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19:00
@Tobias So let $x$ be a nontrivial element of $R^*$. Then consider $A=\begin{pmatrix}x~b\\0~x\end{pmatrix}$ and $B=\begin{pmatrix}x~c\\0~x\end{pmatrix}$. Now how can I show it doesn't have to hold that $xc+bx=xb+cx$?
4
Q: Question regarding Gambler's Ruin

Little RookieConsider a gambling process $(X_n)_{n∈\mathbb{N}}$ on the state space $S = {0, 1, . . . , N}$, with probability $p$, resp. $q$, of moving up, resp. down, at each time step. For $x = 0, 1, . . . , N$, let $τ_x$ denote the first hitting time, $τ_x := \inf\{n ≥ 0 : X_n = x\}$ Let $p_x := P(τ_{x+1} <...

Need serious help here
@ShaVuklia You don't need to have $x$ on all diagonal entries
I never gambled before, maybe I should try.
oh
right
I see it
thanks
\o/
19:13
@ShaVuklia are you still here?
yea I deleted it. I realised I should try other variations
I lied
% Proof 1 at 0.36 (+ 0.01) seconds.
% Length of proof is 41.
% Level of proof is 13.
% Maximum clause weight is 45.
% Given clauses 94.

1 (all a all b all c all d (tu(a,b,c,d) <-> (exists e exists f exists g exists h ((a * e) + (b * g) = 1 & (a * f) + (b * h) = 0 & (c * e) + (d * g) = 0 & (c * f) + (d * h) = 1)))) # label(non_clause).  [assumption].
2 (all a all b all c all d all e all f all g all h (tu(a,b,c,d) & tu(e,f,g,h) -> (a * e) + (b * g) = (e * a) + (f * c) & (a * f) + (b * h) = (e * b) + (f * d) & (c * e) + (d * g) = (g * a) + (h * c) & (c * f) + (d * h) = (g * b) + (h * d))) # l
are you effing serious right now:P
Assumptions:
x+(y+z) = (x+y)+z.
x+0=x.
0+x=x.
x+(-x)=0.
(-x)+x=0.
x+y=y+x.
x*(y*z) = (x*y)*z.
x*(y+z) = (x*y)+(x*z).
(y+z)*x = (y*x)+(z*x).

x*1=x.
all a all b all c all d
(tu(a,b,c,d) <-> exists e exists f exists g exists h (
(a*e)+(b*g)=1 &
(a*f)+(b*h)=0 &
(c*e)+(d*g)=0 &
(c*f)+(d*h)=1)).

all a all b all c all d
all e all f all g all h
((tu(a,b,c,d) & tu(e,f,g,h)) ->
((a*e)+(b*g)=(e*a)+(f*c) &
 (a*f)+(b*h)=(e*b)+(f*d) &
 (c*e)+(d*g)=(g*a)+(h*c) &
 (c*f)+(d*h)=(g*b)+(h*d))).
Goals:
(exists y x*y=1) -> (x=1).
@ShaVuklia :P
lolll wow :P
i'm shook
19:15
LOL
*shooketh
Can we use this program for exam?
shookek
You'll want to show that $xb - bx = 0$ for all $b$, for a fixed $x$, no?
@Daminark Sick, nifty.
19:16
is that directed at me @Steamy? Or leaky?
well I don't know how to show it. I just need to show that the two top-right entries don't match for some matrices
I don't see why I would need $xb-bx=0$, I guess it depends on what matrices you have
Hi @Eric
I hear the storm's pretty bad
@Jasper copyleft infringement!!!
Lol jk I ultra stole those two words anyway
Yo @Eric
19:19
Yah, my parents seem to be hunkered down in a relatively safe place but still scary
@Jasper do you have anything (simple) you want to prove?
Hello chat
@KasmirKhaan hi
@EricSilva Good to hear that they're in safety. Let's see how the storm pans out.
@LeakyNun Hello !:D
19:20
@LeakyNun Nope. =D
Yeah now it's just waiting
Waiting hasn't come to chat for a while. I wonder how her book is coming along...
I also wonder whether the name on that book will be Waiting.
@KasmirKhaan do you want a short proof of (right identity + inverse implies left identity + inverse)?
Btw @Balarka I've been studying a bit of complex stuff and it seems p dope, are you still doing Riemannian things
@LeakyNun You might as well just give it to him, lol.
19:21
@ShaVuklia Yeah, nevermind, I was actually aiming at something you've already proven :P
@LeakyNun Yes I would like that :D, our proof was not that long btw =p
@KasmirKhaan here goes: (cc @Jasper)
============================== PROOF =================================

% -------- Comments from original proof --------
% Proof 1 at 0.01 (+ 0.00) seconds.
% Length of proof is 12.
% Level of proof is 6.
% Maximum clause weight is 11.
% Given clauses 9.

1 0 + x = x # label(non_clause) # label(goal).  [goal].
3 x + (y + z) = (x + y) + z.  [assumption].
4 (x + y) + z = x + (y + z).  [copy(3),flip(a)].
5 x + 0 = x.  [assumption].
6 x + -x = 0.  [assumption].
7 0 + c1 != c1.  [deny(1)].
9A x + y = x + (0 + y).  [para(5(a,1),4(a,1,1))].
@KasmirKhaan Have you finally installed latex? lol
@EricSilva At the moment I am not doing anything; I am thinking I should look at Forster today
@LeakyNun Starred because this is an interesting problem for everyone.
19:23
@Jasper haha yes leaky nun helped me with choosing that :D and many other things =p
@KasmirKhaan What did you choose in the end?
That's where I first learned about covering space theory and stuff in my first year, never got to the second half with the sheaf cohomology and stuff
@LeakyNun What kind of proof is this ?
@Jasper texstudio =p
@KasmirKhaan automated proof
the new brand of proofs
@EricSilva Aha.
What kind of complex stuff are you doing?
19:23
I never seen a proof in this fasion =p
I'm seeing if I can pick that up now from cherns books and papers
@KasmirKhaan Oh OK. So you installed that and what else?
something else wierd
Well I wanted to learn more about characteristic classes so I picked up cherns book on complex manifolds and found some of his papers at Ted's suggestion @Balarka
forgot what it was ><
19:25
@SteamyRoot but can you help me with the rest of the proof? I'm desperate and I want to go to bed:(
@Jasper miktex =p
@EricSilva The font in that book sucks, LOL.
@ShaVuklia amina you have to do some work by yourself -.-
@ShaVuklia It's just a guess, really, but: Suppose that $T^*$ is commutative. Let $x \in R^*$, then $$\begin{pmatrix}x & b\\0&1\end{pmatrix},\begin{pmatrix}1 & 0\\0&x\end{pmatrix} \in T^*$$
New courier is annoying for math yeah, but it's whatever
19:25
@ShaVuklia if they give you the proof you wont learn much
oh shit, I hadn't tried that combination yet
@ShaVuklia I already gave you the entirety of the proof ;)
Your idea of having $1$ and $x$ on the diagonals was great
@KasmirKhaan I think she is asking because she has already tried very hard.
but rather than start with arbitrary $b$ and $c$, it's often easier to start as simple as possible
19:26
@Jasper ah thats fine then :D
@EricSilva Ahh
(if these don't matrices don't work, it's likely some similar-ish matrices will)
@SteamyRoot yea good point
thanks, I hope this works
Also, wut, going to bed at 21:30?
Sounds like you need more coffee
lolll, yea well, I have more to do :P
19:28
I go to bed at 9:30
but I am aiming at going to bed at 22h actually
I never sleep ._.
My most productive hours are between midnight and 2am. RIP sleep.
I keep hearing a sound all night in my house so thats y
streamy those hours are best for sleep
worth like 4 times other hours sleep
19:30
from 9 pm tp 3 am
I'm fine with 5 hours of sleep a night, though
I usually think better after 12 AM
2
@SteamyRoot I used to be like that too, until I lost the ability to have good quality sleep. So now I follow a rigorous sleeping schedule not to drop dead/fail my courses:P
@BalarkaSen Oh wow
19:31
@Mr.Xcoder I prefer not to think at all.
is the statement "the domain has an even number of elements" finitely axiomatizable?
Where do you keep finding these songs/music videos o.O
@Jasper That's completely up to you, but I wouldn't recommend that.
@Steamy loool
that's supplied to me by Mike
Balarka owns youtube, lol.
19:33
@SteamyRoot my favorite music video is this
I don't have 3D goggles though :(
If$$\alpha,\beta$$ are the roots of the equation $$x^2-2x+3=0$$. Then the equation whose roots are $$P= \alpha^3-3\alpha^2+5\alpha-2$$ and $$Q=\beta^3-\beta^2+\beta+5=0$$ is$____________$?

I know that I am supposed to use product of roots(P) and sum of roots(S) property to get:
$$x^2-Sx+P=0$$.

I am working on sum of roots and reached:

$$S= (\alpha+\beta)(\alpha^2+\beta^2-5\alpha-\beta-3)$$

I also know that $\alpha+\beta= 2$ and$\alpha\beta=3$

How do I simplify my result further? Algebra tricks don't strike me instantly so I am stuck.
@SteamyRoot Listen to that shit for long enough and you won't need them
1000% your eyes will be 3d
5
Speaking of 3D goggles I put them on when I watched some movies in the cinema. Actually they can be watched without the goggles too, but they would appear weird.
Not sure if that's a good thing, but okay
19:37
@SteamyRoot I went to bed at 2130 when I was in elementary school. =D
Come on guys, give me something to prove
Collatz
go
@LeakyNun Is your program FOC?
@Jasper first order, right
not sure what C stands for
@LeakyNun Free of charge?
19:38
@SteamyRoot not finitely axiomatizable :(
@Jasper oh, lolllll
yes, it's free
If you prove that, MSE won't have to deal with "is my proof for Collatz correct??!" posts anymore :D
and it does first order logic
@LeakyNun LOL, the only other abbrev I know is FOL for first order logic, lol.
@Leaky Have you considered working on Hilbert's 15th problem :P
@Mr.Xcoder lol come on
19:40
Goldbach's Conjecture is also quite reasonable :P
do Collatz and show up the other person here who thinks its doable :P
(I think they're wrong, but hey)
Did Goldbach ever do a lot of work on it?
@Leaky graph isomorphism in quasipolynomial time
Also I have a question (rather atypically, this is on-topic): Is the complement of any set $A$, just itself subtracted from the universal set? I.E $A^C=U\text{ \\ }A$, for any set $A$
@Mr.Xcoder yes
@Daminark that isn't what I'm referring to lol
19:44
Well, "the" universal set is context-dependent, turns out there is none in general (you gotta play with proper classes), but yeah
Universal set according to Wikibooks: en.wikibooks.org/wiki/Discrete_Mathematics/…
Also on-topic discussion is not allowed and will result in a public hanging
0 is a non negative integer right?
@Faust yes
19:44
@LeakyNun Am I the only one that thinks it is non-sensical. Why does it even exist?
Universal set according to Wikipedia: en.wikipedia.org/wiki/Universal_set
its also a non positive integer yes?
@Mr.Xcoder you aren't. of course the universal set doesn't exist.
@Faust yes
kk ty
If you go by von Neumann/Godel stuff you can make sense of it but you end up losing a bit of the structure of sets at that level.
Basically, there's a universal class which contains all sets. But not all classes. Though that's its own rabbit hole
19:46
as I'm reiterating now, "universal set" in the context of absolute complement != set of all sets.
If you're doing an analysis pset or smth the universe is R
the right word is universe
How do you generally prove that a number is irrational?
Universe is only defined locally :P
Ubuntu has a universe and a multiverse repository, lol.
19:47
Depends what kind of number you're proving irrational.
1 (all x (unit(x) <-> (exists y x * y = 1))) # label(non_clause).  [assumption].
2 (all x all y (unit(x) & unit(y) -> unit(x * y))) # label(non_clause) # label(goal).  [goal].
9 x * (y * z) = (x * y) * z.  [assumption].
10 (x * y) * z = x * (y * z).  [copy(9),flip(a)].
12 1 * x = x.  [assumption].
13 -unit(x) | x * f1(x) = 1.  [clausify(1)].
14 unit(x) | x * y != 1.  [clausify(1)].
15 unit(c1).  [deny(2)].
16 unit(c2).  [deny(2)].
17 -unit(c1 * c2).  [deny(2)].
25 c1 * f1(c1) = 1.  [hyper(13,a,15,a)].
I can prove that sqare root 2 is irrational, lol. That's about it.
Proving $\pi$ irrational is a different problem than proving $\sqrt{2}$ irrational.
2
Puzzle for chat: what is this proving?
@LeakyNun An english word.
19:48
@Mr.Xcoder the method of proof is very context dependent. For roots of non-square integers it's pretty simple, but transcendental numbers are hellish sometimes
@Mr.Xcoder ???
@Dami @Semi Thanks.
@Semi this isn't even sniping anymore
@Mr.Xcoder That Secret isn't the only person who likes to give reams of obscure symbolism :)
5
@LeakyNun I answered what "proving" is.
19:49
there's a beautiful proof that $e$ is irrational
It's basically a 360 noscope
The transcendence of e and pi are proven in some books, eg Jacobson Basic Algebra I.
@Semiclassical bingo
Someone spams stars?
19:50
It follows from the Lindemann Weierstrass theorem.
you can prove irrationality of e using CF's
on the other hand, I think the transcendence of $e+\pi$ is still an open problem
come on guys, give me something to prove
L-W is used for transcendence
@Semiclassical heh, we don't even know if it's irrational.
19:50
of course a much harder fact
@LeakyNun Why do you want to prove something so desperately?
hah, I'd forgotten that
Not desperation as much as boredom
@Mr.Xcoder because I discovered a cool theorem-proving automata
19:50
@LeakyNun Is there a proof for 2>1?
@Abcd define "2", ">", and "1".
I saw that :^)
@LeakyNun Oh ok... Unfortunately I'll never be able to give you something that suits you extreme skills, so I will just abstain.
something something Peano
19:51
@Semiclassical of course I know
I'm asking if he knows
Prove Hilbert's basis theorem
good theorem
brb searching what Hilbert's basis theorem is
oh, I could prove something related to ordered rings
brb searching what rings are
19:52
Prove that for a totally ordered field $(\mathbb{F},+,\cdot,\leq)$, $0 < 1$ ?
it's a gloss, but I come at them as "abstractions of modular arithmetic."
@SteamyRoot that should be doable (I already have a proof in my head)
@SteamyRoot I think you can already prove it for a ring
Probably you can, yeah. It just happens to be one of the first exercises in the Analysis I course at my uni, where it's done for fields (since the authors introduce $\mathbb{Q}$ and $\mathbb{R}$ there)
Guys, let $R$ be a ring, and let $a\in R$. Let $S=\{x\in R\mid ax=xa\}$. I have to show that $S^*=R^*\cap S$. So I only have a problem with one inclusion: let $x\in R^*\cap S$. Then we know that there exists $y,y’\in R$ such that $xy=y’x=1$. Now, it’s good that $x\in S$, but how do I know that these $y,y'\in S$?
Finally got the financial aid this year
19:58
Congrats
[DATA EXPUNGED]
2
BALEETED
@SteamyRoot I can't use any counter-example to debug because any ordered ring must be infinite :(
20:01
[DATA DED]
DEDA
why am i doing this with my life
I am going to study Riemann surfaces
Same
What you want to do with it
Well should I do Riemann surfaces? Idk
20:03
@SteamyRoot that is beautiful
how artful
Oh I remember that
@Daminark jump in if you want
I suspect you'll enjoy it
I'll think about it. Right now I've got so much that I'm saying I should do but little actually being done
mathematically or otherstuffically?
20:08
Mathematically. Like I've got k theory, normal AT, groupsy stuff, commutative algebra, the whole lot
i have found that in those crises it's best to jump into something if you feel good about it than sticking to a plan
Read arbit things
Or at least that's how it works for me
Assumptions:
x+(y+z) = (x+y)+z.
x+0=x.
0+x=x.
x+(-x)=0.
(-x)+x=0.
x+y=y+x.

x*(y*z) = (x*y)*z.
x*1=x.
1*x=x.
0!=1.

x*(y+z) = (x*y)+(x*z).
(y+z)*x = (y*x)+(z*x).

(x<=y & y<=x) -> x=y.
(x<=y & y<=z) -> x<=z.
x<=y | y<=x.
x<=y -> (x+z)<=(y+z).
(0<=x & 0<=y) -> 0<=(x*y).

all x all y (x<y <-> (x<=y & x!=y)).
Goals:
0<1.
Proof:
2 x <= y & y <= z -> x <= z # label(non_clause).  [assumption].
3 x <= y -> x + z <= y + z # label(non_clause).  [assumption].
4 0 <= x & 0 <= y -> 0 <= x * y # label(non_clause).  [assumption].
5 (all x all y (x < y <-> x <= y & x != y)) # label(non_clause).  [assumption].
6 0 < 1 # label(non_clause) # label(goal).  [goal].
7 x + (y + z) = (x + y) + z.  [assumption].
8 (x + y) + z = x + (y + z).  [copy(7),flip(a)].
9 x + 0 = x.  [assumption].
10 0 + x = x.  [assumption].
11 x + -x = 0.  [assumption].
@SteamyRoot ^
It proceeds using -1 just like the standard proof.
Heh, nice, I guess :P
is 0 a composite integer?
@SteamyRoot Wow I don't like memes but I have to give this a star.
20:15
@Faust is your question sensical?
no but im wondering if im asked to consider a fact of all composite integers would that include 0?
no it wouldn't
@LeakyNun is sensical a word?
ok thx\
@BalarkaSen of course it is
20:16
@Faust We just speak of integers at least 2 as prime or composite, usually.
I mean 0 is composite though, everything divides it
Is it?
I know "sensible"
@Leaky! could you help me with that exercise I posted?:P
1 is the sneaky one though
yeah it's not a word :P
20:17
oops :P
As to whether the question makes sense it depends on how you wanna define prime or composite.
@ShaVuklia the prover is timing out
@Daminark if there's a bunch of stuff you want to do prioritize the things you wouldn't learn in class anyway
alright
@Balarka words are all made up tho
20:18
@ShaVuklia you just need to prove that the inverse of $x$ is actually in $S$.
Except the word "meme"
Are you guys trying to confuse me?
@EricSilva bababadalgharaghtakamminarronnkonnbronntonnerronntuonnthunntrovarrhounawnskawnto‌​ohoohoordenenthurnuk
I mean I hate it but it's a word
The (removed) king no longer comes to this chat.
20:19
And yeah I mean, so getting a better grasp on groupsy things is good for rep theory, commutative maybe will come up in AG, but k theory might never come up otherwise, as well as Riemann surfaces
I'm downloading Forster right now
Yeah so do the things u might never see otherwise
Or have fun with life idk
@LeakyNun oh right, that's true
@ShaVuklia for some reason my program is not proving it
lol too bad
20:20
I think $ is used in latex because it was expensive to typeset math back then.
@Balarka good shit
I started reading abt sheaf cohomology this morning and my analysis brain is freaking out
I know only the definitions
@EricSilva It's time to tell your analysis brain to be all rounder brain
the smallest non-commutative ring has 8 elements :O
Not sure what construction you are reading but I like the projective resolution construction
20:22
according to my program
@Jasper I like to think I am but I spent the summer doing analysis with a reading/research group.
@EricSilva I see. Spend the next one doing algebra then, LOL
Probably wont
Ah! I forgot I have a shiny copy of Forster I downloaded directly from Springer using university access once
Cool
That university access swag
The one thing I miss about living on campus
20:25
hehehe
You can still log in using your university library account
Thank goodness for Russian servers.
And have access even off campus
@Daminark ya but I don't need to log in to anything to use libgsn
@Daminark ya but I don't need to log in to anything to use libgen
True but the PDFs are often nicer if you can get them from Springer directly
20:27
Ya true
But idrc so it's nbd
Lmao
But I just buy the books after evaluating them, if I want.
Oh shit my message sent twice when I edited it, that's super weird
The world is full of mysteries.
@Jasper that's how I use PDFs basically, I like physical copies a lot
20:28
it happens if your internet is a little slow
I used to pronounce saliva as sell liver.
I would buy them but it's usually way too expensive to justify so I just stick completely to PDFs or get a library copy
It also helps keep in check how much focus I have, since I can only carry a few books and as a result I don't spread myself as thin
@Leaky I can't do it. I tried the following: $ay'=xyay'=yxay'=yaxy'$, but this doesn't help either:(
@Daminark I actually hate libraries because I can't stand to give books back
20:30
@ShaVuklia what is $y'$?
@Eric Ok, I'm going down Forster rabbithole. I do hope you'll teach me a little comparison geometry someday
It sucks cuz libraries are great things
you're talking about "the" inverse, but I'm talking about a left and a right inverse
but otherwise, good bye differential geometry
so we have: $yx=xy'=1$
20:31
Usually my standard for something being too expensive though ends up being non-zero cost + there exists free alternative
do we know that $y=y'$?
@Balarka boy comparison geometry is very very cool
@ShaVuklia $y=y1=y(xy')=(yx)y'=1y'=y'$
I didn't even buy the 208 book that was like, $10
ohmygod ok
20:32
@Daminark I actually own 3 copies of that book lol
Two different translations and an original
Silverman, non-bastardized translation, and Russian?
I actually heard that there's a more modern Russian edition with more stuff
I really like having original editions even if I can only read old French books
Schlag said he knew Russian since from Austria you could basically see Russia if you squint a little bit, but he needed the English one which was for some reason a translation of the old one
20:35
Can you make anything of the Russian one?
can someone tell me which software was used to do this pictures?
No I can't read Russian at all
I can read a little German bc I know a bunch of medieval Germanic languages but it's really tough going because modern German is a pretty far departure
Lol gg
Luckily most of the people I want to read wrote in French which is basically easy reading
20:39
Yeah I know fuck all French and I have still found it to be readable
Fucking Norman conquest man
Knowing romance languages also helps, they're basically all mutually legible in my experience
I know a few words of Spanish, ish
@Twink Hi!
@Jasper hola
Guten Tag, Bonjour, Buongiorno, Buenos dias, LOL.
20:45
@Balarka I think I've decided what to do about my ridiculously scattered to do list
Add number theory to the mix!
@Daminark Sounds great to my ears!
hey guys. i wonder, how can i calculate $\frac{10!}{3!}\sum _{i=0}^{10-3}\frac{\left(-1\right)^i}{i!}$?
Lolol
lolol is so symmetrical, lol.
'tis true
20:51
Oof friends are starting to send me footage from miami
Has anyone here been to Cancun in Mexico for summer vacation?
@EricSilva how bad is it?
I'm seeing footage of most of the streets underwater, some footage of people's roofs flying off
Oh dear. I didn't know it got the hurrican too.
@BalarkaSen Your part is pretty bad too, the floods, over 1000 died I read.
20:56
@BeginningMath 'tis a finite sum
@Jasper Yeah the northern part of India faced way too much rain this year
@BalarkaSen Are you affected at all?
Nah, I'm reasonably off towards the East
Come on, give me something to prove
Bengal got a lot of floods in the northern bit though
20:58
Heh you didn't find anything to prove yet
@Twink Nobody can say for sure what software is used. But there is a lot of latex packages to do simple graphs like that in latex.
@Leaky ain't nobody got time for that
how would you express D(10,k) in a $\sum$? i know that the result of putting k identical balls in 10 baskets is: $\binomial {10+k-1}{k}. but how can i represent it in a $\sum$ sort of way?

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