Is it possible to add Wikimedia Commons images? Is it possible to add  images in the questions/answers without uploading them on imgur? Since the images on Wikimedia Commons are free, it's a waste of hard drive space to upload them on imgur just to be able to show them in a SE question. I've recently asked a question where I provided links to two Wikimedia Commons images but I would prefer to directly include them in the question - without uploading them on imgur.  If not, then I would like to place a feature request to allow this.
When adding image from Wikimedia Commons, automatically add author/license/link Many Stack Exchange questions and answers embed images from Wikimedia Commons. It is probably the most used image source here. Good news Thanks to their license, all images from Wikimedia Commons can be legally used inside Stack Exchange. Bad news The only requirement is to cite the author/license (), an easy task, but we are totally failing at it, only maybe 1% of Commons images used here are properly cited. That makes our use of these images illegal, and could expose Stack Exchange to various legal threats. Not my problem, you could say, but this is also important because , bringing Stack Exchange closer to closed-data sites like Yahoo! Answers or Experts-Exchange. Proposed solution When using the image addition widget, the widget should detect whether the image's URL has a *.wikimedia.org domain, and if yes:  (optional) Transfer the image to imgur, for perennity reasons. Retrieve the author name and license name from Commons (there is a very reliable API that gives that). Concatenate into a proper citation, maybe something like &quot;*Credit: [&quot; + author + &quot;](&quot; + commonsPageURL + &quot;) , [&quot; + license + &quot;](&quot; + licenseURL + &quot;)*&quot; Insert the citation at the same time as the image's MarkDown.  Benefits:  Remove a large part of Stack Exchange's copyright violations. Makes us a good citizen in the open data world. Remember what happens to sites who disregard ethics (Experts Exchange). Consistent citations. Zero added burden for the users.
Execute command in remote active terminal Suppose you have a terminal emulator (T1) open with a PID of 6350.  From another terminal, type this command (C1):  echo "ls\n" &gt; /proc/6350/fd/0   This writes ls and the new line in T1 but does not execute it. Why?  I also tried using cat|bash with echo "ls\n" &gt; /proc/catid/fd/0 but it is still not executed.  How can I echo the command into another terminal and have the command executed?  possible answer :  $ mkfifo toto; $ bash &lt; toto; $ echo "ls" &gt; toto;   In this case you cannot write anymore directly in the terminal (everything is displayed the same way the command (C1) displayed thing in this terminal. Need to understand why my Schengen visa was refused Whenever we have a changing magnetic flux through a conductor an induced electric field is generated. It is often said that these induced electric field lines are circular in shape.But why is this so?Why are these field lines not elliptical for example? How can this shape be mathematically derived?  Also does this have anything to do with equipotential surfaces?Do such non-conservative electric fields even produce a potential field?If not why? Let $R$ be a commutative ring with 1, we define  $$N(R):=\{ a\in R \mid  \exists k\in \mathbb{N}:a^k=0\}$$  and  $$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$  Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?  I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that. I would like to use  with a video.  With ffmpeg I can export some (.jpeg) images from a video. Can I convert a .jpeg into a valid .tiff or export directly .tiff images from the video with ffmpeg? What happens to Zhang ziyi (Jen) in  when she jumps off the cliff .. she dies? becomes immortal? Just before jumping what does she want Lo to become - a master?  Why does she jump off, guilt? Turning DisplayPort monitor off disables monitor completely I have 2 monitors connected to a NVIDIA GeForce GTX 660:   Left is 1920x1080 via DVI (Samsung SyncMaster 2443BW) Right is 2560x1440 via DisplayPort and is set to Primary (Samsung SyncMaster SA850)   I often leave my computer running overnight (working on long render job, doing backups, etc) so I only power off my monitors manually. However, if I power off the primary monitor, it is disabled completely as if it were unplugged from the graphics card (i.e. icons are moved, open apps are moved and resized to fit on the new primary monitor). If I switch from DP to a DVI cable, this does not occur. It seems to be an issue when using DisplayPort.  Is there a way to disable the monitor from being disabled completely in Windows when I power it off physically?   I have found no relevant settings in either Windows Display settings, nor the NVIDIA control panel software.  Edit #1: I'm using an .  Edit #2: Here is a screenshot of my NVIDIA Control Panel v7.2.710.0:   Edit #3: Here are screenshots of my Monitor configuration:    Clicking on Advanced settings reveals a Generic PnP monitor running at 59 Hz? 60 Hz is the only other option available.    Here is my secondary monitor which is connected via DVI:    Edit #4: If I open the "Screen Resolution" configuration panel and power the monitor off (via the power switch), this is what is shown:    It looks like I disconnected the monitor from the video card. What do the various dpkg flags like 'ii' 'rc' mean? I frequently need to check which packages are installed, and I use the following command:  dpkg -l | grep foo  which gives the following output  Desired=Unknown/Install/Remove/Purge/Hold | Status=Not/Inst/Cfg-files/Unpacked/Failed-cfg/Half-inst/trig-aWait/Trig-pend |/ Err?=(none)/Reinst-required (Status,Err: uppercase=bad) ||/ Name                       Version         Description ii   foo                       &lt;version&gt;         &lt;description&gt;    What does the ii mean?  What other flags are there? How to read the flags? (because the explanation is quite complicated, IMO)   Thanks. can't create a jobs user profile I'm stuck on this page even after linking stackoverflow and github accounts. Will my 320W PSU handle a Asus GTX 750Ti I've got a 320W PSU as far as I know and it's a Chinese PSU because I've seen the Chinese letters on it. I just had a question if I can upgrade my pc to Asus GTX 750Ti 2 GB OC.  My pc specs are:   4GB DDR3 RAM Intel Core 2 Duo E8400 3.0 GHZ 250GB HDD Onboard graphics card Prove $[A,B^n] = nB^{n-1}[A,B]$ I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance. Ubuntu 15.10 is not in Boot Manager I have a problem with Ubuntu 15.10  I have in my computer windows 10. Some days ago i decided to install Ubuntu in partition so i can have both in the same hard drive. The problem is that i cant get into ubuntu, i dont have the grub up so i always go to Windows in default mode. The only way to go into ubuntu is with the CD of installation, when im on the grub of the cd i press Esc, then i text exit in command lines and i go to a boot manager in which i can choose the ubuntu system. Its a strange think because it is the only way i can get into ubuntu, i will like to solve this problem. I try many tutorials already of solving this problem. I didnt get any solution so i will like to ask u before doing another tutorial. Thank u Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.   This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at , and if this could be proven, then the solution would follow immediately. The given differential equation $$\lvert{dy\over{dx}}\rvert\,+\mid{y}\mid+1=0 $$  I don't know how to deal with modulus sign in order to prove the statement  Reference-Different Equations by Shipley L Ross Ch1 Exercise 7(a) I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does. Directing graph such that any outdegree would be at most 2 Let $G=(V,E)$ be a graph. Suppose that for every subgraph $G'=(V',E') , G' \subset G$, we have $|E'| \le 2|V'|$. Show it's possible to direct $G$ such that any outdegree would be at most $2$.   I tried proving it by induction. I've realized that the minimum degree at any subgraph of $G$ is at most 4. I tried removing the vertex $v_0$ with the minimal degree at a certain subgraph from the original graph, claiming that the graph from the remaining vertexes can be directed as required. How can I include the vertex $v_0$ in the new directed graph such that any outdegree would still be at most 2? Thanks This is what I would like to do:  morekeywords={SENÃO, ENTÃO, FAÇA, ATÉ, ENQUANTO, PARA, SE, DECLARA, DECLARE, E, OU, FIM_ENQUANTO, FIM_PARA}   I'm defining a new language, and I would like to use natural language words as keywords.  The problem is with keywords: SENÃO, ENTÃO, FAÇA, ATÉ.  I have solved the first problem, in showing special chars with:   The alsoletter here, doesn't work for me, I got an error when using it: alsoletter={Ç,É,Ã}.  test.tex  \documentclass{article}  \usepackage[brazilian]{babel} \usepackage[T1]{fontenc}  \usepackage[utf8]{inputenc} \usepackage{listingsutf8}  \usepackage{listings}  \lstset{%         inputencoding=utf8,         extendedchars=true,         literate=%         {é}{{\'{e}}}1         {è}{{\`{e}}}1         {ê}{{\^{e}}}1         {ë}{{\¨{e}}}1         {É}{{\'{E}}}1         {Ê}{{\^{E}}}1         {û}{{\^{u}}}1         {ù}{{\`{u}}}1         {â}{{\^{a}}}1         {à}{{\`{a}}}1         {á}{{\'{a}}}1         {ã}{{\~{a}}}1         {Á}{{\'{A}}}1         {Â}{{\^{A}}}1         {Ã}{{\~{A}}}1         {ç}{{\c{c}}}1         {Ç}{{\c{C}}}1         {õ}{{\~{o}}}1         {ó}{{\'{o}}}1         {ô}{{\^{o}}}1         {Õ}{{\~{O}}}1         {Ó}{{\'{O}}}1         {Ô}{{\^{O}}}1         {î}{{\^{i}}}1         {Î}{{\^{I}}}1         {í}{{\'{i}}}1         {Í}{{\~{Í}}}1 }  \lstdefinelanguage{pseudo} { extendedchars=true, alsoletter={_}, morekeywords={ENQUANTO,PARA,SE,SEN\~{A}O,ENTÃO,ATÉ,FA\c{C}A,DECLARA,DECLARE,E,OU,FIM_ENQUANTO,FIM_PARA,FAÇA}, ndkeywords={NUMERO,TEXTO,LEIA,ESCREVA}, sensitive=true, morecomment=[l]{//}, morecomment=[s]{/*}{*/}, morestring=[b]" }  \begin{document} \lstinputlisting[language=pseudo]{foo.pseudo} \end{document}   foo.pseudo  ENQUANTO nota != -1 FAÇA // Comment example  LEIA nota  SE nota &gt;= 7 ENTÃO    ESCREVA "Aprovado"  SENÃO    ESCREVA "Reprovado"    TEST "Algo mais" /* mais  comentário */  FIM_ENQUANTO    I have seen: Tell rate-limited users how long they have to wait before asking a new question New users are limited to one question per an hour, but once they'll hit this limit, all they see is You can only post once every 60 minutes. I think it would be very appreciate to change / extend this message, adding You need to wait next xx minutes. I saw similar solutions on many forums.  Would it be hard to implement something like that?  EDIT: It could be either added as extension to above mentioned warning message or as red-colored reminder above Title field. Or even as an explanation message, preventing user from clicking on a Ask Question button? Anything would be good, to not let user write the question (can be time consuming) only to hit the wall of limit. This can be a little bit annoying or frustrating, don't you think? I thought that the quotation mark (") was simply a type of grouping marker but I'm debugging some NHibernate code and notice that while  SELECT * FROM site WHERE site_id = 3;   Works fine  SELECT * FROM "site" WHERE site_id = 3;   fails with a table or view does not exist error.  What gives? Questions about a commutative ring with exactly three ideals Let $R$ be a commutative ring with identity. Assume that $R$ has exactly three distinct ideals: $\{0\},I, R.$      1) Show that if $a \in R-I$, then $a$ is a unit in $R$.      2) Let $a,b\ne0$ in $I$. Show that $ab=0$.