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00:00
I need to get some shuteye now. I saw your question in main. Feel free to answer it yourself. If no one else does, I will, but only in about 8 hours :-)
G'night!
okay
I will do that thanks again
haha
@JyrkiLahtonen did you see alex argument
It always seems strange whenever authors write rôle instead of role in books. Is there a good reason for this?
It was a traditional way of writing the word. It is mostly died out, but some traditionalists still use it (or were taught that way)
00:15
Does it have some French influence, @CarlMummert?
yes, in French it would still have that accent above the o. Looking at Google NGrams, it looks like the accent use in English peaked in the mid-20th century and has been decreasing since. And the accent version in English was always dwarfed by the non-accent version
That's pretty cool. Do you research math, @CarlMummert?
Guys, I have two questions - is it bad manners if I answer already answered questions, ones that already have a good answer, and thus probably adding little value, if any? And is there anything wrong with writing answers I'm not completely sure about, and later perhaps deleting it if it turns out wrong?
@Khallil yes, I am a logician
I wouldn't suggest doing the last thing you wrote about. As for the first, maybe adding a comment would be better if you won't be adding enough to constitute a separate answer, @Jake1234.
00:21
@Jake1234: adding new answers is OK if you can add something that isn't already covered - new info or a new perspective. If you are just duplicating old answers, people may downvote or complain
Do you mainly dabble in set theory, @CarlMummert?
Alright.
No, I don't know enough set theory to do much research directly in that area. I have done a little in topology, but mostly I work in computability theory and proof theory
@Khallil I wouldn't consider answering, but nobody seems to be answering either, and I've grown to be interested in the question myself - so in some sense I'd like to check whether I've figured out a right answer. I'm not sure I should make a question about it, because that would be a duplicate, so I don't know what to do.
Maybe ask on here, @Jake1234?
I don't know too well to be honest. Carl's answer seems a lot more informed than my two cents.
00:27
I guess that might work, but I think people here aren't very likely to look into time consuming things like checking proofs, unless you someone directly for help.
@Jake1234: if you aren't certain, you can always just mention that in the answer. In my mind the main worry about posting things you aren't too sure of is that someone else might think you are right, even if you aren't, if they know less than you. So you could always add a sentence that your answer is an attempt that might still have flaws
Hm, alright I guess I will do that, thanks!
00:40
the thing i ll never get a proper explanation of, why would some moderator from a totally irrelevant section come and show his ugly peacock tail thus in a way of showin one of those these "forever ambiguous diamand tool" and remove my star from a message.
gd night
00:57
I'm not very well acquainted with proof and computational theory but the wiki entries sound cool.
Did you have many reasons for choosing becoming a logician over an analyst, algebraist etc. @CarlMummert?
I had the unusual history of learning some basics of computability and set theory when I was very young, which probably put me on that path. Also, the job market in computer science was bad around the time that I had to decide whether to get a non-academic job or get a PhD. @Khallil
01:22
Is there a concept of quintessentialism in mathematics? Given a property $P$ that defines a class $C$, is there an element $x \in C$ such that it exemplifies $P$ like no other element?
Or would each element of $C$ be equally exemplary of $P$?
@Axoren: in general, there is no way to pick one just one element from a particular set with some specific rule. In some sets, all the elements "look like" all the others.
huh, it's pretty late here, but happily I finished another problem.
(I should definitely sleep at this time)
@CarlMummert Right, but there is an intuition in some cases, as well as an informal argument for the quintessence of certain members, like the Triangle of the set of Polygons.
I was wondering if someone had formalized it in a generalizable way.
I'm out.
@Axoren: in some cases - I see what you mean by triangles - but for sets in general there is no reason to think that the elements will have any particularly special representative.
01:38
As you can somewhat see what I mean for the set of polygons, then you must somewhat see that there should be a class of "quint-ifiable" sets.
I think the examples you have in mind represent extremes. In the case of a triangle, the fewest edges possible in a polygon. Equilateral triangles are even better.
01:57
@anon does $F_{27}\F_{3}$ has intermediate subfields ?
@L33ter What is the degree of that extension? And what would the existence of an intermediate field imply about the degree?
02:13
@DanielFischer I got it
hello
hello @gansub
Hello Prof Mummert :)
how are you
Doing fine.
02:25
ok
good to know
I have to say, despite my apathy towards the hats up to now
I think I rather like how the mask one goes with my default image
it makes my icon look like a disappointed zordon :P
user174558
03:19
0
Q: Questions which are wrong because situation is not possible

John NashConsider the following question on an elementary school math test paper. The area of a right triangle with integral sides is 5. What is twice its area? Obviously, the answer is 10. But on closer inspection, there is no such right triangle. Primarily, what can be done to avoid such mistak...

user174558
I posted a question for fun. Please give me your answers! Ho ho ho!
user174558
It already has one close vote, LOL
03:54
Hello
I'm new here with a problem some people here may have experience with
I make stupid mistakes in calculus all the time
Errors in algebra, etc.
Have at it.
Math is my favorite subject and it's really fun
especially now that I'm in calc but it is incredibly discouraging to make dumb mistakes all the time
is there a good way to address this
I check my work and sometimes still miss stuff, and I feel like an absolute idiot
Just slow down and make sure every step you make is correct. There is no way to "address" it.
does that eventually turn into doing it right the first time
04:18
@anon around ?
@Portali5t That is for you to find out
user174558
Hello @ted.
Not sure who you are ... May not be so tasteful a name, given the recent death.
@TedShifrin
hi
@TedShifrin could you help me in a question
Hi Karim
04:23
hi @TedShifrin
@TedShifrin Hey!
Hello @Ted!
Can we meet over sometime in which you are free?
Hi Andy, Julian
@TedShifrin I want to show the following
show that every normal field extension E of Q in C that has [E : Q] = 7 is subfield of R
Julian, not sure when I'll be up there ... Sick now, traveling in a week for 3 weeks
Ok, KArim, so?
04:26
@Ted oh. Feel better please :)
is every polynomial of degree 7 has a root ?
is this true ?
You want to think about what normal + containing a complex number tells you.
alright let me ponder for a sec
user174558
@TedShifrin Jasper here, lollollol
I thought so ... Still don't like the name.
user174558
04:34
I will be retiring from this site in a week.
user174558
I already got two downvotes for my question.
retiring may be overdoing it again ...
Is that meant to be facetious @JohnNash
user174558
@Portali5t What is meant to be what?
Were you serious or referencing the downvotes to be sarcastic
Just curious, I don't have much context
user174558
04:36
Well, it is a statement of fact. I got two downvotes. That is all.
No relation to "I will be retiring from this site in a week"
user174558
No relation, lollollol.
so, Karim?
04:54
I am writting down a prove 1 second @TedShifrin
is it true that every polynomial of odd degree must have a root?
I guess this comes from analysis
0
Q: Space of Lipschitz Functions Complete?

Jessy CatConsider the subspace of continuous, real-valued functions on $[0,1]$ that are Lipschitz. Is this subspace complete under the sup norm ($\Vert \cdot \Vert_{\infty} = \sup \{ |f(x)| : x\in S \}$)? I would say yes, since all Lipschitz functions ($d_{Y}(f(x_{1}),f(x_{2}))\leq d_{X}(x_{1},x_{2})$, w...

@TedShifrin
Hello. Is anyone who is good at stats on right now?
Not me, sorry
Like, semi-decent works
just like first year knowledge
I need to know when it is ok to use the least-squares method for curve fitting
05:00
nope, lol, I'm a ninth grader
oh
how much math do you know
Idk, Math is vast and elegant.. And In comparision to that my knowledge tends to zero
Wbu, are you graduate?
limit of Swapnil as (?)
Lol, I was just literally speaking. You sound like a cool math geek like me
grade 11 here
05:04
Great, do you aspire for the IMO?
I just walk chill here
no I want to minor in math
Oh, Major in?
Medicine
its ok you are in grade 9
you know alot of math for your age
sec
Yup, you know, This problem is from a JMO
$Let P=2,3,5,7,11,...P=2,3,5,7,11,... $denote the set of all prime numbers less than $21002100$.Prove that$ ∑p∈P1p<8∑p∈P1p<8,$ where $pp$ represents primes of set PP.
solve it
Without
looking at the answer
05:07
Damn, even I'm in a confusion. Can't anyone provide a elementary solution? That prime number function has nothing to do with a junior olympiad problem. Can you help me?
you asked it
you should know the answer :)
My doubt
you r correct
05:08
It was
in saying that it isnt elemintary
are you familiar with bounds?
Nope :(
I will read it NOW, can u link a source?
There's so much to read, and I'm struggling with other nuisance subjects :(
In mathematics, a function f defined on some set X with real or complex values is called bounded, if the set of its values is bounded. In other words, there exists a real number M such that for all x in X. A function that is not bounded is said to be unbounded. Sometimes, if f(x) ≤ A for all x in X, then the function is said to be bounded above by A. On the other hand, if f(x) ≥ B for all x in X, then the function is said to be bounded below by B. The concept should not be confused with that of a bounded operator. An important special case is a bounded sequence, where X is taken to be the set...
In mathematics, especially in order theory, an upper bound of a subset S of some partially ordered set (K, ≤) is an element of K which is greater than or equal to every element of S. The term lower bound is defined dually as an element of K which is less than or equal to every element of S. A set with an upper bound is said to be bounded from above by that bound, a set with a lower bound is said to be bounded from below by that bound. The terms bounded above (bounded below) are also used in the mathematical literature for sets that have upper (respectively lower) bounds. == ExamplesEdit == 5 is...
What do 9th graders do in the US
05:10
I'm not in the US :(
oh
canada
I hail from India, famous for Ramanujan :P
oh that guy
Eh if you know that much at grade 9, you are set to go
I'll recommend you a book to help
sec
I will be indebted
Is there something similar to $\mathcal P(S)$ for the set of all sequences of a set (using each element at most once)?
Evening @TedShifrin
05:12
It isn't perfect
but it gives some great insight
I'm essentially looking for an Power-Ordered Set of a set.
Thank you very much, It would really help :)
Hi Axoren
@Michael, Did you know that tomorrow in India is celebrated as the National Day of Mathematics?
you want $S^{\Bbb N}$.
05:14
we dont have that in Canada
@SwapnilDas With your math knowledge, you should be decimating school
Oh, bcoz tomoroow is the bday of that guy I mentioned
Oh, I don't attend school half of the day
I hate school
@TedShifrin That's brilliant. It's exactly what I'm looking for. The set of all functions from the $\mathbb N$ onto the set $S$. I didn't make the connection until just now.
@SwapnilDas You need school to become a mathematician
Realistically, I want a subset of that, however.
I want to :P
05:15
Yup, Axoren, that's what sequences are.
It is my aim
Because exists a function $f$ for which it's not defined for $0$
well stop skipping school
I'm looking for a set $S' \subset S^\mathbb N$
To me $\Bbb N$ starts at 1.
05:17
Why @Michael
?
Does $mod$ in this article mean the modulo ? As in the remainder of the division...
@TedShifrin That's irrelevant-ish right now. The problem is that there would be sequences which are undefined at certain intermediate $S_i$
@SwapnilDas Unless you are a God,(250 IQ plus), you need schooling to do mathematics in our time
I'd like to have a notation for contiguous sequences
I don't permanently skip, I go 3/6 days weekly
05:18
why not go every day
Bcoz Indian schools suck
@TedShifrin How can a peace-wise function be continious?
Axoren, I don't understand you.
They ask us to cram math, you knew? @Michael
For example: $f : \{9, 111, 678\} \to \{a, b, c\}$ is such a sequence $f \in S^{\mathbb N}$
There's no $1$st sequence element.
05:20
That's not the defn of sequence, Axoren.
Michael: It All depends on the pieces.
Could you enlighten me? It was my understanding that the reason we could discuss $s_i \in S$ was that there was a function that mapped $i$ to a member of the set the sequence $S$ was constructed from.
Yes, for all $i\in\Bbb N$!
Going forward, let's assume that $S$ is a sequence of alphanumerics.
@TedShifrin I agree that it is true for all $\mathbb N$
@TedShifrin How would you define continious?
But $A^\mathbb N$ is the set of all functions partial and otherwise from $\mathbb N \to A$
05:23
you're now confusing notation for the set and the sequence.
Sorry, we ended up using the same name $S$
Let me correct that
$A$ is the set of alpha numerics.
I agree that $S \in A^\mathbb N$.
I also posit that there exist members of $A^\mathbb N$ exist such that they are not sequences.
Michael, you have a definition in your class.
Which, by definition we are saying must be defined for all $n \in \mathbb N$.
05:25
No, Axoren. By definition every one is a sequence.
$\lim_{x \to c}{f(x)} = f(c).$
That is the one we got
How so? There exists a function $f \in A^\mathbb N$ such that it is not defined for $n \in [1, k]$ for some $k$.
For all $c$ in the domain, Michael.
But some peacewise jump, so f(c) is not equal to limx approaches c
I think I have a flawed understanding of the equation
No, it depends on the example.
No, Axoren. All natural numbers is the domain of every function
05:29
Can you give me an example of a continious peace-wise function @ted
@TedShifrin around ?
so I finished my argument
$f(x)=x$ when $x<1$ and $=x^2$ when $x \ge 1$.
would you like to read it ?
A little busy here, Karim.
05:31
oke I will post on main
or wait until anon comes back
@TedShifrin I must be mistaking this notation, then. I was under the impression that $Y^X$ meant all functions $f : X \to Y$ but it included partial functions.
The idea is very simple,mKarim.
We will argue by contradiction. Suppose E isn’t subfield of R. Since [E : Q] = 7, so this means E has primitive element “a” satisfying a minimum polynomial of degree 7 over Q denote that polynomial by p(x). Every polynomial of odd degree has real root, so in particular every polynomial of degree 7 has real root, since E is normal, so all roots of that polynomial are in E (including the real roots).
No, Axoren, no such thing.
If it does not include partial functions, I run into a different issue. How do I describe the set of finite sequences? Sequences for which there is a last element?
05:33
Do you recommend learning linear algebra before multi variable?
Denote the real root by r we have Q < Q(r) < E with all inclusions being strict, so Q != Q(r) since r isn’t in Q ,since our minimum polynomial is irreducible if it was in Q then it would be irreducible. Also Q(r) != E (since this is by assumption that E isn’t a real field). Since |Gal(E/F)| = [ E : F] = 7 —> Gal(E/F) = Z/7Z,
by galois correspodence of subgroup of Gal(E/F) = Z/7Z and intermediate fields Q < Q(r) < E we have Q correspond to {0} and E correct to Z/7Z, but Z/7Z doesn’t have any subgroup and we showed earlier that Q(r) isn’t equal to Q or E, so the correspodence isn’t a bijection which is contradiction
You are talking about $Y^{P(X)}$.
and that is it
I probably over complicated it
aren't LA and MV entirely independent of each other
not really, @portali.
05:34
@TedShifrin So, the set of all finite sequences of alphanumerics would be $A^{\mathcal P({\mathbb N})}$?
no, axoren.
I typo'd
Is it corrected now?
user174558
All multivariable calculus includes linear algebra in some disguised form.
Oh, right, we're now running into the partial function issue
not true, Jasper
05:35
also do I need a mathjax plugin or something to get rid of the $ I see everywhere
Because now we're using the power set
@Portali5t Yes, but it's less of a "plugin" and more of a bookmarklet
any chance there is a link readily accesible?
You don't need to download anything, just drag something onto your bookmark and push it when you want to read latex in chat
Check the top-starred item on the right
by anon
karim, you never dealt with what it meant to have a complex number in there. And you don't need 7 to be prime. What do you need?
ah there we go, thanks @Axoren
05:37
@Portali5t np
oooh, Karim, you did use it, but where did you use normality? But I do not like your argument depending on 7 being prime.
Prove that 2+2=4
@Michael have you read Russell and Whitehead's Pricipia Mathematica
@TedShifrin Is there a notation for all intervals $[1, k] \subset \mathbb N$ for some k?
@Portali5t No I haven't. I have alot of time over this break. Is it a good book?
05:41
7 being odd
I used normality for first part of the proof
@Michael it is 300 some pages and proves that 1+1 does indeed equal 2
i.e that since E is normal so all roots of that polynomial are in E @TedShifrin
@Michael it takes them 365 pages to do it.
Interesting thing about it is they define exactly what 1 and 2 as well as the addition and equal operator are
@Portali5t Since you know some math books, can you recommend one that discusses sets? I am so bad at that subject, and I hope you can recommend something that goes from 3rd year old basic to first/second year uni
05:45
I have nothing on that subject
I'm in calc BC at the moment
just interested enough to have cursory knowledge
What if the field isn't obtained by adjoining a single root?
@Portali5t Do you know any books that go into first year uni calc?
yeah any single variable calc textbook should do
ie. "calculus of a single variable" by Ron larson and bruce edwards
@Portali5t What exactly does first year uni math consist of?
well it depends
05:47
well, we must have that root to be not rational or it would contradict the fact that our poly is not minimum
if you have taken pre calc and trig then calc 1 and 2 assuming there are semesters
if you have taken calc 1 and get the credit to transfer you would do calc 2 and either 3 or linear algebra
it also depends on the major
If you have a normal extension of $\Bbb Q$ with a complex number in it, what do you know?
@Portali5t What after those courses
a pure math major would take classes on real analysis and other stuff I am not familiar with
probably a course dedicated to differential equations too
@TedShifrin If there isn't, I'll need to create one for my own personal use. I'll probably use $N_{<*} = \{ \mathbb N_{< k} \mid k \in \mathbb N \}$. Can you find anything wrong with this notation? $A^{\mathbb N_{<*}}$
05:49
what grade are you in @Michael
I have finished grade 12 math
im in grade 11 though
that means nothing to me
what is 12th grade math
it will be algebraic
@TedShifrin?
calc 1/some calc 2
@Portali5t It's an odd question to ask if the answer means nothing. :P
05:50
12th grade math is as ambiguous as it gets
well @Michael I'm in the same position
completed calc 1, in calc 2, 11th grade
it says you are getting a bc. What is a bc?
the college boards designation for a year long course that covers calculus 1 and 2
an AP course, since that is what is offered in the US
otherwise it would just be called calc 1 and 2
Oh AP
I did Hl IB
05:53
are you in a different country
Just above you ;)
Canada is a nice place
@Axoren, if only finite sequences, the it's $\bigcup_n A^n$.
Keep going, Karim.
wow topology seems quite hard

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