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00:00
That's not too tough, Nate.
Oh, shit ... the sum of the digits. I missed that.
I have to think.
No, I don't.
Wut
How do you do it, kind wizard.
The product must be divisible by 9.
I'm out for a while, gotta think galois
Night, @EricSilva. I'll look forward to geometry chat during July. :)
But it doesn't tell you how many digits.
00:04
I don't care.
A number (in base 10 expansion) is equivalent mod 9 to the sum of its digits.
So it could be 99, 981, 972, 963, 954, 945, 936, 927
This is casting out 9s, which you've probably heard of.
Hm.
Oh
oh my god
I just realized what I wrote!
What you wrote isn't wrong.
all of those numbers are divisible by 9
00:06
Yup. (This is in the first part of my algebra book. :))
What algebra book!?
Hi Nate.
Is it adequate for somebody like me who has massive gaps!?
Zachary Mix.
My first text was an abstract algebra book ... rebelling at the way algebraists (not Artin) do it.
How have you been, dear sir.
00:06
Tired
Get your blood tested, Zach. I won't be here to keep annoying you with that.
Heh, alright
@Danu by looking at conjugacy classes in SL(2,C) and how their representatives behave as Mobius transformations, we see that every transformation has one of the following cycle types: (i) every point is fixed, (ii) one fixed point and uncountably many ∞-cycles, (iii) two fixed points and uncountably many ∞-cycles, (iv) two fixed points and uncountably many k-cycles for some k>1. If such a transformation stabilizes a finite set, that finite set must be a union of orbits.
Therefore, the permutation of the set of punctures must have one of the following cycle types: 0,1 or 2 fixed points and a bunch of k-cycles for some k. I suspect every such permutation group arises from my aforementioned recipe as the symmetry group of some polyhedron under SU(2) acting on the Riemann sphere by isometries.
Ted did you see my dumb combinatorics problem
Ted, if I buy your book off of Amazon do you get a direct cut of the profits?
00:08
Not if it's a used book, Nate. You don't have the money to spend on that, anyhow.
Heya tern :)
Oh, some of that stuff is in Chapter 8 of my algebra book, tern :P
Ted's books cost more than Ted himself
I think it was because of the publishers
Yes, you're right. But I was thinking something complimentary while in school. I'm pretty sure the multivariable textbook used is James Stewart in second year, I believe.
(Oh hell, that's gonna get starred.)
I think your asterisks are misplaced unless I'm thinking of the wrong word
Nate, after I get back from Europe, send me an email.
00:09
Well not misplaced
Mis-counted?
I don't get that word either.
Unless you elongated it.
OK, so I can't count to 2.
Sure Ted!
Balarka, go to your room.
00:10
cheap who are.
I'm glad the next time I appear the starboard will be far past that.
Teeeeed
Diiiid you seeee the prrooooblem
No, and I'm leaving.
Wow, I'm not gonna talk to you for a whole MONTH now. /s
00:12
Meow mix name an integer other than 1 whose digits have a sum equal to the square root of it's number.
Nate, did you finish the problem with my hint?
Nate, you're as bad as Balarka. Watch those wrong apostrophes.
Hi I have a graphing question and I'm not sure what method I should start with to solve it: The math department needs to schedule 6 classes. 10 students have indicated the courses that they plan to take. What is the minimum number of time periods per week that are needed to offer these courses so that every two classes having a student in common are taught at different times during a day. Two classes having no student in common can be taught at the same time. Each class has 1 lecture per week.
I'm glad you didn't mention me
00:14
I'm fine with apostrophe's.
You're bad, too, Zach. You just escaped because I'm nice.
@TedShifrin I thought we solved it! But it must have to do with the lowest digit losing a number and the second highest digit gaining a number. The lowest number with the property of the sum of the digtis being equal to 18 is 99. Every other number, due to the properties of 9 will be divisible by 9. Since the number is divisible by 9, it cannot be divisible by 3 prime numbers.
I hope I'm close.
What did it say about the 3 primes?
wait
the integer is a product of the 3 primes .
:)
But what about the primes?
00:16
three different primes?
So that's the essential ingredient.
$2\cdot 3\cdot 13$ won't be divisible by $9$.
Oh right
You get it?
because the primes make up the prime factorization
right?
Right.
Okey dokey. I'm out.
00:18
Okay
thanks Ted
I'll send you an email :)
Now that Ted's gone let's have a party
Then you can find out my last name!
I also must go and eat
it was good seeing you Zach
Please come by more often so we can play chess and hang out!
Alrighty
have a good food
Night, all. Back in the next world.
00:25
Hey @Dodsy just saw this message, and Laci is a math/compsci professor at my school, he's had a huge influence on me
He does stuff like combinatorics and computational complexity theory
Also see you @Ted!
Oh I'm back.
Bye ted
@Daminark That's cool
Oh, general question
Sure I have an ocean of knowledge
So, in my class we proved Lagrange multipliers specifically when $f:\mathbb{R}^n\to\mathbb{R}$ and $M = f^{-1}(0)$. Is there some way to extend this to manifolds more generally?
@Balarka Good luck
00:35
@Daminark Given a submanifold $N \subset M$, you want to extremize $f : M \to \Bbb R$ on $N$?
$$\int (1+x)^n \log(x) \textrm{d}x=\frac{(1+x)^{n+1}}{n+1}\log (x) -\frac{1}{n+1}\int \frac{(1+x)^{n+1}}{x}\textrm{d}x=\frac{(1+x)^{n+1}}{n+1}\log (x) -\frac{1}{n+1}\int \frac{(1+x)^{n+1}}{x}\textrm{d}x=\frac{(1+x)^{n+1}}{n+1}\log (x)-\frac{1}{(n+1)(n+2)}\frac{(1+x)^{n+2}}{x}-\frac{1}{(n+1)(n+2)(n+3)}\frac{(1+x)^{n+3}}{x^2}-\frac{2}{(n+1)(n+2)(n+3)(n+4)}\frac{(1+x)^{n+4}}{x^3}-\cdots=\frac{(1+x)^{n+1}}{n+1}\log (x)-\sum_{k=0}^{\infty}\frac{k!n!}{(n+k+1)!}\frac{(1+x)^{n+k+1}}{x^k}$$

Now to figure out how to solve that infinite series...
@Secret Oh my God! How you got that?
You just keep integrating by parts
@Secret I see. The answer is simple, elementary and elegant.
How do you solve:
"Kayla adds the same number to both the numerator and denominator of the fraction $1/10$. Her resultingfraction equals $2/3$ What number did she add to both the numerator and denominator of her original fraction?"

Without brute force?
00:39
(and integration by parts is taught in my high school ,thus it fits the high school part of the challenge)
I have the answer at 17.
The problem is how to solve that infinite series...
@Secret You don't need to go into such stuff. Keep thinking of simple, elementary tools.
Do you make the denominators equal and subtract the smaller numerator from the larger?
You could do the algebra: say she added x to top and bottom, then (1+x)/(10+x) = 2/3.
00:40
so then: $2/3 = 20/30 , 1/10 = 3/30, 20-3 = 17$?
Wait, so you are ok with having an infinite series as an answer?
EDIT: Ok nvm
@Secret No. The answer doesn't contain any infinite series. Just a simple, cute result.
Ah.
benefits of using a hybrid vehicle for day to day driving?
I need to write 500 words and I've been putting it off all day.
Although I'm so behind on my homeowkr that I'll have to pull an all nighter thursday, friday, saturday and sunday.
that's right folks.
@Dodsy Typically always ever try to do the algebra when you have something unknown in the problem that you have to compute.
what a meh essay
@Daminark Anyway, if that was what you wanted to do; sure, if write $N$ locally as level set of some $g$, then you can do the Lagrange multipliers on a coordinate chart. I am not sure if that's what you want.
I have to sleep now.
@Secret It's really late here, the night almost passed. I leave. Don't hurry with the problem.
00:52
I don't really know how to solve for x from that point "(1+x)/(10+x)=2/3"

I suppose I would multiply the whole thing by 3. So (1+x)/(10+x)(3) = 2, and then [3(x+1)]/(x+10) = 2

and then, 3x+3 = 2x + 20, 3x - 2x = 17, 1x = 17, x = 17
Oh yeah, that would've been simpler I suppose.
So, Nate
How are your studies going?
Sigh. terribly.
I have to finish 6 units of work by sunday
so I can study for Functions for a week
and write the exam on the 12th.
because if I get accepted to school I'll have to submit my final marks by august first.
@Balarka in the standard way of doing it, you'd have some known function that you could look at Lagrange multipliers with. Hypothetically, though, on a more general manifold you could have the problem that you could have various charts near the point
So, I have to write 4 exams, and finish a couple hundred hours of work in 5 days.
@Secret @Semiclassical After finishing that version, try another one, slightly different $$\int_0^1 (1+x)^n \log(1-x) \textrm{d}x$$
00:54
They're compatible, sure, but I dunno if you could just do that straight
Good time to go to bed.
Wish me have sweet dreams!
;)
What is the smallest positive integer k such that k/660 can be expressed as a terminating decimal?
33 right?
Yep.
Neat
Five workers together can build a road in 20 days. Suppose every worker works at the same rate. If
three workers work on the road for 10 days before eleven more workers join them, then how long total
will it take to build the road?


15 days right?
Alright I'll stop asking trivial questions.
01:06
1 worker = 4 days / road
3 workers = 4 days / 3 roads
which is
2 days / 1.5 roads
10 days / 7.5 roads
wait huh
______ ignore _____
So what I did was first calculate how much of the road three people could finish
which was 30%
yeah
then I found how long it would take for 14 people would finish a road
and multiplied it by 70%
then added it to the original 10 days.
You can tell I'm terrible at math.
me too
Haha no, you're a genius man.
01:09
OK so one worker would build a road in 100 days
^
I accidentally divided instead of multiplied
So then 3 workers would do it in 100/3 days, so in 10 days they'd do 30% of the work
Now 14 workers would do a road in 100/14 days, but they only need to 70%, so that'd take 5 days
15 days, Nate's right
@Dodsy I'm flattered, but probably not. :P
Thanks Daminark!
What have you been working on?
Ted says you've been doing computer stuff
me?
or dami
01:20
You :P
You Mr. Zach.
Yeah
lots of sleeping too
but most of my programming stuff is in assembly
hides
assembly > all other languages
@MeowMix Depends.
01:27
im kidding
Zee
Zee
Natural Language > assembly
i just like assembly alot
Assembly's great.
It just sucks for certain situations where you have to, say, write a working GUI program in two hours.
Zee
Zee
thats why I program in $ language
But it can't be beat on efficiency unless you want to learn machine language, which...good luck.
Zee
Zee
01:28
Assembly and machine language are the same efficiency
I don't program at all.
I don't even want to take computer science in school..
Oh I didn't even tell you Zach, Queens screwed up my application so now I have to go to UWO -_-
Theoretical computer science is one of the dankest subjects though
Zee
Zee
@Dodsy that's fine , programming is math watered down
Yeah I wouldn't mind theoretical.
Math is cool!
Really, Nate?
That really sucks, sorry about that
01:32
Damn, you guys are still goofing off.
@Zee Not really... I've done a bit of both and compsci asks for a different mindset entirely
Math.SE 2: Return of the Shifrin
We solved a trivial problem, Mr. Shifrin.
The question is, Zach: How many times has a Balarka returned?
Zee
Zee
@Daminark true, CS requires no imagination
01:34
False
Zee
Zee
@Dodsy it's prof shiftin you minions
@Zee raises a skeptical brow
Nate do you want a math question
Knowing how to prove something and knowing how to design an algorithm to find something are different things
At various times
Yeah sure, I probably will butcher it though. I'll try really hard though.
Zee
Zee
01:34
They are the same thing
Especially when you're working around structural obstructions that the system provides
@TedShifrin I emailed him. As most of my emails to professors go, it was four paragraphs of me not knowing what to say or how. But it felt good!
@Fargle: Cool. Did you blame it on me? :)
Zee
Zee
Proofs contain all the germs of thinking
#blameitontedshifrin
01:35
Nate: Do the algebra: $\dfrac{1+n}{10+n}=\dfrac23$, so $3+3n=20+2n$, so ...
1x = 17
:D
@TedShifrin Nah, but I did mention you. Though I think I should see a doctor, because my fists are now ham.
1n*
Not sure I understand that, @Fargle, but ...
Zee
Zee
Why use a computer to study finite processes when you can learn to do things that would take a computer forever to do
01:36
@TedShifrin Everything in the email felt ham-fisted.
Zee
Zee
Sum a series for example
ham-fisted.
Dodsy, here's a little warm-up that should be easy
I didn't mean to make you that uncomfortable, @Fargle. Sorry.
I like it.
Hamliness is godliness
01:37
Haha, no, you're fine!
Find a bijection from $\Bbb Z$ to $\Bbb Q$
Well, sometimes computers can do something people can't
We're good at computing things when there's some niceness to the inputs
I could be emailing a best friend of a decade and a half and feel just as out of place. I just have trouble peopling, but I like to leave my comfort zone now and again.
Seriously, thanks for getting me to do it, @Ted.
Or maybe that's not the best wording
Even if you do math, you eventually have to people.
01:38
@MeowMix from the E of Z to the E of Q?
Zee
Zee
@Daminark computers are no good, just good for watching YouTube and chatting, I would not mind at all if they all disappeared
huh?
That's a statement with a truth value of 0
Oh.
sorry i dont understand
01:39
But, Z is in Q.
@TedShifrin Very true. It's very easy if you don't know the person, I've found. For some reason, the graph of discomfort vs. closeness is a bell curve for me.
Zee
Zee
Oh come on , drop the indoctrination
That is true. :)
Doesn't mean there cant be a bijection between them
I mean, I won't because I'm right :P
Modulo having some random definition of the word "good"
In which case our statements are parsing out differently
Zach, I don't understand either.
Zee
Zee
01:40
@Daminark you eat everything up, you gotta learn to become a rebellious genius like me
@Zee Funnily enough, people said that computers are "just adding machines" a long time ago. I think the fact that it's advanced so far is a demonstration of the power of computing.
huh
did i say something not precise enough?
No, Zach, I was agreeing with you, silly.
:)
Oh, I thought you meant you don't understand my question either
Heh.
01:42
Join my rebellion of not caring what's "good for society" and only studying higher topos theory!
smacks Demonark
So I had to first learn what a bijection was.
A bijection is just where you can "pair" each element together
That sentence sucks.
And also in philosophy where we discuss the metaphysical validity of the ZFC axioms
Zee
Zee
01:43
@Daminark you see, I lied, I don't care what society needs, but the issue is still there, why do you want to do topos theory? couse it's cool right now?
Both for syntax and for math.
You know what else does
Because I don't
See now I can't trust you anymore Zee :'(
And I don't yet have any serious intentions of topos, I just do it for the memes
I will need to think about this problem for some time.
Nobody ruin it for me plex.
Zee
Zee
@Daminark well I lied couse we all lie to ourselves
01:45
Nate do you at least know what it is now
Yes each element in a set is paired to a corresponding element in another set
no element is left unpaired.
I'll need to learn more math before deciding which direction I want to take it, if I had to make as good a guess as I could now, probably something in the direction of algebra, likely enough trying to mash it (and logic) into complexity theory
Good
Now show you can pair up each integer with a rational, uniquely
Well
Find some kind of "pairing"
Nate: I think Zach's question is rather hard, unless you interpret it loosely. You might try to give a function from $\Bbb Z$ onto $\Bbb Q$ without insisting it's one-to-one, or you might try to give a one-to-one function from $\Bbb Q$ into $\Bbb Z$.
(e.g. there are professors at my school who combine algebraic geometry, rep theory, and logic into complexity theory)
Zee
Zee
01:46
@Daminark do you know enough about logic and complexity? Or is it couse they sound sexy?
Oh there's this nice paper I found with an explicit enumeration of $\mathbb{Q}$, as opposed to relying on Cantor-Bernstein
Do you think from $\Bbb Z$ to $\Bbb $Z^2$ would be easier, Ted?
@TedShifrin ah, thanks for the tip.
@Zee using those terms precisely are a bit lewd for me, one time one of those professors subbed in a class I was doing on algorithms in finite groups, and he talked about it some. I liked that a whole lot
Yes, Zach, and even better to restrict to positives.
01:48
Yeah that's what I was thinking
@Dodsy Try just doing $\Bbb N$ to $\Bbb N^2$.
That is, the naturals, to ordered pairs of natural numbers
No, ordered pairs
Logic, I only know that at one point I saw a proof of Ramsey's theorem via ultrafilters
Which I also thought was neat
The plan is to take logic next year, might have to audit complexity theory due to not having much class space
Wait for the Z to Q
could I have done 1/1, 2/1, 3/1, 4/1, etc?
or does the numerator < denominator
That's one-to-one, but far from onto.
ah, i see.
To avoid sounding dull, can someone explain "onto" in this context.
Zee
Zee
01:51
@Daminark i guess we have different tastes, I can't imagine being interested in those things, they are too ... non-visual
This is all a projection (linear algebra intensifies), when I take those classes I'll get back to you on whether they're my thing or not, but right now I'm starting to lean the general direction of algebra and theoretical compsci over, say, analysis
@Dodsy Everything in the codomain is mapped to.
Anyway, enough metamath and back to math itself
I.e., you hit every value in $\Bbb Q$ with your function.
Or, in this case, every rational is paired to some integer.
Yeah.
01:52
at least one
Zee
Zee
@Daminark I'll let you be, math is not a convo but a private activity
But I can't just say, 1 = 1/1, 2= 2/1
Right. Someday I'll be careful and clear.
oh it's a function
I see.
so $f(x)=x/1$ ?
Right, Nate.
01:54
I don't know how I missed that function part.
I guess the word was never used.
No it was, for sure.
@Zee I also don't really agree with that but again, I think that's enough meta for now
Mm.
"Nate: I think Zach's question is rather hard, unless you interpret it loosely. You might try to give a function from ZZ onto QQ without insisting it's one-to-one, or you might try to give a one-to-one function from QQ into ZZ."

-Ted Shifrin 2017
01:55
But yeah so, the idea behind a direct bijection between $\mathbb{N}$ and $\mathbb{Q}_{\ge 0}$ is to say alright, map 0 to 0, then write a tree
Hm, so how does that work, Daminark?
$\frac{1}{1}$ is the starting point, then given $\frac{r}{s}$, the left subtree starts with $\frac{r}{r+s}$ and the right subtree starts with $\frac{r+s}{s}$
Are you quoting me for a reason, Nate?
yeah @Fargle claimed that I was never told it was a function.
I was wrong.
01:57
:)
I also guessed, lol
hey whats up everybody what is this place?
I've yet to read the paper fully, but it turns out this process hits every rational number exactly once, in reduced form
Ohh... Zach was vague in his phrasing, admittedly.
@Davehuff Ostensibly, a math chat room.
01:58
Pinning the blame on me?
@Daminark I was actually about to write $f(x)=x/(x+1)$
Don't give it away, Demonark. Do they give an explicit formula?
You're still only getting a tiny number of rational numbers, Nate.
hm.
Why not, Zach? You always blame me when you can. :)
@Fargle thanks
01:59
Sometimes more ostensibly than others.
A lot of smaller questions get answered here when it is a math chat room.

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