I encountered the error when attempting to transfer source files from a remote Linux web server to a local HP dt1000i Desktop Hard Drive connected to an Ubuntu 12.10 box via USB.  A Samba share is configured for the HP drive to enable the Windows machines on my LAN to access the files. Samba was set up using the built-in graphical folder sharing functionality within Ubuntu. I can modify the files fine in Windows, but problems occur using rsync.  When the error occurred, I was trying to use rsync to backup files from my web space to the HP drive. I executed the rsync command in PuTTY from a Windows 8 machine on my LAN:  rsync -avz --progress --partial -e 'ssh -p 2222' user@www.example.com:/home/user/ /media/HPDesktopDrive/Backup/HostGator/  This resulted in the following error: rsync: failed to set permissions on "/path/to/file.txt": Operation not permitted (1).  I thought the source of the error was the mount options for the HP drive in the /etc/fstab file: both the user_id and group_id options were set to zero so I tried change them to 1000 as per the advice in .  An ls -la of the Samba share reports that the user and group are recursively set to root:root. I tried to address this by executing sudo chown -R mike:mike /media/HPDesktopDrive/, but the files are still set to root:root.  What is the correct command to change the permissions from my Windows machine via PuTTY and will this address the problem?
When I use the -a option as is asked and answered in , I got a lot of "rsync: failed to set permissions on" errors.  rsync: failed to set permissions on "/ata/text/RCS/jvlc,v": Operation not permitted (1) rsync: failed to set permissions on "/ata/text/RCS/jvm,v": Operation not permitted (1) rsync: failed to set permissions on ...   Why is this? The files are normal files with permission of 0664.
Topology on cartesian product and product topology Brown sugar instead of white sugar How to get new answers to old/popular questions read/voted on? If a question has more than one page of answers then a new answer is not likely to get much attention.  I think the stackoverflow engine should give some more weight to newer answers to old questions but I'm not sure how.  Maybe a tweaked UI to get access to newer answers.  NB. In the related questions this  popped up, I hadnt realised either that these tools existed. I'm using ubuntu 13.10 as htpc. To save some power, i have set it to suspend after 30minutes of inactivity. Before entering suspended state, a popup appears to warn me that computer is about to get put to sleep. This is a little annoyance since it appears when i resume the pc later.   How can i disable this notification for good? Java Access Denied The output from the following script \documentclass{article} \usepackage{tikz} \usepackage{graphicx} \begin{document} \begin{tikzpicture}   \node(title)[text width = 0.25\textwidth, text height = 3 cm, draw]{title};   \end{tikzpicture} \end{document}  is  How can I vertically align the text in the node? Specifying the text depth as in \node(title)[text width = 0.25\textwidth, text height = 1cm, text depth = 1 cm, draw]{     title line 1\\     title line 2\\     title line 3};  Does not result in the text being vertically centred. Are the off-world colonies really a "golden land of opportunity"? In Blade Runner, there seems to be quite a  for emigration to the off-world colonies, e.g.,      "A new life awaits you in the off-world colonies, a chance to begin   again in a golden land of opportunity and adventure."   Is there any evidence that people emigrating are actually better off than staying on Earth?  The job descriptions of the replicants Deckard is tasked with "retiring" don't seem consistent with the colonies being wonderful places. If the colonies offered a good life, would they have to do so much advertising? Diplomacy Attacking a Territory as they move Can a player character use the Misty Step spell to bring another character or creature with them when they teleport? I am an Eladrin elf and have the ability to use the misty step spell to teleport with my clothing and items. However, can I grapple a party member (or NPC or other creature) and use the misty step spell to bring them with me when I teleport? Would they have to be willing? Can I get cheaper plane tickets by using a VPN? In a Youtube video by ADVChina, the speaker is recommending using a VPN in China for various reasons, and to sign up using his referral code.  He says there are various practical uses including "You could also go and get cheaper plane tickets by hopping around different countries." and shows a screenshot of a flight website called KAYAK, with the VPN software changing his country to Italy.  Is it possible to get cheaper plane tickets by using a VPN in this way? I've never heard of KAYAK, but if I use something like Skyscanner, is it common for tickets to be priced differently depending on which country the user's IP is from?  Is it because of different taxes in different countries? So could it be illegal to get discounts like this? Or is there some pricing strategy from airlines to charge different amounts depending on the country of purchase? Compute the moment generating function of $Y = X_1X_2 + X_1X_3 + X_2X_3$ Possible Duplicate:          I got two answers for the same question from different users and both of them correct and working perfectly, but I am unable to accept both of them. Why? There may be some questions that can be answered in several ways, but I can't here in Stack Overflow. Why? Java function for arrays like PHP's join()? Center figure that is wider than \textwidth Possible bug in sidebar "question asked:" date Adding newmode with Xrandr - "800x480_60.00" I have 5-inch a 800x480 tft connected to the LVDS connector on a Portwell WADE-8020 mini-ITX embedded system board. I would like to build a XMBC/Kodi box for streming my music from my freenas PC with the tft showing what is been played, I got as fare as this...   sudo xrandr --newmode "800x480_60.00"  29.50  800 824 896 992  480 483 493 500 -hsync +vsync`enter code here   then,  sudo xrandr --addmode LVDS1 800x480_60.00   now that I get the correct resolution, I now have a black band on the top of the screen.  sudo xrandr --output LVDS1 --set "scaling mode"  "Full aspect"   However this did not solve the issue, also I know I need to amend the /bin/sh file with this from googling so I don't need to keep doing this at every boot. But I don't know how to as I'm very new to Linux.  #!/bin/sh xrandr --newmode "800x480_60.00"  29.50  800 824 896 992  480 483 493 500 -hsync +vsync xrandr --addmode LVDS1 800x480_60.00 When to answer vs when to just comment How do you know if something you want to say should be an answer or just a comment?  I assumed that anything that answers the question should be in an "answer" field, and clarifications should be under "comment". But I found this question () in the English Stackexchange site.  The original question asks for words related to the word "chain" and all the answers so far were provided via comment to the original question, instead of in a separate answer field.  Is this the recommended way to answer a question that requires only one-word answers? Or should the comments to the original question be converted to answers instead? Finding total after percentage has been used? Tried my best with the title. Ok, earlier while I was on break from work (I have low level math, and want to be more fond of mathematics)  I work retail, and 20% of taxes are taken out, and I am wanting to find out how much I made before 20% is taken out.  So I did some scribbling, and this is without googling so please be gentle. but lets say I make 2000 dollars per paycheck, so to see how much money I get after taxes I do 2000 * .8 which gives me 1600. I can also do 2000 * .2 which gives me 400 and then do 2000 - 400 (imo I don't like this way since I have a extra unnecessary step)  Anyways, I was pondering how to reverse that to have the answer be 2000, and what I did was 1600 * 1.20 (120%) which gave me 1920, and I thought that is odd so I did 1600 * 1.25 and answer was 2000 exact.  My question is why did I have to add extra 5% (25%) to get my answer? I am sure I did something wrong, and I fairly confident the formula I am using is a big no no.  edit; wow thank you all for your detailed answers. I am starting to like mathematics more and more. Browse an UNC path using Windows CMD without mapping it to a network drive C:\&gt; cd \\somewhere '\\somewhere' CMD does not support UNC paths as current directories.   What I usually do to get around this is to map that directory to a network drive and then I could easily access it from the command prompt.  But is there an easier way on how to get around this? Power set of a finite set is finite proof, no cardinality of power set Prove that $\mathcal{P}(A)$ of a finite set $A$ is finite.      I need to prove this without using the cardinality of $\mathcal{P}(A)$.  This is my attempt, which is by induction: $\newcommand{\lbrac}[1]{\left(#1\right)}$ $\newcommand{\sett}[1]{\left\{#1\right\}}$ $\newcommand{\pset}[1]{\mathcal{P}\lbrac{#1}}$ Proof by induction over cardinality.  If $A = \varnothing$, then $\pset{A} = \sett{\varnothing}$, which is finite.  Suppose that for a set $A$ of cardinality $n$, $\pset{A}$ is finite.  Let $A$ be a set of cardinality $n+1$. $A$ is not empty, since it is of cardinality $n+1 &gt; 0$, and therefore pick some $x\in A$. It follows that $A = \lbrac{A-\sett{x}}\cup\sett{x}$. We claim that $$ \pset{A} = \pset{\lbrac{A - \sett{x}}\cup\sett{x}} = \pset{A - \sett{x}}\cup\sett{C | C\in\pset{A}\wedge x\in C} $$ Suppose $B\in\pset{A} = \pset{\lbrac{A-\sett{x}}\cup{x}}$. Then either $x\in B$ or $x\not\in B$. If $x\in B$, then $x\in B \wedge B\in\pset{A}\Rightarrow B\in C$. If $x\not\in B$ then $B\subseteq A - \sett{x}$.  \newline Either way, $B\in \pset{A - \sett{x}}\cup\sett{C | C\in\pset{A}\wedge x\in C}\Rightarrow \pset{A}\subseteq\pset{A - \sett{x}}\cup\sett{C | C\in\pset{A}\wedge x\in C}$.  Suppose $B\in\pset{A - \sett{x}}\cup\sett{C | C\in\pset{A}\wedge x\in C}$. Then either $B\in\pset{A-\sett{x}}$ or $B\in\sett{C | C\in\pset{A}\wedge x\in C}$ (or both).  If $B\in\pset{A-\sett{x}}$, then $B\subseteq(A-\sett{x})\cup\sett{x} = A$, and if $B\in\sett{C | C\in\pset{A}\wedge x\in C}$, then by definition, $B\subseteq A$. Either way, we get that $B\in\pset{A} \Rightarrow \pset{A - \sett{x}}\cup\sett{C | C\in\pset{A}\wedge x\in C} \subseteq \pset{A}$.  Since we have shown both sets are contained in each other, they are equal.  My goal is to now prove that $\sett{C|C\in\pset{A} \wedge x\in C}$ is finite. I didn't manage to do so without non rigorous claims that could also "prove" the main question.  If this set is finite, and by induction hypothesis $\pset{A - \sett{x}}$ is finite, I'll be done.  Any ideas on how to prove $\sett{C|C\in\pset{A} \wedge x\in C}$ is finite?