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4:00 AM
Hey @Balarka
 
hi chat
 
Assume you contain no constants, then you also can't contain any polynomials whose coefficient of $t^n$ is non-zero, because otherwise taking that derivative barely enough times would kill it, then keep on going, and you see that this invariant subspace is just $0$
So we got that much going for us
 
[Random] Inspired from looking at the way the calculation jobs are arranged:
Sequence of indefinite length $\{a_0,a_1,\cdots a_n\}$, where $n \in \Bbb{N}$ and is unknown
 
@Daminark Note that if you contain a linear polynomial $at + b$ with $a \neq 0$, you have to contain all the constants (because $a$) and also $t$ (subtract an appropriate multiple of $a$ so that you get containment of $at$)
 
not sure what to make of it, other than as long $n$ is unknown, the sequence feels like it is going to go forever
 
4:09 AM
So you get all the linear polynomials.
 
@Secret Can identify that with the polynomial $a_0+a_1 x+\cdots+a_n x^n$.
 
I think the technique will be something like that. Say you have a polynomial $f(x)$ of degree $n$. Write $f(x)$ as $a_n x^n$ plus a linear combination of it's derivatives.
Inductively prove you are containing $x^k$ for all $k \leq n$
Does that make sense?
 
@Semiclassical hmm? but what will the powers of x be, or they are just basis vectors?
 
basis vectors, yeah.
 
I see
 
4:11 AM
I'm thinking of some polynomial ring in the indeterminate $x$.
 
Yeah alright I'm good with this
"Maybe one day I'll learn point-set topology" someone who's giving the algebraic topology talks (not Peter)
 
same
 
tbf I can relate
 
I need to do a lot of things today
but I can't remember what
 
Rip
 
4:27 AM
Pir
 
I'm trying (though will likely fail) to finish by tonight the lecture
 
It's 11:30 pm where you are too, right?
 
Yup
And I've still got quite a good bit left
I'm stuck on this one theorem which the book leaves as an exercise but which I'm gonna include as part of the lecture
 
You have a very flexible definition of "by tonight".
 
That's until I go to bed
:P
 
4:29 AM
Yeah. That's my definition too :P
 
what's the theorem
 
Okay so let's say you have a group $G$ of homeomorphisms on some topological space $X$
You say that $G$ acts minimally on $X$ if for any $x\in X$, $x^G$ is dense
 
So now, take a compact metric space and a group $G$ of homeomorphisms. Show that there's some non-empty, closed, $G$-invariant subset $X'$ on which $G$ acts minimally
I'm thinking it'll be a Zorn argument
 
Ah yes
 
4:32 AM
@Semi do you also define "tomorrow" as, when you wake up? Or the sun rises if you're pulling an all-nighter?
 
Depends on how I feel the next day.
 
Lol
 
tbh if I stay up too late these days I'm usually a wreck the next day.
which means that I effectively start losing days if I get into that cycle
 
Anyway @Balarka, so it's possible that this isn't a Zorn argument, but I think it is, though I'm getting hung up on the execution
 
@Daminark I'm not buying it. Say $G = \Bbb Z/n$ acts on $S^1$ by rotation. What's the $X'$?
 
4:37 AM
I think the nth roots of unity
Note that when I say that if $G$ acts minimally on $X'$, we just ask that orbits are dense in $X'$
 
Ah, so I can just take orbit of any point
 
Why is that closed?
 
it's a bunch of points
finite bunch
 
Oh like in that particular instance? Sure
 
and $x^G$ is not only dense there but the full $X'$
what a dumb example
 
4:39 AM
shrugs
 
About the only thing I can add is that, by the magical power of Google, I came across the following sentence in a preprint: "Choose, by a Zorns lemma argument, a closed G-invariant subset Z of X such that the restriction of the action on Z is minimal."
 
@Semiclassical cheater
 
Okay so we know it's Zorn's
 
@BalarkaSen Eh. It'd be cheating if it said how to do the argument.
 
@Balarka respect Google-fu
 
4:40 AM
@Semiclassical no u cheater
 
All I'm doing is validating that it's the right approach.
 
no u google
google is cheat
 
hi chat
 
respect mah google-fu, son
2
 
google is the most useful skill one can have doing math
5
 
4:41 AM
Well, take the set of closed, $G$-invariant subspaces, we want to define an ordering
Ah this is sneaky, we're asking it to only be dense in itself
 
Is the order defined by $X \leq Y$ iff $\text{cl}(X) \subseteq Y$, or something?
 
These are closed anyway, but my issue is that this ordering doesn't encode minimality at all
 
Is the set $\mathbb{R}$ with respect to any metric is not compact ?
 
@Daminark Do we need to? We can look at the set of $G$-invariant subsets of $X$ and define $S \leq P$ iff $S \subseteq P$
 
Alright, so finite intersection property, Zorn's lemma, etc
You now have some minimal set
How does $G$ act on it?
 
4:52 AM
Cruelest phrase in a text: "It is easy to see that..."
 
and other one may be "it's obvious" :)
 
i cri evrytim
 
(For reference, I'm looking at a question where the asker is the victim of the above: math.stackexchange.com/q/2354417/137524)
I'm annoyed at myself for not knowing how to answer it. Just because I know what Hamiltonian mechanics is doesn't mean I know how to answer questions about sympletic manifolds :/
 
shrugs in Hamiltonian mechanics
 
sorry, internet busted
 
5:02 AM
(I think it may just be Darboux's theorem. But I dunno if that's right.)
 
is that the rekt video again
 
Oh lol should've done that actually
 
you trolling dinosaur
 
Kek
(credits to you for the photo)
 
5:09 AM
@Daminark So in my process I create a closed G-invariant S which contains no other closed G-invariant subset, yes?
 
Yeah
 
I somehow want to say this should be X'
 
Why is it the case that the orbit of any of its points is dense in it?
 
like if x in S, x^G is not dense in S I can take a closure of this in a way to get a closed G-invariant subset of S
but simple cl(x^G) won't do it, because that's not G-invariant
I feel S and X' are pretty close at least
 
Perhaps
I'm tempted to table this problem until office hours
And work on the other stuff
I'm nearly at the stage where I can prove that if $\{S_k\}_{k=1}^n$ is a partition of $\mathbb{Z}$, then some $S_k$ contains arbitrarily long finite arithmetic progressions
 
5:16 AM
That's Dirichlet, isn't it?
 
INTERNT
y u do this to me
 
Nope
It's called Van Waerden's theorem
 
@Semiclassical Eh, Dirichlet says arithmetic progression contains infinitely many primes
I think
 
*Van der Waerden
 
hmm.
yeah, okay.
 
5:18 AM
@Balarka yeah, as long as the jump is coprime to the starting point or someting?
 
right
 
5:40 AM
This book relegates more results to the exercises than it actually proves I'm starting to wonder
Okay not really but like whoa
 
sounds like a good book
 
any answers in the back?
 
Lol nope
 
Hi chat
 
@Daminark the book it replaces relegated everything to an exercise
like it was a list of problems
 
5:58 AM
@Daminark is closure of a $G$-invariant set also $G$-invariant? say $A$ is the set and $z$ a limit point so that there is a sequence $z_k \to z$ with $z_k \in A$. Then for any $g$, $gz_k \to gz$, right? so $gz$ is also a limit point of $A$, isn't it?
if I am not too crazy, that should do it
 
@Eric Does no one want to actually prove some dynamics results or something?
 
the other book was group theory tho
 
Lol anyway I guess this is good practice but like, this is making the lecture take longer than I thought
 
it takes a long time when you've gotten away from the very basic stuff
like it's easier to lecture at the very beginning and graduately gets harder
 
we should get Daminark banned from this chat so he can work on the lectures
 
6:02 AM
I'm working on the lecture, there's a reason I'm spotty on this chat
Just that it's taking me a long time for me to figure out some stuff that ended up being easy so I'm kinda breaking off from the work for occasionally in hopes that I'll figure it out as I get back to it
 
so, huh, i guess what i said means "every orbit is dense" is equivalent to "contains no proper closed G-invariant subset"?
fun
 
So it seems
 
actually, that makes a lot of pictorial sense when i think about it from the foliations point of view
 
i should be working on a lecture rn but rip
hi @Alessandro
 
6:09 AM
Wait you're giving lectures in the second part of the bootcamp too?
 
in our part? yeah
it depends on your instructor
i have to do stuff on rectifiable sets and review some classical stuff on lipschitz functions
i think i could wing it and be fine
 
@Alessandro hey
 
area formula, co-area formula and whatnot
 
@Daminark Do you know an example of a minimal set S of a G-action on X such that orbits of points in X are neither (1) all of S nor (2) dense in X? (maybe forget about it if it's not relevant to your lecture)
 
I see. Do you know how the other professors are operating?
 
6:12 AM
uhh well schlag isnt hands on im p sure, he's just having everyone do their own thing and update him i htink
and marianna is just making them do problems and shit rn i think
idk what silvestre's students are doing
 
@EricSilva Open ones?
 
no
it's like the first week man
 
@BalarkaSen Nothing's coming to mind yet
 
she'll make them do open stuff eventually probably
 
Fair
 
6:20 AM
@Daminark I have to check if I am parsing it correctly (I'm having trouble moving from foliations lingo to this), but here's what I'm trying to say. Consider Z/n acting on S^1 generated by rotating 2pi/n; then S = nth roots of unity is a minimal set, and closure of every point is all of S. Consider Z acting on S^1 generated by rotating 2pi*sqrt(2) angle; then S = S^1 is the only minimal set.
an S which is neither of those (closure of none of the points is all of S, S $\neq$ X) is called an "exceptional minimal set"
 
Also @Eric finally the institutional application is working so I can finish off this aid stuff once and for all
 
these are the most interesting examples of minimal sets imo. if you want i can refer you to read a few pages of Candel-Conlon (the foliations textbook i have been reading) to check out an example. although it's probably not too relevant to whatever you are studying in dynamics
or at least out of syllabus
 
At some point I'd definitely be down for trying out this foliation business
Right now I'm a bit tight
 
sure, sure. the few pages i have in mind are not really parsed in terms of foliations
in particular there's a theorem that every exceptional minimal set is a Cantor set
 
Also I think I've finally worked out the dynamics part of this lecture
Now is the Ramsey theory
@Eric get on fb
 
6:33 AM
3
Q: How to decide $36^\text{th}$ smallest element in max-heap tree of $100$ elements?

Mithlesh UpadhyayConsider a max heap tree with $100$ elements and a node from the same level is chosen randomly among all valid balanced heap trees with $n$ nodes. What is the probability that it is the $36^\text{th}$ smallest element______ . My attempt: (Please, find definition of Binary Heap on Wiki: https...

 
 
2 hours later…
8:42 AM
If $f$ is a continuously differentiable function and $f \rightarrow a$ , for some $a \in \mathbb{R}$ as $x \rightarrow \infty$,then $f^{'}(x) \rightarrow 0$ as $x \rightarrow \infty$ ? is this true or $f^{'}(x) \rightarrow b$ for some $b \in \mathbb{R}$ ? I thought that as $f(x)$ tends to $a$ so it is constant htere after and hence the derivative would be zero as $x \rightarrow \infty$ , is this reasoning correct?
 
@BAYMAX wrong.
 
oh
 
generally the behaviour of $f$ has nothing to do with the behaviour of $f'$
 
but in the definition of $f^{'}$ we have the difference of values of $f$?
 
here $f \to 0$ but $f' \to \pm \infty$
 
8:46 AM
Yes,but still $f^{'}(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$
 
and then?
 
like value of $f^{'}(x)$ depends on nearby points of $x$ ?
 
@BAYMAX so?
 
So I thought that the value of the derivative depends on the value of the function nearby points?
 
@LeakyNun Not really true
 
8:48 AM
well I just provided a constructive counter-example, if you click into my link...
@Astyx $\limsup f' = \infty$ and $\liminf f' = -\infty$ is this better?
 
yes i saw that @LeakyNun
 
It is
 
generally the behaviour of f has nothing to do with the behaviour of f′
or
generally the behaviour of f' has nothing to do with the behaviour of f
 
Funny enough, for that counterexample, even though the limsup and liminf are + and - infinity respectively, you kind of can say that $f'(\infty) = 0$ :P
 
Hey guys
 
8:55 AM
ohi
 
Hi
 
How's it going?
@Astyx you ended up putting ENS as your top choice, right?
 
Yup
 
Nice
 
Perhaps that was a bad idea since I'm now stressed about this week's exams
:p
 
8:58 AM
rehi @Daminark
 
@SteamyRoot $f'(x)=\dfrac{3x^3\cos(x^3)-\sin(x^3)}{x^2}=3x\cos(x^3)-\dfrac{\sin(x^3)}{x^2}$, so why can you kind of say that $f'(\infty)=0$?
 
@Astyx have you already finished them?
 
No, not at all
I have one exam per day until friday
(Actually two on thursday)
 
Oh @Balarka I think I'm finally getting a grip on van der Waerden
@Astyx oh god that's just abusive
 
@LeakyNun Either extend the domain to $S^1 \cong \mathbb{R} \cup \infty$, or (which is probably more common) to the Riemann Sphere $\mathbb{C}_\infty$.
 
9:01 AM
@SteamyRoot I don't see why it would be $0$
 
from the graph it is seen that $f^{'}(\infty) \neq 0$ ?
 
Apply the transformation $z = 1/w$ and look at $w = 0$
You'll end up with a stereotypical example of a function that is differentiable with a discontinuous derivative
 
@Daminark cool, cool
 
Well, modulo understanding what actually happened in the last two sentences, but yeah
 
bye
 
9:04 AM
Now that this is done it might be a good idea to sleep...
 
I advocate that
 
See you @Astyx! And chat at large!
 
9:52 AM
What does it mean by $f_i$'s a are fractions over an algebraic number field $K$?
That $f_i$'s are elements of the type $\frac{a(x)}{b(x)}$ where $a(x)$, $b(x)$ are polynomials in variable $x$ where coefficients of $x$ are elements of $K$?
 
10:28 AM
[Chemistry] The program still refused to work, have been going back and forth with the technicians many times already, fixing as many errors we can find

If after this batch and it still not working, it is possible this error trouble shooting will become so frustrating that even literature review will be comparably more relaxing
There is good news through, one calcualation have worked and it predicts computationally that it got a complex with a rare one carbon bonding mode for that given type of ligand
 
 
1 hour later…
11:30 AM
Any idea why this answer is accumulating downvotes?
https://math.stackexchange.com/questions/2353725/is-it-possible-that-37-12k-for-some-k/2353733#2353733
 
The following question is not the question I am interested in. It is the question that is inspired from this question that I am interested in:
The question is the following:
Consider $a_i,m,l \in \Bbb{N}, m,l$ mixed, and $i \in \Bbb{N}$. Find the conditions given $m$ such that there is a unique sequence $(a_i)$ that satisfy the following:

$$\sum_{i=1}^l a_i =m$$
(We can then easily see that the question that inspired this more general number theory question is the special case where $m=30,l=3$ and $a_i \in S \subset \Bbb{N}$)
 
@Secret But in the inspiring question there was no solution, not a unique one
(unless you allow a box to be empty)
 
I dont understand what you want to do with the sum
ah ok nvm
and im sure that the picture wants to allow things like 19 to be in the box
btw, you can do something really stupid like 19,5 + 7,5 +3
 
@s.harp yeah, you need to either leave a box empty or add commas
 
since you have assume that they want you to "think outside of the box" by being a bit vague with what they mean with using
 
11:41 AM
(right, and there are commas available as the allowed things to fill with)
 
Well, I am curious on how the possible choice of integers will be restricted and in what way as $a_i$ get specified one by one. Rough observations will suggest if the number is big enough ,then you have only few possible choices for the remaining $l-1$ entries.

As for the orignal question that I took from somewhere, pretty much what you guys said as otherwise 3 odds always give an odd number no matter the combination
NB I suspect the general question may have something to do with this:
In number theory, the specialty additive number theory studies subsets of integers and their behavior under addition. More abstractly, the field of "additive number theory" includes the study of abelian groups and commutative semigroups with an operation of addition. Additive number theory has close ties to combinatorial number theory and the geometry of numbers. Two principal objects of study are the sumset of two subsets A and B of elements from an abelian group G, A + B = { a + b : a ∈ ...
though I need to check whether sum of squares fall in this domain
 
user84215
Why does \blacksmiley command not work?
 
what I am suspecting about the general question is that if $m$ is some special numbers such as primes, perfect squares etc., then if $a_1$ is quite large, then there are far fewer possible choices for the remaining $l-1$ entries compared to when $m$ is not special, but that's just a guess, based on how in many domains of maths, symmetries tend to reduce the number of "allowed configurations" or allowed solutions
 
user84215
We can not call packages in MathJax?
 
mathjax is not latex, you cannot use packages
 
12:00 PM
You can use require which is similar
but yeah it's annoying
 
Actually, I thought of a creative solution to the original question:
$$1+9+15=25=30 \mod 5$$
 
$15+13+1$
and one more 'cause you're such a good customer
 
lollol
 
user84215
@AkivaWeinberger Did you answer my question?
 
But yeah three odds can't make an even obviously
@aminliverpool Require is similar to packages I think
but I've only seen it used for one thing (which is the cancel commands)
 
12:05 PM
The general question that is inspired from this is probably not so trivial though. I think we will need to split $m$ in cases where e.g. it is prime, it is square, it is cube etc.
 
user84215
@AkivaWeinberger Could you give an example?
 
In mathematics, arithmetic combinatorics is a field in the intersection of number theory, combinatorics, ergodic theory and harmonic analysis. == Scope == Arithmetic combinatorics is about combinatorial estimates associated with arithmetic operations (addition, subtraction, multiplication, and division). Additive combinatorics is the special case when only the operations of addition and subtraction are involved. Arithmetic combinatorics is explained in Green's review of "Additive Combinatorics" by Tao and Vu. == Important results == === Szemerédi's theorem === Szemerédi's theorem is ...
 
I don't remember
Look up cancel mathjax
It's for strikethroughs
 
user84215
12:27 PM
I want to put a hat over {qwertyuioplkjhjgfdsazxcvb} such that it covers all of the letters. What should I do?
 
$\widehat{qwertyuioplkjhjgfdsazxcvb}$ does not work. The realy wide hats cannot be done via mathjax
https://tex.stackexchange.com/questions/100574/really-wide-hat-symbol
 
user84215
How can I put more than four dots over a letter?
 
$\overset{................................................}{aaaaaaaaaaaaaaaaaa}$
 
user84215
over only one letter
 
$\overset{.....}{a}$
5 seemed to be the limit without them all merging together
 
user84215
12:43 PM
Thanks for your attempts.
 
If $ A \cup B$ connected then it is not necessary $A$ and $B$ are connected. Any simple example for this?
 
set $A = [-2, -1] \cup [1, 2]$ and $B = [-1, 1]$
 
Nvm got one.
Just in time xD
Just a follow up, $A\cup B $ is connected, $A \cap B $ is connected and both non-empty, still A or B not connected.
 
12:59 PM
hello
 
@Mann Still false. Let $A$ be the same as above, and $B = [-1, 1)$
Sorry for the delay, I was away from keyboard.
 
wait a sec is that related to this question I just found?
https://math.stackexchange.com/questions/2354781/operation-on-intervals
 
Both non empty
intersection as well as union
OH wait nvm
my bad.
:|
 
set union is like including all of the elements of set A and set B
like
(omg sorry if my latex is rusty)
$$ A \cup B$$
let A = [1,2,3] and B=[4,5,6]
then the set union would be
[1,2,3,4,5,6]
ow
I'm on earbuds and someone just pinged me
 
@Mann My last example is somewhat cheating, but you can still look at say $A = \{(x, y) \in \Bbb R^2 : x^2 + y^2 \leq 2\}$ and $B = \{(x, y) \in \Bbb R^2 : x^2 + y^2 \geq 1\}$. That's a 2 dimensional generalization.
 
1:04 PM
Just curious, can you prove that mathematical induction works?
 
that's the base case, the k case, and the k+1 case
 
Morning @MikeMiller
 
Yes, but what does it take to prove that induction works?
 
Under ZF you can i think. I don't remember certainly.
 
1:06 PM
@BalarkaSen as long as it works. xD
 
You can prove that it follows from a set being well ordered
 
whoops I didn't study Topology yet x_X
No wonder I couldn't figure out the "connected" term
 
Proof of induction from well-ordering principle?
 
@usukidoll it's fine. xD
 
1:08 PM
OW!
is TOpology hard?
 
@Mann Sorry. I misspoke. $A$ and $B$ are both connected in that example. What I wanted to say is $A = \{(x, y) : 1 \leq x^2 + y^2 < 2\}$ and $B = \{(x, y) : x^2 + y^2 \leq 1\} \cup \{(x, y) : x^2 + y^2 \geq 2\}$
 
@SimplyBeautifulArt proofwiki.org/wiki/…
 
9
Q: Well-Ordering and Mathematical Induction

Alexy VincenzoHere is my attempt to prove the Well-ordering principle, i.e. Any non-empty subset of $\Bbb N$, the set of natural numbers has a minimum element. Proof: Suppose there exists a non-empty subset $S$ of $\Bbb N$ such that $S$ has NO minimum element. Define $A = \left\{n\in \Bbb N : (\forall s\in S...

:p
 
So $B$ is the union of a closed disk of radius $1$ and closed exterior of a disk of radius $2$, and $A$ is the annulus going between
 
How do I prove the well-ordering principle?
 
1:10 PM
 
by induction XD
 
Actually this question is what got me thinking. Why the condition A and B both being open and closed is necessary.
 
@GFauxPas >.<
 
and I forgot the open and closed definitions from analysis nice going self lol
 
In the last two examples I gave, one of $A, B$ are neither closed nor open
 
1:11 PM
@SimplyBeautifulArt you gotta take some axioms as base.
 
But which ones?
 
Exactly, that's why its good. :D
@SimplyBeautifulArt , Most mathematics is founded on ZFC so why not begin with that.
 
and the Axiom of Choice is equivalent to "every set can be well-ordered"
 
Well-ordering principle is not the same as the well-ordering theorem.
 
1:15 PM
Mmm, okay :D
 
right, but, if you want to prove induction works on every set Steamy
no?
though Simply probably would be satisfied with induction on integers
in which case you just need to prove that $<$ is a well ordering on the integers
 
Usually, one takes induction as an axiom, from which the well-ordering principle follows.
Which is way weaker than choice/well-ordering theorem
 
right, but Simkply's question was can you prove induction from other axioms
and then Simply asked "but which ones?" to which Mann suggested ZFC
@SimplyBeautifulArt are you familiar with Peano's axioms?
 
Not much, just a bit
 
user84215
I think it was proved that the axioms of ZFC are independent from each other.
 
1:25 PM
Oh, hello, everyone. I wanted to ask a logic question that wasn't answered in math stack exchange. It is a universal algebra question. What is an explicit identity or set of identities to generate the universally valid equations of the structure $(R,+,*,r)$ where $R$ is the set of real numbers and $r$ is an algebraic constant?
 
1:47 PM
algebraic constant? something like algebraic numbers?
 
$f'(x) = A1 f(x) + B1 f(x+h) + C1 f(x+2h) + D1 f(x+3h)$ then magnitude of truncation error of this scheme is ?
any help ?
 
Ring axioms seemed to be a good candidate for the structure $(\Bbb{R},+,*,r)$ but without any more details on how $r$ behaves, it is unsure if there are other possibilities
I am afraid this does not quite answer my question. I need an identity explicity within the signature $(+,*,r)$ For example, if $r$ was 1, we would need at least the identity $(x * 1) = x$. A natural follow-up question would be the same thing with regard to the signature $(+,-,*,r)$. — user107952 Jul 4 at 17:38
what? you only said $r$ is algebraic, if you are seeking for identities that are specific for each $r$ then you essentially need to run through each cases e.g. if $r$ is 2 and the structure has no 1 nor 0, you would get an extra identity for instance $r$ will be the generator of an ideal
Anyway, I think I have some mistake on my thinking, guess other abstract algebra people will better answer this question
Side note: Is it possible to have a ring like structure involving the reals but with torsion elements?
 
2:03 PM
Is it true that a subgroup $G$ of $GL(n,\mathbb{C})$ is a parabolic subgroup (i.e. contains a Borel subgroup that is, a maximal connected solvable subgroup) if and only if it is simultaneously block-triagonalizable, i.e. there exists a matrix $T$ with $TGT^{-1}$ block-triangular?
 
Yes, I am considering only a single specific algebraic $r$. Every algebraic number is the zero of a polynomial with integer coefficients. In terms of such a polynomial, I need a specific identity. No subtraction allowed, only addition and multiplication.
 
If there is no subtraction, that means there are no negative elements, and thus you will be dealing with $\Bbb{R}_{\geq 0}$ instead of $\Bbb{R}$?
 
No, there are negative elements. It is just that subtraction is not in the signature of the algebra.
My conjecture is that every algebraic number except 0 needs a single identity besides commutativity, associativity, and distributivity of addition and multiplication, and 0 itself needs 2 identities, namely $(x*0)=0$ and $(x+0)=x$
I mean of course there are other identities, but I believe that is a minimal generating set.
 
If the underlying set are the complex numbers, then any polynomial can be factorised into linear factors by the fundamental theorem of algebra, which means if any polynomial contains $r$ as a root, then it must have the factor $(x-r)$.

however, since you are working in the reals, for any algebraic number $r$, there may exists more than one irreducible polynomial in the reals that has $r$ as root. For example, $x^2-4=0$ and $x^2-x+2=0$ will both have a real root $r=2$. That would give you at least two identities to specify $r$ for example
 
Is $L^1([0,1])$ the same banach sapce as $\ell^1(\Bbb N)$?
 
2:21 PM
ok nvm, I realise my examples are faulty, let me see if I can get an irreducible polynomial with a real root
 
For the specific case of r=2, wouldn't the identity $(x*r)=(x+x)$ generate everything? I would be very surprised to see a counterexample.
Just to be clear, I am not allowing constants except r in my equations.
 
hi chat
 
Hi Semi
 
I I have a polynomial $P(x) \in P_n(\Bbb{R})$ such that $\exists y, P(y)=0$, must $(x-y)|P(x)$?
 
$P_n$ being ?
Oh right
 
2:36 PM
the space of polynomials up to degree n
 
Yeah I think so
You write the eculidean division
P = (x-y)Q + R
 
$P(y)=Q(y)(y-x)+R(y)$, yeah.
 
With $\deg R \lt 1$
Then since it values $0$ at $x=y$, it's 0 everywhere
QED
 
ok, so that rules out the existence of irreducible polynomials with algebraic roots that are not factorisable into linear factors with said root
hmmm....
$x=r$ seemed too stupid to be an identity for a structure, but the way to uniquely define $r$ is a solution to its minimal polynomial $x-r=0$...
 
I am still unsure about the status of my conjectures.
 
2:41 PM
$L^p(\Bbb R) = \ell^p(\Bbb N)\otimes L^p([0,1])$
 
Every algebraic number is uniquely determined by a monic polynomial of deg $m$, its minimal polynomial
If $r=\sqrt{2}+\sqrt{3}$ then your conjecture will ran into trouble since its minimal polynomial is $x^4-10x^2+1$ which has constants $-10$ and $1$, but you want a constant free identity
 
So, are those conjectures about 0 and 2 and other algebraic numbers correct? It is a bit tricky because I am not allowing subtraction in the equations, at least not in this particular question.
Oh, I see.
But there could be a way to re-express such a polynomial so that it is constant free except for r.
Like I did with 2.
 
$2$ works because the ideal of even integers are generated by $2$, thus you have this identity $x*2=x+x$ unrelated to its algebraicness
but if you want an identity that is constant free and can uniquely specify algebraic numbers, then you need to find the subset where the minimal polynomials have only powers of x and powers of r in it
That means, you need the set of minimal polynomials for each $r$ where the $x^0$ term can be decomposed into sums or difference of powers of $r$ and powers of x.
and for the first point (existence of a number as sums and difference of powers of r), it is already in number theory territory thus I have no idea
What I am not sure however is whether all algebraic numbers that have minimal polynomials that are not constant (in the context of universal algebra) free can always be expressed as sums, powers and nth roots of algebraic numbers which does
but even then, that's still not what your conjecture is looking for (because you only allow one $r$). So my preliminary conclusion is, there's only a subset of algebraic numbers that will satisfy your conjecture, and its identity is given by its minimal polynomial with terms rearranged such that all - becomes + at both sides
 
Oh man, today's xkcd:
pretty much on point
 
2:59 PM
And one of the most common weird idea (which underlies nearly all quantum mysticism) is the false identification of entanglement with nonlocal signalling
 

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