Mathematics - 2017-06-25 (page 4 of 5)

Astyx

19:00

@Abcd sketchtoy.com/68178191

Abcd

@Astyx Not clear.

Astyx

I have no idea what you're asking for then

Akiva Weinberger

Let $R$ be the ring of casual power series

Abcd

@Astyx See: We have three dimensional graph of time, position and velocity right?

Now, on this graph how do I multiply time with velocity to obtain displacement graphically?

Dodsy

astyx

what math did you self study

you french genius

Astyx

19:07

I'm not a genius (That's so true that I almost wrote "I'm not french" instead :p)

Manifolds mainly, although that's not finished yet

Abcd

@Astyx Are you from France?

Kirill

@Astyx what do you do to let the vector space $L(V)$ to become an algebra? (I do not know algebras). I used to see vectors in vectors spaces, and polynomials in polynomial rings e.g. $K[x]$, but it seems that you have all this together in $L(V)$.

Semiclassical

hi chat

Astyx

Some topology

Yes @Abcd

Akiva Weinberger

@Abcd He is

Sha Vuklia

19:08

I have to consider $f\circ g^{(1)}$

so never mind!

Balarka Sen

@AkivaWeinberger Say $A, B$ are 2x2 matrices with integer entries, with $A, A + B, A + 2B, A + 3B, A + 4B$ all invertible and inverses having integer entries. Prove that $A + 5B$ is also invertible with integer entries.

Kirill

@Semiclassical Greetings

Sha Vuklia

hi @Semi

Astyx

Some number theory and some analysis about measure theory too

Abcd

@Astyx Ok. D'you have any answer to my prev. question?

Hello @Semiclassical

Astyx

19:08

Nothing too advanced really, I'm looking forward to doing this next year

Semiclassical

@BalarkaSen Hmm. Is it true for any $A+nB$ at that point?

Balarka Sen

Yup.

Astyx

I doubt one can do what you're asking for @Abcd

Akiva Weinberger

So this is the same as saying it has determinant 1 @BalarkaSen

Balarka Sen

It's not particularly hard. There's a hint that makes it really obvious.

Semiclassical

19:09

Neat.

Akiva Weinberger

if it's 2x2

Balarka Sen

Yeah.

Well, or -1

Semiclassical

b/c adjugate, yeah.

Abcd

@Astyx So all these calculations are merely theoretical?

Astyx

I don't follow you anymore @Abcd sorry

Dodsy

19:10

Thanks astyx

for your reply.

Abcd

It's alright.

Astyx

@Kiril You see that for $f,g\in \mathcal L(V)$, $\lambda\in \Bbb F$ you have $f+g, \lambda g, f\circ g \in \mathcal L(V)$

Dodsy

sorry I'm busy dying of alergies

Kirill

@Astyx you define composition as a multiplication on the vector space? And then $P(f)$ also lives in $L(V)$?

Semiclassical

Hmm.

Astyx

19:11

Often we forget about the $\circ$ sign and just write $fg$

Note that I'm not being very specific nor formally rigorous

That's right @Kirill

@Dodsy At least die quietly :p

Kirill

@Astyx so there is no polynomial ring on $K$?

Dodsy

haha

will do

Astyx

$K$ ?

Dodsy

:)

Semiclassical

If $A^{-1}\in M_2(\mathbb{Z})$, then $A^{-1}(A+B)=I+A^{-1}B$. So $A^{-1}B$ is invertibe with integer entries as well.

Akiva Weinberger

19:13

We can assume $A=I$, then

Balarka Sen

If I + A is invertible, that does not mean A is invertible

Kirill

@Astyx haven't we set $V$ as a $K$-vectorspace/$K$-algebra?

Semiclassical

Hrm, fair.

It does have integer entries, though.

Balarka Sen

Sure, agreed.

Akiva Weinberger

So we know $I$, $I+A$, …, $I+4A$ are invertible and we want to show that $I+5A$ is

Astyx

19:13

I used $\Bbb F$, but $K$'s good for me

Dodsy

oh

you guys were talking about the Dodsy conjecture?

it is very similar eh?

Akiva Weinberger

since we can do that substitution

Dodsy

shifted up one number

Semiclassical

@Dodsy Yeah.

Dodsy

that's crazy

Akiva Weinberger

19:14

Yeah, it's the same as Collatz @Dodsy

Semiclassical

I mean, it depends on how you write the Collatz map.

Dodsy

that's crazy actually...

Akiva Weinberger

(I think I was the one who noticed it)

Semiclassical

yeah, you were

Dodsy

Wow

Really nuts actually if you think about it

Semiclassical

19:15

eh, it's not entirely shocking in retrospect. if memory serves, the way you got it was by modifying the Collatz map in a specific way

Dodsy

I randomly made up the "dodsy conjecture" (no better name to call it)

right

Astyx

There is a polynomial ring on $K$ @Kirill

Dodsy

I moved the +1 to the other side

and flipped the even and odd

Kirill

@lets use $\mathbb{F}$, $K$ denotes $\textit{Körper}$$\equiv$Field in German. So there is no $\mathbb{F}[x]$

?

Semiclassical

sure. but note that by changing the +1, you made odd stuff into even stuff and vice versa

Dodsy

19:15

exactly

Semiclassical

so that would flip even/odd on its own

Astyx

We also use $K$ in France

Balarka Sen

@AkivaWeinberger Hm, what if I write $C = I + A$ and then write $I + 2A = C + A$, replace it by $I + C^{-1}A$, etc?

Astyx

For Corps

Dodsy

very strange

Semiclassical

19:16

so in doing it a second time you effectively reversed it

Dodsy

right.

Astyx

What do you mean there is no $\Bbb F[X]$ ?

Dodsy

well that could actually lead to some deeper insights into the collatz conjecture, though it remains to be seen.

Astyx

Tomorrow my exams start, not sure how I'm supposed to feel

Semiclassical

still kinda funny that you modified it in just the right way to get collatz back.

Dodsy

19:16

haha yes.

Balarka Sen

Which exams?

SteamyRoot

Ummm... try feeling relaxed?

Dodsy

it was completely by chance.

Astyx

Oral exams for schools I applied to

Dodsy

i am by no means a genius.

Balarka Sen

19:17

Ah ok

Dodsy

I will have to meditate on this for some time.

Kirill

@Astyx you say $P \in \mathcal{L}(V)$. Why not in $\mathbb{F}[x]$?

Astyx

No no no

Semiclassical

pretty sure no one in here is actually a genius, we're just good at temporarily pretending to be such.

Astyx

$P(f) \in \mathcal L(V)$

while $P\in \Bbb F[X]$

I hope I'm not a genius for the sake of humanity

Semiclassical

19:18

Same.

Kirill

@Astyx now perfect. I get it.

Balarka Sen

I am a pretentious hipster. Pretending is what I am good at.

Perturbative

Has anyone read the construction of tensor products of modules in Dummit and Foote?

Semiclassical

@ShaVuklia done with a physics for a bit?

Dodsy

It's interesting that it technically is the collatz conjecture though

really.

this is freaky.

I could be famous!

Astyx

19:20

So you get what the kernel lemma is stating ?

Kirill

@Astyx so the kernel of that endomorphism is a direct sum of the kernels of $P_i(f)$.

Astyx

That's right

Kirill

seems fair

Semiclassical

Just to make explicit what's going on:

Astyx

Now write $P = \prod_{i=1}^p (X-\lambda_i)^{\mu_i}$

Semiclassical

19:21

Collatz says $k\to 3k+1$ if $k$ is odd, and $k\to k/2$ if $k$ is even.

Astyx

Where the $\lambda_i$ are distinct and the $\mu_i$ are integers

Kirill

@Astyx should I interprete lambdas as zeros of the polynomial or take them as eigenvalues?

Semiclassical

note, though, that if k is odd then 3k+1 is even. so I know that the next step would be $3k+1\to(3k+1)/2$.

Astyx

zeros of the polynomials yet

Kirill

ok

Astyx

19:23

Then if $P(f) = 0$ (as an endomorphism), we have $V = \bigoplus_{i=1}^p \ker(f-\lambda_i id)^{\mu_i}$

Semiclassical

So I can equivalently take the Collatz map to be $k\to (3k+1)/2$ if $k$ odd and $k\to k/2$ if $k$ even.

Astyx

Which means the eigenvalues of $f$ are among the $\lambda_i$ (however not every $\lambda_i$ is an eigenvalue in general)

Semiclassical

For the Dodsy map, I instead have (writing $j$ instead of $k$ to make things convenient in a bit) $j\to 3j/2$ if $j$ is even and $j\to (j+1)/2$ if $j$ odd.

Astyx

That last statement is true for the minimal polynomial and the caracteristic polynomial (which cancels $f$ by Cayley Hamilton)

Semiclassical

...I feel like I've mixed something up, though. hang on a bit

Astyx

19:26

Now if a polynomial of which all the roots have multiplicity 1 cancels $f$, then $f$ is diagonalizable since $V = \bigoplus_{i=1}^p \ker (f-\lambda_i id)$

(In particular for $f$ a projector, $X(X-1)$ cancels $f$, so $f$ is diagonalizable)

Kirill

@Astyx there is a basis of $V$ composed of the eigenvectors of $\varphi$?

Astyx

of $f$, but yeah

Kirill

:)

Astyx

At least if the field is algebraically closed

For instance, in $\Bbb R$ you could have an endomorphism of which the minimal polynomial is $X^2+1$

It is therefore not diagonalisable

Semiclassical

Dodsy: 19 -> 10 -> 15 -> 8 -> 12 ->18...
Collatz: 18 -> 9 -> 14 -> 7 -> 11 -> 17...

Astyx

19:29

The rule is : If the minimal polynomial is spearated and has simple roots, you can diagonalize it

If it is separated you can trigonalize it

Semiclassical

If I take $j=k+1$ and plug this into the statement of the Dodsy map, that becomes

Kirill

@Astyx wht does separated means?

Astyx

You can write it as a product of degree 1 polynomials

Not sure about terminology, at all

lattice

Say we have topological spaces $X$ and $Y$, a continuous function $f:X\to Y$, a sheaf $\mathcal{F}$ on $X$ and a presheaf $\mathcal{G}$ on $Y$. Let $f^{-1}\mathcal{G}$ denote the inverse image of $\mathcal{G}$ under $f$, and let $f_*\mathcal{F}$ denote the direct image of $\mathcal{F}$ under $f$.

We said in our lecture that $f^{-1}$ ($f_*$) gives raise to a functor from the category of presheafs on $Y$ to the category of sheafs on $X$ (from the category of sheafs on $X$ to the category of sheafs on $Y$)

Semiclassical

$k+1\to 3(k+1)/2=(3k+1)/2+1$ if $k+1$ is even and $k+1\to (k+2)/2=k/2+1$ if $k+1$ odd

lattice

19:32

I don't really understand how these functors map morphisms

Kirill

@Astyx I have heard this rule only according to the characteristic polynomial

Semiclassical

or, if I shift the input/output down by one for each and write the conditions in terms of $k$ itself:

$k\to (3k+1)/2$ if $k$ odd and $k\to k/2$ if $k$ even. Which is the collatz map.

Astyx

Actually these are equivalences for the minimal polynomial @Kirill

Semiclassical

So Dodsy and Collatz are equivalent.

lattice

and I'm also wondering, are these functors inverse to each other?

Kirill

19:34

@Astyx why $P(f)=0$ implies $V=$[direct sum]?

@Astyx where does the $id$ comes from?

Astyx

What's $\ker P(f)$ then ?

We have $f^0 = id$

$f-\lambda$ wouldn't make any sense

Kirill

@Astyx all the elements that goes to 0 under $f$

@Astyx the set of all these elements

Astyx

$id$ is the unit of the algebra $\mathcal{L}(V)$

@Kirill I meant what is it when $P(f) = 0$ ?

Kirill

@Astyx then $\prod_{i=1}^{p}(f-\lambda_i)^{\mu_i}=0$, but further?

Ivan Ivanov

What do I have to do in order to reduce problem A to NP-complete problem B ?

Astyx

19:37

Are you sure you understand what $P(f)$ is ? @Kirill

Kirill

@Astyx no.

@Astyx it is an element of $L(V)$ first

Astyx

If $f\in\mathcal L(V)$, you agree $f^0, f^1, f^2,\dots$ are all elements of $\mathcal L(V)$

Kirill

@Astyx if it is held by the definition of the algebra, then sure

Astyx

Same for $\lambda f$

No no no

It is just composition

Balarka Sen

@Daminark I'm learning the proof of Whitehead theorem.

Kirill

19:40

@Astyx the composition of linear maps is also linear, yes

Astyx

I'm just saying if you compose two linear maps from $V$ to $V$, you get a linear map from $V$ to $V$

Therefore you can make sense of $a_0 f^0 + a_1 f^1 + \dots + a_n f^n$ right ?

(note $f^0 = id$)

s.harp

Are path spaces always contractible as long as the base is path connected? Ie is Hausdorff important?

Kirill

@Astyx is it like $f \circ 0$ or how can you write it analogous to $f^2=f \circ f$?

Astyx

No, $f\circ 0 = 0$

You just define it as such

Kirill

ok

Astyx

19:42

Just like you define $n^0 = 1$

Kirill

so , it makes sense, yes.

Astyx

You want $f\circ f^n = f^{n+1}$ meaning for $n=0$ $f\circ f^0 = f$

So you'd want $f^0$ to be the identity

Anyway, so if we have $P(f) = 0$, what's $\ker P(f)$ ?

Daminark

@Balarka that only a homotopy equivalence between CW complexes can induce isomorphisms of all the homotopy groups?

Kirill

whole $V$ @Astyx

Astyx

Why ?

Balarka Sen

19:46

@Daminark Yup, weak homotopy equivalence <=> homotopy equivalence for CW complexes.

Kirill

@Astyx as I see there is nodieffference what $v \in V$ I take, as it is always 0. Or - i am totally wrong again.

s.harp

I once mistakenly called whitehead theorem the white-man theorem

Astyx

Yes, we have for all $v\in V$ that $P(f)(v) = 0$

I'll have to go to bed soon

Daminark

Noice

Balarka Sen

wanna jump in? :P

Daminark

19:48

I'm going through Concise rn actually (though I'm still like, on chapter 1)

Oh sure, yeah

Now would be the right time to figure out exactly what a CW complex is

Kirill

@Astyx me too. ok, it is independent from $v$, but why $V$ is a direct sum?

Astyx

Well we stated that $V = \ker P(f) = \bigoplus_{i=1}^p \ker(f-\lambda_i)^{\mu_i}$

Kirill

oh man am I slow

x386

Astyx

Don't say that, it really is quite a conceptual shift from what you're used to do in linear algebra

Balarka Sen

@Daminark Suppose $X$ is a space constructed as follows. Start with a bunch of dots with discrete topology; this is the 0-skeleton $X^0$. Glue intervals so that the endpoints of the intervals are glued to the dots in $X^0$.

The resulting thing is a graph; this is the 1-skeleton $X^1$

Glue disks $D^2$ to $X^1$ by consider loops $\partial D^2 \to X^1$ in $X^1$, and gluing the boundary of $D^2$ along those loops. The resulting thing is a cell-complex; this is the 2-skeleton $X^2$.

Inductively construct $X^{k+1}$ from $X^k$ by gluing $D^{k+1}$'s along maps $\partial D^{k+1} = S^k \to X^k$ (representatives of $\pi_k(X^k)$, I guess you could say).

Kirill

19:55

@Astyx i think i get the plot, but not the details, as I cannot re-tell them myself. I do not see still, why the eigenvalues of $f$ should be among the $\lambda$-s. And why did we use the kernel lemma, actually.

Astyx

Note, I did not prove anything

Balarka Sen

The resulting "CW complex" is the "limit" of the diagram $X^0 \subset X^1 \subset X^2 \subset \cdots$, which is easy to describe: $X = \bigcup_{k \geq 0} X^k$

Kirill

@Astyx for what stood $X$ in $X(X-1)$ by the way?

Astyx

I think the eigenvalue thing is something you can do youself

@Kirill The unknown of the polynomial

Kirill

@Astyx sure, I hope so. As I have learnt a lot of new things with you today, I hope they will accumulate somewhere in the back part of my brain. I still have to use all thes to get the equation from a trace of $\varphi$ to the dimension of the kern.

@Astyx thanks a lot!

Astyx

19:58

Glad to share

Daminark

OK so wait just to be clear about the next level of this construction, we'd get as a 2-skeleton something like, let's say a filled in square? @Balarka

Kirill

@Astyx was it enough to show that the eigenvalues are 1 and 0?

Astyx

No

Balarka Sen

@Daminark Right. In the second step you are filling in various loops in the graph.

Kirill

@Astyx I mean do you think I can show it using the things we have discussed?

Astyx

19:59

You needed to show that we had $V = \ker f \bigoplus \ker (f-id)$

Not sure I understand what you mean by that

Balarka Sen

Here's an exercise, @Daminark: Can you describe the torus as a CW-complex?

Astyx

Bye chat

Balarka Sen

Of course, it'd be a cell complex, because there would not be 3-cells in it.

Bye @Astyx

Kirill

@Astyx you said that the projection has the eigenvalues 1 and 0. I want to show it on the paper. That is it.

@Astyx bye, Astyx. Thank you again!

Paweł Kusz

Hello :) I have a questions :) We have euclidean metric. A is connected $\rightarrow $ A is path connected. It is true?

Daminark

20:02

We can be selective about what we fill in, right? And in that way get a union of a discrete set, a graph, and a 2-skeleton

Balarka Sen

Yeah, of course, fill in a specific set of loops.

As you want

Well, dimension 2 CW complexes are called "cell complexes", rather

in general if $X$ is already a CW complex, the truncations $X^n$ are what we call the $n$-skeletons

Daminark

Oh I thought a cell complex was 2 and up

Balarka Sen

Ah, no. A cell is a disk of dimension 2 :)

n-cell is D^n, which is what you use to fill in stuff in X^(n-1)

Daminark

But alright, so I'm thinking we can do something like, take maybe 8 points?

Use 4 to draw a circle apiece, then connect each point to its corresponding one in the other circle and fill those in

Balarka Sen

That works. Can you give a CW complex structure with 1 vertex?

That is $X^0$ is exactly a singleton? :)

Daminark

20:06

For the torus? Or just in general?

Balarka Sen

For the torus.

Basically I want the simplest possible CW structure.

@Daminark Maybe I should explain this one; think about the fundamental square description of the torus, alright?

Daminark

Maybe thinking about it as a square?

Balarka Sen

damn snipd

Daminark

Sniped

Balarka Sen

sniped!

Daminark

20:09

Darn

Balarka Sen

lol

Daminark

But yeah you take the square, roll it into a tube, and then roll that into a torus, I'm just trying to think of how to formalize that precisely

Do we have to have just a graph for $X^1$ or can it be a multigraph?

Balarka Sen

Formalizing, as in? The square description can be formalized by saying $T^2$ is a quotient space of $[0, 1]^2$ by an appropriate relation.

Oh, it can be a multigraph.

That means you have edges with the same endpoints, right? Like loops starting and ending at the same vertex?

Daminark

Yeah

Because then you can sorta draw a square, but then double each edge and that should do it

Balarka Sen

Yup. It's basically a bunch of points with edges going between them in whatever way possible.

Not sure if I understand that.

Daminark

20:13

Wait I dunno if that works now

Ivan Ivanov

Hello can someone clarify how to prove that a problem X is reduced to NP-hard problem Y ?

Balarka Sen

@Daminark Here's a small hint. Suppose you take the square, identify pairs of opposite edges in the same orientation. The "interior" of the square (the open bit) is an open disk. What is the "boundary" of the square, as a subspace of the torus?

What do you get from the boundary of the square after all the relevant identifications?

Daminark

So wait are we taking just a raw square?

Balarka Sen

Yes, just a $[0, 1]^2$

Daminark

Oh weird, I thought you needed more than that

Balarka Sen

20:18

You actually don't. The answer to my question should explain why.

This puzzled me when I was thinking about CW complexes actually.

Daminark

But alright, so now we're giving it the orientation where the vertical edges both point, say up, and the horizontal ones to the right, as opposed to the normal orientation

Balarka Sen

Ya

Daminark

Oh god so we've upgraded to digraphs. Well, will the boundary give you like, two circles on the torus which intersect at a cross or smth?

Balarka Sen

Yes! A longitude + meridian on the torus.

Which is a figure 8

Daminark

Wait what? I don't see the figure 8 yet

Oh I mean

I guess yeah if you allow it to move off the torus

Sure

Balarka Sen

20:22

Yup.

So torus is a 2-cell (interior of the square) glued to the figure 8

And this, folks, is precisely a CW decompositon of $T^2$

Daminark

Wow, this is bizarre

Balarka Sen

What confused me a lot in my Hatcher Ch 0 days: 2-cell has boundary $\partial D^2 = S^1$. What is the loop $S^1 \to "\infty"$ that we are gluing the 2-cell along?

Can you explicitly describe this loop?

Daminark

O lawd

That's... one way of saying it, I suppose

Balarka Sen

lol

BTW, $S^2$ has a dumb CW complex structure. So does $S^n$.

(with a single point as 0-skeleton)

Daminark

I'd say take two points, draw an equator, then fill both sides?

That's the first easy one

Balarka Sen

20:29

There's a dumber one.

Daminark

Oh right wait a second I forgot you're allowed to have a loop

Balarka Sen

:)

Daminark

I wonder how much you'd lose if you forced it to strictly be a graph

No loops, no multiple edges

Balarka Sen

Not much, I don't think. If you have a loop, you can just draw a point somewhere in the middle to get no loop there. So any graph is a no-loop graph.

=> any CW complex can be given a CW decomposition with no loops in $X^1$

So like, you are just marking off a point on the loop to get two vertices there, and two edges are glued to the two vertices like this: ()

Daminark

That checks out

Balarka Sen

20:34

So, did you get the CW complex structure on $S^n$ with a single vertex/singleton as 0-skeleton?

John Jack

Typhon

Correct me if I am wrong but when solving differential equations of the form y' + yQ(x) = P(x) the coefficient to multiply by both sides is the product integral of Q, right? I could've sworn it was... and have thought that for some time. Now I'm doubting that assertion?

Daminark

Take a point, draw a loop, fill in that loop twice

Balarka Sen

@Daminark Interesting. That works; there is a still more simple one :)

Pick a point on $S^2$. Call it $p$. What is $S^2 - p$?

Daminark

Doesn't seem like a CW complex at all

Typhon

20:38

Be back in a bit.

Balarka Sen

Well it's homeomorphic to a very well known object.

Daminark

An open ball

Balarka Sen

2-d ball, or disk, rather. But yep.

So you get a CW decomposition of $S^2$ by taking $X^0 = \{p\}$, and gluing a closed disk $D^2$ to it by the constant map $\partial D^2 = S^1 \to \{p\}$

Daminark

Well, I mean you still need to have a loop if you want to fill in a cell complex?

Oh wait a second you're allowed to glue things over multiple dimensions?

Balarka Sen

Yes [said in Dumbledore's voice]

Daminark

20:41

Jeez you guys are being nice to yourselves

Sha Vuklia

@Semi haha yea two more weeks of math now. I will redo my E&M test tho, so I will do physics next week on Tuesday.

Daminark

I was under the impression that it was very rigid, like if you could only stick on a cell complex to a 1-skeleton

Balarka Sen

That's why this CW decomposition of $S^2$ is dumb. $X^0 = X^1$.

The 0-skeleton is the same as 1-skeleton, because there are no 1-cells.

@Daminark Right, you will soon realize that you can even glue very very high dimensional cells to n-skeletons for n very very small even by non-constant maps

That's the yoga of the Hopf map: that's a map $S^3 \to S^2$ which is not nullhomotopic.

Daminark

Okay that's just cheating

shrug

And lol I remember Neves went through blood constructing that

Balarka Sen

Don't worry though, I am not teaching you definitions in full rigor. We're trying to understand the class of CW complexes via examples.

hahah

In any case, similar for $S^n$. $S^n = D^{n+1}/\partial D^n$, so you can just say $\{p\} = X^0 = X^1 = \cdots = X^{n-1} \subset X^n = S^n$ where the $n$-cell is glued by the constant map $\partial D^n \to X^0 = \{p\}$ sending everything to $p$.

Basically it only has cells in dimension 0 and dimension n.

Sha Vuklia

20:56

Guys, question about a "Chinese Remainder Theorem" Algorithm. So say we have $n_1,\dots,n_t$ which are relatively prime, and $a_1,\dots,a_t$. Say we let $a=a_1$. We want to find $x$ such that
$$
xn_1\equiv 1\mod n_2.
$$
Then we get
$$
a’=a+(a_2-a)xn_1.
$$
So now we have that $a’\equiv a_i\mod n_i$ for $i\in\{1,2\}$. Now my book says that we could also replace the factor $(a_2-a)x$ by its remainder after division of $n_2$. I don’t see why this is the case. Anyone any idea?

What I do see is that

replacing $(a_2-a)x$ by anything won't affect the congruence with $n_1$

so.. oh

right I can't delete it anymore

I see it now

Daminark

21:25

@Sha (removed)

Balarka Sen

Should I sleep, do more algebraic topology, or watch a movie?

Just win baby

sleep

and dream about algebraic topology :P

Daminark

Dream that you're watching a movie that Peter May made

Balarka Sen

I'm not watching a Fifty Shades movie

Daminark

Not a movie about Peter May

A movie he made

Perturbative

21:35

I laughed a bit too hard

Just win baby

at?

Perturbative

The previous two messages

Just win baby

:-D

laughing is good

Daminark

@Perturbative :P

Zee

I hate the word perturbative

Daminark

21:48

I suppose we can perturb the word and get another one which you like?

Just win baby

lol

Zee

Only if the end justifies the means

Perturbative

@Zee Eh, I'm too lazy to change my username

Daminark

@Zee You are the end in that case

We are treating you not merely as a means to end, but as an end in yourself

Zee

Filthy kantian

Dodsy

21:55

Dami how have you been?

We don't talk enough as of late.

I am sure you are very busy though

Perturbative

lol @Zee

Zee

@Dodsy you sound like my ex

Dodsy

oh...interesting @Zee

Daminark

@Zee Kek

@Dodsy True, it's been a while, I'm on the chat a bit less

Dodsy

eh me too dami

Zee

21:57

He's also less inclined to argue with me, which is making this place boring...

Daminark

Lol, at some point we'll have a showdown yet again, fear not

Kant's name shall never be besmirched under my watch

Zee

Besmirched, wtf does that even mean

Daminark

Like, damaged reputation, I guess

(Damage from the previous height of having perfectly correct moral philosophy :P)

Transcript for

$Mathematics$ Mathematics