Suppose R is a commutative ring, not necessarily with an identity and only have two ideals the trivial ones. Also R is not equal to the zero ring.
If there is a r in the ring s.t. Rr=R show that $\{s\in R: sr=0\}=\{0\}$
Since Rr= R so there is some say a s.t. ar=r and I claim that a is the identity. Because if we can take any other k\in R then k=br, and so ka=bra=bar=br=k since the Ring is commutative. Then since we have an identity, the ring is commutative and doesn´t have proper ideals, then is a field and so doesn´t have divisors of zero
@JoseAntonio I love your insight in observing that ar=r implies a=1. You start to derail after proving that (though it is still a valid path). Simply note that a=a'r for some a', so sr=0 implies s=sa=sa'r=sra'=0.
@Will Actually 10.337 pages a day, including appendices and intro's, so even less, but I don't want to bother finding actual pages(3773 including everything)
I would suggest picking a really cool result from one of the books and challenging yourself to understand that within a certain time period. That would be much more fulfilling for me.
@KarlKronenfeld I really enjoy doing the textbooks, but if I am not consistent with doing some work on them every day or two, then I just give up on them. It is to motivate consistent work :)
another problem is that it is extremely difficult to measure your own understanding, since one often measures it too early on compared to when one should do this sort of test.
Periodically reviewing an idea or method leads to its retention, as opposed to just learning it once.
That's perhaps a justification for my suggestion of doing this one interesting result at a time, instead of one page at a time.
I interrogate myself constantly, if I don't understand absolutely everything on the page perfectly or I can't answer one of my own questions, then I stay where I am
@Karl Kronenfeld Thanks so much for your help. Sorry I was listening a sarah vaughan
The last part says the following: Let R be a commutative ring, not necessarily with identity and whose only ideals are R and \{0\}, and also R is not the trivial ring. Show that R is either a field or (R,+) is isomorphic to Z/pZ for a p prime and the product of any two elements is always zero.
The first part is trivial if it has a 1, it follows that is a field as is commutative and doesn't have proper ideals. But if not have a 1, then we claim that the product of any two elements is always 0. If not, so let rs\not=o, then (r) it must be entire Ring, i.e., Rr=R and so there is a\in R a s.t., ar=r and is easy to see that this a must be 1, contradiction.
Since the product of all the elements is zero and the Ring is not the zero ring, so for a non zero r, Rr=\{o\}. So let \langle r \rangle is a subgroup of (R, +) generated by r is clearly an ideal an so there must be the entire group R. SO there is a n s.t. nr=0. (From here I'm not completely sure how to show that n must be prime) but from here and the homomorphism Z \to (R,+)=\langle r \rangle ,sending n \to nr
is clear that nZ is the kernel and by the isomorphism thm. it follows that Z/nZ and (R,+) are isomorphic.
to show that n there must be prime maybe this works if not, so n is composite so n= ab and then the order of ar must be b but generates (R,+), whose order is n. I´m not sure.
@JoseAntonio I've never seen Pugh's book (which book is this?), but I liked Rudin. Many of the exercises proved things beyond the scope of the text and following them gave a much deeper understanding of the material.
Also the two volumes of analysis of Terry Tao are terrific, the explanation are amazing, and incredible rigorous
@robjohn What is your opinion of Dudley´s book? It's what my professor uses. The proofs are really beautiful but is sometimes so difficult. His proof for example of Arzela Ascoli using ultrafilters was too difficult to follow, for me.
@BalarkaSen The set contains $1$ because you get there by repeated applications of $\lfloor\sqrt{n}\rfloor$. Then it must contain all powers of $4$ by repeated applications of $4n$. Given any $k$, there is an $m$ large enough so that $\left(\frac{k+1}{k}\right)^{2^m}\ge4$. Then there must be some power of $4$ between $k^{2^m}$ and $(k+1)^{2^m}$, so repeatedly applying $\lfloor\sqrt{n}\rfloor$ to that power of $4$ will get you to $k$.
@robjohn Yes, that's it. Just multiply the endpoints $\log_2(k)$ and $\log_2(k)$ by some large power of 2 such that the difference of the floors are at least one.
@robjohn Our study at this moment is about the smallest $c = c(n)$ for which there is perfect power between $[n^{c(n)}, (n+1)^{c(n)}]$. This also relates to the number of perfect powers in $[n^k, (n+1)^k]$ for some $k$, which can be heuristically answered conditionally on abc-conjecture.
Why would you possibly know lots of $13$ and $14$ year olds who know much more than you do? That makes no sense, I don't know a single person who is $19$,$20$ here that could answer a question about Weierstrass's elliptic functions
I was wondering something : Consider the modular equation $X(5)$. Wiki claims that the corresponding Riemannsurface is a sphere with 12 points removed. Any idea how to actually prove that? @ccorn
@ccorn Haha OK. I was just wondering if you can describe them by sheets or whatnots. For examples, those punctures can be topologically transformed into copies of C by z |-> 1/z and all what remains then is to determine the order of the flipping of each of the sheets that will wrap those slits.
The question asker over here has made a comment asking for more references on the classical theory of elliptic functions. But he has already listed the good ones. And he cannot read the german Weber's Lehrbuch der Algebra. He seems to be moving toward modular equations, but prefers fumbling with $k$ and $K$, and generally seems to like ~1900 literature. If you have suggestions, please drop a comment there.
@Chris'ssis It is quite obvious that $\displaystyle \sum_{k = 2}^\infty \frac{k^s}{k^{2s} - 1} > \zeta(2s) - 1$ but that factor of $3\sqrt{3}/2$ is bug-a-boo. What interests me is that $3\sqrt{3}/2$ is of geometric interest, so I'd like to know how it came in there.
$(\forall x)(\forall y)(\exists z)(y \geq x \Rightarrow y=x+z)$
@Chris'ssis Help me with trivial thing above? I imagine it is because I am tired, but I can't see if this is true or false
I always negate things to test if they are true or false, and $(\exists x)(\exists y)(\forall z)(y \ne x+z \Rightarrow y\lt x)$ seemed true, but so did the original statement
Oh sorry forgot to say, $x,y,z\in\mathbb{N}$ with $0$
@DanielFischer How's it going? Could you please assist with a question I posted. Any insight would be helpful. I have posted two proposed solutions. The post is here.
@Alex Whether the limit exists or not depends on the used definition of $\lim_{x\to 0} f(x)$. I don't think using the sequence criterion is helpful here, though.
@DanielFischer I'm assuming the usual definition of the limit at point for a function defined on $\mathbb{R}$. Cant the result I am using from wikipedia not be used the way I used it in proposed answer 1 since x_{n} = x for some n?
@Alex One uses $(\forall x) (x\in D(f) \land 0 < \lvert x-a\rvert < \delta \implies \dotsc)$, the other is identical except it uses $x\in D(f) \land \lvert x-a\rvert < \delta$. The second definition includes the value of $f$ at $a$ if $a\in D(f)$, and then the only possible value of the limit is $f(a)$ [if the codomain is a Hausdorff space]. The limit exists for the first, but not for the second definition.
@DanielFischer I understand how they differ. The first one I guess would be the one more commonly used in say first year calculus (in Stewarts book). I wasn't aware of the second definition.
@DanielFischer The result quoted from wikipedia that I included in my post, do you know how to interpret that? Is it saying that we can only apply it if x_{n} is not equal to x for any n?
@Alex Federico Poloni and mjh apparently also were not aware of it. But if the task is to show that the limit doesn't exist, that must be the definition used, or the exercise is incorrect. [It would be different if you were to show that $\lim\limits_{x\to 0, x \in\mathbb{C}} f(x)$ doesn't exist. That limit exists in neither definition.]
It's really sad because the first tme it happenned to me was when I was 5 and my neighbors convinced me a meteorite would come and we would all die in a year, after that my mind became screwed
@DanielFischer So would my second proof also be okay where I consider two separate sequences rather that two subsequences of the same sequence as in the first proposed proof?
@DanielFischer Okay thanks. I just want to confirm that the reason why the limit does not exist for the second definition is that both 1 and 0 satisfy it.
@TedShifrin Not much in my analysis class...Our first midterm is in November because we're moving so slowly, and our second midterm is a few days before our final exam, lol
@Alex No, neither $0$ nor $1$ satisfy the condition. There are points arbitrarily close to $0$ [including the option of being $0$] with $\lvert f(x) - 1\rvert > \frac{1}{2}$, and points with $\lvert f(x) - 0\rvert > \frac{1}{2}$.
@TedShifrin Peano Systems, recursively constructing $\mathbb{N}$, defining addition/substraction, etc. and we're just about to construct $\mathbb{Z}$ and $\mathbb{Q}$
My favorite induction problem, @nabla (which can be done other ways) is this: Prove that no matter what $n+1$ integers you pick between $1$ and $2n$, there must always be a pair where one divides the other.
I also want to practice with you, but later because now I have a calculus test. I have been studying completely alone in the library for weeks, and I have negleted all the other aspects of my life, which I think is what is causing my mental health problems
@DanielFischer If you have a chance could you explain why the 'if and only if result' I quoted from wikipedia is relaxed to hold even if x_n = x for n if I consider the second definition of limit of a function that you gave.
@DanielFischer I'm not following the reasoning. Would I be right in saying that if using the first definition of limit of a function, that wiki result only applies to sequences where x_n \neq x for any n?
@Alex For the first definition of the limit, you are not allowed to look at $f(x)$ at all [if $f$ is not defined at $x$, both definitions coincide], so you must restrict to sequences avoiding $x$.
@DanielFischer I'm starting to see the light, but its still quite dim. So the 'iff result' as I stated in the post is stated for the case where we are using the first definition of the limit of a function(whether f is defined at x or not ) or if we are using second definition of limit of function where x is not defined. But if using the second definition and f is defined at x then there is no restriction...Is that the idea?
Right, @Alex. The second definition of limit corresponds to using all sequences in $D(f)$ converging to $x$, and the first definition corresponds to using all sequences in $D(f)\setminus \{x\}$ converging to $x$.
Hi, from yesterday I'm with the following exercise and I'd appreciate If someone could check with a I have so far. Thanks: Let $R$ be a commutative ring, not necessarily with identity, and whose only ideals are $R$ and $\{0\}$, and $R$ is not the trivial ring. Show that $R$ is either a field or $(R,+)$ is isomorphic to $\mathb{Z}/p\mathbb{Z}$ (which I denote for simplicity $Z_p$) for $p$ a prime numeber and the product of any two elements is always zero,
The first part is completely trivial because if it has a $1$, it follows immediately that is a field (as is commutative and doesn't have proper ideals). But if not have a $1$. First I shall show that the product of any two elements is always $0$. If not, let $rs\not=o$ then $(r)$ (the ideal generated by $r$) it must be entire Ring, i.e., $Rr=R$ and so there must be some $a\in R$ a s.t., $ar=r$ and is easy to see that this $a$ must be the $1$, contradiction.
Since the product of all the elements is zero and the Ring is not the zero ring, so, for a non zero $r$, $Rr=\{o\}$. So let \langle r \rangle$ is a subgroup of $(R, +)$ generated by $r$, and is clearly an ideal an so there must be the entire group $R$. So there is a $n$ s.t. $nr=0$ and we can choose the least such $n$. (From here I'm not completely sure how to show that $n$ must be prime) but from here and the homomorphism $Z \to (R,+)=\langle r \rangle$, sending $n \mapsto nr$
is clear that $nZ$ is the kernel and by the isomorphism thm, it follows that $Z_n$ and $(R,+)$ are isomorphic.
To conclude only we have to show that $n$ in the above argument must be a prime. Well I'm not completely sure but if not, so is composite $n=ab$, and also the subgroup generated by $ar$ is an ideal and has to be the entire ring. But this is a contradiction, since the order of $ar$ is $b$ but $R= \langle r \rangle$ has order $n$.
@Alizter Probably must be taken as a principal value integral, $$\lim_{\varepsilon\downarrow 0} \int_\varepsilon^{\pi-\varepsilon}\lfloor \cot x\rfloor\,dx.$$
