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/sci/ - Science & Math


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9239141 No.9239141 [Reply] [Original]

What are you studying today, /mg/?

http://www.lmfdb.org/universe
>The top half of the diagram is based on the Langlands program, which predicts that any motivic object corresponds to an automorphic object via their L-functions.

Previous thread >>9223313

>> No.9239177

I don't really understand how to correctly translate a natural language sentence into propositional logic

>The definition for a bijective function is that the function is both injective as surjective

should it be
[math]I \wedge S \implies B[/math]
Or
[math]I \wedge S \equiv B[/math]

>> No.9239181

>>9239177
the second

>> No.9239183

>>9239181
but why?

>> No.9239208

>>9239183
If it is injective and surjective, it is a bijective function.

>> No.9239215

How do integrals work? Why can we take the antiderivative of a function f and then have that F(b) minus F(a) is the area of f between those points? I know it’s true but why? Is there an explanation for dumb me who has only done calculus? I know you’re adding infinitesimal rectangles but how does this relate to the antiderivative?

>> No.9239227

How do I solve for d, where p is a large prime?

[-1 = 5^d] mod p

Thanks from terrible-at-crypto anon

>> No.9239294

how do I prove that (3/2)^n > n by mathematical induction, help

>> No.9239296

>>9239215
That is just what the fundamental theorem of calculus says. If you want to know why then read the proof.

>> No.9239303

>>9239294
prove the base case then prove the induction step

>> No.9239318

>>9239227

Euler's theorem, and then since if p is odd, p-1 even, so there is a square root of 1 which is -1, because otherwise the order would be too small.

>>9239294

Show that 3/2 > 1, and that the left hand changes by a factor higher than 1 each n, and the right hand changes by exactly 1, so therefore the LHS must stay larger.

>> No.9239328

>>9239296
Thanks I didn’t see it said that too.

>> No.9239333

>>9239296
>>9239328
Thanks I read https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning and I understand it better now!

>> No.9239352
File: 145 KB, 759x639, bullshit.png [View same] [iqdb] [saucenao] [google]
9239352

Can we agree this proof is shit? In (ii) there is no reason for W(a) != S.

Glossary: I is the set of ideals, W(a) is the initial interval determined by a.

Dugundji's topology btw.

>> No.9239356

>>9239352
>usage of classical logic
>Can we agree this proof is shit?
Absolutely.

>> No.9239358

Can anyone give me an example of two metrics which induce the same topology but are not equivalent?
Wikipedia claims this is possible but doesn't give an example.

>> No.9239369

>>9239358
Inducing the same topology is by definition being equivalent. Do you mean that inequality using two constants? If so I have a tip: boundness of a metric is not an equivalence invariant.

>> No.9239376

>>9239356
>muh constructionism
Let the big boys talk.

>> No.9239382

>>9239376
>muh
>constructionism
Way to expose your newness.

>> No.9239386

>>9239382
ESL m8.

>> No.9239397

>>9239386
Disbelief.

>> No.9239431

>>9239382
>Way to expose your newness.
Oh, the irony...

>> No.9239441

>>9239431
This makes no sense in context. Are you mentally ill?

>> No.9239442

>>9239441
>This makes no sense in context. Are you mentally ill?
It makes perfect sense in context. Are you mentally ill?

>> No.9239454

>>9239442
How truly pathetic.
*yawn*

>> No.9239456

>>9239454
Which part of the irony confused you?

>> No.9239464

>>9239456
Maybe you are confused.

>> No.9239497

>>9239464
>Maybe you are confused.
It's possible, but I doubt it.

Meanwhile it's very clear your 'newness' comment was hilariously ironic.

>> No.9239502

>>9239497
You are daydreaming.

>> No.9239520

>>9239502
>You are daydreaming.
How so?

>> No.9239569

>>9239520
Unlock your heart, anon.

>> No.9239631

>>9239569
>Unlock your heart, anon.
What do you mean?

>> No.9239650

>>9239631
A-anon, I...

>> No.9239653

>>9239650
>A-anon, I...
Are you ok?

Why are you stuttering?

>> No.9239697

>>9239352
Stop bickering and answer me faggots. Am I crazy or Dugundji can actually make mistakes?

>> No.9239709

>>9239697
>faggots
Why the homophobia?

>> No.9239723

>>9239709
Too much /fit/ anon, I'm sorry for your feelings.

Nvm, fixed the proof, Dugundji was lazy as fuck. If you are in the (ii) case then $S=W(b) = \cup_{a<b} W(a)$ and these $W(a)$ are strictly contained on $S$.

>> No.9239726

>>9239709
>>>/r/eddit/

>> No.9239744

>>9239726
>>>>/r/eddit/
What about reddit?

>> No.9239749
File: 10 KB, 1366x156, Capture.png [View same] [iqdb] [saucenao] [google]
9239749

>>9239726
huh?

>> No.9239844

>>9239744
It's your home.

>> No.9239853

>>9239844
>It's your home.
What do you mean?

>> No.9240841

>>9240483
might be time to consider this has more to do with your personal experience. Clearly there are many people who do active research in it, even in major branches of math. I know some people from JHU, and they have a very active CT community, almost entirely applied (number theory, homotopy theory, geometry, etc.). All the people I met from there were very familiar with CT (Emily Riehl is there), but also with their field of study, and understood the importance of both moving forward.

Don't let the CS fanboys who've only read Awodey ruin a great tool and field for you.

>> No.9240992
File: 56 KB, 335x459, how I feel.jpg [View same] [iqdb] [saucenao] [google]
9240992

I'm trying to define some basic algebra concepts in Category Theory but I'm running into trouble with formalisms.

The category Set is defined as usual:
Objects: sets
Morphisms: set functions


If [math]G[/math] is a group then we can define it as a category [math]\mathcal{G}[/math]:
Objects: Single object [math]G[/math]
Morphisms: [math]g:G\to G[/math] for each group element so that the identity element is the identity morphisms on [math]G[/math] and composition is (right) multiplication in the group and all morphisms are isomorphisms (yielding inverses).

Note that [math]\mathrm{Aut}_\mathcal{G}(G)[/math] yields our group under composition.


Let [math]\mathcal{U}:\mathcal{G}\to\mathrm{Set}[/math] be the functor that sends the group to its underlying set in the category [math]\mathrm{Set}[/math].


A group action where [math]G[/math] acts on a set [math]R[/math] is defined as a functor
[math]\mathcal{A}:\mathcal{G}\to\mathrm{Set}[/math]
[math]\mathcal{A}:G\mapsto R[/math]
[math]\mathcal{A}:g\mapsto f[/math] where [math]f\in \mathrm{Aut}_\mathrm{Set}(R)[/math]

Note: Automorphisms on a set are permutations. So [math]\mathrm{Aut}_\mathrm{Set}(R)
[/math] yields the symmetric group [math]S_{|R|}[/math] under composition.


A group action where [math]G[/math] acts on itself is defined as a functor
[math]\mathcal{A}:\mathcal{G}\to\mathrm{Set}[/math]
[math]\mathcal{A}:G\mapsto\mathcal{U}(G)[/math]
[math]\mathcal{A}:g\mapsto f[/math] where [math]f\in \mathrm{Aut}_\mathrm{Set}(\mathcal{U}(G))[/math]

For instance, the group [math]S_3[/math] acting on itself is analogous to taking [math]S_3[/math] and embedding it in [math]S_6[/math] (since [math]S_3[/math] has [math]6[/math] elements then [math]\mathrm{Aut}_\mathrm{Set}(\mathcal{U}(S_3))[/math] is [math]S_6[/math] under composition). Unfortunately, this makes it difficult to define an action that does conjugation (as it would require multiplying elements of [math]S_3[/math] with elements of [math]S_6[/math]). Any help?

>> No.9240997

>>9239358
let d a metric

define a new metric d' = d/(1+d)

Same topology, but all distances are bounded by 1 in d'

>> No.9241070

Can somebody post challenging probability problems?

>> No.9241076

>>9241070
What's the probability that all non-trivial zeros of the Riemann zeta function have real part 1/2?

>> No.9241484
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9241484

>>9241070

>> No.9241486
File: 7 KB, 462x57, 3b91627e24f0135cb585a1a6c07c87af.png [View same] [iqdb] [saucenao] [google]
9241486

>>9241070
i also have the solutions, have fun anon

>> No.9241889

>>9241076
1/2
>>9241484
1/2
>>9241486
1/2
either it is or it isnt
probability is easy

>> No.9241901

>>9241889
Can someone calculate the probability than any of these answers is within some margin of error?

>> No.9241930

Hello, i have a linear algebra question. Can someone help me understand the relationship between the trace of an n x n matrix A, and the eigenvalues of matrix A

>> No.9241945

>>9241930
sum of eigenvalues is equal to the trace of the matrix

>> No.9241949

>>9241945
Ok I see, what about the eigenvectors for A

>> No.9241950

>>9241889
There is a 100% chance you are pissing me off

>> No.9241973

>>9241949
what about them?

>> No.9241981
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9241981

When they say "vanishes" do they mean goes to 0 or becomes discontinuous?

>> No.9241982

>>9241973
well I would to find the eigenvalues of A using only the determinant and trace.

>> No.9241998

>>9241982
(not the same guy)
Can you recall the definition of eigenvalue and eigenvector off the top of your head?

>> No.9242000

>>9241981
zero
that theorem is called L'Hospital's rule

>> No.9242020

>>9241998
idk. i know the determinant is 0

>> No.9242027

>>9240992
>he fell for the category theory meme

>> No.9242054

>>9242020
You Always should read the theory from you textbook before trying to do excercises/examples.
An eigenvector v is a (non zero) vector which A only scales by a factor of λ (it can't rotate it or do something else). λ is called the eigenvalue corresponding to v. In more compact notation Av=λv, or you can also write it this way (A-λI)v=0.

If A had an eigenvalue λ, then you would have (A-λI)v=0 for some non-zero v.
This happens if and only if A-λv is non invertible.
Which happens if and only if the determinant of A-λv is 0.

Thus λ is an eigenvalue if and only if det(A-λv)=0.
Therefore, you can find all the eigenvalues from this equation.

Now, say you solved the system and you found the eigenvalues. Take one of the eigenvalues, let's say λ.
The eigenvectors v corresponding to λ must, by definition, satisfy the following:
Av=λv, or in other words (A-λΙ)v=0. That's a linear system that you have to solve. The solutions are the eigenvectors for λ.

Watch this:
https://www.youtube.com/watch?v=PFDu9oVAE-g

>> No.9242063

>>9242054
I just need to figure out how to find the eigenvalues of a 3x3 matrix given using the determinant and trace. But i understand eigenvalues and eigenvectors now

>> No.9242100

>>9242063
The determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues.
You probably have more info about the matrix (e.g. you already know on eigevalue) which will allow you to just solve for the eigenvalues.

>> No.9242404

>>9242027
>>he fell for the category theory meme
I'm not a "he".

>> No.9242434

What are some hidden properties of finite abelian groups?

Stuck on a proof

>> No.9242489

>>9242404
Ignore this baitfag. There are many men and women who research category theory. Here is Eugenia Cheng, she does higher dimensional category theory and makes retard-proof youtube videos explaining basic category theory concepts.
https://www.youtube.com/watch?v=g47V6qxKQNU

>>9242027
I do pure category theory. It's super useful and interesting. Algebra nerds constantly misuse algebra terminology and notation while claiming that Algebra can be approached through category theory. I thought I'd give it a try and I'm finding it's actually quite hard. I suspect that the solution here may require some overly complicated (and somewhat contrived) approach where I construct something in the category of groups capturing the notion of conjugation and then send it out to the category of sets.

>> No.9242493

>>9242489
>misuse category theory terminology and notation
fixed

>> No.9242598

How do I prove something that's too obivous but still written as a standalone question

>> No.9242611

>>9242598
Write the statement formally (i.e. via it's formal definition) and derive it explicitly.

>> No.9242798

>>9242598
If a statement is too obvious to the point you can't write a proof about it then you either don't understand it and you fooled yourself, or you don't understand some new machinery that is used to prove the statement.

>> No.9242814

>>9239141
trying to understand Knuths abstraction of the Josephus Problem.

so pure mental masturbation basically

>> No.9242825

>>9242027
>meme
>>>/r/eddit/

>> No.9242870

>>9242493
I know "natural" is a big one, but what are some more?

>> No.9242912
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9242912

>>9242887
any takers?

>> No.9242928
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9242928

I'm supposed to graph this to see the set of feasible solutions, and then use the slopes, but how am I supposed to graph this when the constraints are variables? Did I set this up incorrectly?

x_1 - # cakes sold
x_2 - # Pastries sold
b_1 - lbs of flour avail.
b_2 - lbs of sugar avail.

max 40x_1 + 9x_2
subject to
10x_1 + 3x_2 <= b_1
15x_1 + 2x_2 <= b_2

>> No.9242946

Is calc 2 harder than 1 and 3 or am i just retarded? i breezed through calc 1 but in calc 2 every exam takes more time then i have to complete it (ie im too fucking slow despite knowing the material) and i get a shitty grade. I think if i grind enough i can mange to get a B- overall.
should i just give up on taking any consecutive calc courses and change to a shitty non-engineer major?
should i give up on my dreams?

>> No.9242949
File: 48 KB, 208x136, 1508226757089.png [View same] [iqdb] [saucenao] [google]
9242949

>>9242946
>shitty non-engineer major
Is there anything even shittier?

>> No.9242956

>>9242949
>is there anything even shittier?
not sure if you're saying that engineering is shitty or everything else is, but thanks for reading my entire post and replying with a dumb joke and completley ignoring my question

>> No.9242959

>>9242956
>completley ignoring my question
I didn't. I'm implying that engineering is mainly for retards so you shouldn't be too worried.

>> No.9243224

>>9242959
>I'm implying that engineering is mainly for retards so you shouldn't be too worried
if you havent taken calculus and your reply says nothing about calculus then obviously your post was neither an answer to my question or a meaningfull statement

>> No.9243247

>>9243224
>your post was neither an answer to my question or a meaningfull statement
It's actually both, think about if for a while and you should realize.

>> No.9243256

>>9243247
so i have suffered defeat at the hands of a N.E.E.T. larping as someone who has left his parents house and gotten an education thus interacting with people like a normal human would, but for some reason can only contribute via insult rather than any explaination of any kind or length.
why are you on /sci/? do you visit the advice board and tell people their retarded while you wait on mommy to bring you tendies? please stop wasting air and join the military or something before you turn 30

>> No.9243258

>>9242949
>>9242959
>>9243247
You seem hostile. Are you hostile?

>> No.9243264

>>9243256
I didn't insult you though, I was simply stating a basic fact about you and most engineers. You didn't seem to realize it so I decided to help.
>why are you on /sci/?
I'm not on /sci/, I'm on /mg/. I couldn't care less about this board as a whole.
>>9243258
Not at all.

>> No.9243271

>>9243264
>simply stating a basic fact
Nah, you're misusing a meme term as an excuse for excess hostility. Nobody who comes to /sci/ uses that meme that way. Someone should probably ban you for being cancer at some point.

>> No.9243279

>>9243271
What are you even trying to explain? It seems like you have difficulty communicating.

>> No.9243281

>>9243271
>meme
see >>9242825

>> No.9243307

>>9242000
I think you mispelled that there name friend

>> No.9243314

>>9239141
this is shit

>> No.9243324

>>9243307
o with a hat in French means that there was an s after it, but it stopped being pronounced after time passed.

>> No.9243348

>>9243314
>this is shit
What do you mean?

>> No.9243356

>>9239177

The first one isn't wrong, it's just incomplete. The second one is just

[math](I \wedge S) \leftrightarrow B[/math] which is equivalent to the conjunction of [math](I \wedge S) \rightarrow B[/math] and [math]B \rightarrow (I \wedge S)[/math].

>> No.9243367

>>9243314

goto the website its the fucking tits

>> No.9243375

>>9243367

this website is fucking awesome

http://www.lmfdb.org/EllipticCurve/Q/150/c/1

its everything i've been dreaming about once i started learning algebra, full classification, tables of properties

>> No.9243397

Soooo...Stewart or Spivak?

>> No.9243404
File: 38 KB, 695x324, CONFINF1.png [View same] [iqdb] [saucenao] [google]
9243404

>>9239141
I would like to take a full semester course in every word in that diagram. Too bad my only option for getting the money to pay for it would have made it impossible to attend the classes.

>> No.9243411

>>9243404
>I would like to take a full semester course in every word in that diagram. Too bad my only option for getting the money to pay for it would have made it impossible to attend the classes.
Just read some books m8

>> No.9243429

>>9243397
Neither, since the subject is trash.

>> No.9243469

>>9243281
>>9243271
>>9243264
thanks for the shitfest guys, but has anyone here actually taken calculus 1 2 and 3 and is capable of iliciting a legitimate response to my question?

>> No.9243471

>>9243397
Dunno about Stewart, but Spivak is decent.
Pirate them both.

>> No.9243733

For natural numbers, how can I show 2^(n^2) is greater than or equal to n! (factorial)?

>> No.9243742

>>9243733
https://en.wikipedia.org/wiki/Mathematical_induction

>> No.9243750

>>9243733
Assume that [math]0 = 1[/math], then it follows trivially.

>> No.9243765

>>9243742
>>9243750
I figured it out, thank you anons

>> No.9243774

>>9242404
tits

>> No.9243775

>>9242912
yeah, could do with some for all epsilon>0 etc though, and a written contradiction

>> No.9243802

>>9239141
how long would it take to start from elementary math and grasp/understand each subject continuously up until calculus? i.e., given you aren't a brainlet.
idc if anyone believes me, as I know it's custom on this board to assume brainlet status for any anon who posts, but I believe myself to be in alright condition to understand any material, given I have time the to study it, which I do.
This question does apply to me; I won't mention details despite them being considerably relevant to the inquiry, just please answer how quickly you think an able minded person could accomplish this feat.

>> No.9243824

>>9243802
It's trivial if you aren't a brainlet. That's a big "if" right there.
Calculus is garbage anyway, so you shouldn't bother.

>> No.9243853

>>9241484
I cant latex but its = sum of (-1)^(i+1) * 1/i! for i running from 1 to n

>> No.9243860

Based on the statements below, let E = "Engine is good", G = “Gearbox is good”, and S = “Steering is good”. Convert the three data from the feedback system below into propositional logic notations using E, G, S and logical connectives.

- Engine provides feedback that Gearbox is faulty while Steering is good.
- Gearbox provides feedback that Engine is good if and only if Gearbox is good.
- Steering provides feedback that at least one of the other two parts is faulty.

>> No.9243873

>>9243853
no it's [eqn]\sum_{i=1}^n \frac{(-1)^{n+1}}{n!}[/eqn]

>> No.9243874

In linear algebra we are studying bilinear forms and today we proved that the product of two linear forms is a bilinear form.

I have two questions about this that my professor did not know how to answer at the moment, and googling is not helping

1) If I have a bilinear form, is it always possible to factorize it as the product of two linear forms?

2) Is the representation of a bilinear form as the product of two linear forms unique?

In short, what I am asking is if linear forms are like prime numbers for multilinear forms or what's the deal here? Are there bilinear forms that do not have a "prime" factorization? And are "prime" factorizations unique?

>> No.9243875

How do I keep vector calc interesting?

I feel like because it’s a generalization of calc 3, that I’m sort of glazing over the course without really taking in everything that’s being done.

Sort of like psyching myself out by thinking it’s easier than it may be.

Given, the course work is not hard, but does anyone struggle with this, and how do you combat it?

I’m just worried the course will get to the point where content will not be so obvious solely based on prior knowledge in calculus and linear algebra.

>> No.9243877

>>9243824
I just need an estimation given preferably in months

>> No.9243879

>>9243877
1-2 months in the worst case scenario.

>> No.9243887

>>9243860
Not entirely sure what is meant by "provides feedback that" but I'll give it my best shot:
[math] (\neg G \wedge S) \Leftrightarrow E \\
(G \Leftrightarrow E) \Leftrightarrow G \\
(\neg E \vee \neg G) \Leftrightarrow S
[/math]
It could also be
[math] E \Rightarrow (\neg G \wedge S) \\
(G \Rightarrow E) \Leftrightarrow G \\
S \Rightarrow (\neg E \vee \neg G)
[/math]

>> No.9243892

>>9243887
>provides feedback that
as in "shows that"?

>> No.9243902

>>9243874
>1) If I have a bilinear form, is it always possible to factorize it as the product of two linear forms?
yes iff it has rank 1
https://math.stackexchange.com/questions/2226229/rank-1-bilinear-form-is-a-product-of-two-linear-functionals-on-a-finite-dimens

>2) Is the representation of a bilinear form as the product of two linear forms unique?
I think if B(x,y)=L(x)K(y) then you probablyhave B(x,y)=(c*L(x))(1/c * K(y)) for any nonzero c

>> No.9243919

>>9243902
>https://math.stackexchange.com/questions/2226229/rank-1-bilinear-form-is-a-product-of-two-linear-functionals-on-a-finite-dimens

Nice. We have not even gotten to define the rank of a bilinear form but I see it is just (naturally) defined as the rank of the matrix of the bilinear form, which we just defined today. That's cool.

Do you have some examples for bilinear forms that are not of rank 1?

>>9243902
>I think if B(x,y)=L(x)K(y) then you probablyhave B(x,y)=(c*L(x))(1/c * K(y)) for any nonzero c

I see, I had not thought of that. But I think this is not really what I mean. This is trivial. L(x) and c*L(x) are not linear independent so technically this is a trivial non-unique factorization. Maybe there is unique factorization up to linear independence? Do you know anything about this?

>> No.9244013

I am bored.
Gimme some good shit about geometry (problems, articles, books, whatever)

>> No.9244082

How do you guys convince yourselves to study for a good couple of hours when you could play video games instead?

>> No.9244091

>>9244082
By convincing myself that I am not a hedonistic monkey, and do the thing that will help me progress in life.

>> No.9244099

>>9244082
One question:
Do there exist twenty-one consecutive integers each of which is divisible by one or more of the primes 2,3,5,7,11,13?

The moment you hear that question you should immediately feel the urge to grab a piece of paper and start thinking about it. These are the kinds of questions that are so nice they are actually better than video games and maybe sex.

Anyways, if any of you care. You can easily prove that indeed they exist and there are infinitely many and that if you have a computer then you can find one.

>> No.9244103

>>9244099
Oh man I barely understand the question.
Also my induction method is shit.

>> No.9244105

>>9244103
It is not induction. A solution is 9440.

>> No.9244107

>>9244103
>>9244105
By that, I mean that a solution is the 21 consecutive integers starting from 9440.

>> No.9244134

>>9243879
Thanks.

>> No.9244143

>>9244099
>prove that there are infinitely many
>infinitely many
No such thing.

>> No.9244146

>>9244143
>The Chinese remainder theorem does not exist
How do you even do math without infinity mate?

>> No.9244150

>>9244099
>The moment you hear that question you should immediately feel the urge to grab a piece of paper and start thinking about it
I definitely would if I were a number-theory subhuman, sadly I'm not though.

>> No.9244155

>>9244146
>The Chinese remainder theorem does not exist
It seems like you're new to mathematics, a theorem itself can be correct (and still exist) even if it assumes false statements.
>without "infinity"
I don't understand what you're saying. It doesn't sound rigorous, whatever it is.

>> No.9244159

>>9244155
>doesn't even know elementary theorems
>"It seems like you're new to mathematics"
ironic

>> No.9244164

>>9244159
>doesn't even know elementary theorems
What are you even talking about? Do you have some sort of mental disability? I guess it's to be expected from an engineer.

>> No.9244165

>>9244164
>What are you even talking about?
you not understanding the chinese remainder theorem

>Do you have some sort of mental disability?
no I'm fortunately just not a freshman like you

>> No.9244167

>>9244099
>>9244155
>infinity
>>>/lit/
We demand rigor here.

>> No.9244171

>>9244167
>We
Who?

>> No.9244175

>>9244165
>the chinese remainder theorem
Can be trivially reduced to assuming that [math]x \in \varnothing[/math]. A collection of "all" "integers" can't be shown to exist to begin with.
>not a freshman
That's pretty sad. I would understand your condition if you were still young, but I guess it's too late at this point.

>> No.9244177

>>9244150
>I definitely would if I were a number-theory subhuman
To be fair, you have to have a very high IQ to understand number theory.

>> No.9244179

>>9244177
Indeed, understanding such a trivial and retarded field requires super-human intelligence.

>> No.9244184

>>9244171
Mathematicians. Take your emotional arguments to >>>/lit/

>> No.9244197

>>9244175
>Can be trivially reduced to assuming that x∈O. A collection of "all" "integers" can't be shown to exist to begin with.
so this is the brainpower of freshmen...

>That's pretty sad. I would understand your condition if you were still young, but I guess it's too late at this point.
if you can't even understand the chinese remainder theorem you should stick to whatever they teach in the high school/freshman classes you're in until you're old enough to at least understand a proof

>> No.9244214

>>9244184
>Mathematicians
You're obviously not one based on your posts

>> No.9244223

>>9244214
I am one. Why would my post suggest otherwise?

>> No.9244228

>>9244223
Because you're not even aware of basic theorems that literally every student learns

Unless by mathematician you mean something different than what everyone else means

>> No.9244230

>>9244197
The Chinese remainder "theorem" assumes the existence of "integers" (which can't be shown to exist).

>> No.9244232

>>9244228
>theorems
You seem to be an idiot. No theorems have been mentioned in the latest posts of this thread.

>> No.9244234

>>9244230
the chinese remainder theorem doesn't assume the existence of the integers at all, try again when you've at least made it into university

>> No.9244238

>>9244179
>trivial
Please find me a formula that generates infinitely many twin primes

>> No.9244239

>>9244232
>You seem to be an idiot. No theorems have been mentioned in the latest posts of this thread.
You seem to be an idiot. No where in my post did I imply that theorems have been mentioned in the latest posts of this thread.

>> No.9244244

>>9244239
You're claiming I'm not aware of some "theorems". What are those?

>> No.9244249

>>9244238
>Please find me a formula that generates infinitely many twin primes
>infinitely many
Your goal reduces to finding an element of the empty set.

>> No.9244250

>>9244244
>What are those?
the chinese remainder theorem

>> No.9244259

>>9244249
Okay, find me a formula that generates the first 1 billion twin primes

>> No.9244260

>>9244013
>>9244013
pls

>> No.9244264

if the set of all axioms form a category then what are the morphisms?

>> No.9244270

>>9244250
How am I not "aware" of it? Is me being not aware of it a theorem?

>> No.9244273

>>9244264
STOP ASKING QUESTIONS JUST BELIEVE THAT CATEGORY THEORY WORKS AS A FOUNDATION PLEASE DO NOT DIG DEEPER STOP

>> No.9244275

>>9244259
The property of being prime can't be shown to be well-defined for large integers.

>> No.9244278

>>9244264
>all axioms
All axioms of what?

>> No.9244280

Lemma: You can't prove that ZFC is inconsistent or consistent.
(known fact)

Theorem: ZFC is consistent
Proof: Suppose ZFC has an inconsistent statement. Then I could use that inconsistent statement to prove that ZFC is inconsistent, contradicting the lemma. Therefore ZFC is consistent.

>> No.9244282

>>9244273
Retard detected.

>> No.9244286

>>9244282
Yes! I am so retarded! Lets talk about how retarded I am lol. Who wants to talk category theory? NO ONE. Who cares! Lets just throw banter around haha stop digging deeper haha I am so dumb!

>> No.9244289

>>9244280
>Lemma: You can't prove that ZFC is inconsistent or consistent.
>You can't prove
In what system? It's trivially provable if your system has LEM.

>> No.9244291

>>9244289
>In what system?
In ZFC itself.

>> No.9244294

>>9244286
>Who wants to talk category theory?
Apparently you do, but your retarded mind for some reason thinks that "the category of all axioms" (if it exists or even is well-defined (which it isn't)) has anything to do with category theory being used as a foundation.

>> No.9244296

>>9244294
Yes haha I am so retarded. Lets keep talking about how retarded I am.

>> No.9244298

>>9244291
Lemma: ZFC proves [math]Con(ZFC) \lor \neg Con(ZFC)[/math].
Proof:
Trivial. Apply LEM.
[math]\blacksquare[/math]

>> No.9244299

>>9244270
>How am I not "aware" of it?
See all your posts above.

>Is me being not aware of it a theorem?
No, it's a fact.

>> No.9244300

>>9244273
>JUST BELIEVE THAT CATEGORY THEORY WORKS AS A FOUNDATION
but it doesn't, and it's irrelevant to most of mathematics anyway

>> No.9244301

>>9244296
At least you realize that yourself.

>> No.9244303

>>9244298
>autism
You are proving that one or the other is true. But which one is it? I proved that it is consistent.

>> No.9244304

>>9244278
>All axioms of what?
Of mathematics.

>> No.9244305

>>9244275
>"large" integers
I don't understand what you're saying. It doesn't sound rigorous, whatever it is.

>> No.9244308

>>9244300
https://www.youtube.com/watch?v=huHY5ViHRYw
NO NO NO NO NOOOOOOOOOOO
NOOOOOOOOOOOO
STOP
STOP

PLEASE LETS JUST TALK ABOUT HOW RETARDED I AM THANK YOU

>> No.9244310

>>9244303
>But which one is it?
ZFC is non-constructive, so you won't know simply from my proof.
>I proved
Your assumption is false, you can prove anything you like.

>> No.9244311

>>9244308
>video games
>number theory
We need to euthanize these vermin.

>> No.9244312

>>9244304
What do you even mean by this? Is there a formal definition?

>> No.9244314

>>9244310
>Your assumption is false, you can prove anything you like.
I am just assuming Godel's incompleteness theorems.

>> No.9244316

>>9244312
>Is there a formal definition?
Of what?

>> No.9244319

>>9244311
Who the fuck mentioned number theory?
Also yes, lets tak about number theory. FORGET ABOUT CATEGORY THEORY NO NEED TO DISCUSS THAT IT JUST WORKS!

>> No.9244320

>>9244314
You "proved" the negation of the lemma given in >>9244298 so your assumptions contain a contradiction.

>> No.9244321
File: 126 KB, 960x720, Roberts-KurtGodel.jpg [View same] [iqdb] [saucenao] [google]
9244321

>>9244314
>Godel's incompleteness theorems.
This is what I was getting at with >>9244264
since Godel's "incompleteness" theorems is proved only for sets of axioms, maybe we can get around it by using a category of axioms instead

>> No.9244324

>>9244316
Of "all axioms of mathematics".

>> No.9244326

>>9244319
I couldn't care less about category theory, I just despise vermin such as yourself.

>> No.9244327

>>9244324
it's the construction of a category \mathcal{C} such that if \mathfrak{A} is an axiom then \mathfrak{A}\in Ob(\mathcal{C})

i just want to know what the morphisms will be

>> No.9244329

>>9244326
But I am a simple man. I am not a number theory man. I am a category theory man (but we should not talk about it)

>> No.9244333

>>9244327
>the construction of a category
It's not a construction of a category if you didn't specify what the morphisms are.

>> No.9244337

>>9244321
I think you mean "proper class" instead of category.
Look up compacity theorem: anything provable with infinitely many axioms is provable with a finite number of them.

>> No.9244338

>>9244333
do you know how to read, brainlet? that's why I'm asking what the morphisms are

>> No.9244339

>>9244329
>I am a category theory man
Didn't know they used number theory. Seems like they will have to perish as well.

>> No.9244340

>>9244337
>I think you mean "proper class" instead of category.
no, I mean category, I'm not sure what makesyou think I'dwrite something I don'tmean

>> No.9244344

>>9244338
>that's why I'm asking what the morphisms are
The identity morphisms.

>> No.9244346

>>9244339
But dude, everything is numbers. For example, just look at your screen. How many numbers can you see? Your post number, my post number, everyone else's post numbers. Those numbers on the corner counting how many posters are here, the 10-second auto refresh counter, etc.

Numbers are everything! Why do you hate them?

>> No.9244348

>>9244338
>that's why I'm asking what the morphisms are
>are
What morphisms are you talking about? There is no category yet.

>> No.9244349

>>9244348
>What morphisms are you talking about? There is no category yet.
theones in the category ofaxioms

do they noteven teach brainlets how to read anymore?

>> No.9244350

>>9244349
>category ofaxioms
Prove that such a category exists.

>> No.9244351

>>9244350
i don'tknow why you're asking me irrelevant questionsi don't care about, i'm only concernedabout themorphisms

>> No.9244352

>>9244346
>numbers
>video games
Aw dude you have aids, poz my neg hole please.

>> No.9244354
File: 1.80 MB, 1202x910, math and physics.png [View same] [iqdb] [saucenao] [google]
9244354

Threadly reminder to work with physicists.

>> No.9244355

>>9244351
see >>9244344

>> No.9244357

>>9244355
yes but those are in every category so it'sirrelevant to my question, that's not the answer

>> No.9244358

Somehow I'm not surprised the dog-eating subhuman's friend is one of the shitposters.

>> No.9244360

>>9244357
The only morphisms in your category are the identity morphisms.

>> No.9244361

>>9244360
>The only morphisms in your category are the identity morphisms.
whats' theproof?

>> No.9244362

>>9244314
No.
You wrote

>Lemma: You can't prove that ZFC is inconsistent or consistent.

What Godel theorem says is *if zfc is consistent* then it cannot prove that it is consistent (nor inconsistent).

There's always the possibility that ZFC is inconsistent and proves everything

>> No.9244363

>>9244361
It's included in the proof of your category being a category, look it up.

>> No.9244365

>>9244363
>It's included in the proof of your category being a category, look it up.
which book? i tried leinsters basiccategory theorybook but he doesnt go over the categoryof axioms

>> No.9244367

>>9244362
>then it cannot prove that it is consistent (nor inconsistent)
See lemma >>9244298
It's trivially provable.

>> No.9244369

>>9244365
Use the book you learned about the category of axioms from.

>> No.9244371

>>9244369
i told you its not in the leinsterbook

>> No.9244374

>>9244367
Your joke doesn't even work anymore the way I phrased it...

>> No.9244377

https://en.wikipedia.org/wiki/Axiomatic_system
>In mathematics, an axiomatic system is any set of axioms

how can an axiomatic system be a set of axioms if the sets require axioms?

>> No.9244380

>>9244374
What joke?

>> No.9244385

>>9244358
>dog-eating subhuman's
Why do wh*tes put dogs on a pedestal?

>> No.9244391

>>9244374
It's not a joke, brainlet.

>> No.9244393

>>9244380
Yes

>> No.9244396

Keep posting, subhuman. I can tell even your unrelated posts apart now.

>> No.9244398

>>9244371
On what page does his book mention the category of axioms?

>> No.9244399

>>9244396
>I can tell even your unrelated posts apart now.
Why wouldn't you be able to tell unrelated posts them apart?

>> No.9244400

>>9244396
which posts?

>> No.9244401

>>9244398
>On what page does his book mention the category of axioms?
if it did i wouldn'tbe askning what the morphismsare you fucking retard

>> No.9244403

Your friend is too talkative for his own good.

>> No.9244405

>>9244377
Whenever you do logic you are studying an target theory using an ambient theory. If you pick ZFC for both you have an ambient set of axioms that define the target sets.

>> No.9244406

>>9244401
You said you learned about it from his book right here
>>9244371

>> No.9244408

>>9244405
why bother with a 'target theory'?

>> No.9244410

>>9244406
do you not knowwhat the word 'not' means retard?>>9244406
>You said you learned about it from his book right here

>> No.9244416

>>9244410
Where did you learn about this category from?

>> No.9244417

>>9244408
Why bother with groups ? rings ? topological spaces ? numbers ?
Logic is just about studying theories, which are mathematical objects just like the ones above.
When you study ZFC you have to study it in some ambient mathematical framework, just like when you study groups.

>> No.9244420

>>9244416
>Where did you learn about this category from?
/mg/

>> No.9244421

>>9244408

>>9244416

>>9244417
idk it's real good for cooking dogs

>> No.9244424

>>9244417
>Why bother with groups ? rings ? topological spaces ? numbers ?
most of those are actually useful but not topological spaces

>> No.9244427
File: 13 KB, 220x165, 220px-Gaegogi-01.jpg [View same] [iqdb] [saucenao] [google]
9244427

>>9244421
Wha'ts the most mathematical way to prepare dog?

>> No.9244428

>>9244300
>>JUST BELIEVE THAT CATEGORY THEORY WORKS AS A FOUNDATION
>but it doesn't, and it's irrelevant to most of mathematics anyway
Is this true?

>> No.9244429

if the set of all axioms form a category then what are the morphisms?

>> No.9244430

>>9244424
>actually useful
In the grand scheme of things we're all just a speck of dust in the universe

>> No.9244433

What's the latest genuinely exciting theorem proved?

>> No.9244434

>>9244427
idk i use a physical way of cooking dogs

>> No.9244435

>>9244430
>In the grand scheme of things we're all just a speck of dust in the universe
Speak for yourself.

>> No.9244437

>>9244429
There are no morphisms, it's a set, you said it yourself. A set doesn't have morphisms.

>> No.9244440

>>9244424
true... you can't use them for cooking dogs

>> No.9244441

general topology makes me want to kill myself
two months into a semester of it and the entire course is nothing but endless definition spam

>> No.9244442

>>9244437
>There are no morphisms, it's a set
a set is trivially a category.

>> No.9244443

>>9244437
>There are no morphisms, it's a set, you said it yourself. A set doesn't have morphisms.
i'm asking aboute category theory not set theory

>> No.9244445

>>9244441
>general topology makes me want to kill myself
Do some category theory first then go back to it and it will make more sense.

>> No.9244446

>>9244443
prove that it exists.

>> No.9244451

>>9244440
Could you prove this assertion in ZFC using a locally small category of axioms ?

>> No.9244452

>>9244446
i'm not going to prove to you taht category theory exists just stop embarassing yourself stupid brainlet

>> No.9244454

>>9244442
Which doesn't have morphisms

>> No.9244455

>>9244454
trivially let every arrow be the identity.

>> No.9244456

>>9244454
all categories have morphisms brainlet

/mg/ really needs skill-testing captchas

>> No.9244458

>>9244451
no zfc is not locally dog-closed so i can't actually

>> No.9244461

Is it weird to tell people you dreamt about them?

>> No.9244462

>>9244458
can you do a canine compactification to make it locally dog closed?

>> No.9244463

>>9244456
>skill test
You go first: what is the initial object in the category of categories ? How many morphisms does it have ?

>> No.9244464

>>9244463
you don't even know the definition of a category brainlet, why are you proposing skill testing questions about things you don't even know a clue about?

>> No.9244467

>>9244464
WRONG. Please leave /sci/ immediately

>> No.9244468

>>9244462
nah the co-canine functor isn't representable

>> No.9244470

>>9244467
if I was wrong then you would know the definition of a category, but you don't ass seen here >>9244454

therefore i'm right, and so skill-testing questions would do the job they're supposed to by filtering brainletslike you from the discussion

>> No.9244471

>>9244463
>You go first: what is the initial object in the category of categories ?
it's isomorphic to the terminal object in Dog

>> No.9244476

>>9244471
what are the morphisms in [math] \textbf{Dog} [/math]?

>> No.9244478

>>9244470
>>what is the initial object in Cat?
>you don't even know the definition of a category brainlet
You claim this is the right answer to my question ?

>> No.9244479

>>9244478
>You claim this is the right answer to my question ?
no, weher did I claim that? you can do your own homework brainlet, if you can solve thatyou'll at least finally know what the definition ofa categpry is

>> No.9244483

>>9244476
the co-morphisms in [math]\textbf{Cat}[/math]

>> No.9244490

>>9244479
>weher did I claim that?
when you claimed you weren't wrong

>> No.9244492

>>9244490
that's a double negation

>> No.9244495

>>9244490
>when you claimed you weren't wrong
i wasn't, see my prof here >>9244470

>> No.9244499

>>9244492
You claim to not be wrong but you don't claim to be right ? So you reject excluded middle ? I respect that.

>> No.9244504

>>9244437
The approriate way to turn a set S into a category is by letting Ob(C) be a single point and then Hom(C)=S

>> No.9244506

>>9244499
>reject excluded middle
i accept it's double negation.

>> No.9244507

>>9244495
Your proof uses proof by contradiction. You can't use it if you choose to reject excluded middle. They are logically equivalent.

>> No.9244510

>>9244504
How do you define composition

>> No.9244511

>>9244510
>How do you define composition
w/ axiom of choice, which led to a contradiction when you construct the category of axioms

>> No.9244514

Hey guys it's 12:00am, have an algorithms exam next october 25th.
Convince me to stop playing Fallout to study a little.

>> No.9244515

>>9244507
>You can't use it if you choose to reject excluded middle.
but excluded middle is an axiom in the category, how can you reject it?

>> No.9244523

>>9244515
I think you just have to keep the full subcategory whose objects are the remaining axioms.

>> No.9244526

>>9239177
Whenever you have a sentence that says
>term is defined as ..
>The definition of term is ..
>If ... then we say it is term
>etc..
it is the same thing as saying
>A thing is term if and only if ...

For your example:
>A function is bijective if and only if it is both injective and surjective.

In general, definitions are always if and only if (biconditional). I have never seen an example when this is not the case though I have seen many textbooks only write "if" instead of "iff".

>> No.9244533

>>9244526
What are the morphisms in the category of all definitions

>> No.9244536

what ifwe consider the category where if X isa proposition then [math] X \in Ob(thecategory)?[/math] what does it mean?

>> No.9244538

>>9239141
>jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon jargon

>> No.9244539

>>9244538
>muh jargon!

>> No.9244541

>>9244536
Hom(X,Y) would be the set of all proofs that X implies Y, I think we're on to something anon

>> No.9244544

>>9244541
is the composition rule just concatenation of proofs?

>> No.9244555

>>9244544
Morally yes with some modus ponens inserted.
Isn't the 'and' operator a cartesian product ? And 'or' is a coproduct ?

>> No.9244568

>>9244538
>this expression indicates a non-mathematicians confusion and lack of understanding.

>> No.9244572
File: 913 KB, 500x450, 1508303002126.gif [View same] [iqdb] [saucenao] [google]
9244572

Is Azusa a category? What are the morphisms?

>> No.9244574

>>9244572
>Is Azusa a category? What are the morphisms?
Category theory may not be ready for such problems.

>> No.9244575

>>9244264
>>9244555
>>9244544
>>9244541
>>9244536
You can define a category where your objects are logical statements and your morphisms are proofs of entailment.

If [math]P[/math] and [math]Q[/math] are statements (objects in the category), then [math]a:P\to Q[/math] is a proof that P entails Q.

Conjunction = products in the category
Disjunction = coproducts
Implication = exponential objects
Bottom (falsity) = initial object
Top = terminal object

Negation and axioms are handled through special morphisms and exponential objects to/from top and bottom.

In general, any category can be interpreted as a logic (though the type of logic may vary depending on the limits the category supports), we call such interpretations the "internal logic" of a category.

>>9244300
>>9244273
Ignore these brainlets and their chain of responses.

>>9244428
No, but imho there are things that category theory does not seem suited to doing (and all the textbooks out there will just handwave it away whenever it becomes contrived or inconvenient). At least that's been my experience here: >>9240992


>>9244437
>>9244454
Why the fuck are you people listening to this brainlet.

>>9244504
>>9244443
>>9244442
Morphisms and objects are defined in the category. In the category set your morphisms are functions from sets to sets. However there are lots of other categories one could define where your objects are sets and your morphisms are something different (eg. [math]A\to B[/math] iff [math]A\subseteq B[/math].

>>9244533
Kill yourself, moron.

>>9242870
All of them. Just look at all the brainlets ITT.

>> No.9244581

>>9244575
>>>9244264
>>>9244555
>>>9244544
>>>9244541
>>>9244536
>You can define a category where your objects are logical statements and your morphisms are proofs of entailment.
>If P and Q are statements (objects in the category), then a:P→Q is a proof that P entails Q.
>Conjunction = products in the category
>Disjunction = coproducts
>Implication = exponential objects
>Bottom (falsity) = initial object
>Top = terminal object
>Negation and axioms are handled through special morphisms and exponential objects to/from top and bottom.
>In general, any category can be interpreted as a logic (though the type of logic may vary depending on the limits the category supports), we call such interpretations the "internal logic" of a category.
>>>9244300
>>>9244273
>Ignore these brainlets and their chain of responses.
>>>9244428
>No, but imho there are things that category theory does not seem suited to doing (and all the textbooks out there will just handwave it away whenever it becomes contrived or inconvenient). At least that's been my experience here: >>9240992
>>>9244437
>>>9244454
>Why the fuck are you people listening to this brainlet.
>>>9244504
>>>9244443
>>>9244442
>Morphisms and objects are defined in the category. In the category set your morphisms are functions from sets to sets. However there are lots of other categories one could define where your objects are sets and your morphisms are something different (eg. A→B iff A⊆B.
>>>9244533
>Kill yourself, moron.
>>>9242870
>All of them. Just look at all the brainlets ITT.
thanks

>> No.9244590

>>9244575
Can you help with >>9244572

>> No.9244599

>>9244575
>Why the fuck are you people listening to this brainlet.
Nobody is listening to him, we don't listen to subhumans here and anyone who doesn't know basic category theory is one.

>> No.9244608

>>9244599
>basic category theory
Redundant.

>> No.9244619

>>9244608
Not really, it's a technical term.

>> No.9244623

>>9244619
>Not really
There is no "advanced" category theory.

>> No.9244624

>>9244623
There is "higher" category theory, which is again a technical term. Not that a subhuman would know.

>> No.9244627

>>9244624
>There is "higher" category theory, which is again a technical term.
Higher category theory is a subset of basic category theory.

>> No.9244629

>>9244627
>subset
wew lad

>> No.9244630
File: 2.42 MB, 320x240, 1462228321315.gif [View same] [iqdb] [saucenao] [google]
9244630

>> No.9244632

>>9244629
>wew lad
I'm not a "lad".

>> No.9244641

>>9244629
must have typo'd 'subcategory'

>> No.9244656

>>9244632
Fuck off to >>>/t/aiwan and >>>/r/dogs/

>> No.9244662

>>9244627
>subset
I don't believe in sets.

>> No.9244663

>>9244662
>"believe"
Not math.

>> No.9244682

>>9244663
Spotted the freshman.

>> No.9244686

>>9244682
>Spotted the freshman.
I'm not a "man".

>> No.9244687

>>9244682
lol... the irony

>> No.9244689

>>9244686
Fuck off to >>>/t/aiwan and >>>/r/dogs/
>>9244687
>lmao
Fuck off to >>>/r/physics/ it's a good place for dog-eaters

>> No.9244693

>>9244689
>>lmao
Who are you quoting?

>> No.9244697

>>9244693
i'm quoting a dog i just butchered

>> No.9244699

is the death penalty an acceptable solution to dog-eating/physics? i think it is

>> No.9244700

>>9244699
What's wrong with eating dogs?

>> No.9244730

>>9244700
Don't you know anything?

>One winter, when our people were starving, a young man and his family were camped by themselves as they searched for food. The wolves found the family and appeared to them as young men bringing fresh meat to the lodge. The wolves took this family with them, showing the man how to cooperate with other people when he hunted buffalo and other animals. The wolves introduced the people to the other animals in their world. The human beings learned that animals with hooves and horns were all right to eat, but that animals with paws and claws should be left alone.

>The wolves disappeared in the spring, but we still see them in the sky as makoiyohsokoyi, the Wolf Trail (the Milky Way). These stars constantly remind us of how we should live together.

>> No.9244743

>>9244730
>>One winter, when our people were starving, a young man and his family were camped by themselves as they searched for food. The wolves found the family and appeared to them as young men bringing fresh meat to the lodge. The wolves took this family with them, showing the man how to cooperate with other people when he hunted buffalo and other animals. The wolves introduced the people to the other animals in their world. The human beings learned that animals with hooves and horns were all right to eat, but that animals with paws and claws should be left alone.
>>The wolves disappeared in the spring, but we still see them in the sky as makoiyohsokoyi, the Wolf Trail (the Milky Way). These stars constantly remind us of how we should live together.
>>>/lit/

>> No.9244745
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9244745

>>9244743
hoofman detected

>> No.9244753

There's nothing mathematical about putting dogs on a pedestal instead of eating them like any other meat.

>> No.9244880

>>9244753
i dont care about mathematics lmao
eating dogs is very physical its studied in string's theory and physical Intuitions help out with cooking them

>> No.9245045
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9245045

What are some good authors for combinatorics besides Stanley?

>> No.9245156
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9245156

anybody familiar with model theory, could you briefly explain quantifier elimination and methods to prove that a theory has it, how to prove nullstellensatz, omitting type theorem and all that 'pre morley model theory'? I'm curious about those topics but don't have enouh time to study the details right now

>> No.9245417
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9245417

These threads are so bad nowadays. They were never good, but they weren't this bad either. At least I can give lectures on algebraic topology and category theory to my stuffed animals.

>> No.9245453

>>9245417
Answer this: >>9240992

>> No.9245490
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9245490

>>9245453
For a group [math]G[/math] and any element [math]g \in G[/math], you can define conjugation using this element by [math]\varphi_g \colon G \to G[/math] with [math]\forall h\in G: \varphi_g (h)= ghg^{-1}[/math], as you already seem to know (since you mentioned the embedding), but this can also be done the same way if you consider endofunctors of this sort of a groupoid of one object. This will then give rise to a category whose object is the group category and arrows are these conjugation morphisms, a subcategory of the category with all fully faithful endofunctors, both of which are groups in the categorical sense. I'll think about how to define a general action. It could simply be an endomorphism of the arrow class of the object your group is acting on (for example the (subgroups of the) Möbius group acting on the upper half plane can be thought of mapping the identity morphism to other homeomorphisms), but I'm not 100% sure yet.

Did this help?

>> No.9245551

Where did I go wrong in my proof?

Let R be a relation on a set X that is both symmetric and transitive. Let a, b ∈ X and suppose that aRb; then by symmetry, bRa. Then since aRb and bRa, transitivity means that aRa. Therefore every element of X is related to itself, and R must also be reflexive.

>> No.9245574

>>9245551
>Where did I go wrong in my proof?
here:
>Therefore every element of X is related to itself, and R must also be reflexive.

>> No.9245578
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9245578

>>9245551
Nothing guarantees there is a [math]b[/math] for every [math]a[/math] satisfying [math]aRb[/math].

>> No.9245584

>>9245574
>>9245578
Yep I see. Now that I think about it, you can have relations that are symmetric and transitive but not reflexive, so the entire premise of my proof is false anyways

>> No.9245586

>>9245490
I was thinking something similar would be required for dealing with conjugation. I think it would have to be done in one of two ways.

Either as
>endofunctors on the group category as you described
or as
>natural transformations on underlying set functors (from the group category to Set) somehow.

Both ways seem like crude fixes but at this point I'm pretty sure there isn't any other more straightforward way (I've looked through a bunch of textbooks and as soon as they reach the topic of the conjugation action of a group onto itself they quietly abandon category theory and do things in a handwavy way that mixes groups (not as tuples) and sets.

>It could simply be an endomorphism of the arrow class of the object your group is acting on
I'm honestly not sure what you mean here.

>for example the (subgroups of the) Möbius group acting on the upper half plane can be thought of mapping the identity morphism to other homeomorphisms), but I'm not 100% sure yet.
This has me more bewildered. I've only briefly studied mobious transformations and wasn't aware they formed a group. Moreover you're talking about homeomorphisms so I guess you're talking about functors from that group to the category of topological spaces?

You have been helpful. At least it seems there isn't some obvious answer I've missed and (I think) we seem to be approaching it along similar lines.

>>9245551
It is not necessarily the case that a is related to b. For an example, let [math]A=\{a\}[/math], [math]B=\{b\}[/math], [math]R=\emptyset[/math]. Then [math]R\subseteq A\times B[/math] satisfying the definition of a relation. Moreover, the sentences:
[math]aRb\Rightarrow bRa[/math]
[math]aRb\land bRc \Rightarrow aRc[/math]
are both vacuously true. So, you have transitivity and symmetry. However, since there are some things that aren't related to anything (since [math]R[/math] is empty) then your relation is not reflexive.

>> No.9245692

>>9245417
Answer this: >>9244572

>> No.9245696
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9245696

>>9245586
>Both ways seem like crude fixes but at this point I'm pretty sure there isn't any other more straightforward way (I've looked through a bunch of textbooks and as soon as they reach the topic of the conjugation action of a group onto itself they quietly abandon category theory and do things in a handwavy way that mixes groups (not as tuples) and sets.
Yes, I have encountered the same.

>I'm honestly not sure what you mean here.
I was meaning something like this: if a group [math]G[/math] acts on some object [math]X[/math] in a category, then the action could maybe be expressed as a function [math]\varphi_g \colon \text{Hom}(X, X)\to\text{Hom}(X, X)[/math].

>This has me more bewildered. I've only briefly studied mobious transformations and wasn't aware they formed a group. Moreover you're talking about homeomorphisms so I guess you're talking about functors from that group to the category of topological spaces?
In hyperbolic geometry, you have this group [math]\mathbf{Möb}(\mathbb{H})=\{ z\mapsto \frac{az + b}{cz + d}\ |\ ad -bc >0\}[/math], and this is indeed a group. It's action is quite easy: [math]\gamma .x = \gamma(x)[/math], but this can be thought of as [math]1_{\mathbb{H}} \mapsto \gamma[/math], or actually just the composition of the Möbius transfromations from the left if I remember correctly. This is just one example, but it's good to have something concrete to motivate the abstract.

>> No.9245735

>>9245417
Physishits and their friends have infested this place, what else did you expect?

>> No.9245830

>>9245417
>These threads are so bad nowadays. They were never good, but they weren't this bad either.
What's so bad about them?

>> No.9246089

Pretty brainlet question, but I'm wondering if there is a specific name for computing the limit of fairly complicated rational functions as x approaches infinity, by approximating the behaviour of the function at extreme x values with another rational function that only contains the dominant terms? I don't know what this is actually called.

>> No.9246189
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9246189

>>9246089
>infinity
No such thing.

>> No.9246216

>>9246189
*zeno's laterally*