Is $G$ a group under those conditions?
Right identity and Right inverse implies a group
Filter subscription email is not being delivered (June 2019 - Feb 2020) I have created a custom filter in  and specified that I would like to get email updates:  However, I never got a confirmation mail nor any notifications based on my filter. I tried the &quot;resent confirmation mail&quot; button on the right hand sidebar:  But again nothing happens. The mail address should be fine  I used it for all kinds of notifications from Stack Exchange. For example the rep recalculation email arrived just fine on this address.  I retyped it several times to make sure there really isn't any typo  I double checked that it did not end up in the spam filter  I don't think it is a problem with my particular email provider, users with other mail providers seem to have the same problem, see     This is possibly related to , but this claims to be &quot;status complete&quot;.   EDIT: Although the &quot;status-completed&quot; tag was added, the problem still persists  EDIT 2: As of Feb 2020, it finally works I declared random list using TikZ package and can get items randomly. But at same time I need to get all items in sequential order. How can I do it? Is it possible print \mylist items declared in noncomma variat?  \documentclass[a4paper,14pt]{extreport} \usepackage{tikz}   \def\mylist{{one}{two}{three}{four}{five}}   \pgfmathdeclarerandomlist{mynum}{\mylist} %Define the list   \begin{document} \pgfmathrandomitem{\mynum}{mynum} \pgfmathresult~\mynum\\  \pgfmathrandomitem{\mynum}{mynum} \pgfmathresult~\mynum\\  \pgfmathrandomitem{\mynum}{mynum} \pgfmathresult~\mynum\\  \pgfmathrandomitem{\mynum}{mynum} \pgfmathresult~\mynum\\  \pgfmathrandomitem{\mynum}{mynum} \pgfmathresult~\mynum\\ \end{document} What are Vertex and Pixel shaders? Whenever a question pops up on SO about Java synchronization, some people are very eager to point out that synchronized(this) should be avoided. Instead, they claim, a lock on a private reference is to be preferred.  Some of the given reasons are:    (very popular this one, also has an "accidentally" variant)     Other people, including me, argue that synchronized(this) is an idiom that is used a lot (also in Java libraries), is safe and well understood. It should not be avoided because you have a bug and you don't have a clue of what is going on in your multithreaded program. In other words: if it is applicable, then use it.  I am interested in seeing some real-world examples (no foobar stuff) where avoiding a lock on this is preferable when synchronized(this) would also do the job.  Therefore: should you always avoid synchronized(this) and replace it with a lock on a private reference?      Some further info (updated as answers are given):   we are talking about instance synchronization both implicit (synchronized methods) and explicit form of synchronized(this) are considered if you quote Bloch or other authorities on the subject, don't leave out the parts you don't like (e.g. Effective Java, item on Thread Safety: Typically it is the lock on the instance itself, but there are exceptions.) if you need granularity in your locking other than synchronized(this) provides, then synchronized(this) is not applicable so that's not the issue Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective How can I proof this:     Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective, i.e., there are $A,B \in \mathbb{R}^2$ such that $A \neq B$ and $f(A) = f(B)$ (Hint: show that given $(a,b) \in \mathbb{R}^2$ in every open neighbourhood of $(a,b)$ there are points $(c,d),(c_1,d_1)$ where ${\partial}_1f(c,d) \neq 0$ or ${\partial}_1f(c_1,d_1) \neq 0$).   The problem is that I have no idea how to use the hint.  Thanks for all your comments. Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero How do we show that equality holds in the triangle inequality $|a+b|=|a|+|b|$ iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the three conditions happens. Is there a string math evaluator in .NET? If I have a string with a valid math expression such as:  String s = "1 + 2 * 7";   Is there a built in library/function in .NET that will parse and evaluate that expression for me and return the result? In this case 15. Why would Louise telling the Chinese commander about his wife's last words change his mind? In the future in Arrival, the commander says that he changed his mind about going to war with the aliens because Louise called his private number and told him his wife's dying words. My question is what did Louise say to him (like in English) and why did it change his mind? Why is King Tommen Baratheon the "First of his Name" rather than at least third? Monty hall problem extended. How does the wording "for each" work? Some cards in magic has something happen for each {something}. For instance:  makes you gain 2 life for each creature you control. Tapping  adds one green mana to your mana pool, and you gain 1 life for each non-land card revealed.  How is this life gained: all at once or "one bit" at a time? Say I tap Selvala and 1 land, and 2 non-lands are revealed: do I gain 2 life, or do I gain 1 life two times? Would  have one or two +1/+1 counters placed on it if I had it in play at that point? I have a dataset with a large number of Yes/No responses. Can I use principal components (PCA) or any other data reduction analyses (such as factor analysis) for this type of data? Please advise how I go about doing this using SPSS. Got a bunch of .mp4 videos (info shows H.264 Video + AAC Audio) that are chapters of a larger video.  I'd like to merge them, but not sure which is the best way.  I don't have Quicktime Pro, but I'll get it if that's the best way to do this.  Edit: Just to clariify this is Mac OSX 10.6 I'm on. I came across the phrase, “suck it up and go” in the columnist’s answer to a question from a reader of Carolyn Hax's column in Washington Post’s “Lifestyle” section (July 2nd).  The Q&amp;A titled “Ever the aunt, never the godmom” begins with the following sentence:     “Dear Carolyn: We just got an invitation to the baptism of my seventh nephew on my husband’s side. Once again we were not considered as godparents. I felt snubbed. -- Is this a legitimate snub? Should I talk to my sister-in-law about it?”    and Carolyn answers:     “Boycotting the baptism in a wounded huff would be petty, yes, and that’s the main reason to suck it up and go - but you also need to go because nothing would undermine your own purpose more tidily than staying home.”   I checked the meaning of “suck it up and go” in English Japanese dictionaries at hand only to find one of them shows “suck it and see” meaning “to try” as a British and  Australian usage. Online dictionaries register “suck it (up) and “go suck,” but not “suck it up and go.”   What does “suck it up and go” mean? Is it an American version of “suck it and see”? Any Set with Associativity, Left Identity, Left Inverse is a Group Related Link:    Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra     I will present my proof (distinct from those in the link) for critique and then ask my question.  $G$ is a set and $ \times $ is an associative binary operation.  Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$.  Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.  $a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$  Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.  As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.  For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.  Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.  In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly.  Specifically, the identity in the second axiom is not well-defined.     Let $(G, *)$ be a semi-group. Suppose    1. $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;    2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group?    This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? –  Derek Holt Sep 17 '11 at 15:31   Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,  when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} =  e_1, aa^{-1}_2 = e_2$?  I think my wording eliminated the ambiguity.  It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes. Im asked to prove that a polynomial of an odd degree has a root. Im going to use IVT to prove this, but I'm wondering if I can use the assumption that an odd degree polynomial has either two cases:  $$ \lim_{x\to\infty}h(x)=+\infty \\ \lim_{x\to-\infty}h(x)=-\infty \\ $$ $$and$$ $$ \lim_{x\to\infty}h(x)=-\infty \\ \lim_{x\to-\infty}h(x)=+\infty \\ $$ Then by this there exists numbers $a,b$ such that $$ h(a)&lt;0&lt;h(b) $$ $$or$$ $$ h(b)&lt;0&lt;h(a) $$ Which I think would prove that odd degreed $h(x)$ has at least a root. Or is there a more convincing way to show that there are numbers in $h(X)$ greater than zero and less than zero?  edit: (sorry about the bad limit notation) How can I achieve the correct garden tap orientation when threading it in? I'm installing an external tap using a wall plate like this one:    And a tap like this one:    This might be a stupid question, but if I screw the tap into the wall plate it will end up at a certain angle. I can install the wall plate with that angle and the tap will end up with the correct orientation (i.e. pointing down). But what if I want to put another tap in later that has its thread cut slightly differently? How do I prevent it ending up upside-down or whatever? Are all the threads cut exactly the same in these taps?  This seems like a stupid design. If I know $\| f_n - f \|_{L^p(\mathbb{R})} \to 0$ as $n \to \infty$, do I know that $\lim_{n \to \infty}f_n(x) = f(x)$ for almost every $x$? I have a really strange situation here. My PC works fine, at least in most cases, but there's one thing that I can't deal with. When I try to copy a file from my pendrive, everything is ok -- I got 16-19M/s , it works pretty well. But when I try to copy something to the same pendrive, my PC freezes. The mouse pointer stops moving for a sec or two, then it moves a little bit and it stops again. When something is playing, for example, in Amarok, the sound acts like a machine gun. The speed jumps from 500K/s to 15M/s, average 8M/s. This occurs only when I'm copying something to a pendrive. When the process of copying is done, everything backs to normal.   I tried everything -- other pendrive, a different USB port on front panel or those ports from back, I even changed the USB pins on motherboard (front panel), but no matter where I put my USB stick, it's always the same. I tried different filesystem -- fat32, ext4. I have no problem with the device on Windows, on my laptop. It has to be my PC or something in my system. I have no idea what to look for. I'm using Debian testing with standalone Openbox. My PC is kind of old -- Pentium D 3GHz, 1GiB of RAM, 1,5TB WD Green disk. If you have something that would help me to solve this issue, I'd be glad to hear that.  I don't know what else info I should provide, but if you need something, just ask, I'll update this post as soon as possible.  I tried to reproduce this problem on ubuntu 13.04 live cd. I mounted my encrypted partition + encrypted swap and connected my pendrive to a usb port. Next I tried to start some apps, and now I have ~820MiB in RAM and about 400MiB in SWAP. There's no problem with copying, no freezing at all, everything is as it should be. So, it looks like it's a fault of the system, but where exactly? What would cause such a weird behavior? Software was blocked from opening because it is not from an identified developer I went to run LibreOffice on my OS X 10.9 and it would not let me install it. It said it was from an unidentified developer. I'm aware of the security features of the Mac and went to System Preferences to allow the software to run.   When I opened this setting I saw a button that said, '"LibreOffice" was blocked from opening because it is not from an identified developer.' and there was a button next to it that said "Open Anyway". I clicked this button and nothing happened. I clicked it multiple times, like a hundred, out of frustration. Is there something else I need to do? Could someone verify this is not a bug?   UPDATE: I've included the photo below of what my System Preferences screen looked like and the button I pushed that didn't work.  UPDATE II: So I tried opening the application again and went into the system preferences and pressed the "Open Anyway" button again and nothing happened at first but a few minutes later while I was typing an email something flashed across my screen, possibly another dialog (or more likely a chewpacabra), but it quickly disappeared and the next thing I know LibreOffice launched!