12:01 AM
well in that case, gcd of the a,b parts of each basis vector entry then. But you can always scale a prime back down to 1 and then to whatever other prime anyway
the way you said scalar multiple of the other made it seem like they just had to be coprime or something

12:17 AM
Hello community, I am doing this exercise and I can't figure out the mistake. This is the exercise: abs(2/(x+1)) < 4. In the case "2/(x+1) < 0", I have replaced abs(2/(x+1)) with "-2/(x+1)" to obtain "-2/(x+1) < 4".
Is that substitution correct ? Doing that, I can't obtain the correct result, that's why I am wondering.

@EE18 you are talking about the field of rational functions, which is however not what the excerpt you are quoting is about
the algebra you're getting at does make sense, but the excerpt you are quoting is not sweeping anything under the rug, it's simply about something different
the function $x\mapsto p(x)/q(x)$ makes perfect sense as-is (the domain, of course, being the complement of the zero set of $q$)

12:45 AM
Mathematics, in essence, is just about manipulating numbers, isn't it?

12:55 AM
@user402514 No?
I haven't seen a number in twenty years...

1:09 AM
Structuralism is a theory in the philosophy of mathematics that holds that mathematical theories describe structures of mathematical objects. Mathematical objects are exhaustively defined by their place in such structures. Consequently, structuralism maintains that mathematical objects do not possess any intrinsic properties but are defined by their external relations in a system. For instance, structuralism holds that the number 1 is exhaustively defined by being the successor of 0 in the structure of the theory of natural numbers. By generalization of this example, any natural number is defined...
In the philosophy of mathematics, formalism is the view that holds that statements of mathematics and logic can be considered to be statements about the consequences of the manipulation of strings (alphanumeric sequences of symbols, usually as equations) using established manipulation rules. A central idea of formalism "is that mathematics is not a body of propositions representing an abstract sector of reality, but is much more akin to a game, bringing with it no more commitment to an ontology of objects or properties than ludo or chess." According to formalism, the truths expressed in logic and...
@user402514 these might be of interest

@user402514 not at all
Its a common misconception by non-mathematicians
Mathematics in essence is about proofs

In the center of mathematical interests are proofs. Mainly those about common mathematical structures
Not even theorems, just proofs themselves

> The essence of mathematics lies in its freedom.

1:23 AM
Proofs and proving things
Computing integrals etc. that's a really small part of mathematics
Well sorry not small... I guess I should say uh. Its not essential if it doesn't bring any new results
Say you want to compute integral of 1/x. That's boring and no one cares

@Jakobian It's hard to distinguish logic from math in that sense

@Thorgott Understood now. I guess it is fine as is, though I suspect they are doing some sweeping because they have introduced $K(X)$ with precisely this sort of notation at this point
But I see what you mean that there's no need to even know about $K(X)$ in order to answer this question
They do say "rational function" after all no? Implying that every rational function can be written $r = p/q$ which is the sort of sweeping alluded to?
Anyway, appreciate you clarifying that I was roughly speaking on the right track re: the algebra. Will record that in my notes...and now I get to actually solve the problem now that work's done ;)

@EE18 that's a different matter then
@EE18 no matter what you think a rational function is exactly, this statement holds true definitionally

Engaging in a bit of pedantry, we'd have to clarify that by $p,q$ we here mean their embeddings in $K(X)$, no? And 1/q really means the inverse in $K(X)$ of $q$'s embedding?

the only potential confusion is distinguish being a rational function as an algebraic thingy or as an actual function on some subset of $K$, but this is not relevant to the exercise at hand

1:36 AM
Yes, that is well taken

@EE18 sure, that is to be understood automatically

Well, on second thought, there is some sort of relevance insofar as we start with the algebraic thingy $r$ and then we speak of its evaluation at $x_n$, which is to say we are tacitly using the homomorphism into the field of functions $K^K$ right?

it does not seem to me as if we start with an algebraic notion, but I'm not privy to the context nor the author's intent
also, there is no such homomorphism as you can only evaluate a rational function at a point where the denominator (in lowest terms) does not vanish

$K(X)$ has been defined as the minimal field of fractions obtained from $K[X]$, where the latter is the subspace of $K[[X]] = K^\Bbb N$
Hmm, I thought maybe there was a way to restrict out such cases but I guess not
OK, will mark all this down and now get down to the analysis

@EE18 sure, but I don't know if the exercise intends to evoke that notion

1:44 AM
I guess we would indeed have to define evaluation of a rational function in the manner I described then if there's no hope of using the homomorphism which does exist for $K[X]$
Seems kind of wonky/not as elegant because it's choosing an element of the equivalence class rather than the "usual" function on the equivalence class itself induced by a function on each element of the class which is compatible
That's what I was searching for, not sure if that made snse^

the result does not depend on the choice
but this is also tangential to the exercise, because you are there given two polynomials
and all we do is evaluate both of them separately and then take the quotient of the resulting numbers

For sure, I usually just like to write something extra in my notes if I feel like more can be said based on what I've learned to that point
also helps me recall the notions i saw earlier
You're saying that I can show that for any $\langle p,q \rangle,$\langle p',q' \rangle \in r$I have$p(x_n)/q(x_n) = p'(x_n)/q'(x_n) $? Hmm, I can't recall now what made me so focus so fastidiously on the p,q in lowest terms this morning then Bc from Enderton that sort of compatibility would've been my first instinct you need to put it in lowest terms to read off what the domain is on which the rational function is defined for example,$\frac{x^2}{x}=\frac{x}{1}$, but in the former expression, you do not see that evaluation at$x=0$is legitimate Ah so my claim is certainly not true above given your counterexample but if$\frac{p}{q}=\frac{p'}{q'}$and$q(a)\neq0\neq q'(a)$, then$\frac{p(a)}{q(a)}=\frac{p'(a)}{q'(a)}$1:52 AM Got it. But the proper way an algebra book would speak about "evaluating" rational functions would be indeed to single out lowest terms? not necessarily, but it is one thing you could do you can also use the second claim I just made and say "a rational function can be evaluated at$a$if it can be written as$p/q$with$q(a)\neq0$and is this case the value is defined to be$p(a)/q(a)$" (in other contexts, it is also convenient to consider a rational function to be defined everywhere by allowing it to take the value$\infty$) I see, as in defining the evaluation of$r$as saying if there exists$p,q$with$q(a) \neq 0$then$r(a) = p(a)/q(a)$, and there is no issue with this definition because you've shown that for any$p',q'$with this property we get the same thing i.e. there is no "choice" in the technical sense yeah, it's well-defined merci as always thorgott :) no problem 2:11 AM BTW, how would you make sense of some (finitely many)$x_n$being such that$q(x_n) = 0$Would it strictly be abusive to speak of the limit of this sequence since we haven't even defined a sequence (i.e. we'd have to "delete" things) @Thorgott A thought as I begin to do the exercise. In order to get a bound it may be easiest if i can use the same$p,q$at every$n$right. In which case I should use the pair in lowest terms (I guess it's a theorem that if there exists a$q$with$q(x)n) \neq 0$then certainly the monic$q'$in lowest terms obeys$q'(x_n) \neq 0$) 5 hours later… 7:07 AM when I'm well rested, I feel way less in the department of fluidity of thought Is there a term for the relation on group elements that says a and b are equivalent if there is a group automorphism mapping one to the other? Clearly if a and b are conjugate they are equivalent in this way because there is an inner automorphism. But there might also be an outer automorphism even if a and b are not conjugate. Homogeneous I guess At least that is the term for topological spaces when there is a homeomorphism mapping a to b @SoumikMukherjee that's not right I mean the space is called homogeneous, not the elements though 3 If$G$is a group, we can say$g$is automorphism-conjugate to$f$if there is a group automorphism$\alpha : G \to G$such that$g = \alpha(f)$. This is an equivalence relation. Is there a standard name for this equivalence relation? Is it a well-studied notion? Does it have some importance som... 7:19 AM @leslietownes Thank you kindly. so... equivalent then I've seen a paper on group theory just use plain equivalent, without automorphically I think in general they also just call them equivalent elements, without adding that they are equivalent automorphically, unless it might be unclear from context 2 hours later… 9:03 AM The one on left knows that if he tries to jump, then I'll catch him :) 9:15 AM They follow me when I go outside. And when I get out of their sight during their following, they meow. 😅 9:41 AM > Let$U\subset \mathbb R$be open and suppose that$f:U\times E\to\mathbb R$satisfies > (i)$x\mapsto f(t,x)$is integrable for all$t$> (ii)$t\mapsto f(t,x)$is differentiable for all$x$> (iii) for some integrable function$g$, for all$x\in E$and all$t\in U$, $$\left|\frac{\partial f}{\partial t}(t,x)\right|\leq g(x).$$ > Then the function$x\mapsto \frac{\partial f}{\partial t}(t,x)$is integrable for all$t$. Moreover, the function$F:U\to\mathbb R$, defined by$F(t)=\int_E f(t,x)\, \mu(dx),$is differentiable and $$\frac{d}{dt}F(t)=\int_E \frac{\partial f}{\partial t}(t,x)\,\mu(dx).$$ Why is$x\mapsto \frac{\partial f}{\partial t}(t,x)$the limit of measurable functions? 1 hour later… 10:46 AM ok, I think I understand, we have by assumption$x\mapsto f(t,x)$is integrable for all$t$, hence measurable for all$t$. So $$c_n(x)=\frac{f(t+h_n,x)-f(t,x)}{h_n}$$ is measurable for all$t$and its pointwise limit is$(\partial f/\partial t)(t,x)$. 1 hour later… 11:58 AM @EE18 as I mentioned before, a "sequence" should really be allowed to start at any integer and not just at$0$and where a sequence starts is totally immaterial to any question pertaining to its convergence. for large enough$n$,$q(x_n)\neq0$(why?) and so you can speak about the sequence from that point onward and hence its convergence. @EE18 I don't think there is any need to put the fraction in lowest terms 12:30 PM is there an algorithm which generates udecidable statements in an axiomatic system 1 hour later… 2:00 PM 1 hour later… 3:15 PM$\mathfrak{C}$(ignore this, just had to test it) 4:11 PM hi sup @Koro nice 4:28 PM i have a question Given the linear transformation$f: \mathbb{R}^3 \to \mathbb{R}^2$defined as follows:$f(x,y,z) = (x-y+z, y-2z)$. To find the associated matrix, i can either use a standard basis, consisting of the vectors$(1,0,0)$,$(0,1,0)$, and$(0,0,1)$, and calculate$f$applied to each basis vector. Alternatively, i can directly observe the expression of the transformation$f(x,y,z) = (1x-1y+1z, 0x +1y-2z)$. The resulting matrix is $$A = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \end{pmatrix}.$$ These approaches are correct, right? yes. "directly observing the expression" of the transformation is implicit use of the standard basis @Thorgott I know this is useless to think about, but what would it mean to "speak about" the sequence from that point onwards. Are we strictly speaking about a subsequence thereof? I guess we could choose "speaking of" this to mean "a subsequence beginning past the finitely many zero terms, or a subsequence deleting all zero terms" etc. as required and that's your point? correct @Thorgott OK, I will continue to think then. My proof strategy was going to be related to that to elaborate on this, if you think of a map from R^3 to R^3 as a function of "(x,y,z)" i.e. where the variables are the coefficients x(1,0,0) + y(0,1,0) + z(0,0,1) in the expression of the input in terms of the standard basis, and write the output as a 3-tuple of numbers, i.e. again as a linear combination of the standard basis, it is not surprising that you can "read off" the matrix of f with respect to the standard basis from those entries of that 3-tuple 4:36 PM @leslietownes @SineoftheTime thx! @EE18 we went over this already... @EE18 it is not reasonable to talk about a subsequence of a sequence that technically speaking isn't defined in the first place and, in general, subsequence is a much more flexible notion all I'm proposing is that the sequence is only defined for sufficiently large values of$n$and that's all that is needed to discuss any matter relating to convergence can i ask the last rhing wait the points were listed like this 1. determine the dimension and a basis of the ker$f$; 2. determine the dimension and a basis of$Im f$Is it okay if i first determine the dimension and base of the$Im$? Because then to find the dimension of the ker I can do (dim starting space - dim of the$Im$), is that ok or do I have to use another method so first find the dim and a base of the ker$f$? @psie I suppose you mean a limit of measurable functions. And what you say after explains this, indeed pizza: mathematically you could do that in any order, but if that is really a question about what a specific instructor/evaluator would require in written work, if they posed the question that way, it is a question that only that person can answer 4:50 PM @leslietownes Can the procedure I wrote be considered correct? sometimes the order in which related exercises are given is not significant, i.e., they're ordered simply because it's basically impossible to convey more than one thing in natural human language without somehow ordering them. sometimes the ordering is significant because it suggests a framing for how an instructor expects you to approach a problem in terms of the tools that you are expected/allowed to use. @leslietownes yes, I think they are set in order by how the teacher expects me to approach a problem pizza: in life, nobody really cares whether or how you determine the dimensions of anything. if you asked most people to determine the dimensions of the kernel and the image of a linear map, without specifying more, nobody would care what order you solved those problems in. i can't imagine why anyone would care about what order you solved them in, and there is no general reason why you would do one first and then the other there's no "natural" order to that that i can think of no i mean if you think the ordering has some kind of message from the teacher hidden in it, the best way to reveal that message is to ask them nobody here can know that 4:55 PM is this correct: to find the dimension of the ker I can do dim starting space - dim$Im$? there's a background theorem here, which i expect you know, which tells you that if you calculate either one of those, you can determine the other one okay, yes, dim(ker f) + dim (im f) = dim (the domain of f) that is a question that a random person on the internet who does not know your teacher's mind can answer :) i understand , thanks! 2 hours later… 6:37 PM what is the analogue of the term basis in linear algebra to something in abstract algebra? null space as a subspace of a linear map is like kernel as an ideal/normal subgp of a hom. between rings/groups respectively a basis seems to be like the generator of a cycle group, which is not necessarily unique the dimension of which is kind of like the number of groups in a direct sum/product representation? idk probably not obliv because there is some freedom in defining what you want in an "analogue" of a basis, there is more than one conceivable answer here, not all of which will carry a whole lot of the linear algebra understanding over to some more general setting What about for a matrix? if$T: V \to W$maps n dim vector space V to m dim vector space W, we have M(T) an m by n matrix many abstract algebra books discuss the classification of finitely generated abelian groups, which are examples of "modules" over the ring Z of integers, and there is a notion of "basis" in this setting represented by some non unique basis vectors one natural setting would be "how much linear algebra can you do with the scalars coming from some [commutative?] ring that might not be a field," and that's basically the study of modules over a [commutative] ring 6:47 PM @leslietownes this intro abstract algebra book didn't cover modules but from what I can gather they're like group actions on a ring or something? and you run into matrices and things there although having them is not always as useful as you might expect fraleigh is an example of an abstract algebra book that would at least introduce the notion of 'basis' appropriate to modules over Z ight I'll check that out artin would be a more sophisticated treatment that does the whole thing (including i think at least some canonical forms of linear maps over modules, where the coefficient ring is nice enough) what's an ectension field? is it the same thing as a field extension? smells like UK 7:07 PM @Obliv Looks like a typo? There, it is spelled "extension". Where did that TOC come from? 7:38 PM looks like an international edition of fraleigh odd that the copy editors would bother editing the TOC, let alone by introducing errors 8:04 PM hi 9:07 PM Not to mention omitting chapter 8 it looks like did fraleigh make another algebra text? I don't know if this is the one you were talking about or a grad text version I don't immediately see modules covered I'm really liking fraleigh though not by that name. in the chapter on what fraleigh calls "advanced group theory," there is a discussion of free abelian groups and finitely generated abelian groups, and the notion of 'basis' is introduced for those things. abelian groups are Z-modules fraleigh is making a pedagogical choice there not to run off to "oh, let's just do everything in terms of abstract modules." i'm mixed on the book as a whole, but i think that's the right choice. i wouldn't recommend that anybody go off and learn about modules before being really solid on abstract linear algebra, which fraleigh isn't assuming (he covers the very basics in his discussions of fields) every factor/quotient ring is an ideal right Every factor/quotient ring is an ideal which is also the kernel of a ring hom. ? Idk, I like his textbook font & style but I can't give it a good assessment from the brief skimming I'm doing @Obliv literally the LEAST important things... but I like axler for the same reason. It's visually clear not just a dense page of words like some grad texts i've seen (also friedberg) don't forget the binding 9:22 PM I disagree, if it's in braille how could I read it effectively? I think visual clarity is pretty important @leslietownes the binding is actually really important. Hatcher's book really fails on that one. I also like these little historical notes he puts in @Obliv That isn't a choice of fine and style, though. some printings of royden's real analysis book (3rd ed, it is now in a 4th) had really bad binding. my own copy had pages falling out, something wrong with the glue. so i went to the library, and both of their copies had missing pages @leslietownes Probably people stealing the pages they need to fix their own copies. 9:26 PM not enough horses in the glue Oh stop.. It could be he put too much horse in the glue mixture. I don't get this, what are they saying exactly? That$F$is a subfield of$E$? Why are there$F(x),F(y)$as two separate branches in$\mathbb{R}$and how does that combine to make$\mathbb{C}$They are giving two examples Those rhs fields are not related to lhs fields yeah. maybe better set as two diagrams, with different spacing between them maybe put them on either sides of the paranormal distribution 9:34 PM hehe @leslietownes u mean like that? I still don't get it @Obliv Which part? "A field E is an extension field of a field$F$if$F\leq E$" \leq means subfield is the only qualification really just to be a subfield? 9:40 PM obliv: to make the remark clearer, the diagram involving C, R, and Q has no logical relation whatsoever to the other diagram. F(x,y) and F(x) and F(y) and F are not intended to relate to the C, R, and Q in the other diagram @Obliv yes Ohh I see so the Q,R,C thing is a tower of fields the right side is just some other subfield diagram to illustrate the extensions I.. can't be arsed to redo the meme just pretend the tower is on the left and the other one is on the right side @Obliv Also note that C is not a finite extension of Q, but F(x,y) is a finite extension of F, so you can't relate them. Honestly don't know what F(x),F(y),F(x,y) even mean F is a field but F(x) looks like a function.. Sorry to interrupt, can I ask you a question? If I have a matrix of dimension 3, the number of the eigenvectors is less than the dimension of the matrix, can i Say that the matrix is not diagonalizable?? 9:45 PM pizza: if by 'number of eigenvectors' you mean something like 'largest number of linearly independent eigenvectors,' yes @Obliv F[x] is the polynomial ring, F(x) is the fraction field of that ring what is a dimension of a matrix? Shouldn't it be m by n format @Obliv yes 3 rows 3 columns obliv: he's talking eigenvectors so implicitly it's a square matrix with m = n. can you guess what the common value is intended to be? For example 9:46 PM @SoumikMukherjee what is a fraction field pizza: recall the kind of annoying fact that if v is an eigenvector of a matrix, so is kv for any nonzero scalar, so (when your field of scalars is infinite) any matrix with an eigenvector will always have infinitely many of them. the 'count' you want to do is a count of linearly independent eigenvectors It isn't obvious to me, btw, that a polynomial ring is a field extension of a field When an earthquake happens the fields get fractured, those are called fraction field not to mention a polynomial ring isn't a field Just kidding 9:48 PM I.. I don't know what to believe anymore In abstract algebra, the field of fractions of an integral domain is the smallest field in which it can be embedded. The construction of the field of fractions is modeled on the relationship between the integral domain of integers and the field of rational numbers. Intuitively, it consists of ratios between integral domain elements. The field of fractions of an integral domain R {\displaystyle R} is sometimes denoted by Frac ⁡ ( R ) {\displaystyle \operatorname {Frac} (R)} or... @Obliv The polynomial ring itself is not an extension, because it's a ring and not a field Reminds me of a skit where these guys visit a hospital for the mentally ill (for lack of a better word) to pick up their friend and someone who looks like a doctor greets them and offers to guide them. After a while, the doctor just starts making strange noises and skips away. Another "doctor" catches up to them and apologizes for that patient. @SoumikMukherjee is the gist that we define a field of rational functions like$F(x) = \frac{P(x)}{Q(x)}\$ or something

I found the clip haha

@leslietownes Now I'm on the phone and so I was checking an exercise and I saw this: emathhelp.net/en/calculators/linear-algebra/…
and It gave me this answer: Since the number of the eigenvectors is less than the dimension of the matrix, the matrix is ​​not diagonalizable.
I'm going anyway, see you tomorrow!

10:13 PM
time to stop being lazy :p
tfw u trace a definition/theorem only to be led to another definition/theorem
so I have to duplicate the pdf multiple times so I don't lose my page :(