If X is finite, and A is a subset of X, is A cofinite? Or is cofiniteness only defined for infinite sets?

The definition of co-finite as I understand, for universal set $U$ we have $A$ is cofinite if and only if $A^c$ is finite.

If $X$ is our universal set in this case, and if $X$ is finite, you have $A^c = X\backslash A$ and therefore $A^c\subset X$. In particular this implies that $|A^c|\leq |X|<\infty$ by assumption.

So, yes, for any finite universal set, all subsets are finite and cofinite.

completely unrelated topic: I remember reading something a lot time ago on the subject of "tracing needles along paths." Given any continuous f(x), it is always possible to move a needle from one end of the graph to another, with the requirement that both endpoints of the needle be on the graph at all times. I've tried to Google this to no avail, so I don't actually know if it is a real thing or not.

@PhiNotPi: wtf

Define "needle" and "endpoints of the needle." You mean to say like a sewing needle being laid flat on the piece of paper, with the spot you put the thread in and the pointy tip both touching the graph representation of the function (the curve)?

Okay, I guess the way I explained it made no sense?

He means the needles can be bent @JM

Even with a rigid needle., yes you can. It should follow directly from the definition of continuity.

I'll try to explain what I am talking about some other way, I guess....

This is literally something I randomly remembered but certain haven't seen anything about in a very long time......

MS Paint to the rescue.

How do I show that (z+w)*(1+(conjugate z*w)) is a real number given the fact that |z| = |w| = 1 and that zw does not equal -1?

$(z+w)\cdot (1+\overline{zw}) = z + w + z\cdot \overline{zw} + w\cdot \overline{zw} = z + w + |z|^2\overline{w}+\overline{z}|w|^2$

oh wait, i messed up

^ Possibly helpful MS Paint drawing

@Mathy it winds up being then equal to $z+\overline{z}+w+\overline{w}$

The idea being that not matter how crazy the curve is, a needle can work its way through it, even if it requires backtracking to pass through the tighter curves.

How did you get that? @JMoravitz?

and yea, @PhiNotPi the one on the left is a function, the one on the right isn't.

I used z = a+bi , w = c+di but I got the messiest expression

and again, it should follow from the definition of continuity

@Mathy note first that $z\cdot \overline{z} = |z|^2$, and also that $\overline{zw} = \overline{z}\cdot \overline{w}$

I don't see how it follows from continuity. Perhaps a bit of background: I'm a high school student in AP Calc. I've only ever seen the epsilon-delta definition of continuity.

And yes, I know the thing on the right isn't a function.

@PhiNotPi Try approaching it from the contrapositive. Suppose instead that you couldn't do it. Then there exists some point at which there is not somewhere at radius $r$ away from the point on the graph.

@Jmoravitz: Got it now. Thanks.

hmm.. I'm beginning to wonder though whether it will work in certain pathological cases, such as the cantor function (the devil's staircase), where it is technically continuous, but clearly skips values.

Morning

Hi mike

How're ya?

Doin fine personally. My students have a test tomorrow. Basic set theory and counting methods (combinations, permutations, the like)

I wish them luck.

I wish you luck.

That too.

Well, they seem to be understanding., or at least the ones who are talkative seem to be.

It's always the quiet ones...

I have had around a third of the class show up sometime this month to office hours, so thats a good sign

Curiously, my office hours are usually empty. I'd like to take that as a sign that my discussions are all they need to succeed, which is totally the correct interpretation.

<the moment you realize you're the only person in the room who isn't a teacher>

Under some definition of teacher.

Is $(M \otimes N)/(P \otimes Q) \cong M/P \otimes N/Q$ for $M, N$ $R$-modules, $P \leq M$ and $Q \leq N$?

Indeed @MikeMiller @PhiNotPi, I'm technically a teaching assistant, but I do have my own class I'm responsible for.

@SamuelY No. Pick $M=N=\Bbb Q, P=Q=\Bbb Z$, all as abelian groups = $\Bbb Z$-modules.

(That this is a counterexample involves some calculation, of course, but it's a good exercise.)

hmmmmm

so we're left with $(\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q})/\mathbb{Z}$, I guess

Hello!

hi

@cand hw are you?

good

@SamuelYusim You should be able to compute $\Bbb Q \otimes \Bbb Q$. It's a group you know well.

cool

I guess it's probably just $\mathbb{Q}$

and I guess the other side is 0 so there we go

well, guessing's no good. but your guesses are correct, so you oughtta prove them.

alright well take $a/b \otimes c/d \in \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$. then we can write $a/b \otimes c/d = 1/b \otimes ac/d = 1/b \otimes bac/bd = 1 \otimes ac/bd$ so there we go with that

I don't see how that proves it. You need to use the definition of the tensor product somewhere beyond just rearranging everything

eg show that every bilinear map factors through $\Bbb Q$ etc

(every $m \otimes n \in \Bbb Z_p \otimes \Bbb Z_q$ is of the form $1 \otimes a$, but that doesn't show that the above group is isomorphic to $\Bbb Z_q$)

well I guess I can do that. Just take a bilinear map $f$ from $\mathbb{Q} \times \mathbb{Q}$ to some module $P$ and notice that $f(a/b, c/d) = f(1, ac/bd)$ by the calculation above, and so we can uniquely identify the map $f$ with one from $\mathbb{Q}$ to $P$, say $\hat{f}$, such that $\hat{f}(p/q) = f(1, p/q)$.

Mhm.

And then it's easier to show that the RHS of your equation way back above was zero.

this time I really can just show any arbitrary pure tensor is 0, right

yeah

because the group is generated by them, if they're all 0, so's the group

good, I like it when things work the way I think they do

it's great when that happens

I'll suppress all the necessary $+\mathbb{Z}$'s or overlines because they're annoying to type out but here goes: $a/b \otimes c/d = 1/b \otimes ac/d = 1 \otimes ac/bd = 0 \otimes ac/bd = 0$

Da

and so now I've successfully proven myself wrong, hurray

anyway I'm trying to come up with a way to finish off my proof of right exactness of tensor products and was hoping this would do it, but it seems not

so like imagine I have $M \to N \to P \to 0$ with $f: M \to N$ and $g: N \to P$, and I take $M \otimes C \to N \otimes C \to P \otimes C \to 0$ with the obvious choices of $f \otimes \text{id}_C: M \otimes C \to N \otimes C$ and $g \otimes \text{id}_C: N \otimes C \to P \otimes C$.

mhm

to show this is exact at $N \otimes C$ my plan was essentially to barehand it by first showing $\im(f \otimes \text{id}_C) = \im(f) \otimes C$ and $\ker(g \otimes \id_C) = \ker(g) \otimes C$ but this second equality is a real mess

oh, I guess I can't use that command, that's annoying

you do need to use that $g$ is surjective, of course

(otherwise you'd have just shown that tensoring is exact, not just right exact)

yeah but when I took this route I ended up needing the "fact" that if an irreducible sum of pure tensors is 0 then each summand must be 0

and so I said screw it and thought about using the first isomorphism theorem instead somehow but I dunno about that anymore either

I don't remember how to prove this off the top of my head, but in general, I think you're relying too much on elements

The best strategy is often to think in terms of universal properties and such

so how can I use the universal property here?

I need $\varphi: \ker g \times C \to \ker(g \otimes \text{id}_C)$ which is gonna be all kinds of messed up, isn't it?

also why is there a built-in command for kernel but not for image? did anyone think this through?

@SamuelYusim Don't use the universal property.

You want to use the tensor functor is adjoint to the hom functor.

we haven't learned what a functor is in class and I don't really have a great grasp of the concept myself so that feels like a bad idea

although I've seen the trick you're talking about and I like it, I don't have the machinery.

You don't need to know the word functor for what he's suggesting

I think the point is this: do you know that $A \to B \to C \to 0$ is exact iff for any $R$-mod $M$, $0 \to \hom(A,R) \to \hom(B,R) \to \hom(C,R)$ is exact?

if so, use that $\hom(A,\hom(B,C)) = \hom(A \otimes B, C)$.

no

I don't know that

@SamuelYusim Now you know. =)

I recommend you prove that first.

Then it is easy cake.

this is the proof i remember learning now.

That proof is perfect.

I feel like this might be going against the spirit of the problem as it was given though

Don't be stubborn.

shaddup pedro

Shuts up.

well like, I don't want to be the asshole who shows up to his class with some super high powered answer to a question like this as if bragging to the prof about how smart he is

@SamuelYusim LOL.

we've got a couple of those people at my school already

You don't need to brag, and this is not high powered.

@SamuelYusim you might try to do it the way you think was intended and this way

hi

@SamuelYusim The fact that a sequence is exact iff the hom sequence is exact for every $D$ is actually very important. It is easy to prove, but very important.

can I chat here about early math?

@gemma Sure.

You are talking about functions, are they like functions?

Functors sorry spell check

Sorry?

Oh.

Yes, they are "functions" between structures called categories.

My spell check things they are lol

You can edit your messages for up to 5 minutes.

But perhaps you're on a phone, and in that case you need to go to the menu and click "edit last".

@Pedro your easy to prove

?

The functor wiki page makes it seem impossible to learn lol

every wikipedia article about category theory is absurdly dense but really you need to know abstract algebra before you'd ever want to know any of that stuff anyway

I have dummit and foote so I will be there soon. What's after category theory?

@gemma What do you mean?

professorship @gemma

@gemma: On a more serious note, are you in college?

I mean arithmetic->Algebra->Abstract Algebra->Category Theory->?

@Don no why?

I know someone named Gemma.

@Don I won't be in college for many years lol

the idea isn't that you learn one thing so you can get to the next

I think that is exactly the idea lol

in life as well as math

Ah young one, I used to be the same way. Life is not linear...

I believe you gain abstract capacities when you are in your teen years.

not at all. where do set theory and logic come in if you already know arithmetic?

if you were going in order surely those would come first, if not something like homotopy type theory

They are in a different linear progression ;P

This is the linear progression for algebra stuffs

I don't know lol

I don't even know what homotopy type theory is

you probably shouldn't

mathematics is rarely taught (or made) in any sort of linear progression. some things just have other things as logical prerequisites.

Does anything have category theory as a logical prerequisite?

sure

What?

is it

algebraic geometry would be the next thing in your chain, I suppose

Arith->Alge->Abst Alg-> Cat The->Alg Geo. ok anything more?

lol

will this take years?

Also is there another name for Cat theory that starts with an A? the inconsistency is bugging me!

lots of people only see what comes after the basics of abstract algebra once they've already gone through 4 years of university

oh that's not good

I'd also put linear algebra somewhere in there

but If I do 6 hours a day for a few years ill probably get it

probably you'd need to do it at the same time as abstract algebra

@Samuel so what do you think about lately

You just want to break my linearity!

think about?

@gemma: that's what linear algebra is for, isn't it?

:PPP

I suspect you think about things

above you were thinking about some commutative algebra

oh, sure

I wanted to learn math so people stop thinking I am dumb lol

I'm thinking about set theory, combinatorics, field/galois theory, commutative algebra, and linear programming lately. not coincidentally these are all courses I'm taking

A friend of one of my brothers is in mensa and has an IQ of 154 and he said Math will make me think better. so I got dummit and foote and I will do 6 hours a day from now on

haha

christ knows why I decided to do this to myself but now I'm locked in

I've faith

it's all fun though so I guess it's fine

but in general, yes, if you study too much at once you leave with none

how about you?

bye I will come back after I know the first 30 pages, you'll see!

dummit and foote is hard and the first 30 pages are particularly hard

especially since they expect that it's review for you

i will do them within a week!

alephsub0.files.wordpress.com/2014/02/135_course_notes_f12.pdf try this first?

for someone who cares so much about prerequisites you sure are intent to skip them, @gemma

Fine I'll do 100 pages of that one within a week then!

I'll finish this one is 30 days of 6 hours a day I believe, and then D&F

best of luck with that, I don't think it's something I could've read before I was maybe 17

I'm trying to prove that given $a,b,c\in\mathbb{Z}$. If $a^2+b^2=c^2$, then $a$ or $b$ is even.

by contradiction

so $a$ and $b$ are both odd and $a^2+b^2=c^2$

then it's trivial to argue $c$ is even, but I'm having trouble constructing a contradiction

hint: modular arithmetic

oh I didn't consider it, I'll look at it again

@SamuelYusim i'm learning gauge theory and kirby calculus now, which are both methods to study the (differential) topology of 4-manifolds. on the side I'm trying to pick up some PDE theory.

sound pretty cool

I claim they are

the only thing I know about 4-manifolds is that they're way more messed up than n-manifolds for any n not equal to 4

they get a bad rap

the difference between the smooth category and continuous category is much different there, and the failure of h-cobordism is rough, but otherwise they're pretty well behaved (if you restrict to simply connected)

is that good?

oh wait i screwed up f is still even

pardon the list of adjectives, but topological, closed, simply connected 4-manifolds have a very nice classification

you don't have to use foil if you want to do it nicely

@gemma

the interesting questions, then: which (adjectives) 4-manifolds are smoothable? (open. the last remaining question here is the widely believed 11/8 conjecture, which is written down on the above wikipedia page), and for any given (adjectives) smoothable 4-manifold, can we classify all its smooth structures? (pretty much wide open. we don't know the answer for the 4-sphere.)

i think there are some manifolds we actually know the answer for, maybe

but not many

what is foil?

hi

sorry, you don't need to square any binomials

it's just a stupid phrase we're taught in school here for doing that

i don't know what a binomial is either lol

@SamuelYusim I didn't see any obvious way modular arithmetic helps me show $c$ is anything other than even

can someone tell me where this came from

hell (typographically speaking)

also @MikeMiller I need to admit I don't know any topology beyond the definition of a topological space, do you'll have to forgive me for not really understanding

sp*

so*******

no forgiveness needed

when do i learn topology?

once you know calculus and set theory

so beforeish abstract alge?

there is no order

prerequisite wise there is?

@SamuelYusim What construction were you thinking of?

as I haven't found anything more novel than $1+1\equiv 0\mod 2$, as before

@GBeau Look at it mod 4

okie

yeah do that

$a^2+b^2\equiv 2\mod 4$ but $c$ is even so $c^2\equiv 0\mod 4$ ?

$2\not\equiv 0\mod 4$

aye

squares are either 0 or 1 mod 4

man the first guy who thought of that must have been pretty smart

thanks

I suppose the easiest contradiction to $\nexists a,b\in\mathbb{Z}, 21a+30b=1$ then is over modulo $3$ >.<

(this is my writing), I'm making a logically correct proposed contradiction, right?:

as in, nothing negative :<

you haven't shown anything yet but I suppose that's one way to go

right I was just checking the method (the question is "prove by contradiction" and just a list statements)

alternatively just prove it constructively by noting first that x/2 < x if x is positive

I was going to let $y=\frac{x}{2}$ to contradict $y\geq x$

which is the same as the constructive proof, honestly :P

yep

in fact you don't even need to know x/2 < x for that proof

elaborate?

just say $x/2 \geq x$, as you did and use some algebra: $x \geq 2x$ so $0 \geq x$, contradicting positivity of $x$.

oh I see

that's more rigorous anyway

I wonder how long I would have been stuck on the 21a+30b=1 problem without being framed by the modular problem beforehand

I quickly understood it as untrue from gcd theorems

(as that would imply gcd(21,30)=1, which isn't true)

but didn't want to invoke that knowledge to solve it (but the modular idea is really the same thing)

Morning, @Ted

@SamuelYusim There's no way both of these problems are here by coincidence

Hey. Quick Question: would this be a good book for entertainment and a little overview of category theory? fef.ogu.edu.tr/matbil/eilgaz/kategori.pdf

It is for high schoolers like me!

@JulianRachman Why do you insist on reading category theory? I mean, it's not that reading it is something bad for your health or anything, but there are lots of things to learn before category theory.

Personally, I think it'd do you good to read Tom Apostol's Mathematical Analysis.

I know. I am taking a course on Analysis at a college

@Pedro i am sorry if I am thinking about these things but I was referred this book

I'm running out of ideas that don't involve listing a bunch of even and odd cases D;

rip length of this proof

@GBeau You can type TeX here.

>.<

See \LaTeX in chat on the right.

@GBeau Expand your denominators so you get an equation of the form $x^2+y^2=3z^2$, where $(x,y,z) = 1$. Take everything mod 4. Note that the RHS is either 0 or 3; if it's 0, both of the LHS terms are divisible by 2, contradicting $(x,y,z) = 1$. But 3 cannot be written as the sum of two squares (mod 4).

I guess it's easier to just consider the cases of $a^2+b^2=3c^2$ which is shown to be untrue by the "modulo 4" comparison in every case except all of them even

Aye.

and the last case contradicts reduced fraction

but not trivially

if all 3 are even, then a or d is even and c or b is even and b or d is even

$\implies$ at least one of the fractions isn't reduced

Working too hard. If you have an integer soln to $a^2+b^2=3c^2$, you have one such that $(a,b,c) = 1$. (Otherwise just divide out by the gcd.) So assume this and derive the contradiction as above.

what do you mean by that notation $(a,b,c)=1$

$(a,b,c)$ is shorthand for $\gcd(a,b,c)$. You'll probably run into it a lot.

oh

Sorry for not clarifying - I got annoyed at the same notation the first time I ran into it, and I've forgotten my own annoyance :)

@MikeMiller feels weird to write out

apparently my professor had suggested to do the problem modulo 3 but I didn't see it

it seems like more work that way

is there an easier way to disprove "For every $n\in\mathbb{N}$, $2n^2-4n+31$ is prime." without finding a counter-example

well, it's not terribly difficult to come up with a counterexample; just set $n=31$

what is your intuition for the counter-example, then?

typo. whenever you see something like that, just plug in the constant term, or some multiple of the constant term

I mean, I could take the number and run, but what made you choose that?

oh, more gcd stuff :<

the resulting thing, if nonzero and not the constant term itself, will be divisible by the constant term and something else (and hence not prime)

just that the thing immediately factors 31(2*31-4+1)

right :<

"Suppose $a,b\in\mathbb Z$. If $a\mid b$ and $b\mid a$, then $a=b$."

I guess not true, because $a=-b$ is possibility

@MikeMiller @r9m James Grime and another whose name I can't remember right now, are actual mathematicians. The former is a public speaker on behalf of a project from the University of Cambridge.

And in fact, you can see here they're all mathrmaticians, physicists, science writers and similars. After all, apart from those one or two videos, they're really enjoyable. Even on the very same matter, Edward Frenkel was great in another video, and Grime himself did a good job on his personal channel.

@bourbaki do you read Bourbaki?

@Chris'ssis hi

@VincenzoOliva you know Dr james grime

He's the guy who does the group theory video on youtube

Yes and also Numberphile

is For $n\in\mathbb Z$, $2\mid n^3\iff 2\mid n$. true?

there doesn't seem to be an obvious counter-example :s

not sure how to prove the forward case

I have noticed that completely empty rooms started appearing here in chat. (No messages, just the name of the room.) Like here, here or here.

It seems somewhat strange. The rooms which are imported from comments contain some messages. I don't think that the rooms I mentioned above were created automatically.

@GBeau For any prime we know that if $p\mid ab$ then $p\mid a$ or $p\mid b$. This is called Euclid's lemma.

So for any prime $p$ we have $p\mid n^3\Rightarrow p\mid n$.

And 2 is prime number.

But for the number 2 you can prove it also without knowing anything about prime numbers.

Is there necessarily a counter example then for, say, the same thing with $4$

Simply show that if $n$ is even then $n^3$ is even; if $n$ is odd, then $n^3$ is odd.

(Generally, for small numbers, considering all possible cases is doable.)

Oh, I wasn't sure why that was sufficient but it's because we have $2$

oh, right

just like $4\mid n^2$ with the same counter-example

hi

@SayanChattopadhyay hi

u like number theory@user153330

@SayanChattopadhyay yes and no

If yes then have a look at what i have proposed

@SayanChattopadhyay where?

on the main site

@SayanChattopadhyay let me see ....

the question is that can the cube of every perfect number be expressed as the sum of three cubes

@SayanChattopadhyay \o/ you already have a complete answer

I want the proof and it does not contain that

@SayanChattopadhyay the proof would require some analytic musings, too far from my reach ...

so do u agree that answer dosent dosent contain the proof

@SayanChattopadhyay no, just some interesting computations

what do u mean

@SayanChattopadhyay computations = calculations

No does it contain proof

hi @huy

hi

@Huy hi hi

@VincenzoOliva: Why do you even care what others think about James Grime and other people from Numberphile? If you think you want to learn from them and/or their videos, do as you wish.

@huy u dont think numberphile is good

@SayanChattopadhyay: I've only seen one video and I found it rather terrible.

Which one

@SayanChattopadhyay: The one about $1+2+3+\dots = -\frac{1}{12}$.

Oh that one........then see one more that all triangles are equilateral

@SayanChattopadhyay: No, thank you.

Fine

@SayanChattopadhyay: If I want to waste my time I'd rather browse reddit.

Do u know some place to send my question

No.

Fine

@Huy: What made you think I care about what others think about them, and even more, that I want or don't want to learn from them? If you read the whole conversation you'll see I actually critisized them in the first place. I put in evidence that they are mathematicians (and other stuff) because Mike Miller seemed skeptic about it. Nonetheless, I do like them in general, especially Edward Frenkel, whose book is in my shelf, and I certainly do as I wish.

ahahahahahahaha I don't think they thought about this statistic before it was quoted: "Every year since 1950, the number of American children gunned down has doubled." (quoted from 1994)

the mathematical ignorance to reach that is astounding

@VincenzoOliva i watch numberphile amd like matt parker the most

@GBeau Obviously, the number was $0$.

:)

@GBeau Hmm, what made him say that?

@ABeautifulMind him referring to @DanielFischer or the author of the quote?

@GBeau The author.

@ABeautifulMind it was quoted for a sociology thesis, but I assume the source incorrectly rephrased it from somewhere else

@SayanChattopadhyay Yeah, he's great too.

@SayanChattopadhyay Numberphile is the most idiotest thing ever.

Greetings