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00:15
Could be worse I suppose!
I need to go to bed. Good night all.
hello all
anyone around?
hi there :)
hey
whats up
does anyone here understand 2nd order Peano arithmetic, or perhaps just the arithmetic hierarchy, who may be able to offer me some guidance?
00:27
not really but I can be a rubber duck
heh
which means?
hhh
hhh
Suppose a problem "Find a factor for the term 1+y in a problem such as 'x+xy'". Here for example x(1+y) so the asked factor is x. What is the name for this "asked factor"?
I heard that 2nd order PA has only one model, the "true natural numbers".. yet there is an axiomatic extension which is not inconsistent yet as no model
PA with a new symbol $\infty$ and the axioms $\infty \not = Z$, $\infty \not = SZ$, $\infty \not = SSZ$, ...
its consistent because any proof of false has to use finitely many axioms, but any finite portion of the axioms are modelled by the standard model
but it doesn't have a model...
so second order is very weird
@hhh "common factor"
00:32
aha
@caveman: Yeah, so it's sort of an artificial construction.
Interesting, I suppose.
I was only really aware of the first result.
That's actually part of what I'm concerned with (though I'm already acquainted with it well enough by now).
have you proved it?
@caveman: So now I'm trying to find a Sigma_1-axiomatizable deductive system for PA^2
well, Dedekind proved it, but I found a nice proof in a book and read over it :)
lol
I didn't realize that went back as far as Dedekind
yeah. apparently the original Peano Arithmetic was 2nd order though, so less surprising then perhaps
hhh
hhh
@anon 1*2+3, what is the common factor of 2?
00:38
so. are you familiar with Sigma_1 formulas?
hhh
hhh
1?
they're easily enough defined anyway
even if the arithmetical heirarchy is a little more complicated.
hhh
hhh
1*(2+3)=1*5, common factor of 1 is 5? Right?
they're just of the form (quantifier) (bounded quantifier) (quantifier-free formula)
where the first two are optional i believe
00:42
hmm
@hhh that question is (a) of a different type than the other one and (b) doesn't make sense. in a sum a1+a2+..., the collection of summands {a1,a2,...} has a common factor which is the gcd of the terms. in the case of 1*2+3, you don't say "common factor of 2," nor do you say "common factor of 1 is 5": you say the common factor of 1*2 and 3 is 1 (since gcd(1*2,3)=1). note that the common factor of a sum depends on how it's presented, e.g. the c.f. in 1*2+3*2 is 2 but the c.f. in 2*2+2*2 is 4.
hhh
hhh
Another example 1+1+1+1+1+1+1+1 -- so the gcf=1?
@hhh $\gcd(1,1,1,....1)=1$, yes
hhh
hhh
-
What is this factor called?
3*4*5*5 with respect to 5 should return 3*4
3*4*5*5 with respect to 4 should return 3*5*5
3*4*5*5 with respect to 3 should return 4*5*5
(B) This factor called?

3*4*5*5 with respect to 5 should return 3*4
3*4*5*5 with respect to 4 should return 3*5
3*4*5*5 with respect to 3 should return 4*5
(C) This factor called?
:(
who cares
00:55
2 years later...
01:09
...2 years earlier
01:43
@κρανίοπεριπολία I made an animation again:
0
A: Folded Paper Shape TikZ

Karl's studentsJust 4 fun with PSTricks. \documentclass[pstricks,border=12pt]{standalone} \usepackage{pst-eucl,multido,fp} \FPset\Width{4.00}% paper width \FPset\Height{6.00}% paper height \FPset\Step{0.50}% interline skip \FPeval\Lines{round(Height/Step-1:0)}% number of lines \def\X{2}% abscissa of the t...

02:03
@κρανίοπεριπολία Thanks!
02:22
The last message was posted 3 years ago.
02:34
The first message will be posted 3 years from now.
This message will never be posted.
@anon Hi, anon! I miss your other gravatar. But this one is nice...does it depend on mood? this one seems a little more "light-hearted".
yes, it depends on mood. I have like 10 old gravatars, not sure which one you miss.
@anon I think the preceding this (most recent). I've been switching a bit, too...but not drastically... :-)
03:48
This is not a message that was never posted.
The last message was posted 5 decades ago.
@Karl'sstudents Again?
@GustavoBandeira Yes. again and again.
@Karl'sstudents What are you?
04:04
@GustavoBandeira I am just a free human being without a job.
Mainly I am learning TeX and PSTricks.
Why you learn that?
@GustavoBandeira graduated from high school only :D
Why don't we dominate the world?
@GustavoBandeira Because there is no purpose. :D
Of course there's a purpose.
But, considering it doesn't. Why don't we dominate all banana plantations?
banana playstations.
I'm looking some old photos of me. I was so beautiful. =/
04:15
Keep lookin' playa...
What is interesting is that I thought I was ugly at the time.
I know how to solve the problem: Until now I have evidence that I'll find myself beautiful in the future, by knowing that I know that I'm beautiful now.
$[3,1]=[3,1]'[3,1],+[3,2]'[3,2],$
@κρανίοπεριπολία KrakaKarmenElektra
@Ethan Hello!
hello
What do you study? Everytime you seems to be asking something hard, dunno.
04:21
you first
lol
what do u study
$\Huge\text{(removed)}$
Oh, I am dumb. I'm reviewing algebra basics and I'm going through algebra, naive set theory, logic and calculus now.
oh
I study math
But I know almost nothing because I neglected math studies while I was younger.
elementary
number theory
04:23
Elementary?
i don't really know any analysis
Why do you like number theory?
@κρανίοπεριπολία Kraka Kaka Cabbage
C'mon. Don't annoy him.
This is the reason for studying NT?
I didn't at first, but after I did it for a while I started to like it
Nice. I feel some curiosity on NT.
04:27
XD
I remember the time when I thought that NT was only about prime numbers.
Oh, Ethan. Is NT a disliked field?
I'm not sure, but I guess people hate NT, Geometry and set theory.
From what I know, and I know very very very little
It's mostly based on the number of "I hate (insert_field_here)" and "I love (insert_field_here)" I've read in chat.
Yes.
NT is used in cryptography, right?
I shouldn't talk anymore, il make myself look stupid, because of my lack of knowledge
Pft. I'm making myself look stupid for a lot of time.
I think for the Rsa cryptosystem
04:30
But I really don't care, it's good to chat with people about math.
yeah number theory and crypto are best buds
I don't know enough about cryptography to really judge its use
infact I don't really know anything about cryptography
Just the rsa algorithm
lol
rsa is basically unbreakable but super slow
The older term for number theory is arithmetic. By the early twentieth century, it had been superseded by "number theory".[note 2] (The word "arithmetic" is used by the general public to mean "elementary calculations"; it has also acquired other meanings in mathematical logic, as in Peano arithmetic, and computer science, as in floating point arithmetic.)
The use of the term arithmetic for number theory regained some ground in the second half of the 20th century, arguably in part due to French influence.[note 3] In particular, arithmetical is preferred as an adjective to number-theoretic.
Oh, it was called arithmetic.
most crypto research nowadays is in multivariate crypto
04:32
Now it kinda make sense in my head.
I sometimes find somthing I think is interesting, then study it for a couple of days only to find it is trivially obvious
Its the feeling you get after manipulating a bunch of shit only to find the original thing is equal to itself
I don't know anything though
A Red Indian Thought He Might Eat Tobacco In Church.
I have only been studying math for about 2 years
04:35
But you're studying it formally? You're going to math school?
Im studying it on my own time
Nice.
Same to me.
How do you solve your doubts? Only MSE?
I havn't studied anything really for a couple of months just screwing around with the knowledge I already have looking at stuff I think is interesting
I have trouble reading stuff cause of my ocd
@Ethan Hey there... just got back from proctoring a final at UCLA
@robjohn where?
@Ethan where was the exam?
ye
@Ethan In Powell
@robjohn do you know people in the math department?
04:40
@Ethan There are still some people who used to be there when I was on the faculty there, but no students.
@GustavoBandeira It is not always true. Poincare's conjecture proof is not obvious even though we know the answer.
@Ethan doesn't sound familiar
@robjohn don blasius?
@Ethan I've heard the name, but I don't think he was there that long
nvm lol, oh well, hopefully I can run into you sometime
not that id even know it lol
@robjohn they have finals past 8?
@GustavoBandeira what kind of math did you do 2 day
04:50
@Ethan Well. Not much, I'm just trying to solve aditional doubts on this:
1
Q: Doubts on the truth table of $\models$

Gustavo BandeiraI'm reading Shawn Hedman's A First Course in Logic: An Introduction to Model Theory, Proof Theory, Computability, and Complexity: Definition 1.18 Formula $G$ is a consequence of formula $F$ if for every assignment $A$, if $A\models F$ then $A\models G$. We denote this by $F\models G$. ...

When I doubt my knowledge on something or am not totally sure something is true, my feat start to cringe
Haha
It's not hard. It's basic in logic.
defintion 1.18, i asked about like 2 hours ago
I think
this is how I worded it, I don't understand the logic notation
If the truth of statement a implies the truth of statement b
Do you know truth tables?
and the truth of statement b implies the truth of statement a
then statement a conveys the same info as statement b
04:55
Hello everyone!
@GustavoBandeira no what are those
I am thinking, if in the dearth of cohomology tools, what would be left in number-theory?
what are cohomology tools
@Ethan In boolean algebra, you have certain operations.
Their arity is 2, and the only numbers allowed are 0 and 1
(Not has arity 1)
04:59
@GustavoBandeira what are they used for? are the validity of logical statements ever that complicated that in order to phrase them rigorously you need to introduce all this notation
@Ethan Actually you need to memorize only a few of them: Not, Nand, And, Or, Implies, etc.
@Ethan The tools using homological algebra in number theory, such duality theorems, Herbrand quotients, etc.
Sorry for the wrong tag.
@Ethan These gates are the tools with which we build our computers.
I'm reading this book, I have already built the ALU (Arithmetical Logical Unit), look.
mathematicians also want $\forall$ and $\exists$ in their work. yes, being familiar with propositional logic is extremely useful for anyone who wants to do serious math. I'd recommend it as mandatory even.
@anon For me?
05:04
lol
I learned some of that notation from linear algebra
@GustavoBandeira have you designed digital circuits before
@anon Yep. I've read about it here.
@GustavoBandeira Would you be able to explain what first-order logic is in a nutshell? I head of a weaker notion of isomorphism between algebraic structure is elementary equivalence, where two objects (say, groups) share the same set of true first-order statements that can be said about them - in a sense, this says they are approximately the same, at the resolution of "first-order" descriptions, and I'd like to know what this means.
@κρανίοπεριπολία I am here!(raising hands
05:08
@anon Sorry. I still don't know about that, I'm stuck in the problem from my last question and I'm too afraid of speculate and give wrong information.
@GustavoBandeira which answer?
I guess there's one on logic.
12
A: Mathematical subjects you wish you learned earlier

andyvn22I wish I'd learned logic much, much earlier. Obviously young students couldn't handle much depth, but at least a basic introduction to a few concepts would be nice. Just understanding the concept of axioms and deductive rules would put all of math into some perspective. When I finally understood ...

I was trying to emphasize "anyone" in anon's recommendation:
propositional logic is extremely useful for **anyone** who wants to do serious math.
I see. Sorry for the misunderstanding.
05:11
@Ethan I've only learned a bit of electronics when I was younger, most of the instruction to build the circuits is given in the book.
06:01
(removed)
This message will be (removed) every 5 minutes...
This content was removed because it violates our Terms of Service.
what does $Lt_{n \to \infty} (a_n)^{1/n}$ mean?
Is this supposed to be $\lim_{n \to \infty} (a_n)^{1/n}$?
@Eric I think so.
@Karl'sstudents These terms of service were removed because they violate our users.
06:12
@GustavoBandeira :-)
Whoever was violated read the Terms of Service.
06:30
Mathematics was born to make your life better!
We were all born to try to make mathematics better.
07:28
No.
What the heck.
I hate fatalism.
07:44
What's that got to do with fate?
We don't have to die trying.
Just make a choice to try or not...
nice try^
We were all born to decide if we want to try and make mathematics better.
@Ethan The last final ended at 6 PM
@GustavoBandeira They wish...
08:25
@robjohn Can you contribute to my last question? I guess you might know something about it.
user19161
Hey @GustavoBandeira nice new pic!
@JasperLoy Yep. Recursion.
user19161
@GustavoBandeira How did you make it?
I just used GIMP, I copied me, inverted the image, paste some copies on the other side and reduced the size og these instances.
user19161
Ah, GIMP is too complicated for a banana like me.
08:40
says the banana that purportedly switches out distros like his daily wardrobes
and distributes his razor sharp wit with uncanny ease
09:39
@κρανίοπεριπολία And there is a math teacher with a radar gun at the side of the road
@GustavoBandeira You mean this question?
@robjohn Yes.
@JasperLoy: Let's see the TeX.SX chat room: chat.stackexchange.com/rooms/41/tex-latex-and-friends
@GustavoBandeira I'm sorry, but I perhaps don't understand the question fully, but I don't see an area where any tool (computers included) might not be useful
@robjohn Oh. Now I get what did tried to say.
The definition for "useful" - which implies usefulness only in some circumstances - is poorly defined.
09:54
@GustavoBandeira well, that's good, because I thought Did's comments were non-constructive
@κρανίοπεριπολία I have another alternate.
Greetings
@robjohn: is this limit familiar to you? $\lim_{x\to\infty} \int_0^x \frac{\log(1+t)}{t} \mathrm{d}t - \frac{\log^2(1+x)}{2} = \frac{\pi^2}{6}$ I evaluated it in 2 different ways, one involves dilogarithm identities and the other one(rather hard) involves mutivariable calculus. I was wondering if we can avoid the dilogarithm identities and the multiple integrals.
10:18
@Chris'ssisterandpals that is $$\int_0^\infty\frac{\log(1+t)}{t(1+t)}\mathrm{d}t$$
@robjohn: right. A very good point.
10:33
Since
$$\zeta(x) \Gamma(x)=\int_0^{\infty} \frac{y^{x-1}}{e^y-1} \mathrm{d}y$$
$$\int_0^{\infty} \frac{y}{e^y-1} \mathrm{d}y=\zeta(2) \Gamma(2)=\frac{\pi^2}{6}$$
and the last integral is obtained by letting $1+t=e^y$
$$
s=\frac1{1+t}\Rightarrow\mathrm{d}t=-\frac{\mathrm{d}s}{s^2}
$$
$$
\begin{align}
&\lim_{x\to\infty}\int_0^x\frac{\log(1+t)}{t}\mathrm{d}t-\frac{\log^2(1+x)}{2}\\
&=\int_0^\infty\frac{\log(1+t)}{t(1+t)}\mathrm{d}t\\
&=\int_0^\infty\frac{\log(1+t)}{(1+t-1)(1+t)}\mathrm{d}t\\
&=\int_0^\infty\frac{\log(1+t)}{(1+t)^2}\left(1+\frac1{1+t}+\frac1{(1+t)^2}+\frac1{(1+t)^3}+\dots\right)\mathrm{d}t\\
&=-\int_0^1\log(s)(1+s+s^2+s^3+\dots)\mathrm{d}s\\
&=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\dots\\
@robjohn: your way is far nicer :-)
@Chris'ssisterandpals It is just a disguised version of the proof that $\int_0^\infty\frac{t^{x-1}\,\mathrm{d}t}{e^t-1}=\Gamma(x)\zeta(x)$
@robjohn: sure, but you did things elementarily
@robjohn: thank you for your feedback. I go now to continue my work on some problems.
@Chris'ssisterandpals good luck :-)
 
1 hour later…
11:50
Hello. Is there any body out there?
Don't know if you can hear me.
@PeterTamaroff I can hear you
(well, see what you write)
@PeterTamaroff In space, no one can hear you scream...
12:35
0
Q: Number theory $a/d$

pourjourSuppose in $\mathbb{N}^** \mathbb{N}^*$ the equation $(E): x^2+y^2+xy-13x=0$. We set $x=ad$ and $y=bd$ and $d=PGCD(x,y)$. How can I prove that $a/d$?

ah any help please?
0
Q: Number theory $a/d$

pourjourSuppose in $\mathbb{N}^** \mathbb{N}^*$ the equation $(E): x^2+y^2+xy-13x=0$. We set $x=ad$ and $y=bd$ and $d=PGCD(x,y)$. How can I prove that $a/d$?

ah any help please?
@pourjour $a/d$ is not a statement, so there is nothing that can be proved
@TobiasKildetoft u can read like that $d=ka / k \in \mathbb{N}$
@pourjour no, I can't
@TobiasKildetoft is there any other symbol
@pourjour you mean to write that $d$ divides $a$?
that is written as $d\mid a$
12:53
@TobiasKildetoft $a|d$
user19161
@robjohn What about the person himself?
13:11
@JasperLoy hmm good question
oh thats so depressing
some students are not smart enough to unterstand what to do in a multiple choice test
@DominicMichaelis you mean not smart enough to answer the question or not smart enough to be able to fill in the answer?
It was Put a "F" if the statement is false and a "T" if the statement is true
and some made a cross
ohh, that's bad
@TobiasKildetoft any idea abour the last problem
1
Q: Number theory $a|d$

pourjourSuppose in $\mathbb{N}^** \mathbb{N}^*$ the equation $(E): x^2+y^2+xy-13x=0$. We set $x=ad$ and $y=bd$ and $d=GCD(x,y)$. How can I prove that $a|d$?

13:24
@tobias
@rob
i wanna show tahat a lin transf. on a vector space V of dim = n is nilpotent iff $f^n\equiv 0$
Now, if f^n =o
It is nilpotent
@peter do you know jordan ?
@dominic no
now suppose f^n =/= 0
And take a basis
ah bad luck :(
It is clear f^k =/= 0 for k<n
So lets show it is tfue fir k>n
@JasperLoy just quoting Aliens
@PeterTamaroff how do you define nilpotent?
13:28
that is, for each k>n, there exists a basis vector fir which f^k(v_i)=/=0
i think we can use induction, and the division algorithm i think
@rob there exists an s for which f^s=0 identically
@PeterTamaroff Ah, so you are wondering about showing that $n$ is sufficient.
@PeterTamaroff the easiest way is using Jordan Normal Form
13:43
how to prove $n^7 \equiv n[42]$
@pourjour: 42 = 2 * 3 * 7, so it's enough to prove n^7 = n mod 2, n^7 = n mod 3 and n^7 = n mod 7.
@m.k. thanks
no problem
@robjohn Hey
I changed my display name :)
@PeterTamaroff: If a $n \times n$ matrix a satisfies A^k = 0, then its minimal polynomial is of the form x^t. Here t <= n since the minimial polynomial has degree at most n by Cayley-Hamilton. Does this work?
13:56
@PeterTamaroff Prove this: A linear map $T$ is nilpotent iff there is a basis for $V$ such that the matrix for $T$ is upper triangular with all zeros on the diagonal.
@JasperLoy I changed my display name
@PeterTamaroff got it?
14:18
look anon is bill dubuqe math.stackexchange.com/a/337641/58512
hehe
it does look better than the usual cursive latex
Ok ben. I'll try that
14:44
@林伟轩 So I see
Another joins the ranks of the encrypted ;-D
(removed)
@robjohn How did you ping me?
How many ways can I map {1,2,...,n} to {1,2,3} so that the range has two elements? Would this be 3C(2n,n) since there are 3 pairs of length 2 in {1,2,3}?
I'm starting to doubt that it is a combination, and I doubt my parameters (2n, n)
14:50
@林伟轩 Lin?
@Karl'sstudents yes?
@林伟轩 Lin Wei?
@Karl'sstudents what are you asking?
@林伟轩 copy and paste?
@林伟轩 Spelling
14:51
@Karl'sstudents huh?
@林伟轩 how to spell your display name?
that's it
@林伟轩 what do you mean? The same way I usually do...
@robjohn Like with you I would click r for your name to come up
Using "reply to this message"
14:53
with mine
"If d divides all the terms but one in an integer equation, it must divide the last"
is that a rule
@林伟轩 that doesn't link to the previous comment
posted 15s ago – permalink
reply to this message

star as interesting
flag as spam/offensive

flag for moderator
@pourjour, yes, if aX = Y then a|Y
@κρανίοπεριπολία that takes too long
14:54
@pourjour, so e.g. aA + Y + aB = aC implies a|Y since we have a(A+B-C) = -Y
:8633991
Click there...
@caveman so can we conclude from this $a^2d+b^2d+abd-13a=0 \Rightarrow a|d$ or $13 |d$
14:58
@Karl'sstudents @ 你 会说 中文 吗?
@Karl'sstudents 你 是 中 国 人?
@pourjour, you definitely get a|d, I don't know what you mean 13/d
@pourjour, you can also conclude d|13a
@robjohn True using the LHS takes one extra click.
@κρανίοπεριπολία yep
@伟轩 Only Jasper Loy knows.
@Karl'sstudents 你 是 哪 儿 来 的?
14:59
[About this question](http://math.stackexchange.com/questions/315073/how-can-commuting-with-frobenius-imply-the-order-of-an-element-in-the-inertia-gr)
Maybe commuting with the Frobenius implies that the representation is one-dimensional, hence lying in the ground field, thus the order statement?

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