I was getting trolled or were people arguing?

I cant even think about doing math after getting that grade back, I was pretty certain I got a B on the test

@Jordan Where?

@Jordan What grade did you get?

62%

Basically I am going to fail the class

@Jordan Isn't that a good grade?

Like a $C$

62%| like $\frac{62}{100}$

I know, but it is a passing grade!

Well in the US you have to get an A or B or you pretty much are failing

If you get less than a B you can never get into a good university

Sounds like an athlete :P

@Jordan A Fields medal is one of the greatest honors in mathematics.

Never heard of it

besides he probably got poor grades at a really good school

Yeah looks like he was already at a good school

I am just failing high school math

well most people have no problems getting good grades

like those douchebags infront of me, one got 100 percent on the test the other got 95 percent

they really like math, a lot, they are probably just good and work hard

I am really not good at anything, I just have to work really hard for poor results

Hi @Jordan, you seem to be highly unmotivated. Without motivation, it's unlikely to be successful. I've been following your questions, and I can clearly see good progress. Perhaps it's slower than you like, but it's definitely there.

Maybe but school doesn't move at my speed then so what do I do? Do I just keep retaking classes until I am out of money? Honestly this just feels so pointless, I feel like I am not smart enough for school but I keep doing it because I don't want to do anything else.

If you think that something is pointless, you're unlikely to do well in it. Also, it's true that people have different abilities, but that's only part of the equation. It takes a lot of hard work, even for those most gifted, to achieve good results.

I put in the hours but I never get results, I just wish at least once I would get results, but I never do

not even once

Seeing your questions, you're clearly more capable of evaluating integrals now. Your course seems to be moving very fast though. It's covering a lot of material.

I am going to withdraw from this class

@PeterTamaroff next sem I die

@BenjaminLim LOL why?

supervisor wants to cover 240 pages of hardcore several variables @PeterTamaroff

If I get a D in this class I will never be able to transfer out of my shitty school

@BenjaminLim Strap up people! Here we go!

I actually had a nightmare all night last night that I was going insane because of math, I think it is happening

@Jordan Stop it.

Just stop.

@PeterTamaroff I think I might die!!

@PeterTamaroff Algebraic topology is already looking insane

That is easy for you to say Peter, you have things so easy

You are 18 and you are already in your third year of college math

@Jordan I don't. You don't know what I've done to get where I am.

@Jordan I'm in first year, BTW.

@BenjaminLim I imagine!

I want to learn some Point Set from Rudin.

@Jordan Do you think we've all had it easy? I don't exactly go around telling people how bloody brutal last semester was for me. Do you think @PeterTamaroff is where he is now without all the bloody hard work? Wake up Jordan STFU and stop complaining. Everyone has had to work their arse off to get anywhere at all. Just stop the complaining, shut up.

@PeterTamaroff nice. You're becoming a real mathematician

@BenjaminLim I have easily +1000 pages of notes and theory handwritten by me, by myself, studied by myself from books and other similar sources. I know what I know because of me, and the (sometimes not so little) help I have been given. Now I want to stop this rant none sense.

@PeterTamaroff exactly.

People are getting put off, the room is getting empty. Darn.

@BenjaminLim Know that feel, bro.

@JasperLoy 1400 sides is crazy

@BenjaminLim I'll show you a pic.

@JasperLoy Well, that, I meant that.

@PeterTamaroff start with munkres

@BenjaminLim I have it. I just need some extra time. I'll get a two month vacation july-august, so I'll get on 5th gear there!! Expect many questions in main!

Yes, I came here to see what the hubbub was all about.

It seems to have sorted itself out, so I'll be off now.

Good day.

@PeterTamaroff where is the pic?

you can show it on facebook

@BenjaminLim Give me a sec

@JasperLoy: has there been a lot of flagging here?

I was working on questions and missed the fun, I guess.

@robjohn well we all know what was the reason for all the flags

@BenjaminLim I couldn't find the Sd adapter.

@JasperLoy He said "math faggots". I flagged that one.

huh???

@JasperLoy I think that is offensive.

@BenjaminLim OK, it worked.

@PeterTamaroff huh?

@BenjaminLim The image. I got it.

for all you biebermaniacs out there, justin bieber has released a new album

oh shit

@Eugene Save yourself!

@PeterTamaroff where is the image?

@JasperLoy what are you doing in the islam room?

@BenjaminLim Uploading...

@JasperLoy cody simpson is hot

@JasperLoy youtube.com/watch?v=evTrDL7Hfvc&feature=plcp

The two first are analysis.

The little one is number theory.

@PeterTamaroff ....

@Eugene What?

@PeterTamaroff you're a lot more organised than me

this is legal proof of your copyright violation.

All I have is a big pile of papers

@Eugene That is handwritten, silly.

those are notes?

@Eugene What do you call "notes"?

@PeterTamaroff Wow, did you print out all 56,657 MSE questions?

@JasperLoy i just flagged you

@BillDubuque Hahahaha, it is handwritten Bill.

@JasperLoy welcome

@BillDubuque Jordan accused Ben and me that we got it "easy". I object with that.

@JasperLoy maybe that'll teach you not to call someone "silly"

@BillDubuque I can't stand the complaining

@BenjaminLim Honestly, I can't either.

@BenjaminLim Isn't it possible to block users on chat, so you don't see their posts?

@BillDubuque yes it is

you open their chat.SE profile then click ignore this user.

@JasperLoy rhetorical means doesn't need an answer but doesn't prohibit me from giving one.

done.

@JasperLoy Yess.

@JasperLoy so you don't subscribe to the "i'm rubber and you're glue" philosophy?

or rather i'm a non-cohesive polymerized tree sap?

@Jasper Have you never used usenet newsgroups? Most newsreaders have filtering capabailities. They prove very handy for large diverse groups.

@JasperLoy @Eugene One should love allbeings.

@BenjaminLim did you have salad today?

NO, even though it's been de-criminalised in the ACT

@JasperLoy it can be found here

@Eugene love all beings

@BenjaminLim i don't think i believe you

why?

One should strive to love all beings in the universe

given a function continuously differentiable over $]a,b[$, does $f(x_0)=0\implies f'(x_0)=0$

@Eugene Hater.

@BenjaminLim Obivously that is asymptotically possible.

@leo Nou.

@PeterTamaroff yes I just realized that

@PeterTamaroff once you teach your first calc class you'll sing a different tune

Unless $x_0$ is a root of multiplicty $\leq 2 $ and $f$ is apolynomial, for example.

$x^2+x$ does not

@Eugene Hahahaha lol.

@leo In general, if $(x-x_0)^k$ divides $f$ then $(x-x_0)^{k-1}$ divides $f'$.

@Eugene We had a cooler tune: "Rebota rebota y a vos the explota".

@leo a polynomial $f(x)$ has a multiple root iff the $\textrm{gcd}(f, f') \neq 1$

This is true in all characteristics

@PeterTamaroff yes

@BenjaminLim in all fields?

@leo no. not in really small ones.

@leo Yes the characteristic is irrelevant

@Eugene Hhahahaha

@Eugene It's true in characteristic $p$.

"Only for large values of $\pi$" beats them all.

@BenjaminLim what about THE TRIVIAL FIELD

by the way there are no field of characteristic 6, right?

ie $F = \{0\}$

@leo that depends on how drunk you are.

@Eugene you're trolling $\{0\}$ is not a field

@leo all fields have either characteristic 0 or prime characteristic

@BenjaminLim while i was joking initially, actually it is a field. most people require rings to have unity but there's the weird rngs.

@Eugene I see. Then almost always no these days

@leo oh that's kind of sad.

@BenjaminLim then there are no vector space with exactly 6 elements, right?

Any help is appreciated.

well the joke is that rings without unity are called rngs.

@Eugene yes indeed. I'm about to finish the period

@leo no

@BenjaminLim no what?

They really call them that.

it is false?

@leo there is no vector space with 6 elements

@JasperLoy well... that's debatable...

Also rigs, which are rings without inverses.

such a vector space must necessarily be over a finite field

of prime characteristic

from the fact that there are 6 elements

6 is not a power of any prime number

so there we go

@BenjaminLim is the reason of that the characteristic of the possible field?

i thought rngs were actually called pseudo-rings.

yes

@BenjaminLim That's why we mvoe on to octonions?

so I'm right :-)

@PeterTamaroff dunno

@PeterTamaroff ?

@leo why?

@BenjaminLim some time ago I was trying to approach that ´problem with group theory

ah

@Eugene $\Bbb C$ then $\Bbb Q$ then $\Bbb O$

@PeterTamaroff i know what the octonions are...

@JasperLoy yes that why i said they are jokingly called rngs..

@Eugene I'm jsut saing that there is nothing in between... I know you know

@PeterTamaroff because some characteristics of vector spaces are not possible?

I was trying to prove that such a space is isomorphic to $Z_6$, then see $Z_6$ as a vector space over some field $Z_p$ with $p$ prime and then arrieve to a contradiction

But I fail

@Eugene Beats me, it's what Ben is saying.

@PeterTamaroff hay una inclusión que es fácil

there is an answer in there.

@leo In where?

@PeterTamaroff essentially the quaternions is an extension of the complex numbers. the motivation is very geometrical. it's mostly used in physics though. so if $\Bbb{C}$ is like $\Bbb{R}^2$ then $\Bbb{Q}$ is like $\Bbb{R}^4$.

@Eugene I remember when I got a complex numbers text and it said "\Bbb C" has the topology of $\Bbb R^2$ and I was like "Whaaaaaaat?"

@Bill Would it be senseless to prove that $\tau (n) \leq 2 \sqrt(n)$ by induction?

I'm pretty sure I can see why it is true, but I'm not sure if I can make a elementary argument.

It seems if $d$ is a divisor of $n$, $d \leq \sqrt n$ equality only of $d^2 =n$, i.e. $n$ is a perfect square.

Clearly there are stronger bounds, but that is a "necessary" bound.

@PeterTamaroff $\tau$ Euler's totient function?

@leo No, that is $\varphi$ or $\phi$. $\tau$ counts the number of divisors of $n$.

Simply $\tau(n) = \sum_{d\mid n} d^0$

In terms of the exponents of the prime factors, i.e. of $k_i$ in each $p_i^{k_i}$ of $n$, $\tau(n) = \prod_{m=1}^r (k_i+1)$

Surely if $d|n$ and $d<\sqrt{n}$, we have $\frac{n}{d}|n$ and $\frac{n}{d}>\sqrt{n}$.

It just occurs to me that people here might enjoy my new pentagon-themed Tumblr blog, Fuck Yeah Pentagons!.

I prefer hexagons.

@anon Yes.

@PeterTamaroff in that case $\varphi(n)=\sum_{d|n}\varphi(n)$ as well :-)

@anon There is already a Fuck Yeah Hexagons! blog.

@leo ?????

fuckyeahhexagons.tumblr.com

@MarkDominus I prefer that to $\tau(n) = \sum_{d \mid n} 1$

It is a worthy effort, and I enjoy it. But I do prefer pentagons.

I prefer divisor-sigma, $\sigma_s(n):=\sum_{d|n}d^s$.

There is also a HexagonBot on twitter that retweets everything anyone says about hexagons. I tried following it for a while, but there was too much volume.

@anon Well, sure $\tau(n) = \sigma_0(n)$... =)

@anon Any help on the problem?

problem?

I need to prove $\tau(n) \leq 2 \sqrt n$

By noting that if $d \mid n$ then $d \leq \sqrt n $ or $\frac n d \leq \sqrt n$

@Eugene Hey.

i don't have any interesting elliptic curve questions to ask anymore

oh well.

i'm thinking of offering a bounty

@Eugene For what?

Just split the divisors of $n$ into two groups, that are in bijection with each other (except you'll have to be careful about whether $n$ is a perfect square). One of them is $\le\sqrt{n}$ in cardinality, so...

for this

Or you could just bump it when you see Emerton active on the front page.

@anon Oh, right!

holy crap 9 flags pending

everybody panic

@anon I had one. I don't see 9.

@anon that is a good idea. lol.

@PeterTamaroff Either at 10k or 20k you get a tools bar that lets normal users see flags that have been raised on the mainsite that haven't had action taken on them.

@leo $n|n$....

@anon Wait. $\leq n$ in cardinality since $\operatorname{card}= \frac{\tau(n)} 2 \sqrt n$?

@anon $n|n$?

@PeterTamaroff One of the sets will be a subset of those numbers in $\{1,\cdots,[\sqrt{n}]\}$, or somesuch. Not sure where in the world you're getting that expression.

@leo Yes, I'm telling you that $n$ divides $n$, so that your equation reads $\varphi(n)=\text{blah}+\varphi(n)$. You sure it's right?

@anon no

@anon The numer of divisors is $\tau(n)$, since halve of them $\leq \sqrt n$ the amount of them should be $\leq \frac{\tau (n)}{2}\sqrt n$ for an even amount of divisors.

Why in the world are you multiplying $\tau(n)/2$ with $\sqrt{n}$?

I don't know! AAHHHHHHHHH

Split the divisors into two subsets, A and B. Prove that |A|=|B|, though you'll need an exception for when n is a perfect square. Then since $A\subseteq \{1,\cdots,[\sqrt{n}]\}$, we get $|A|\le\sqrt{n}$, hence $|B|\le\sqrt{n}$, hence adding things together...

it should be $n=\sum_{d|n}\varphi(n)$, my apologies

Imma get some food

@leo yes, that's correct

@anon I don't know what I was doing!!!

@anon enjoy. Me too

@anon Does the notation

@PeterTamaroff does the notation what?

@anon I'm typing wait

${\rm A} = \{ d : d\mid n, d \leq \sqrt n\}$ and ${\rm B} = \{ d : d\mid n, \frac{n}{d} \leq \sqrt n\}$ make sense?

To write the proof.

Do be careful of whether or not $n$ is a perfect square. If it is, you will have $A\cap B=\{\sqrt{n}\}$. But that works.

@anon Oh, OK. I'll make that clear.

But I mean. I'm starting like this:

Suppose $d \mid n$. Then $d \leq \sqrt n $ or $\frac n d \leq \sqrt n$. Group the divisors of $n$ into

${\rm A} = \{ d : d\mid n, d \leq \sqrt n\}$ and ${\rm B} = \{ d : d\mid n, \frac{n}{d} \leq \sqrt n\}$

Any divisor of $n$ not a squareis to be $\leq \sqrt n$ so that ${\rm A} \subseteq \{1,2,\dots, [\sqrt n]\}$ and ${\rm B} \subseteq \{1,2,\dots, [\sqrt n]\}$

But since there is a bijection from $\rm A$ to $\rm B$ ($d \mapsto \frac{n}{d}$), we have $|\rm A| = |\rm B|$.

Actually you don't need to care about whether n is a perfect square because you can use $|A\cup B|\le |A|+|B|$ in the end. Also note that "not a perfect square" and "squarefree" don't mean the same thing.

Hhee yes that was slip.

Thus the totality of the $d$s, $\tau(n)$ is to be $\leq{\rm |A|+|B|}=2\sqrt n$. $\blacktriangle$.

I'm not sure why you're saying "any divisor of $n$ not a square is to be $\le\sqrt{n}$" ..., nor do I get your $B\subseteq\cdots$ statement...

@anon $n$, $n$ not a square,

the other is wrong.

What?

Sorry, I didn't pay attention to that, it was copy paste.

@anon "Any divisor of $n$, $n$ not a perfect square, is to be"

@anon Actually $B \subseteq \{[\sqrt n ], \cdots, 2 [\sqrt n ]\}$ right?

But that's still not true, a divisor of a not-perfect-square $n$ can surely be $\ge\sqrt{n}$, nor do you need this to be true. Just say that $A\cup B$ is the set of all divisors, and $|A|=|B|$, and $|A|\le \sqrt{n}$.

@anon OK

@PeterTamaroff Why end it at $2[\sqrt{n}]$? Not that what $B$ is a subset of is needed for anything.

@anon That was ad hoc. I got it. I have to think things more thoroughly next time. My bad.

@PeterTamaroff @BenjaminLim Honestly I feel like you probably do work hard, but I work harder and I have far less to show for it. So yes in a way you do have it easy.

@Jordan Really? You haven't event apologized yet, and come back with that?

Apologized for what?

@Jordan Never mind.

@anon Would one say $d \mapsto n/d$ bijects $A$ with $B$?

Yes. Or "is a bijection $A\to B$." Interestingly it's also a bijection $B\to A$.

@anon He, yes!

@anon Ok, I wrote the proof down. My "brain pressure" went down.

@anon That $\tau(n) \equiv 1 \mod 2 \Leftrightarrow n =a^2$ is kind trivial, right?

That's a relative term, but on our level I'd say yes.

@anon I'll take that as a compliment. I hope I didn't fall too much from the last one, I really was saying some nonesense!

That's nothing compared to Eugene and me discussing Iyengar's problem one night. He was a trainwreck!

@anon Hahahaha Iyengar's problem, let me google taht.

Also, I figured out the problem I posed yesterday in chat was utterly trivial. (Though I figured out how to amend it to make it nontrivial, though it requires the group to be infinite.)

@PeterTamaroff Iyengar is a user on MSE.

We were discussing one of his questions. (That he later deleted and then reposted.)

@anon What was his problem? (That explain google failing! Phew!)

@anon That's cool! Making something non trivial!!!

Actually it looks like he deleted his second version of the question too.

@anon Well, what was it about?

Lemme see if it's in Recently Deleted (10k+ toolbar!) for a screenshot or summary.

Okay it's not there, but I'm pretty sure I have a cap of the first version on my hd.

https://i.sstatic.net/teDfU.png

@anon Interesting question.

I don't like Diophantine equations, and wasn't really interested by it.

@anon I see.

What problem?

Is it evident that if $p$ is odd then $\frac{p^{k+1}-1}{p-1}$ is odd only when $k$ is even?

The one in the imgur pic I just linked, Frank.

Maybe a hard problem.

by geo sum formula it's $\equiv 1+kp\bmod 2$, so yes

@anon Ha!

Very keen observation.

note that $x\equiv x^2\equiv x^3\equiv\cdots\bmod2$.

@anon Yep.

Well, $a\equiv b\pmod 2$ is a conventional notation.

don't tell me you're turning into a Michael Hardy

@FrankScience Why the parenthesis?

@anon He is an MH.