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12:17 AM
I've constructed a solution for $x_{k_n}$.
For each $x_{k_n}$ and for all Cauchy sequences $\{y_i\}$ in $E$ with $y_i\to x_{k_n}$ in $X$, we have $g(y_i)\to g(x_{k_n})$, again from the definition. For each $n$ pick $y_{i_n}$ such that $d_X(y_{i_n}, x_{k_n})<1/n$. Letting $n\to\infty$, we know that $d_Y(y_{i_n}, p) < \varepsilon$ for any given $\varepsilon>0$. From the uniform continuity of the first function $f$, it follows that $g(y_{i_n})\to g(p)$, but then $g(x_{k_n})\to g(p)$.
Do you think it is logically valid?
Forget subsequences, and just deal with a sequence. Minimize notation to make things as clear as possible.
Your $y$ will need the $n_k$ index, too. Now think about my Cantor diagonal comment.
Do you say that I need it here for this to work or that I should think about it as another cleaner method?
Your notation is not sufficient. You need a sequence of $y$s for each $x_{k_n}$. First, simplify by making $x$ just a sequence, not a subsequence. Then for each $x_k$, you have $y_{k,i}\to x_k$. I'm suggesting that you look at $y_{k,k}$.
Ah, double indices, how I could forget them here.
12:38 AM
what's the difference between an R-algebra and an R-module, practically speaking
if you can do something with one of them, can't you do the same with the other?
To show that it is continuous, let $p\in X$. Assume $\{x_n\}$ is a sequence in $X$ such that $x_n\to p$ and $x_n\neq p$.
For each $x_n$ and for all Cauchy sequences $\{y_i\}$ in $E$ with $y_i\to x_n$ in $X$, we have $g(y_i)\to g(x_n)$, again from the definition.
Let $\{y_{n,i}\}$ stand for some such sequence corresponding to each $x_n$. Looking at $y_{n,n}$ and letting $n\to\infty$ we have $y_{n,n}\to p$, and by uniform continuity of the first function, we have $g(y_{n,n})\to g(p)$, which means that $g(x_n)\to g(p)$.
Here is full.
WTF is an $R$-algebra? I know what a $K$-algebra is.
i'm no expert but I think it's when you take $R$ to be a ring instead of a field
my question is the same for a $K$-algebra, what's the difference with a vector space?
I know that. What is an algebra ?
12:53 AM
you take a ring homomorphism $\phi:A \to B$, then $B$ is an A-algebra
the isomorphism theorems give you that $\phi(A)$ is a subring in $B$
which makes me think that $B$ behaves exactly like a vector space wrt. $\phi(A)$
Tell me the difference among ring, vector space, and algebra.
a ring is, wrt. addition, an additive abelian group, and wrt. to multiplication it has associativity and distributivity
vector space has the same additive abelian group structure, but it has a scalar multiplication $S\times V \to V$ operation instead of the ring's $R \times R \to R$
an A-algebra is a ring $B$ s.t. you accompany it with a homomorphism $\phi: A \to B$
i'm talking commutative unital rings here, i don't know anything about any other sort of ring
No, that is not the definition of an algebra. It’s typically a ring with an additional multiplication by a.
No, that is not the definition of an algebra. It’s typically a ring with an additional multiplication by a field. The ring is a generalization,
Polynomials over $k$ form a $k$-algebra,
i'm having trouble distinguishing the vector space multiplication from the k-algebra multiplication
1:08 AM
It’s a vector space with an internal multiplication as well.
Alternative viewpoint.
oh, so you can multiply "scalarless" vectors directly?
I often hear *morphism words, even out of algebra context. Do they initially teach undergrads those morphism things in first algebra course?
they're the noteworthy functions between algebraic structures
Did you first hear them when you started self-studying algebra?
I mean, their definitions mentioned.
you'll hear isomorphism in linear algebra
1:13 AM
@noballpointpen yes
Howdy, leslie.
@noballpointpen not limited to algebra! It’s just a word for mapping.
Homeomorphism, diffeomorphism.
But where undergrads hear them (their definitions) first time?
@shintuku you can consider it as a ring with $K$-vector space structure
hello. to clarify my response, i think the average US undergrad, at least someone who went to high school in the US, would first hear 'morphism' words in an undergraduate algebra class, and maybe not their linear algebra class.
1:15 AM
at ted: have found the definition defining the additional multiplication. i wasn't actually understanding that it was a distinct multiplication that coexisted with scalar multiplication
Stop with algebra as the center of the universe.
I remember in linear algebra, the base field is usually denoted by $F$ but outside linear algebra, suddenly it's denoted by $k$.
@shintuku With a ring as the object, with extra scalar multiplication, that is clear.
@onepotatotwopotato Nah.
@leslietownes Unless they took topology first.
At least Hoffman Kunze uses $F$
1:17 AM
Ok, thanks everyone.
ted: what a weird school that would be. what is it, hogwarts?
Probably it would be someone who enrolled already with experience in math.
@noballpointpen Nah.
@leslietownes I took algebra and point-set simultaneously!
Oh, and I learned diffeomorphism before topology !
1:34 AM
Hey all, is this statement true? I can only seem to find it on this website…I have been trying to look for this theorem in Hall’s intro to lie theory text to no avail
ted: at hogwarts
Yup. Go swim with ducks.
Or if anyone knows of a lie theory text that may have something to say about wherher the adjoin action of a lie matrix group on its lie algebra commutes with the exponential mapping :) that would be helpful
@noballpointpen the first time i got an actual definition was my second proof based math course: abstract algebra I. In my linear course the notion of isomorphic vector spaces was mentioned but not defined. And then the first time i heard of the word morphism verbatim is in a category theory text
goose: math.stackexchange.com/questions/1173056/… there seem to be various versions of this on MSE.
1:45 AM
Ah! I just came across such a post hehe. The title was ostensibly unrelated to the commuting business I was interested in. Hm i bet it is somewhere in Hall then
2:38 AM
Hi @TedShifrin. If you remember, in our previous interaction, I presented an alternative argument for the proof of the second case, where the countable sets are not all disjoint , which was some hours back.
Although you mentioned that my approach is quite equivalent to yours, I was still left with some lingering doubts about the validity of my proof as I feel, your comments weren't too clear as it never mentioned about the validity of my arguments but only mentioned the approach is much equivalent to yours.I understand that exposition plays a crucial role in conveying mathematical ideas effectively, and I aim to improve the presentation of my proof.
However, I am much curious about your insight in assessing the soundness of my arguments, or better say to give your judgements about the validity of my arguments/ proof for the case of the theorem, when, the countable sets are not all disjoint. To be precise, I want to know whether my proof is "correct" or not.
Add on: I apologize if one of my previous comments were unclear.
@TedShifrin If you think my arguments are correct. Then please consider posting it as an answer to the original post. I will be more than glad to accept it.
2:54 AM
Link please.
Q: If $A_n$ is a countable set for each $n ∈ \Bbb N,$ then $\cup_{n=1}^{\infty}A_n,$ is countable.

Thomas FinleyI was recently, going through a proof of the following theorem, which states that: If $A_n$ is a countable set for each $n ∈ \Bbb N,$ then $\cup_{n=1}^{\infty}A_n,$ is countable. [Definitions that are givem in the book: Countable Set: A set A is countable iff $\exists $ a bijective mapping $f:...

3:42 AM
@TedShifrin Nevermind, I got my validation and now I know, that my proof's valid. Thanks for your cooperation !
consider that we are only working with matrix lie groups. Let $Y \in \mathfrak{g}$ and $X \in G$. hm so this result is a distinct result from "matrix exponentiation commutes with conjugation"?
because as i understand it the LHS will be some power series involving nested commutators and it is at least not obvious of how to turn that expression into some sort of exponentiated conjugation of $Y$
1 hour later…
5:11 AM
those matrix groups are all a big lie
6:02 AM
Let $f \in C[0, 1]$ . Then construct a sequence (g_n) \subset C[0, 1] such that $g_n(0) =0$ and $g_n\to f$ in the integral norm ( $\|f\|_1=\int_{0}^{1}|f(t) | dt$)
Such sequence clearly exists.
Because $\{g\in C[0, 1] : g(0) =0\}$ is the kernel of the unbounded linear functional $\phi_0 (g) =g(0) $.
what do you mean by that :0 @copper.hat
you can be pretty explicit about this. consider e.g. h_n given by h_n(0) = 0, h_n(1/n) = 1, and h_n(1) = 1, and having a graph that is line segments in between. h_n f is 0 at 0 and h_n f goes to f in || ||_1. you can show that ||h_n f - f||_1 <= ||f||_oo/n, for example, where || ||_oo is the sup norm.
@leslietownes Thanks.
So I have proved that the quotient space X/M is a normed space iff M is a closed linear subspace of the normed space X.
M closed linear subspace iff $\|x+M\|=d(x, M) =\inf\{\|x-m\|:m\in M\}$ is a norm on X/M
Reverse implication: Suppose M( linear subspace of X) is not closed. Then choose $x\in cl(M) \setminus M$ .
Then $d(x, M) =0$
6:20 AM
is there a closed form of the adjoint action of the lie group on the lie algebra for matrix lie groups? I.e., a closed form for $ad^{-1}_X$
i am trying to see if i can evalate this formula without deferring to the power series for the middle fraction; since the power series is a bit intractable :P
sorry I mean to say *inverse of the adjoint action...
7:00 AM
@SillyGoose I don't know if this answer helps.
7:17 AM
It does use the power series, however.
It's hard to avoid the "Power" of the power series...
1 hour later…
8:24 AM
@AlessandroCodenotti So I was browsing a theorem about how having a function $h\in C(X)$ unbounded on a set $E\subseteq X$ implies that $E$ contains a $C$-embedded copy of $\mathbb{N}$. I didn't realize it at first, but that copy of $\mathbb{N}$ is closed, which I proved explicitly. But funny thing is that there's an approach using more theory, one can just prove that realcompact $C$-embedded subsets are closed.
maybe it's an analogue of compact $C^*$-embedded subsets being closed, lol
actually I guess it is, since all compacts are $C^*$-embedded
If I learned game theory can I be better at games?
8:49 AM
This is a very beautiful answer math.stackexchange.com/q/1023681/977780
for high-rep users only :)
9:22 AM
Q: The set of all sums of 3 primes covers almost all of $2\Bbb{Z}+1$ (result of Helfgott), what can one say about $\Bbb{P} - \Bbb{P} = 2\Bbb{Z}$ problem?

D Left Adjoint to UBy Helfgott's ternary Goldbach conjecture we have that $\psi : \Bbb{P}^3 \twoheadrightarrow (2\Bbb{N}+1)\setminus \{1,2,3,5\}$ given by $\psi(p,q,r) = p + q + r$ is surjective. Which is to say every odd integer greater than $5$ is a sum of exactly $3$ prime numbers. Now consider $\Bbb{P}' := \pm...

9:39 AM
I don't get why this got downvoted
looks interesting
10:04 AM
@Jakobian upvote dot chit den mon
I completely rewroted per my readers' request
you guys are my readers
I'm da writer
In fact, downvote attracts people's attention more than zero vote.
A downvote is like carrion signal to the vultures
10:37 AM
Does Bessel's inequality in functional analysis imply that any inner product space can be continuously embedded into the sequence space l2?
(with countable orthonormal basis)
Well, more so Parseval's identity
it implies that there's an isometry from your inner product space to $l^2$
Well, I thought that what I wrote is a consequence of Bessel's inequality. Is this correct?
Oh I see, Parseval is needed to get isometry
ah, yeah I suppose Bessel's inequality still shows continuity, but I don't see any advantage in using it
Just continuity doesn't need completeness I think
it wouldn't be an embedding then
oh, you mean completeness of inner product space
10:47 AM
Parseval's identity doesn't require completeness
all spaces are assumed to be complete
that's a weird thing to say...
11:11 AM
Yeah, it's like saying you assume axiom of choice. Why assume something trivially true?
11:32 AM
Trivially true 🐼
11:45 AM
of course but well-ordering principle is definitely false.
12:26 PM
Just to be sure: If $S$ is an $R$-algebra and $M$ is $RG$-module (group ring), then $S\otimes_RM$ is an $RG$-module by $rg\cdot (s\otimes m) = rs\otimes gm$, correct?
$S,R$ are all commutative unital
12:45 PM
I mean left action (and the action map is additively extended). It's noncommutative.
Ah we can just suppress left anyway
what is the meaning of the curved arrow when defining a function? usually it is
f : X \rightarrow Y
But sometimes i see, f: X rightarrow y but with a little curve like a cane on the end
what that means depends on context, and whoever is using it should ideally be defining it somewhere before they do that. it is often used to denote functions that are one-to-one, or inclusions, or "embeddings" (in some sense), or more generally are monomorphisms in some category of interest
In quantum mechanics, bra–ket notation, or Dirac notation, is used ubiquitously to denote quantum states. The notation uses angle brackets, ⟨ {\displaystyle \langle } and ⟩ {\displaystyle \rangle } , and a vertical bar | {\displaystyle |} , to construct "bras" and "kets". A ket is of the form | v ⟩ {\displaystyle |v\rangle } . Mathematically it denotes a vector,...
Hi leslie, glad to see you again
If you scroll down, where they define the embedding "phi" they use that arrow between the hilbert space and its dual
So they mean that function is a one to one?
1:09 PM
Actually, even an isomorphism
1:41 PM
@冥王Hades theoretically, yes.
> Coursera offers 420 Game Theory courses from top universities
Am I the only one hating the new layout to display the score of the questions and answers ?
We'll see how many 🌠s that gets.
@冥王Hades you can become better at obliterating the iraqi army
@SillyGoose:-) just a play on the word Lie (as in group)
2:38 PM
Can you have a solution to a PDE on $\Bbb R^n$ that can be expressed as a product of 1 dim. solutions?
$\Phi(u,s_1,s_2,\cdot\cdot\cdot,s_n)=\Phi(u,s_1)\cdot \Phi(u,s_2)\cdot\cdot\cdot \Phi(u,s_n).$
I often see solutions expressed as infinite series, as linear combinations of solutions
@Mad well... yes, although the wikipedia's link to "embedding" doesn't exactly make that clear, and the text of the entry itself doesn't say exactly how much structure they intend to convey by the use of the curvy arrow.
although this is a good example of how a lot of readings of the curvy arrow can simply be read as "\to" i.e. you may not need to know the precise sense of in which the notation is used to understand e.g. the definition of the map itself, which may clue you in as to other properties
2:54 PM
May 20 at 16:59, by one potato two potato
Now the claim is there exists a unique maximal ideal $\mathfrak{m}$ of $R$ such that $M\simeq R/\mathfrak{m}$ as an $R$-module.
This was about simple $R$-module, which is more or less straightforward.
anyone here familiar with the braket notation?
I can translate a bit
if A is a linear operator on hilbert space, then the dual of A(v) is $\phi_v \circ A^* $
with$ \phi_v = <v,->$
Prove an ideal has at least one generator. Since an ideal I of R is a subgroup of the additive group of R, and this latter group is cyclic, we have I cyclic group. Since x is an element of I if and only if it is an element of the additive cyclic group I, x is generated by some member of I.
does anyone believe me?
why is that statement true?
In braket notation_:
the dual of$ X \vert a > is <a \vert X^*$
book: Sakurai
@shintuku problem: is an ideal with two generators really a cyclic group?
3:08 PM
What can I do if my institution doesn't provide access to a certain journal, and there is no preprint on arxiv?
I dont understand what you are saying
ok that works thanks
3:32 PM
The following question was downvoted. Although I cannot know for sure, I'd like suggestions on why, please.
Q: Is $\pi$ almost surely normal? A reference request.

ShaunNB: This is a reference-request question and so there's not much context I can give. It is claimed here that $\pi$ is almost surely normal; that is, the probability that $\pi$ is normal is one. I haven't found a reference for this anywhere. If it is true, please would you provide one? Context: I ...

@shintuku the additive group of a ring doesn't need to be cyclic
@Shaun To be somewhat coarse: The claim you are asking about is somebody talking out of their ass.
Read the downvoter's mind!
3:50 PM
Is there a definition for residue map in commutative algebra?
@JaakkoSeppälä the canonical homomorphism between a ring and a quotient ring?
Is it true that $\lim_{x \to 0} f(x) = 0 \implies \lim_{x \to 0} f\left(x\left(\frac{1}{x}-\lfloor\frac{1}{x}\rfloor \right)\right)=0$?
I think it is true, because $\left|x\left(\frac{1}{x}-\lfloor \frac{1}{x}\rfloor \right)\right|=\left|x\left\{\frac{1}{x}\right\}\right| \le |x|$ and so, given $\epsilon>0$, from $\lim_{x \to 0} f(x)=0$ there exists $\delta>0$ such that $|x|<\delta \implies \left|x\left(\frac{1}{x}-\lfloor \frac{1}{x}\rfloor \right)\right|<\delta \implies \left|f\left(x\left(\frac{1}{x}-\lfloor\frac{1}{x}\rfloor \right)\right)\right|<\epsilon$
4:05 PM
za: careful now. consider something like f(x) = 0 if x is nonzero and f(0) = 1. note that lim x to 0 f(x) is 0. but if you let e.g. x_n = 1/n then f(x_n (1/x_n - floor(1/x_n))) = 1 for all n (and with more thought, lim x to 0 f(x) does not exist).
there's that crap about in the limit statement, there's delta such that (begin highlight in bright red) 0 < (end highlight) |x| < delta implies some other stuff.
@Shaun take my approach. i've been downvoted because i don't understand what i'm asking about. instead of wondering why, i relentlessly hate the downvoters
shin: another thing you can do is just go and randomly downvote someone else. there's this concept that if you have low self esteem, making someone else feel bad for no reason gives a temporary high that for a brief moment is almost like having self esteem.
or randomly downvote five other people.
an eye, arm, and a leg for a bruised ego?
i have the eternal, just and immutable right of the underdog, i do not doubt myself
sry that's just the pain of the downvote speaking
4:21 PM
@leslietownes so the mistake in my reasoning was that there are real values of $x$ in each neighborhood of $0$ that makes $x\left(\frac{1}{x}-\lfloor \frac{1}{x}\rfloor \right)=0$ and so the implication given by the limit hypothesis does not hold?
@s.harp I did ask whether the claim is true, though, so why is that an issue?
@shintuku Same, but I try my best to get feedback to better my future posts, no matter how asinine that feedback is.
@onepotatotwopotato I wish I could! But, yeah, I know I can't tell for sure; I'm just looking for suggestions.
za: given epsilon, there's delta such that if 0 < |t| < delta, then |f(t)| < epsilon. if |x| < delta, then you can definitely deduce |x(1/x - floor(1/x)| < delta too. but unless you also know that 0 < |x(1/x - floor(1/x)|, you wouldn't be able to deduce that |f(x(1/x - floor(1/x))| < epsilon from anything we've written.
shaun: a limitation of this approach is that the people who will give you suggestions are most likely not the people who are downvoting you. they might help you "better" your question, but maybe not in the direction of "avoiding downvotes," because they don't know how those people think any more than you do.
@leslietownes thank you very much for your help!
5:22 PM
@JaakkoSeppälä currently reading hungerford, i just stumbled upon his name for it, he calls it the canonical projection or canonical epimorphism, fullstop
Q: Help me please proving the theorem below

KottoamatsukamiI have been trying to prove the theorem for several days now, but I still can't, so please help me. The formulation of the theorem is as follows: $$ $$ Prove that $\forall f\in C(K): $ if $f(K)$ is bounded in $\mathbb{R}$, then it follows that K is a compact. $$ $$ $K$ is a subset of $\mathbb{R}$...

Interesting question :)
A topological space $(X, \tau) such that \forall f\in C(X, \Bbb R) is bounded is called pseudo-compact space.
Compact implies pseudo compact.
Converse need not be true ( ex:infinite particular point space)
However in a metrizable topological space pseudo-compact implies compact.
5:45 PM
@SouravGhosh why such example?
it has pretty bad separation properties
If you find this interesting: Prove that $X$ is pseudocompact iff for all $f\in C(X) = C(X; \mathbb{R})$, $f[X]$ is closed.
An example of a space with good separation properties that's pseudocompact is $\omega_1$
It's $T_5$, pseudocompact, not compact. Any function from $C(\omega_1)$ is eventually constant.
@Jakobian I will try to prove.
@Jakobian First uncountable ordinal with the order topology.
@Jakobian Suppose $X $ is pseudo compact and $f\in C(X, \Bbb R) $ . Claim: $f[X]\subset \Bbb R$ is closed.
Let $f[X]$ is not closed. Then $\exists y\in Cl(f[X] ) \setminus f[X]$
Now define $g:X\to\Bbb R $ by $g(x) =\frac{1}{f(x) -y}$
Then $g$ is continuous but not bounded.
As $\exists (x_n) \subset X$ such that $f(x_n) \to y$
6:10 PM
solving system of equation :
For $a+2b+3c=0$ and $5a-2b+7c=0$, I am told that $\frac{a}{20}=\frac{b}{8}=\frac{c}{-12}$. It seems to me that they got denominator for $a$, i.e., 20 by ignoring $a$, and taking determinant of matrix formed by coefficients of $b,c$.
But I can't see why it should hold.
6:25 PM
Consider a real analytic manifold $(M,g_1,\mathfrak{A})$ with metric $g_1$ and real analytic foliation, $\mathfrak{A}.$ And consider $(N,g_2,\mathfrak{B})$ with analytic diffeomorphism $f:M \to N,$ for $\dim(M)=\dim(N).$ Take the foliation $\mathfrak{B}$ and $g_1$ and form the object $(O,g_1,\mathfrak{B}),$ where $O$ is a properly embedded submanifold in $M.$ Due to the embedding, $O$ inherits the induced metric $g_1.$
Give an example s.t. $(O,g_1,\mathfrak{B})$ solves $xu_{xx}=tu_{t}$ and $(N,g_2,\mathfrak{A})$ solves the corresponding equation for the metric $g_2.$
@Silent Find the null space of $A=\begin{pmatrix } 1&2&3\\5&-2&7\end{pmatrix}$
6:39 PM
@SouravGhosh sir I see that I need to find null space, possibly by row reducing. But I can't understand why the same task can be done by simply taking determinants while ignoring one of the three variables at a time
6:59 PM
@Shaun To make the example more extreme: You would get downvoted for asking what the guy in this picture is talking about i.redd.it/y9dk1prvwf2b1.jpg . I think in this case its easier to understand why: the downvoter interprets the question as distracting noise. I think for your question the sentiment is similar, its kind of an unhappy surprise to click on the link and then you see somebody talking ****. (I didn't downvote your question, so I guess this is just reading tea leaves)
deleted false attempt
@Shaun Although going through your questions and answers it seems more likely that you have a hater since random totally innocuous answers are downvoted
$F(\alpha)$ is a finite field extension implies $[\exists N \in \mathbb N$ s.t. $\forall n \in \mathbb N$, $n \geq N \implies \alpha^n \in F(\alpha)$ is in the span of $\{1, \alpha, \cdots, \alpha^N\}]$, this proposition clearly follows from the definition of a finite field extension. Does the converse hold?
7:17 PM
@Silent $a+2b+3c=0$ and $5a-2b+7c=0$ : 2D plane.
@Silent Solution of (1) and (2) is a line.
Line of intersection is parallel to (1 2 3) × (5 -2 7)
and the line passes through the origin.
Compute the cross product of $r_1=i+2j+3k $ and $r_2=5i-2j+7k$
Then the equation of the st. line in symmetric form ( passes though (0, 0,0) and direction cosines (20, 8,-12)) is given by a/20=b/8=c/-12
@shintuku Suppose the converse hypothesis, with $n \geq N$. then $\alpha^n = p$, a polynomial in that basis, so $0 = \alpha^n - p$, so $\alpha$ is algebraic.
thanks discord
7:35 PM
$\alpha\in F(\alpha) $
Then $\alpha^n\in F(\alpha) $ for all n\in \Bbb N
If $\alpha^n\in \span\{1, \alpha, \ldots , \alpha^N\}$ for some $N$ then [F(\alpha): F]\le N$
Experiencing major brain overload right now
hm, that's not obvious directly to me, unless we take the route of saying $\alpha^n \in \text{span}\{1,\alpha, \cdots, \alpha^N\} \implies \alpha$ algebraic $\implies [F(\alpha):F] \le N$
at sourav
Correct. Then degree of the minimal polynomial of $\alpha$ over $F$ is at most $N$.
7:44 PM
oh ok
@Jakobian I like the particular point space,a simple topological space with many weird properties.
It is a non KC space. We can construct a non compact space (infinite particular point space) with compact dense subset.
I don't like it because it's really bad separation properties-wise
that's more of a preference, but I like working with (at least) Tychonoff spaces
I don't mind working with spaces with weaker separation properties of course, but I prefer that they do have strong separation properties when they can
8:09 PM
8:34 PM
To prove irrationality why integrate?
Q: To prove irrationality, why integrate?

Timothy ChowI have been reading David Angell's lovely book, Irrationality and Transcendence in Number Theory, which has given me some fresh insights even with some of the easier proofs. But the book reminds me of something that I've long been puzzled by, which the book hasn't cleared up for me. In the book'...

MSE also had a post about niven's proof. math.stackexchange.com/questions/4051354/…
IMVHO those proofs are pretty good examples of rabbits out of hats, even to the extent that people can come around later and come up with just-so stories about where they 'came from." i don't mean this pejoratively, it is fine for rabbits to pop out of hats, and coming up with maybe-not-correct stories about where something could have come from is a good way of finding new math.
ban magic from math
For the proof of irrationality of $e$ does this at all hinge on the fact that the integration of f(x)e^x dx is expressible in terms of $e$?
for a suitably chosen $f$
banish math sorcerers, chain them to mana-seals
I mean that the integral evaluates to something in terms of $e$
8:45 PM
@leslietownes Munchkin and I would prefer ducks out of hats.
like i can generally understand various proofs of irrationality/transcendentality that use continued fractions at the level of individual steps, or maybe one or two higher level thought processes, but i don't see the big picture, or even know if there is one. i don't think there's any 'continued fractions approach' that yields useful information a whole lot of the time. any numbers you like have continued fraction expansions and integral formulas, and the concepts are still hard.
I found the words in the title of this paper pretty funny : mathos.unios.hr/mc/index.php/mc/article/view/621
"Thats gold Jerry, gold!"
heh. do you know if that's a topic of general interest? or is it one of those things where people create a 'field' out of some goofy corner of math and every paper in the field only cites like 10 other papers in the 'field,' 5 from the same authors.
furthermore, is it an algebraically closed field?
very important i know, pls respond
Its not of interest, the definition is artifical in that it has nothign to do wiht the golden ratio and basically amounts to choosing two complementary distributions of the tangent bundle, also the main result of the paper is completely wrong
9:00 PM
ah, i had my suspicions when i looked at the list of papers cited, although i was somewhat disappointed that some of them did not have 'golden' in the title.
if i had stayed in academia i would have made an effort to publish papers like that in as many fields as i could find. with a view toward having a lot of exotic journal names on my CV.
for more on "fuzzy" quantum pseudo-golden manifolds, see my paper in the transnistrian journal of mathematics
I think people gossip a lot, if you try this nobody will hire you because you will get a reputation for publishing garbage
yes, but it will be exotic garbage
oh yeah, in this plan i only do it after tenure.
i think the authors of a lot of that stuff know it is garbage, but may be working within academic systems where there is some kind of pro forma pressure to produce X published articles per annum or the government takes your housing away, or something.
so you want papers? i'll give you papers. i think that's the logic.
9:04 PM
I mean in this case the reviewer is also at fault for not recognizing that the theorem is incorrect
reviewers may be under similar pressures. would not surprise me if the journal has a publication fee that can only be paid in currencies other than ones used where the editors live.
Look at the many words run together in the abstract. This has to be phony.
So digging a little deeper, this seems to be the initial papers about Golden geometry: https://www.sciencedirect.com/science/article/pii/S0960077908001690
its published in "Chaos Solitons & Fractals", which had a huge scandal since the editor El Naschie published like 300 of his own papers in that journal without peer review. This particular paper cites 4 articles by El Naschie, which is really kind of suspicious.
when it comes to the math I still don't see what in what way this "golden structure" is different from what the authors call an "almost product structure", which is the same as a
sigh. better form is to have your own journal for that. should have done "El Naschie Notions"
9:41 PM
So much chaos I’ve missed …
9:58 PM
eh, what the heck. i'm in. when do we begin killing apostates?
10:11 PM
I am an atheist in all regards.
what if we dropped phi_t down to 2/5? that's the best i can do.
10:23 PM
I think people associate atheism with many negative things such as no purpose in life
maybe it's a negative outlook, or leftover from religious way of thinking
a person looks for a reasonable explanation of the world around them, at the same time trying to keep their sanity intact, I feel like atheism lacks a little in the latter regard if not approached carefully enough
for example, why would life need a purpose
10:42 PM
The reason why $\alpha$ algebraic implies $F(\alpha)$ is a finite field extension is because the minimal polynomial gives a finite basis for $\alpha^n$ with any $n$, right?
Your sentence makes no sense, shin.
Jakobian: Your thesis that one needs religion to have a valid life is laughable. Most war and violence we can thank religion for.
i.e. i mean that if $\alpha^n + \cdots k_0 = 0$ is the minimal polynomial satisfied by $\alpha$, then $\alpha^i = \frac{-\alpha^n - \dots - k_0}{k_i}$
therefore, $F(\alpha)$ is a vector space with a finite basis
No. The equation says that $1, \alpha, \dots, \alpha^{n-1}$ give a basis for $F(\alpha)$ as $F$-vector space.
shin: not specifically what you've just said, but as a background vibe, the ability to write some power of alpha in terms of lower powers of alpha shows you that any power of alpha (or polynomial in alpha) is in the span of a fixed, finite set of powers of alpha, and more generally that inverses of polynomials in alpha are also in there, and any rational function of alpha is in there too.
I don't think I said anything like that, not sure what you mean. I think you misunderstood my comment, I was saying what I think a lot of religious people may think and misunderstand
10:50 PM
at ted: we wrote at the same time, is my last sentence the same as what you've stated?
in the sense i didn't clarify it was specifically an $F$-vector space?
at leslie: yeah i think this is what i'm trying to get at
No, you are not seeing that all powers starting with $n$ can be rewritten in terms of the lower terms.
oh right, we also have $\alpha^j = \alpha^N + \alpha^i$, where $i$ is smaller than $n$ and $N$ is a multiple of $n$
No. This is just not right in any general way.
Write out some simple examples.
Like $x^3 + x + 4=0$.
11:00 PM
@Jakobian i understood this. even non-religious people sometimes feel a kind of guilt or emptiness about it, even if they did not grow up in a tradition, and sometimes the source of that is a feeling of a lack of structure, or that they're not buying into something they 'should' be buying into.
ted with his examples.
Buying into religion for hatred and discrimination, for example?
Oh, that remark was disconnected?
haha, ted, don't be afraid to share what you really think about religion. penny for your thoughts.
"ted with his examples" was in connection with the polynomial algebra.
shin: you should really work with ted's example.
Well, looking at most of the red states in the US gives beautiful examples.
And Israel, too. And Serbia, and ….
muzzles self
well, in my remark above, i was thinking of the many people i know who are religiously atheists but also culturally catholic or jewish, and for some reason feel "bad" about not being more religious. like, they've internalized a stigma about it.
if you want to give an example of exclusive or, "culturally catholic or jewish," applied to a person, might be a good one.
Clearly I don’t feel bad. Yes, I am culturally Jewish, but even my parents were not raised religiously.
11:08 PM
my wife was raised catholic and part of her envies people who have kept up with it, even if they're people that she does not otherwise agree with or even respect very much.
the algebra of complex numbers is maybe too simple of an example of how you get a finite dimensional vector space out of an algebraic element.
am working on it
I wanted an equation with more terms so shin might unravel his mess.
yeah i was very wrong, will come up with answer soon
I think your wife’s guilt is part of the whole religious premise.
i wonder if there's a market for a book that does calculational linear algebra, like, computational homework exercise stuff, but in the context of the theory of algebraic extensions of Q. it's kind of a shame that a lot of the basic arguments are so concrete, but by the time you get to them, it's in books that assume you know that stuff already and don't put in big matrices or systems of equations in them anymore.
which is not to say that i want to see a week's worth of computing polynomial resultants or anything like that. just a taste of the 19th century would be enough.
11:13 PM
it's the groebner basis stuff no?
i wasn't even thinking that, but yeah, similar. in that vein, i saw that fraleigh put a chapter on groebner bases into the newer editions of his fairly computational abstract algebra book.
or whoever 'fraleigh' is now. whoever owns the fraleigh cinematic universe.
fraleigh the idea vs fraleigh the man
@TedShifrin munchkin is going to grow up with guilt about not spending more time at the duck pond.
ted, you might like this. there's a male mallard duck who, for whatever reason, has been appearing on the roof of our neighbor's house across the street every night. in the same place. he just stands there, sometimes for half an hour or more. occasionally he quacks.
haha "you can't just say perchance." my wife sometimes shares with me the greatest hits of attempts at transitional or summary phrases at the ends of paragraphs from her students essays. you very much can just say perchance.
11:18 PM
@leslietownes Probably waiting for Munchkin!
there he is.
Yup, he wants Munchkin.
the other night we were kicking a soccer ball back and forth until it got dark and the duck just stood there like that, watching us. it was kind of creepy
Not quite Harold and Maude creepy.
thankfully not, although that does reminds me of the time we encountered the corpse of a heron at the duck pond. lots of questions that day.
11:27 PM
Wonderful movie!
yeah, it's great. one of our first VHS rentals from the library after we got a VCR. i don't know why that was on VHS in the 1980s, let alone at our library, but it was.
edit: removing a contradiction
I wonder if it’s anywhere these days.
Shin … WTH are you doing?
wait i almost have it
Do my example.
11:43 PM
ok your example has $x^{10} = -47x^2-48x+48$. am now trying the general case again
what I was doing was fine
supposing $\alpha^n + \cdots + k_0 = 0$ is our monic polynomial satisfied by $\alpha$, then we have, for $j>n$, the following: $\alpha^j=\alpha^{qn+r}=(-k_{n-1}\alpha^{n-1}-\cdots-k_0)^q\alpha^r = (-k_{n-1}\alpha^{n-1+r/q}-\cdots-k_0\alpha^{r/q})^q$ from which we see that the degree of the polynomial will be at most $\alpha^{q(n-1)+r}$
all we're arguing for is that the resulting polynomial has smaller degree than $\alpha^j$
What the hell are you doing?
We want a field extension of degree $3$.
we're showing any power $j$ of $\alpha$ that is greater than $n$ must be expressible by a smaller power than $j$
supposing our minimal polynomial is of degree $n$
You’re doing the wrong argument.
argh my bad
let me reread
Given the minimal polynomial $f$, an arbitrary polynomial $p=qf+r$, so $p(\alpha) = r(\alpha)$, and that is a polynomial of degree ….
11:54 PM
oh, that's a nicer argument, the polynomial must have smaller degree than $f$
It’s the right argument.
we get that from the proof of the division algorithm
hm i see
Not the proof. The algorithm.
It also proves concretely that you have multiplicative inverses.
very cool, thanks for the help

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