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12:08 AM
@MikeMiller Yeah, cigarettes are not the way to go. As Erdos has shown, it's all about methamphetamines. /s
 
12:23 AM
@Zee I also am a smoker.
@MikeMiller I don't think cigarettes improve my mathematical ability.
I need a friend.
 
 
2 hours later…
2:48 AM
Today in "Nuking a potato from orbit: proof edition" is proving that the degree of composition of functions is the product of degrees via forms
 
I prefer "nuking with a potato from orbit"
or I would, if I knew how to do it
 
nuking the orbit with a potato would be very good
 
Orbiting a potato with a nuke?
 
That's just dangerous
 
Anyway you get the idea, probably a very high powered tool to do the job but why not?
 
2:52 AM
Um, I don't get why that's a nuke. It's perfectly reasonable to define degree of a map as the number the integral of the top dimensional form is multiplied by when pulled back.
Indeed, in algebraic topology, you define degree of a map $f : M \to N$ to be $f^*(1)$ where $f^*:H^n(N) \to H^n(M)$ is the map induced on top cohomology (de Rham in smooth world), and both groups are identified with $\Bbb Z$.
 
I mean given that we defined it via intersection number
 
Sure, if it's defined as the cardinality of a generic preimage that's obvious
But then it's nonobvious why that's equivalent to the version I mentioned.
 
I mean yeah, we did prove that
This is what I'm saying, using the degree theorem to prove this statement when it's possible to prove it outright via intersection numbers is probably not what Neves was going for
 
i actually usually think of it in terms of forms
 
Lol
I mean in fairness forms just seem like a cool way to think about stuff
 
3:01 AM
I know you'll hate me for saying this but I can't help but thinking of them as a tool for computation
a very cool tool which saves a lot of work, sure, but i never think about them if i have a geometric alternative
I mean this is probably a personal failing
 
shrugs
I haven't gotten deep enough into that side of the world to say anything
Mostly I've liked that they give a nice way of looking at various things which I've appreciated
 
i mean @Balarka maybe you just haven't done the problems that really are natural in forms :P
 
Degree, Cauchy integral theorem
 
like lots of things carry the same data as forms, but sometimes things really are natural in that language
 
@Eric You're probably right.
 
3:09 AM
[Random] It is not known whether $\tau$ has special numerical properties unique to it but not $\pi$

More generally, it is not known whether given any real number $x$, whether $nx$ with integers $n$ can have new properties that are not found in the base itself
 
Is that so?
@Secret
 
3:35 AM
Actually I am not sure if what I said is correct, but I think it holds for multiples of $\pi$ I guess, but not so much for integers (e.g. 0 and 1 are very different, 3 and 2 are different because one is odd and one is even (and even numbers are part of an ideal of the integer ring
so 2, 0 has properties that 3 does not have, and 1 has properties unique to it not shared by other integers
 
Hmm
 
Hi,Is there anything that does symbolic calculations?Like $a = \alpha x + \beta y$ and $b = \alpha x - \beta y$ then when i enter $a.b$ it must return me the correct result as $a.b = \alpha^{2}x^{2} - \beta^{2}y^{2}$ ?
 
3:58 AM
"...to my best knowledge, the book was not dedicated to erotic problems of people in outer space... As Solaris' author I shall allow myself to repeat that I only wanted to create a vision of a human encounter with something that certainly exists, in a mighty manner perhaps, but cannot be reduced to human concepts, ideas or images. This is why the book was entitled "Solaris" and not "Love in Outer Space"."
lel
 
4:31 AM
@BAYMAX Mathematica allows you to define symbolic operators
 
4:50 AM
Thanks@Secret
 
5:05 AM
An initial condition of a pde is $u(x,0)=c$, c is a constant, after I solved the pde, then $u(x,t)=c+kt$, right? I did something completely incorrect
 
5:36 AM
Hmm. ChatJax doesn't seem to be working for me.
 
i don't know how t =o use Chatlax at all.
 
 
2 hours later…
7:40 AM
@TedShifrin How do you feel about eigenvalues of the Laplacian?
 
don't know. have never studied that.
chargrin
 
8:57 AM
soooo I got my F-1 Visa for the US confirmed today
 
congrats
 
i only know academic matters
outside schools I can't find my milieu
I did well in schools
after leaving school, I flounder in the wilderness
 
then return
 
I went to my alma mater to visit my professor in graduate school, and he blamed me for not having a paid job.
 
how is that your fault?
(in his opinion, of course :-)
 
9:11 AM
he said it's the responsiblity of an adult to have a paid job
 
its your choice, no?
 
I am a master gaduate in theoretical physics with research field in gravitation. I can't find a job requiring my expertise.
 
return to university and do research
 
there is no job opening to master graduate in theoretical physics at all
theoretical physics only has job opportunity for PhD graduates
 
you can be a lecturer
 
9:18 AM
@MikeMiller I doubt I can help with the little we did in the PDE course, but on which space of functions are you considering the Laplacian?
 
SBM
9:35 AM
Hello
 
SBM
How's everyone?
 
 
1 hour later…
10:46 AM
Hi all
@SBM I'm good, I have an exam tomorrow though.
Hey, I've just asked a question on this question after a really long time. Check it out!
0
Q: How must have Liu Xin calculated $\pi$?

NickI know there won't be a definite answer to this question because of a lack of required historical evidence but there is so much we know of Liu Xin and of that time period. In that time, to get to an approximation 3.154 was a phenomenal feat. To this day, no one knows how Liu Xin did it but we ca...

 
SBM
All the best @Nick
 
@SBM Aww, thanks. It's math.
Maybe you guys can help me study.
 
SBM
OK
 
Is there a formula or algorithm for calculating the motion such that the distance from an object is changing at a constant rate, but the speed of the movement is variable.
 
11:03 AM
@Benjamin You can try to derive one.
 
@Benjamin One could take, for example, $v(t) = (\cos (t^2), \sin (t^2))$. This always lies on the unit circle, but its speed is $|2t|$.
 
I meant:
if in a compact space every convergent subsequence of a sequence converges to the same limit then does it imply that the sequence itself converges?
 
@Fargle Okay, that was the sort of thing I was thinking, but I wanted to see about other possible equations other than a circle.
 
11:34 AM
Hm, I think this could go wrong in a compact but not sequentially compact space where you have a sequence without converging subsequences
 
SBM
Oh
 
Happy Towel Day everyone!
 
@SteamyRoot I never could get the hang of Thursdays.
 
why there is no smiley here? are mathematicians usually so serious that they do use smileys?
don't
actually I don't use smileys often.
 
12:00 PM
Happy towel day @SteamyRoot
 
SBM
towel day?
:}
 
Towel Day is celebrated every year on 25 May as a tribute to the author Douglas Adams by his fans. On this day, fans openly carry a towel with them, as described in Adams' The Hitchhiker's Guide to the Galaxy or share their folded animal towels to demonstrate their appreciation for the books and the author. The commemoration was first held 25 May 2001, two weeks after Adams' death on 11 May. == Origin == The importance of the towel was introduced in The Hitchhiker's Guide to the Galaxy original radio series in 1978. The follow-up book explained the importance of towels in The Hitchhiker's Guide...
 
Don't have to carry a towel / if you don't leave the house
insert dude-tapping-a-finger-to-his-temple meme
 
Hi chat
 
hi chat
 
SBM
12:13 PM
Hey @Astyx and @user314159
 
@SohamChowdhury like this?
 
@Balarka: $\alpha_\ast$ from yesterday is actually a functor, and a pretty nice one
iirc it has both left and right adjoints
@user314159 ye
 
12:38 PM
how do you describe your English language skills in school application?
 
SBM
?
pardon?
 
Fluent etc
 
Beginner, intermediate, advanced, fluent
 
when you apply for graduate school, there is a column inquiring your English language skills.
 
12:41 PM
Native speaker is used also.
 
What's your native language, out of curiosity?
 
I am not a native English speaker.
Chinese
 
Is fluent better than advanced ?
 
Yes.
 
Fluent means you're the same as a native speaker.
Advanced means you're really good, but not necessarily fluent.
 
12:43 PM
Weird. I would say I'm fluent, but not advanced
 
(Modulo that weird thing where you put a space before question marks)
 
As in, a native speaker does not necessarily have a very rich vocabulary etc
 
I can speak understandable English to foreingers.
 
Lol, you mean like this ?
 
12:45 PM
I know 40000 vocabularies.
 
@CaptainBohemian What about when English speakers speak to you?
 
I find it easier to read, and it's usual typography in France (as you probably guessed). I guess it's just a habit
 
@CaptainBohemian You mean you know 40000 words?
 
I can understand standard English spoken by foreigners even they speak very fast.
 
I recently found out about this website that might be helpful, by the way: cambly.com/english
@CaptainBohemian *Even if (or even when)
 
12:48 PM
but if the speaker pronounces not standardly, like Japaneses or Koreans, I would have difficulties in understanding them.
 
@Soham Yeah, pulling back coefficients is a functor over the slice category of modules or something
 
> …like Japanese people or Koreans, I would have difficulty understanding them (or I would have a hard time understanding them)
(Unlike "Korean", the word "Japanese" cannot function as a noun.)
 
@BalarkaSen "restriction of scalars"
 
(Also, the first part of the sentence sounds awkward, but I don't know how I would fix it.)
 
@Soham Ah yeah that's the name. Thanks
 
12:50 PM
Japaneses
 
"people from Korea or Japan"?
 
That's probably better, yeah
 
I always think about $R-Mod$ as $Vect/B$, and restriction of scalars is the same thing as pulling back bundles over different bases
 
Or "Japanese or Korean people"
 
SBM
oh
 
12:51 PM
extension of scalars is $M \otimes_A B$ with $i : A \to B$ given
 
That's a functor A-Mod to B-Mod right
 
$- \otimes_A B : {\rm Mod}_A \to {\rm Mod}_B$
 
"If the person doesn't pronounce in a standard way" might be better
 
well, actually that should be $i^*B$, not $B$. this functor is written $i_!$.
 
sure sure
 
12:53 PM
Your English is really good. I'm picking out small details, which means that everything else is good. @CaptainBohemian
 
hm, do I understand what extension of scalars does vector bundle level?
 
@AkivaWeinberger really? then how do you refer people of nationality Japan?
 
and, naturally, all these operations work on sheaves
 
"Japanese people"
 
@Captain Japanese is an adjective, not a noun. Korean is both.
 
12:54 PM
You can also use "the Japanese", but it kinda sounds weird
 
so the bundles case is okay too
in the interest of not linking to nLab pages:
In mathematics, Grothendieck's six operations, named after Alexander Grothendieck, is a formalism in homological algebra. It originally sprang from the relations in étale cohomology that arise from a morphism of schemes f : X → Y. The basic insight was that many of the elementary facts relating cohomology on X and Y were formal consequences of a small number of axioms. These axioms hold in many cases completely unrelated to the original context, and therefore the formal consequences also hold. The six operations formalism has since been shown to apply to contexts such as D-modules on algebraic...
 
@Soham yeah but what's the picture?
 
so troublesome. why is there no single word coined to refer people of nationality Japan?
 
the problem is that I only know ~one nontrivial bundle
the torus
 
"the Japanese" means the people in general I think
Not the individuals
 
12:56 PM
the torus is a trivial bundle, actually.
but you know the moebius strip
 
mobius strip
yeah
oh, sorry, I forgot what trivial bundle meant
 
@Soham you should learn a few things from Hatcher's VBKT
 
i was thinking "product with $\Bbb R^n$". silly me, nothing special about $\Bbb R$
 
it's pretty good and doesn't require much prior knowledge
 
yes, now is definitely the time for me to do that
huh
 
12:57 PM
Some nationality words are both nouns and adjectives ("Korean," "Russian," etc.). Some nationalities have different words for the nouns and the adjectives ("Swede/Swedish"). Some nationalities have a word for the adjective and no word for the noun. I don't know why.
 
@TedShifrin Heating up (although I'm probably way too late :p)
 
i guess trivial budle means a bundle which can be covered by a single coordinate frame without singularities
 
@SohamChowdhury yeah that's the trivial vector bundle
torus is a trivial S^1 bundle over S^1
'cuz S^1 x S^1
 
when I say "bundle", I don't mean "circle bundle"
anyway.
 
@AkivaWeinberger what about Chinese? Is Chinese people=Chineses?
 
1:02 PM
gotcha but torus is not a vector bundle over anything
 
right
@BalarkaSen would it be accessible to me? I guess I could take a long hard look at it then
 
the first few pages are def accessible. let me look again
 
huh, I don't remember how to put a manifold structure on $TM$.
 
oh yeah I had hard time understanding that at first when i learnt that it was a thing.
so let $U$ be a chart on $M$
 
Japanese adj.
Of or relating to Japan or its people, language, or culture.
n. pl. Japanese
1.
a. A native or inhabitant of Japan.
b. A person of Japanese ancestry.
2. The Japonic language of the Japanese, written in kana and Chinese characters.
 
1:06 PM
then what is $TU \subset TM$?
 
was a bit fiddly
 
(hint: what does a chart mean?)
 
just $U$ would work, I guess
maybe using the chart map
 
mhm
 
I think writing $TU$ is kinda circular, since before then the notion only extends to the "germs" (so $T_xM$)
(strictly speaking)
@BalarkaSen ...?
 
1:09 PM
Hm. by $TU$ I just mean union over all the $T_pM$'s where $p \in U$. Is there anything circular with that?
 
oh, then not at all
 
@SohamChowdhury Yeah I am pretty sure it's accessible for a large portion of the first section
 
okay, I'll be going now. let's go through the construction later.
 
sure.
 
I'll try to think hard about gluing the $TU$s together
probably the transition maps on $U_i$ can lift somehow to give transition maps between $TU_i$
 
1:11 PM
That's not a bad way to think about them :) Vector bundle is just a bunch of trivial bundles glued togather anyway
Yeah exactly
 
and charts are the same ones, morally
 
you should be able to make "somehow" rigorous :)
(btw I hope you see why $TU$ is diffeom to R^N for some N)
 
hi chat
 
Hi Semi
 
Semi hi
 
1:14 PM
well, I am stupid. I somehow convinced myself that a person had asked the same question twice when they actually totally didn't.
 
Semi stupid? :P
 
nothing semi about it
I'm just glad I can retract flags
(i'd still say it's a bad question, but the flag was entirely unwarrented)
Here's a question, motivated by this one: math.stackexchange.com/q/2296197/137524
Is $a^0=1$ if $a\neq 0$ true by definition for real $a$, or is it something one proves?
 
I guess you'd want $a^na^m = a^{n+m}$ in any case
 
Yeah.
 
So if $a$ is invertible, you have $a^{-n}a^n = a^0 = 1$
But it's just a notation really
 
1:24 PM
I guess one could say that it follows from the requirement that $a^0 a^n = a^{0+n}=a^n$.
 
@Semiclassical I have a doubt in a probability question try to answer:A person can score 0,1,2,3,4,56 per run in an over, probability that in an over a total run of 30 can be obtained.
 
But the fact that it's a requirement does not mean it's something one proves
 
Right.
This really comes down to how exponentiation is defined.
 
Agreed
 
@Semiclassical please answer my simple doubt;
 
1:26 PM
@satyatech "...per run in an over" what?
 
Your sentences don't make much sense
 
Means in a ball he can score 0,1,2...
 
I think there's a language barrier here.
Your sentences don't make sense in English.
 
You probably should use subjects
 
*in an over *should not be used after "per run " my mistake ..as it makes the sentence confusing@my mistake
Let me write the question again;
( A cricketer can take runs in 0,,1,2,3,4,5,6 .So let's say he plays a over ,what is the probability that he will score exactly 30 runs in the over?
)
@Semiclassical got it now..
 
1:31 PM
Oh, cricket. This would explain why I don't recognize the terminology.
 
What's an over ? Is the probability distribution uniform ?
 
(I'm an American, and we're typically ignorant of cricket)
 
The probability of scoring 0=probability of scoring 1= bla bla bla..
A over consists of 6 balls ,so you get 6 chances ...
 
Ah. So you have 6 chances to score runs, and on each attempt you can score 0-6 runs with uniform probability.
 
Yes,,,
 
1:34 PM
(a pity it's not 1-6, or I could say "oh it's just dice")
 
Ah it would have been die thrown 6 times(but it's not)
 
or 6 dice thrown at the same time, yeah.
aaaanyways.
 
Just pretend it's 1-7 and say "oh it's just the (seven-sided) dice"
 
What's the doubt?
@Astyx heh, and subtract 6 at the end :P
 
Exactly :p
 
1:37 PM
Now ,I cannot use children's way of counting no. of ways in which X(1)+X(2)+X(3)+X(4)+X(5)+X(6)=30,rather I should use a formula..
 
Well, probably you should start by finding the total number of possible outcomes.
Do you know how to do that?
 
I just forgot how to do that,does it use binomial distribution?
Or multinomial theorem
 
Probably multinomial, since you need to say how many of their attempts give 0 runs, how many give 1, etc and these should add to 6.
 
1:40 PM
Do you know about generating series ?
 
No,but I can do after a bit hint...
 
Problem with the multinomial approach, though, is that the counting is pretty painful as far as I can tell.
Though there may be no way around it...
 
Then should I ask in the real place ...To get how many people attempt it??
 
Maybe.
 
In different ways?
 
1:43 PM
The route I can see is via generating functions as well.
 
Wait 3-4mins and then you will see my question going online,,check the main site ...And try there please..
Ok bye
 
mmkay.
 
Well generating functions only save you the hassle of explaining why the result is what we claim it to be, but it's still multinomials
 
Right.
 
That is the coefficient of degree $30$ in ${1\over 7^6} (1+X +\dots + X^6)^6$
 
1:53 PM
ya
 
Which for the record is $462/7^6$ unless Mathematica betrayed me
 
A Gaussian approximation doesn't do badly either.
Yeah, that's what I get too
Going that route, I think the thing to do is write that polynomial as $$\frac{1}{7^6}(1-X^7)^6(1-X)^{-6}=\frac{1}{7^6}(1-6X^7+\binom{6}{2}X^{14}+\cdot‌​s)(1+6 X+\cdots)$$
bah, \cdots not \cdot s
So that'll give an alternating sum.
 
Oh yeah nice
 
still pretty annoying, though
 
So you only need the terms of degree 30, 23, 16, 9, and 2 of the last parenthesis
 
2:04 PM
Right.
if memory serves the $k$-th coefficient will be $\binom{5+k}{k}$
 
Is it not $-6\choose k$ ? (or maybe that's the same thing?)
 
Pretty sure it's the same. I always have to go back and check to be sure, though.
 
$f : B(0,1) \to B(0,1) $ , we know $||f(x) - x || \le1 $ i need to prove there is $x_0$ s.t $f(x_0) = 0$ . i know that there is $x_1$ s.t $f(x_1) = x_1$ , but not sure how to continue, someone see the answer?
 
Up to a minus sign it's the same
 
it is in $\Bbb R \ ^ 2 $
 
2:08 PM
But yours is right
So we're left computing $\sum_{k=0}^5 {6\choose k}{5+k\choose k}$
 
$$\binom{-6}{k}=\frac{(-6)!}{k!(-6-k)!}=\frac{(-6)(-6-1)\cdots(-6-k+1)}{k!}=(-1)‌​^k \frac{6(7)\cdots(5+k)}{k!}=(-1)^k \frac{(5+k)!}{5! k!}=(-1)^k \binom{5+k}{k}$$ phew
also, that doesn't look right.
one doesn't need 5C0, 6C1, etc.
one needs specifically 35C30,28C23,...
plus, it'd still be an alternating series due to the signs from the X^7 sum
 
Guys ,it's pretty difficult to survive in this world ,you can see my question here and understand why it's so long there...
0
Q: A Simple interesting probability problem;

satyatechQ) A cricketer can score 0,1,2,3,4,5,6 runs per ball.So ,he plays an over ,what is the probability that he can score exactly 30 runs in that over ?(Assume that P(throwing a ball )= constant and the event of throwing a ball is an independent event having equally likely outcomes, similarly P(s...

 
Should be: $$\sum_{k=0}^4 (-1)^k \binom{6}{k}\binom{5+30-7k}{30-7k}=\sum_{k=0}^4 (-1)^k \binom{6}{k}\binom{35-7k}{5}$$ @Astyx
 
Oh yeah I was being silly (again)
 
Though I guess one doesn't need to worry about the range of $k$, since the first binomial coefficient vanishes for $k<0$ and the second for $k>4$
 
2:18 PM
Weirdly $$\sum_{k=0}^5 (-1)^k{6\choose k}{5+k\choose k} = - \sum_{k=0}^4 (-1)^k {6\choose k}{35-7k\choose 5}$$
 
I guess that's just a coincidence
 
i'm not awake enough to understand why atm
 
It's even weirder considering the sum bounds are not the same
 
If the $p$-value is $0.04$, that means we can have a $96$% confidence interval at best, right?
 
2:21 PM
With that we pass from probability theory into stats where I can't say anything.
Trying to figure out how to answer this without making it trivial: math.stackexchange.com/questions/2296290/…
 
someone tried my question ? :P
 
It basically just boils down to "complete the square and do the remaining gaussian integral in cartesian coordinates"
 
it was given in a Game Theory course as an "extra" question, and it seems interesting
 
@Liad I assume $f$ is continuous ?
 
yea
 
2:24 PM
Which question?
Oh, the one above
 
ya
 
it is in $\Bbb R \ ^ 2 $ ,
 
What's $B(0,1)$? I'm forgetting the notation
 
unit ball
inside the closed unit ball
 
Ball of radius 1 centered on $0$
 
2:25 PM
of course it is.
 
:P
and we know there is $x_0$ s.t $f(x_0) = x_0$ , im not sure why there is $c$ s.t $f(c) = 0$
 
So the image of $f(x)$ is within 1 of $x$ and is inside the unit ball.
 
The image is connected
 
and there's some fixed point.
 
yea
the fixed point is not given in the ex. it is because the unit ball is compact and convex
 
2:26 PM
The image of the circle is connected
 
hm.
 
However the only point which is in $B(x,1)$ for all $x$ in the circle is $0$
 
okay, i'm officially brain dead right now and need to take a nap
 
hehe :P good night
 
See ya @Semi, sleep well
 
2:28 PM
@Astyx im not sure i follow , why this implies a point that goes to $0$
 
cya
 
(the f*-ness of my schedule is something like ...10 centi-balarkas?)
 
lol
 
@Liad With some topology I believe you can show that the image of the ball does not have "holes" in it, and with what I just said that 0 is surrounded by points of the image of the ball (that is the image of the circle)
 
might be a bit higher, but I haven't gone full-balarka yet :P
 
2:30 PM
Never go full Balarka
 
@Liad Don't rely too much on what I'm saying, it's not 100% true
 
Am I allowed to ask an algebra question? Given is a group G and a subgroup H with order n. G acts on G/H by $g\circ xH=(gh)\circ H$. Let $phi$ be the associated homomorphism $G \to S(G/H)$. I have to prove that $#Im \phi $ is at least n. I thought I could use one of the isomorphism theorems by realising that Ker phi is contained in H and therefore $Im \phi \sim G/Ker(phi)$. How to continue?
 
@Astyx ok, it is in Game Theory, so there are students who dont know basic topology there :P
 
I don't know much about game theory, what is it about ?
Oh I know
 
2:34 PM
this particular question is not really about game theory , more calculus :P
it is about games generally
:)
 
Was it asked to prove there was a fixed point ?
 
i thought defining $g(x) = ||f(x)-x||$ we have $im(g) \subset [0,1]$ , and there is $c\in [0,1]$ s.t $g(c)= 0$ , and to try somehow to show that there is $w$ s.t $g(w) = ||w||$
no, the fixed point i came up with
it was asked to prove there is a point $x_0$ s.t $f(x_0) = 0$
the $c$ i wrote is in $B(0,1)$
 
Yep
The solution I can come up with involves quite a lot of topology
That is consider $f_1:t\in[0,\pi]\mapsto f(e^{it})$ and $f_2:t\in[0,\pi]\mapsto f(e^{-it})$
These are homotopic unless I'm mistaken
And on goes on the right of $0$ while the other goes on the left of $0$
Actually I meant $e^{it}$, not $f(e^{it})$ (same for $e^{-it}$)
 
2:52 PM
yea i thought something is a bit wrong when you said one goes on the right and one on the left, because we dont know how $f$ act
 
Yes we do, this still holds
We know the image of $i$ is in the upper half of the complexe plane
 
i meant acts on the unit cevtros
 
While the one of $-i$ is in the bottom half
 
yea this we know
 
(I hope you don't mind using complex notation instead ?)
 
2:54 PM
its find :)
how do you continue from here?
 
Compose $f$ and the homotopy
And suppose it never goes to 0
Let $h$ be the homotopy, take $z(x)$ be a point of minimal norm in $f(h([0, \pi], \{x\})$
And find a contradiction from there
There is probably a much easier way of doing it
I have to go now, bye !
 
bye
 

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