@AkivaWeinberger this one. With a being of lesser intellect :p
Now, there must be another element $y$ (other than $x$). We can take $u$, $v$ such that $-1<ux-vy<1$.
how good is that?
And, those elements would be infinite in number.
So, if not cyclic, then all the elements are going to be rational. (if my above claim is true at all). Now, the field of fractions of $\mathbb{Z}$ is $\mathbb{Q}$.
Now, we can proceed with "fractional ideals"
Finally we apply "prime avoidance"
@BalarkaSen don't know how is this going to work out
I don't actually officially start learning this language until, you know, school starts, but I'm glad I'm learning some of this outside of a classroom because there's no way I could do this under time pressure
It's just so different from English
Also there's a really annoying pronunciation detail
The word ga is pronounced with the "ng" sound like in "sing", so it's really more like nga
and I keep on saying na by accident but that's a different word
> However, /ɡ/ is further complicated by its variant realization as a velar nasal [ŋ]. Standard Japanese speakers can be categorized into 3 groups (A, B, C), which will be explained below. If a speaker pronounces a given word consistently with the allophone [ŋ] (i.e. a B-speaker), that speaker will never have [ɣ] as an allophone in that same word. If a speaker varies between [ŋ] and [ɡ] (i.e. an A-speaker) or is generally consistent in using [ɡ] (i.e. a C-speaker), then the velar fricative [ɣ] is always another possible allophone in fast speech.
@AkivaWeinberger. Let $T=\{a_1, a_2, ..., a_m\}$ be a complete enumeration of the set. And suppose, $H$ is not cyclic. Let $a_t$ be the element closest to $0$ (>0) and for some $h \in H$, $h \notin <a_t>$ (we choose $-h$, if $h<0$). Now, $p a_t \leq h < (p+1) a_t$ for some positive integer $p$. Now, $0 \leq h- p a_t <a_t<1$, and $h-pa_t$ is not in $T$. So, h- pa_t$ must be $0$...
@BalarkaSen !?
@AlessandroCodenotti I read your proof. Didn't understand much.
@AkivaWeinberger, This particular problem can be solved using the statement I was trying to prove beforehand.
@SubhasisBiswas I think this is very much flawed since I am using ordering
@SubhasisBiswas Sometimes easy to state questions have hard answers. This might be one of those cases, or maybe there is a proof not using geometric group theory that I can't see.
@AlessandroCodenotti $Z =<1>=<-1>$. A cyclic group has only cyclic subgroups. Suppose, $<a> \subset Z$ be of finite order. $n a = 1$. (n being its order). This implies that $na(1)=1$, i.e. $Z$ becomes a group of finite order.
Hey, @cis. It's called "foot of the altitude" as written here. Took a bit of googling myself to be honest with you. Here's the link: https://en.wikipedia.org/wiki/Triangle#Points,_lines,_and_circles_associated_with_a_triangle
Anyways, I'd like to ask something myself. I'm currently studying linear programming and combinatorial optimization and I'm trying to grasp some concepts, but I'm struggling a bit with a kind of intuitional understanding of some things.
Specifically, I'm currently at the "Hyperplane Separation Theorem". Now, I'm mostly fine with the proof, but there's an essential step I feel I don't fully understand. At one point you choose a point from each polygon (say $x \in X, y \in Y$), such that these have the least distance of all such pairs. Using these two points you're trying to construct a hype…
I have a dumb question. I have a space $X$ such that $H^i(X, \Bbb Q) = \Bbb Q$ if $i = n$ and $0$ if $i \neq n$. Is it at all clear that $H_i(X, \Bbb Z)$ should behave similarly, degreewise?
But $\text{Hom}(A, \Bbb Q) = \Bbb Q$ and $\text{Ext}(A, \Bbb Q) = 0$ doesn't quite force $A = \Bbb Q$, right? Any cyclic group will also satisfy these two.
manjul bhargava helped draft an elaborate educational reform policy recently but it's clear the government implementation will have nothing to do with it...
Oh and there was a guy from Calcutta university who came to give MMath interview. He himself admitted he had no clue even about compactness... I wonder how he got selected...
btw how much time you have on a week on average for self study (ie excluding the curriculum work which everyone else has to follow) ? ISI course structure seemed pretty rigid in general
@BalarkaSen so they "enjoy college life" by spending time on facebook/silly jokes/watching movies ?
any advice for how to utilize it (i.e how to eke out time to do maths) ? I'm very bad in utilizing time and I get distracted easily online (on hangouts/MSE chat/FB etc)
Hello. Vague question, but why is having compact support particularly good (say in the context of integration theory), and good in the sense that perhaps some nice theorems not apply, or they are well behaved in some sense (or however else you want to interpret it)
oh cool, thanks. Do you sometimes feel depressed/temporarily uninterested in math when you're stuck on a problem for a long time without any progress ?
Say that $g:\Bbb R^d\to \Bbb R$ is continuous with compact support and let $g_h(x)=g(x-h)$.
Then it is claimed that $$\int_{\Bbb R^d} |g_h(x)-g(x)| dx\to 0$$ as $h\to 0$.
By continuity it holds that $|g_h(x)-g(x)|\to 0$ as $h\to 0$. Then for any $|h|<\delta$ we have $|g_h(x)-g(x)|<\epsilon$ so that $\int_{\Bbb R^d} |g_h(x)-g(x)| dx \leq \int_{\text{supp}(g_h-g)} \epsilon dx$
@cdt Idk I suppose I move on to something else for a while, then come back, sleeping on the problem, the usual schtick. It's good to acknowledge that I'm a stupid person in general, and that it shouldn't contradict with efforts to understand something I don't understand
I was just using that by continuity for any $\epsilon>0$ there exists a $\delta>0$ such that whenever $|(x-h)-x|<\delta$ (i.e. $|h|<\delta$) then $|g(x-h)-g(x)|=|g_h(x)-g(x)|<\epsilon$
I think you are right that compactness of the support is not necessary though. I can always take a large compact set containing the support (if it's bounded - that is crucial), in which everything is happening and everything is uniformly continuous
Boundedness is crucial, and not because of volume, you can have unbounded subsets with finite volume (think of union of the intervals $[a_n, a_{n+1}]$ where $a_n$ is the $n$-th partial sum of $\sum 1/k^2$) but you can have a function which is $x^2$ on those intervals and vanishes on some open neighborhood of the union of those intervals
That's not uniformly continuous on the support! So yikes!
I've always seen "continuous with compact support" refer to the version of support where you take a closure in the definition. Closedness of the support is not an issue, boundedness is the important part
If I have understood it correctly, the elements of the subalgebra are ${\Bbb C}$-linear combinations of the form
$$a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n},$$
where $m_1,m_2,\ldots,m_n$ are integers.
I'm not particularly fond of Dummit and Foote.
The exercises seem to fall into three categories: way too easy, way too hard, or way too specific to really gain anything from them.
That last category is a bit ill-defined, but oh well.
I don't know of any deep connection. If a function is absolutely improperly Riemann-integrable, it is Lebesuge-integrable. This does not hold in the non-absolutely convergent case (e.g. $\sin(x)/x$). And being Lebesgue-integrable generally does not imply being Riemann-integrable.
@cdt too often. Then I start nagging about it in this chat room because I can't do it irl. It's not a good thing. Two or three days ago a girl in kolkata committed suicide because she was too obsessed with getting into ISI.
I searched for semiprimes of the form of $n^n+n!$, where $n \in \Bbb{N}$, for a range of $n \le 2 \times 10^4$ on PARI/GP and found semiprimes of the form $n^n+n!$ only for $n=2, 3, 7$.
We can write $n^n+n!$ as:
$$n^n+n!=n(n^{n-1}+(n-1)!)$$
Therefore we can also alternatively look for primes of ...
I need to check if I'm over- or under-thinking something here. An integral domain is called almost-dedekind if its localizations at primes are all discrete valuation rings. An integral domain is called Prufer if its localizations are all valuation rings. Then to find an example of something that is Prufer but not almost-Dedekind, any non-discrete valuation domain should work... right? Please excuse my paranoia.
If I have understood it correctly, the elements of the subalgebra are ${\Bbb C}$-linear combinations of the form
$$a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n},$$
where $m_1,m_2,\ldots,m_n$ are integers.
