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12:14 AM
Hey @Akiva!
 
How's it going?
 
Good
My brother's making a fifteen-minute film
*directing
 
He directed one before, yeah?
 
There's like a huge crew
@BalarkaSen Yeah
This is his final project for school.
 
12:16 AM
Hey @BalarkaSen thanks again for helping me with chem.
 
(He's a senior at NYU's film school)
 
no problem
 
Although I'm too drunk to even work on it.
 
@Akiva Gotcha
 
Not drunk I guess.
 
12:17 AM
The last one was eight minutes; this will be fifteen. It's insane the amount of work that goes into it.
 
Films are cool.
I used to want to be a director.
Must be a hard course, Jim Morrison took film I think.
:)
 
a lot of technical work goes into actually making a film yeah
 
And lots of all-nighters
Last Tuesday night, Wednesday night, and Thursday night, as well as tonight
 
Heya DogAteMy :)
 
12:31 AM
Come on baby light my fire. China set the night on fiii-iiire.
I know that's not the words, but it's funnier that way.
0
Q: $x$ in a Euclidean Domain is a unit if and only if $h(x) = h(1)$

ALannisterLet $R$ be a Euclidean domain with function $h$ satisfying the condition of the following definition: Definition: An integral domain $R$ is a Euclidean Domain if there exists a function $h: R\setminus \{0 \} \to \mathbb{Z}_{\geq 0}$ satisfying the following conditions: $h(xy) \geq h(x...

 
I remember watching your brother's last opus, DogAteMy. Looking forward to this one!
 
Hi @Akiva
 
Hi
Wanna hear something that includes my voice but also lots of other voices?
(I don't have any solos)
Also some of it is in Hebrew
 
12:47 AM
Wanna play some BP?
 
Oh, my mom has a solo in "Avinu"
(This is my first time listening to this)
@MeowMix I have a paper to work on, sorry
 
Oh, good luck with that.
When do you have spring break?
 
Now
But the paper is late
 
Oh
 
@ALannister: Part of what you posted for the first part of the proof seems wrong to me. I don't get how you saw that $h(x)\ge h(1)$. It has nothing to do with the Euclidean algorithm.
 
1:07 AM
@TedShifrin I'm not surprised that part isn't correct honestly
 
$h(x\cdot 1)=h(x)$.
 
Right.
 
But, as Will Jagy suggested, this does not look like the usual definition.
 
@TedShifrin I get that a lot
 
LOL
 
1:10 AM
Usually the definition is just the second part.
 
Right.
 
At least that's what I've gathered from looking online. Lemme check Hungerford. Sometimes he steals things from there.
 
But using your first condition, what I just typed does the other inequality.
 
Are you sure that's enough? Isn't a multiplicative norm needed?
 
It's not customary in the definition, @Alessandro.
Whoa. Way past your bedtime!
 
1:13 AM
Yup @TedShifrin it's in Hungerford.
It's in the definition of a Euclidean Ring.
 
This must be one of those things like whether we assume $1\in R$. :)
 
Yeah, what I meant isn't a multiplicative norm needed to have $x$ invertible iff $h(x)=1$?
 
But the exercise is $h(x)=h(1)$ here, @Alessandro.
 
Ah, wait, you have $h(1)$ there, hmm, I'm not sure, nevermind
 
And note, as I've hinted, that $h(1)\le h(x)$ follows trivially.
 
1:15 AM
@TedShifrin well, we're definitely assuming that $R$ is a ring with unity
 
Yeah, @ALannister, I got that :)
 
If you have a copy of Hungerford, it's Definition 3.8 on p. 139
 
I don't.
 
Meh, you're better off.
trying to be funny and failing
 
I got rid of thousands of dollars of books when I retired.
But I never owned that one :)
 
1:18 AM
I prefer Dummit & Foote
But we don't use a text for this class.
 
Anyhow, back to your question. I just fixed the first part.
 
What do you mean you fixed it?
 
I think the second part is stupidly easy.
 
You edited it? B/c it's not showing up on my end.
 
No, I just typed it up here ^^^^.
You proved that if $xy=1$, then $h(x)\le h(1)$. I'm saying $h(x)\ge h(1)$ is automatic.
 
1:19 AM
It doesn't follow from the second part?
 
I never do edits to people's posts.
Nope, nothing to do with the second part.
 
What's the reason for it being automatic?
 
It's always true that for $x\ne 0$, $h(x)\ge h(1)$.
 
Er... okay.
Is it incorrect though to say that if $yx = 1$, then, $yx + 1 = 2$, and so $r=1$, and then from the second part, where $h(r) < h(x)$ we get $h(1) < h(x)$?
Incorrect or just unnecessary?
 
What are you dividing by what?
 
1:23 AM
I guess I'm dividing $y$ by $x$ in this case
 
No, you're definitely not doing that.
 
Hm...
Yeah, I'm dividing x by 2
 
Other way around.
 
Correct. Sorry bout that.
 
I guess then you can conclude it, but you don't need that if you have the first property.
 
1:25 AM
Anyway, why is it always true that for $x \neq 0$, $h(x) \geq h(1)$?
 
Because $h(1)\le h(1\cdot x)$ for any $x\ne 0$.
 
As long as it isn't logically not true
Oooh, duuuuh! Of course!
Okay, now for the and only if part.
 
Now I think the other part of the problem is equally silly.
 
How do I go from $h(1) = h(x)$ to x is a unit?
@TedShifrin they usually are, but it often takes a while for whippersnapping novices like me to see that.
 
Oh, maybe I don't see it after all.
Now let's use what you did before.
 
1:29 AM
Wha? The remainder stuff?
 
If you divide $1$ by $x$, no matter what $y$ you try, you never get $xy=1$.
So for every $y$ there is $r\ne 0$ so that $xy+r=1$.
 
Huh?
 
Be more specific :)
 
LOL
Isn't $xy = 1$ for some $y$ exactly what we're trying to show here?
 
Precisely. I'm assuming $x$ is not a unit (doing contrapositive).
 
1:32 AM
Yeah, I just realized this as I was typing
 
So are we OK?
 
I'm trying to recopy what I did that was good, add stuff that was better, and figure that out all at the same time listening to my husband asking me if we should add the vision plan to our insurance.
 
LOL
Being blind is fine.
 
And multitasking stresses me out.
I've never known any blind mathematicians.
 
OH, some amazing visual stuff has been done by blind mathematicians. A blind French mathematician figured out how to turn the sphere inside out (which a theorem of Smale's said was possible), but no sighted person had seen how to do it explicitly.
 
1:34 AM
Dang.
Are there braille symbols for all the operators and stuff, then?
 
Good question. I dunno.
 
Because I just think writing stuff out would be harder if you had to do it in braille.
It's not like you can easily erase.
 
easier now with TeX
Anyhow ... back to where I was.
 
@Ted I watched the "turn a sphere inside out" video about a dozen times.
 
Pretty amazing, isn't it? That was Bernard Morin who figured that out.
When I was a grad student at Berkeley, there were models up in the coffee room of the process (made by Charles Pugh).
@ALannister: You here?
 
1:40 AM
@Ted What is it called when you like deform it like that
 
I think I understand Smale's proof better than the video.
 
isotopy
 
@TedShifrin yes I am. Catching up...
 
OK. So think about dividing $1$ by $x$. You have to get a nonzero remainder. What does that tell you?
 
I'm thinking
 
1:46 AM
So like, what's the topological difference between a normal sphere and an inside out one?
 
None, Zach :)
But if you think about it yourself, when you try to pass the sphere through itself and turn it inside-out, you have points where tangent vectors collapse to 0.
 
Well, by the definition, @TedShifrin it means that $h(r) < h(x)$...
 
The theorem says you can do it without that ever happening.
Right, @ALannister, and what do we know (from what I said earlier) about $h(r)$?
 
OK, so it doesn't work for a circle, what about $n$-spheres in general?
 
Whoa, Zach. Say again?
 
1:48 AM
$h(r) > h(1)$?
And then by transitivity...
 
Can you turn any spheres in higher dimensions inside out?
 
You can try @Zach but thaey won't like it ;P
 
@ALannister: We only know $h(r)\ge h(1)$, but that's good enough.
@MikeM: Smale's result was just for the 2-sphere? I've forgotten. Is this easy in high dimensions for the usual reasons?
 
Yay. But that only tells us that we get only one side of the inequality. Do we know that we absolutely cannot get the other side?
 
Huh?
 
1:51 AM
This tells us that $h(1) < h(x)$. But have we shown that when $x$ is not a unit then we absolutely cannot have also that $h(x) < h(1)$?
It seems to me the proof won't be complete unless we do that.
 
So we assumed $x$ was NOT a unit and proved $h(x)>h(1)$. By contrapositive, if $h(x)=h(1)$, then $x$ IS a unit. (Remember that we can't have $h(x)<h(1)$.
Oh, so what you typed is what I just reminded you of. What did we already prove holds in general always?
 
Oh, because of the def.
Awesome.
 
not definition, but my proof of that part of the first part.
 
Oh yesh.
 
You got it?
 
1:53 AM
Well, @TedShifrin it was nice doing bidness with you
I think so.
 
LOL, OK, I'm going to make/eat dinner.
 
Please eat, pee, sleep.
 
Have fun making/eating dinner
 
Maybe Mike will answer my question later, Zach.
LOL, @ALannister. Silly.
Way early where I am.
 
I need to sleep
Some time
Not now though
 
1:54 AM
Zach: I guess it can never work in odd dimensions. I'm not sure about higher evens.
 
Husband went to bed.
Star Trek's on
 
@ALannister Are you gonna dip his hand in warm water?
 
No because I have to sleep in that bed too
And that would be messy
It's the salt monster episode! Awesome
 
2:11 AM
@ALannister Euler was blind later in life, I think
And I vaguely remember reading a Wikipedia article on Braille for mathematical symbols once
 
LOL
I laughed at this:
"Statistically, the average human has one testicle"
 
Anyone here know of a fast algorithm for finding a minimal edge cover for an acyclic, directed graph? A brute-force solution I hacked together takes several minutes for a 460-node graph
Brute-force here as in taking what I believe to be exponential time
 
2:36 AM
@Ted: I really just meant Hirsch-Smale. You can only invert spheres in dimensions 0,2,6. The reason I don't like the video proof is I don't understand why it's special to 2 dimensions
 
Hello!
 
Hey @Dodsy
 
Hey Zach!
 
Did you think about the problem with the convex sets?
 
Not much, honestly, probably tomorrow.
Find two differentiable convex functions from R→RR→R whose graphs intersect countably many times
By countably I suppose you mean infinitely?
 
2:40 AM
Yeah, sorry
 
That's alright, I was just clarifying!
I looked at a couple of your chess games, you are sincerely a good chess player.
 
Oh, they're up? That's embarassing.
Would you liike to play another game? I have the time for a 30min. one if you'd like.
 
Well, I took the liberty of looking at a couple games. If you look at my recent loss, it's very embarssing.
Sure!
 
I'm just a little tired, so, I'll probably make a few blunders.
 
I make mistakes too often.
And, I've drank 12 bottles of beer and a tall boy of gin and something....
 
2:45 AM
Perhaps a bit too much.
Only a little though. :P
 
In a mathematicians apology G.H Hardy talks about chess, actually.
He says that chess is a form of mathematics.
My internet is terrible right now..
:/
 
Is he the British guy who invited Ramanujan to England?
 
Yes!
He is famously quoted as saying "mathematics is a young mans game"
 
I knew I heard that name somewhere.
 
and "discovering" ramanujan.
That is why I am quite envious of you.
 
2:47 AM
OK, so go to the live chess site.
 
I wish I had discovered my love for mathematics at your age.
I'm there!
Can you send me the request, I have no Idea what I'm doing.
 
Sorry, I tried but I entered the wrong game.
Sent.
 
No problem!
 
3:21 AM
Hey everyone, I've been wondering, what exactly is the deal with Fourier/harmonic analysis? I've only seen it in the context of being convenient in physics, but is there something... I dunno, deeper about the subject?
 
I have a book on the subject that I never finished
puts on mental "to do list"
 
Yes there is @Daminark
aware of area perimeter theorem ?like using Fourier analysis we can connect a relationship between perimeter and area.
 
@Akiva Which subject?
 
Fourier series
 
Good game meowmix
 
3:27 AM
"Fourier Series" by Georgi P. Tolstov
 
You too, Nate
 
And @Baymax I have not heard of the theorem by that name, at least
 
I'm guessing it's something to do with the circle being optimal
when you optimize area with a fixed perimeter
 
@AkivaWeinberger Did you hear about my convexity problem?
 
Oh, I think I've heard of that as the isoperimetric inequality or something
 
3:28 AM
@MeowMix Which one?
 
I'll just state it
I don't know the answer
 
You really are a good chess player, I thought you had it there many times!
 
yeah youa re right@Daminark
 
Don't forget that I studied the game for a couple years.
 
sorry for wrong name.
 
3:28 AM
And that's p cool @Akiva, the last chapter in my analysis book is about Fourier analysis on locally compact abelian groups
 
I must go though meow mix.
 
Topological groups? @Daminark
 
Don't worry about it @Baymax, chances are this name of it exists, just that I didn't hear of it as such
 
@Dodsy Don't flatter me :P. I'm not too good, and your skill definitely shows
 
I'll talk to you tomorrow about that hyperbola problem (the one that you posed to me)
Nah, I am not a very good player!
You have a lot of potential.
I have reached my limit.
 
3:29 AM
@AkivaWeinberger So, take some non-convex differentiable contour
 
Sky is the limit.
 
And take the union of all it's tangents
Then take it's complement
Is that set always convex?
 
$\lim_{n\to\infty}f(n)=\rm sky$
 
Furthermore, is that the smallest convex set which your contour contains?
 
@BAYMAX I wish! I've studied thousands of games. Read over 30 books. I cannot improve anymore in chess. Others have talent for the game, I do not.
 
3:30 AM
Yup @Akiva
 
@MeowMix For that last bit: Clearly not, as points are convex...
 
yes@Akiva
 
Largest******
 
I think meowmix could improve greatly and shows promise, I am at my peak. I don't have the passion or talent for the game.
Talk to you tomorrow Zach!
It was fun playing with you and chatting!
 
You as well, have a nice night :]
 
3:31 AM
Maybe next im we make the game longer so we can discuss maths.
 
oh@Dodsy , you will definitely improve , practice and effort!
 
no i'm tho.
 
@AkivaWeinberger So that's a little food for thought.
 
It doesn't do everything just on groups in the abstract, it talks about Fourier analysis on the circle, on $(\mathbb{R},+)$, and (in typical Sally style) p-adics
 
3:32 AM
@MeowMix What was the hyperbola problem that Dodsy mentioned?
 
Find with proof, all integer solutions to 4x^2 - y^2 = 15
also please do not ruin it for me.
 
Oh, it wasn't hard, just a generic lattice problem. Find the integer solutions of $4x^2 - y^2 = 15$
@Nate there's a clever and sneaky trick here that makes it trivial.
 
I think I see it
 
Yes
 
Once I leave you guys can discuss it!
I will be working on it tomorrow Zach, before starting chemistry for the day.
then tomorrow, we will discuss my conclusions.
 
3:34 AM
Alright, good luck with it :]
 
Thanks!
See you tomorrow! We don't have to play another game of chess, but I appreciate your time!
 
It was my pleasure.
@AkivaWeinberger I assume you just factored it?
 
So clearly you factor the difference of two squares
Yeah
$(2x-y)(2x+y)=15$
 
Mhm, then both factors must be integers
 
which means we have several cases to check, starting from $(-15)(-1)$ to $(15)(1)$
 
3:35 AM
so you look at the integer factors of $15$
$1,15$ and $3,5$
 
$(-15)(-1)$ gives me $(x,y)=(-4,7)$.
 
positives/negatives dont matter
 
Right, I just realized
 
so just $(\pm 4, \pm 7)$
 
And $(\pm2,\pm1)$.
Yeah?
 
3:37 AM
Yep.
 
Arright, cool
 
I added in the "with proof" part so that he wouldn't try to brute-force it
 
ha nice @Fawad
 
@MeowMix A much harder (and more interesting) one is $x^2-2y^2=1$.
 
No solutions I believe
 
3:38 AM
A starting point is $(3,2)$.
 
Wait
Wrong one
 
And $(7,5)$, as well.
 
Aren't these called Pell's equations or something?
 
It is, yeah
I wasn't sure if you'd heard of it
 
How to solve $(x-1)(x-3)(x-5)(x-7)+7=0$?
 
3:40 AM
What tools do I need to solve this one?
Would I know it?
 
@MeowMix You would
I mean, it's far from obvious
but one starting point would be the same as the other one (difference of two squares)
@Fawad Integer solutions?
 
Well, that works
 
A starting point would be to look at the cases where each is negative/positive
There must be an odd number of negatives so either $x-1, x-3, x-5$ are all positive or just $x-1$
so $x \in [1,3] \cup [5,7]$
 
Wolfram Alpha just multiplies it out and factors it
It ends up being the product of two quadratics.
 
3:48 AM
Also, look at just $(x-1)(x-3)(x-5)(x-7)$. It has a line of symmetry about $x = 4$
This suggests that there are 2 in each subset, symmetrical to eachother
I guess that was obvious though
 
Oh, that's a point, we should probably substitute $x\mapsto x+4$ then
so $(x+3)(x+1)(x-1)(x-3)+7=0$
or $(x^2-9)(x^2-1)+7=0$
which becomes a quadratic in $x^2$, right?
$x^4-10x^2+16=0$
$(x^2-8)(x^2-2)=0$
which becomes $x=\pm2\sqrt2$ or $x=\pm\sqrt2$. But remember — we did that substitution.
 
@AkivaWeinberger this is something my maths teacher used yesterday , to get fractions polynomial
 
So, to fix it, do $x\mapsto x-4$, which gives us $x-4=\pm2\sqrt2$ and $x-4=\pm\sqrt2$
Which gives us our final answer: $x=4\pm2\sqrt2$ and $x=4\pm\sqrt2$.
 
@AkivaWeinberger shouldnt take $x+4=y$ and get quadratic in $y$,then get $x=y-4$ ? It will be less confusing
 
Yeah, sorry
Same thing, less confusing
@Fawad Well, no, you want $x=y+4$ and $y=x-4$.
 

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