Well, then I suppose you cannot answer the question.
What you have learned is that a set $G$ together with the operation $\star : G\times G \to G : (a,b) \mapsto a$ either has no identity, or is the trivial group.
If you don't know anything else about $G$, that is the best you can say.
That being said, I am pretty sure that Neel's example began with the assumption that $G = \mathbb{Z}$, which gives you more information.
If e is an identity of (".", ℤ) , then for all x ∈ ℤ, x = x.e = e.x = e x = e Let x be 0 and 1. 0 = e 1 = e 0 = 1 Contradiction Thus (".", ℤ) doesn't have an identity. Thus (".", ℤ) is not a group.
Rather, if there is an identity $e$ for the operation $a\star b := a$, then $x = x\star e = e\star x = e$ for any $x \in \mathbb{Z}$. That is, $\mathbb{Z} = \{e\}. $ But $\mathbb{Z}$ contains more than one distinct element. Contradiction.
Let $f:\mathbb R\to \mathbb R$ be a continuous function such that $\int_0^\infty f(x) dx$ exists. Then there are four options: a)if $\lim_{n\to \infty} f(x)$ exists then it is must be equal to $0$. b) the limit in a) must exist and equal to $0$, c) if f is non negative then limit in a) must exist. d) if f is differentiable, then $\lim_{n\to \infty} f'(x)$ must exist and equal to $0$.
@Xander: I was thinking whether I should post on mse, because I wanted to show attempt also in one place.
So here, I think that option a) is correct. Because let $x_n:=\int_0^n f(x) dx$, then the sequence $(x_n)$ is convergent. By Fundamental theorem of calculus, the function $\int_o^u f(x) dx$ is differentiable for all $u$ so by LMVT, there exists a $c_n \in (n,n+1)$ such that $x_{n+1}-x_n= f(c_n)$. Taking limit on both sides, it follows that $\lim f(c_n)=0$
If ∃y ∈ ℤ ∀x ∈ ℤ (e.x = x.e = x)
Let e ∈ ℤ such that ∀x ∈ ℤ (e.x = x.e = x)
∀x ∈ ℤ (e.x = x.e = x)
Given a ∈ G
∀x ∈ ℤ (e.x = x.e = x)
e.a = a.e = a
∀x,y ∈ ℤ (x.y = x)
e.a = e
e = e.a = a.e = a
e = a
∀x ∈ G (e = x)
e = 0
e = 1
0 = 1
⊥
¬∃y ∈ ℤ ∀x ∈ ℤ (e.x = x.e = x)
Thus (".", ℤ) doesn't have an identity.
Thus (".", ℤ) is not a group.
@XanderHenderson Is this proof understandable to you?
b) is not correct. Counterexample can be constructed as follows: Let's consider a triangle OAB of height $\frac 12$ and base of length 1 with $O$ as origin, I call this triangle T1. Then, I inductively define $Tn$ as the triangle which has base length equal to half of base length of $T_{n-1}$. Also, suppose all Tn's have the same height that is 1/2. These Tn's define a function g with domain $[0,\infty)$ and range =$[0,1/2]$. This function can be extended to $f:R\to R$ by setting f(x)=0
on x<0 and $f(x):=g(x)$ on $x\ge 0$.
It can be shown that $\int_0^\infty f(x) dx=\lim_{n\to \infty}\sum_{i=1}^n (\frac 12)^i=1$ but clearly $\lim_{x\to \infty} f(x)$ does not exist.
c) is false by b)
d) I couldn't come up with a counterexample nor could I prove it.
Also, I'm not sure if my understanding for b) is correct.
@Prithubiswas what is G? And how does that differ from $\mathbb Z$. I think you meant $\mathbb Z$ there.
@Prithubiswas First off, I don't know why you are writing it in that format. Most working mathematicians (other than, like, formal proofs people) try to write in English. When I am reading an argument, I prefer that things be written in words as much as possible, with notation used only when words are going to cause problems or create ambiguity.
Second, you have $e=0$ and $e=1$. This makes no sense to me.
Again, there is a simple, direct argument: If $a\star b = a$ for all $a,b\in \mathbb{Z}$ and this operation has an identity element $e$, then $x = e$, which implies that $\mathbb{Z} = \{e\}$, i.e. the set with one element. But $\mathbb{Z}$ contains more than one element.
@XanderHenderson Not sure what being a "formal proof people" means. If it's a question in math logic, or in logic, we see that due to the contradiction, the proof is complete. But I agree that in other classes, words help immensely, to make a proof flow better.
@XanderHenderson I get it, and agree, save for what seems to happen on this site, where folks struggling with English, feel more confident in using the "universal language of math". But thanks for clarifying.
Consider the group $\Bbb R/\Bbb Q$. That is, $\Bbb R/{\sim}$ where $x\sim y$ iff $x-y\in\Bbb Q$.
Apparently, you need the axiom of choice to prove that that has a total order
That is, you cannot constructively define a relation $\prec$ on $\Bbb R/\Bbb Q$ with $x\nprec x$ (irreflexivity), $x\prec y\land y\prec z\Rightarrow x\prec z$ (transitivity), and $x\prec y\lor x=y\lor y\prec x$ (connectivity)
@XanderHenderson hey... that's how I make my living. I wrote and maintain the formal system that is used by UCLA and some other places to teach symbolic logic.
@AkivaWeinberger In the system we use, Certain exercises are answered to justify Theorems that can then be used. Students can also cite theorems they have written and proven themselves in other proofs.
Yeah, smoothing out corners is a common approach (and an important one, for example approximating a piecewise-smooth curve by a smooth one).
I haven't read through all the moderator self-nomination stuff. Is there anyone running who thinks that there is a rôle for those of us who like to comment interactively and engage in socratic education (as opposed to posting textbook proofs for people to copy into their homework)?
@XanderHenderson I think the proof will be valid if I replace the "e" in the first and last line of my proof with "y". I think that was the big issue (is it?)
i'm having trouble with hatcher, page 151, namely the homology group of klein bottle pi.math.cornell.edu/~hatcher/AT/AT.pdf. How in the world does $\Phi(1)=(2,-2)$ and what in the world does that have to do with "since the boundary circle of a M¨obius band wraps twice around the core circle."
i'm not sure. I don't know how to understand the generators geometrically or how to get them @TedShifrin. I know $A\cap B$ is homomotpy equivalent to $\mathbb{S}^1$
@Prithubiswas I don't enjoy reading proofs which are presented purely symbolically. I am not the person to proofread your purely symbolic proof. I have already told you how I would present the proof. If you want to know how to write it in a completely formal or symbolic manner, ask someone else.
@amWhy Sure---but perhaps my point would better have been made by saying "most work-a-day mathematicians write in natural language, not in the language of formal proofs". I, personally, have no desire to proofread a proof which is written purely symbolically. Someone working in the field of formal proofs might, on the other hand, be interested.
@TedShifrin Yeah sometimes I do see collaboration and people talking which genuinely feels good, but there are lots of antisocial people and just people who are not interested in sharing and listening (which is a general fact I guess)
@monoidal Here's a suggestion. Put a blue dot on the boundary and attach it with a string to a red dot on the central circle. As the blue dot goes around the entire boundary, what does the red dot do?
@Sayan The best thing to do is to have a seminar with several other mathematicians where you go through a paper of common interest. Interacting on that often leads to joint research work later on.
I just think as math people, or atleast the community I have been around, we put "the problem/idea" way ahead of ourselves. This sometimes goes to the point where you lose your self identity and all you are is a husk of theorems/ideas which looks down upon anyone and everything who maybe wants not to be one dimensional
OK, there are different ways to do that, @fin. One is to use the theorem about invertibility and non-singularity. One is to use just nonsingularity so show that the only solution of $Ax=0$ is the trivial solution.
Okay. I see. Now i'm starting to understand things. Why do we get the map $\Phi(1)=(2,-2)$. Isnt the map supposed to be $\Phi: H_1(A\cap B)\rightarrow H_1(A)\oplus H_1(B)$. I know theyre isomorphic to $\mathbb{Z}$ and $\mathbb{Z}^2$ respectively, 1 is the generator for $\mathbb{Z}$ but where does $(2,-2)$ come from?
@TedShifrin Maybe but for instance not to make things weird, being in the LGBTQ community and being open about it (personlly) has made life kind of difficult to not be judged around in the Indian math community. This identity is thought of as an accessory which I put around and becomes a justification whenever I have mathematical shortcomings
Yeah it's frustrating. Anyway, I do enjoy math, but I don't know till when I can do it sustainably. Hopefully I can understand a few things until then.
@fin I think a lot of the progress that we'd made in this country has been reversed by the enabling of the overt white supremacy and hatred thanks to the last administration. And it is not going away. There have been anti LBGT trends most places in the world.
@fin Back to the math. If you were going to pursue more mathematics, you would encounter invertibility more generally in groups and rings. This is a very natural algebraic notion, just like inverse functions. Singularity/non-singularity is just talking about rank of a square matrix.
@monoidal You're just starting on computing homology groups, I guess. But think about a torus and a torus with two holes. Those might be less confusing than the Klein bottle.
@fin So did you figure out a proof of that question with $AB$ and $B$? Can you write a formula for $A$ in terms of those two matrices $AB$ and $B$?
So my understanding is this, $\Phi:H_1(A\cap B)\cong \mathbb{Z} \rightarrow H_1(A)\oplus H_1(B)\cong \mathbb{Z}^2$. hence $H_1(A\cap B)$ is free abelian of rank $1$ and $H_1(A)\oplus H_1(B)$ is free abelian of rank $2$. So as a map $\mathbb{Z}\rightarrow \mathbb{Z}^2$. $A\cap B$ is the boundary of the mobius band, the boundary circle is homologous to the integer linear combination $\sum_i\sigma_i$ of the generators. Since the boundary circle wraps itself twice around the core circle
Hmm. So $\Phi:H_1(A\cap B)\mathbb{Z}\rightarrow H_1(A)\oplus H_1(B)\mathbb{Z}^2$. We need to calculate the matrix representation of it as a map $\mathbb{Z}\rightarrow \mathbb{Z}^2$. Let $[\sigma]\in H_1(A\cap B)$ be a generator of $H_1(A\cap B)$ then $\Phi([\sigma])=([\sigma]_{A},[\sigma]_{B})$ where $_$ denotes the target of the path being in $A$ or $B$
I'm probably very tired, But how do I relate $\sigma$ to the core circle and how do I use the fact that every closed loop is homologous to a $\mathbb{Z}-linear$ combinations?
It's supposed to help humans (eventually, one day). Think of it as a virtual library of books, but theorems are encoded in a graphical language
It has a much easier user experience once all the core libraries are fleshed out
Just drawing arrows essentially. If you surmise, there are only 4 or so atomic operations in Category Theory diagramming world. The theorems combine these operations into diagram rules, but each edit to the diagram is just adding/deleting an arrow/object
Desktop software has to be complicated for the advanced users yet streamlined / simple enough for new users
So right now essentially all I have going is a fancy unicode-based mathematical diagram editor (sort of like Quiver, but without the LaTeX / TikzCD output for now). But eventually I'm going to becoding diagram look-up and rule application
We have linear map from $R[x]\to End(V)$ ($R[x]-polynomials$)
For elements in $End(V)$ we have composition. I am reading notes and it says that
we have
$(h_1 \times h_2)(f) $ = $h_1(f) \circ h_2(f)$ and if you want to derive you can easily derive it from $f_1 \circ(f_2+f_3)=(f_1\circ f_2)+(f_1\ci...
This lacking details. What type of spaces are $R$ and $V$ and, most importantly, what's the linear map $R[x]\rightarrow\mathrm{End}(V)$ that you say you have? And in which space do $h_1,h_2$ and $f$ live? And what does the $\times$ denote?
What you mean by what type of spaces?$R[x]$ denotes polynomials over real numbers.We have $\beta_f$ linear map from polynomial of real numbers to $EndV$ but now I understand that I gave extra information.$h_1,h_2 \in R[x]$,$f_1,f_2,f_3 \in End(V)$
@anakhro Thanks for the link! I'm going to cover the history and also the proof. I recently learned that it was independently discovered by two different mathematicians. So I'm going to look into if there are two different proofs and choose which to cover for the sake of time.
In Conway's book on complex analysis, he identifies the extended complex plan $\Bbb{C}_\infty$ with the sphere $S = \{(x_1,x_2,x_3) \mid x_1^2 + x_2^2 + x_3^2 = 1\}$ and $\Bbb{C}$ with $\{(x_1,x_2,0) \mid x_1,x_2 \in \Bbb{R}\}$. However, he says something I don't quite understand. He says that for each point $z$ in $\Bbb{C}$, the straight line in $\Bbb{R}^3$ through $z$ and the northpole $N= (0,0,1)$ intersects the sphere at exactly one point $Z \neq N$.
He says that if $|z| > 1$, then $Z$ is in the northern hemisphere; if $|z| < 1$, then $Z$ is in the southern hemisphere. Doesn't $|z| < 1$ correspond to being inside the sphere? And if so, I don't see how the straight line from $z$ to $N$ intersects the sphere at any other point besides $N$.
Let alone intersecting some point on the southern hemisphere.
$\lim_{n\to\infty}\sqrt[n]{1+\cos2n}$
It took me and my tutor 2 days to find this limit, we believe that this function has no limit or if it has, we are not intellectual enough to find it. Hope someone can help us.
So if $n_k$ be indices such that $\cos n_k\to -1$ then $\cos 2n_k$ being a subsequence of $\cos n_k$ converges to $-1$ and now what can be said about $\lim_{k\to \infty} (1+\cos 2n_k)^{\frac 1{n_k}}$
This can be fixed, I think. I made a mistake here. I plan to use $\lim|x_n|^\frac 1n=\lim \frac{x_{n+1}}{x_n}$ if limit on RHS exists.
$\frac{|\cos^2 (n_k+1)|}{|\cos^2 n_k|}=2\frac {(\cos n_k\cos 1 -\sin n_k\sin 1)^2}{2\cos^2 n_k}=2\frac {(\cos n_k\cos 1 -\sin n_k\sin 1)^2}{1+\cos 2n_k}$ and $\cos 2n_k$ is a subsequence of $\cos n_k$ so $\lim \cos 2n_k=0$ hence the limit is $\sim 2(\cos n_k\cos 1 -\sin n_k\sin 1)^2\to 2\sin^2 1<1$ so by ratio test, the limit exists.
no that's wrong.
Or may be not completely. I think that we can conclude: $|\cos^2 n_k|^{\frac 1{n_k}}\to 2 \sin ^2 1$
Now, we are almost done. There exist a subsequence $\cos n_j$ such that $\cos n_j\to 1$, it follows that $(1+\cos 2n_j)^{\frac 1{n_j}}=e^{\frac 1n \log (1+\cos 2n_j)}\to 1$
@LeakyNun and professor Ted, is this okay?
Hi Leslie!
@Koro never mind, that's wrong. It is just not true that $2\sin^2 1<1$. I fed it into calculator which understood 1 as 1 degree hence the wrong conclusion.
An item in diablo 2 has usually a fingerprint which is 4 bytes big. So that would be 2^32 possibilities if I'm right. What is the formula to calculate the neccessary items to have a P% chance to have 2 or more items with the same fingerprint? This is just the birthday paradoxon with different numbers or?
$\lim_{n\to\infty}\sqrt[n]{1+\cos2n}$
It took me and my tutor 2 days to find this limit, we believe that this function has no limit or if it has, we are not intellectual enough to find it. Hope someone can help us.
