in n$ and the case $m=n$ ?

Yes I think this makes sense.
I am not really up to speed with this. But it seemed alright to me what you wrote.

11:21 PM
Aha! You're right! I'll have to think for a sec about how to physically construct square roots of lengths, but that completes @MaryStar's problem.

Let me know if you're getting stuck.

hi. I'm looking for examples about modular aritmetics similiar to this question:find the all solutions of $x^4=6(mod\quad 23)$. could anyone help me

I don't think factoring is helpful.

Wait
It would be finding roots of $x^4 - 6$ in $\mathbb{Z}_{23}$.
Ugh. That would be a pain for large fields though.

Factoring is harder than finding linear factors :P

I'm fairly certain there's number theory I don't know yet that makes that problem easier :P

11:27 PM
One way to think about it, @emmett, is to consider $6+23k$ for integers $k$, and see which ones are fourth powers of integers.
Alternatively, for $x=1,\dots,11$, compute their fourth powers mod $23$.

Yuck! @TedShifrin

LOL

Brute force :(
I guess it is convenient that one does not have to check $1-24$ though.

Here's something to ponder, @Kaj, @emmett: How many solutions does $x^4=1$ have in $\Bbb Z_{23}$?

@quid Now I tried to prove the other direction. That's what I did:
$m' \leq n' \rightarrow m' \in n' \lor m'=n'$
If $m' \in n'$ then $m \cup \{m\} \in n \cup \{n\} \rightarrow m \cup \{m\} \subset n \cup \{n\} \rightarrow m \subset n \cup \{n\} \wedge \{m\} \subset n \cup \{n\} \rightarrow (m \subset n \lor n \in m) \wedge (m \in n \cup \{n\})$
Is it right so far? How could we continue?

11:30 PM
@emmett: We don't know how much algebra you know.

Hmmm. Well, we have the homomorphism of $\mathbb{Z}_{23}^\times$ into itself defined by $x \rightarrow x^4$...

Right, @Kaj. So ?

We just need to figure out the kernel...

Right, and part of that would be easier if you knew if $-1$ were a square.

@TedShifrin i found its solutions as $x_1=9(mod 23)$ and $x_2=-9(mod 23)$
need more to practice :)

11:31 PM
Oh yes. One sec, I think I have this.

How did you get it, @emmett? And are we sure this is it?

It has only $2$ solutions, because $4$ does not divide $22$.

Cool ... for $1$ it has two solutions, hence for $6$ it has either $0$ or $2$ solutions.

Very cool!

But I know no way other than brute force to determine one solution for $6$.
@Kaj: I'll let you explain what you figured out to @emmett ... Still not sure what he knows. I sometimes assign stuff like this right at the beginning of the course, when it's just for brute force.

11:33 PM
@evinda I think the penultimate \in should be reversed.

I think a good background question first would be whether @emmett is taking algebra or number theory.

No. actually I guess you might have wanted an = there.

I'm not really sure what a number theory course covers :/

@TedShifrin $6$ is a square, hence $6^{11} \equiv 1 \pmod{23}$, and $6^{12} \equiv 6 \pmod{23}$. So $6^6 \mod 23$ does it.

@quid
If we have $m \subset n \cup \{n\}$ do we conclude that $m \subset n \lor n \in m$ OR $m \subset n \lor m \in n$ ? :/

11:38 PM
Promotion: the room formerly known as Jury Duty (now "Reopen? Undelete? Close? Delete?") seeks voters willing to consider "cold cases". Mostly reopening right now.
2

@Fundamental Hello Care Bear.

@evinda the former. Sorry I guess I said something wrong earlier.

So is that what I have written so far right, @quid ?

Yes. I think so.

@quid Do you have an idea how we could continue?

11:46 PM
@TedShifrin is this wrong? first found a primitive root for mod 23 which is 5. and compute $ind_5(6)=18$ and solved $4y\equiv18(mod \Phi(23))$ found its solutions as $y_t=(10+11t)(mod22);t=0,1$ sorry for slow writing because of the mobile

I will go to sleep now.. I will be on again tomorrow @quid
Good night!!! @quid @DanielFischer @JasperLoy @KajHansen @robjohn @ArthurFischer

Good night @evinda. Sorry I was a bit slow today.

@KajHansen At the point where you say to apply the Pythagorean theorem do you mean that we have to use $\sin^2 \theta + \cos^2 \theta=1$ ?

No problem :) @quid

In a sense @MaryStar
Draw a right triangle with $e^{2i\pi}$ at the end of the hypotenuse.

11:55 PM
and the other two sides are the imainary and real parts of $e^{2 \pi i/p}$ ?? @KajHansen

Exactly @MaryStar. Now relate the real part, which we've denoted $a$, with the imaginary part.

@evinda good night! sleep well

Let b be the imaginary part... Then $a^2+b^2=e^{4 \pi i/p}$ ?? @KajHansen

Certainly, but the length of the hypotenuse is $1$, since that value is a root of unity @MaryStar. So it gets a lot simpler.