Let $L,M$, two subspaces of $V$ where $L,M\ne V$. Prove $L\cup M \ne V$   My Try:    Obliviously, $L\cup M \subseteq V$. It's left to show $L\cup M \subsetneq V$. $L,M$ have bases as subspaces. Let: $$l = \sum\limits_{i=1}^{j} = \alpha_il_i$$  $$m = \sum\limits_{i=1}^{k} = \beta_im_i$$  Now, I want to claim that $v=l+m$, is of course $\in V$ but $\notin L\cup M$  Is that how am I suppose to do that? How to proceed then?  Thanks.
Let $L,M$ two subspaces of the vector space, $V$ such that both $L,M \ne V$.   Prove: $L\cup M \ne V$.     I think this is a case of a proof by contradiction. Lets assume $L \cup M = V$.   Hence, $$\dim(V) = \dim(L) + \dim(M) - \dim(L\cap M)$$  How to proceed?
I would like to configure my Ubuntu 12.04 server to synchronize its clock with a specific NTP server. Where to set this up? I've just installed Ubuntu 16.04 LTS on dual boot. I've looked for many solutions and tried them all but Chrome never starts; the logo appears on the left but no window opens.  I downloaded the deb file on google chrome website and installed it through Ubuntu Software, but it doesn't start.   I have tried installing it through the command line with   sudo dpkg -i --force-depends google-chrome-stable_current_amd64.deb sudo apt-get install -f   I have checked with uname -a and my system is 64 bit. I have also tried gdebi with the same result.   My repository universe is up to date.  I only have one display in settings>displays> so I can't turn off built-in display nor check the mirror display (I read it could be the problem as I also have a graphic card but it seems it is not in my case).  Here is what I get in the command line when I try to run Chrome from it ( the message is the same, whatever the method I used to install Chrome):  ~$ google-chrome [11831:11831:1122/114854.195473:ERROR:process_singleton_posix.cc(433)] readlink failed: Permission denied (13) [11831:11831:1122/114854.195493:ERROR:process_singleton_posix.cc(256)] readlink(/home/basil/.config/google-chrome/SingletonLock) failed: Permission denied (13) [11831:11831:1122/114854.195508:ERROR:process_singleton_posix.cc(256)] readlink(/home/basil/.config/google-chrome/SingletonLock) failed: Permission denied (13) [11831:11831:1122/114854.195515:ERROR:process_singleton_posix.cc(280)] Failed to create /home/basil/.config/google-chrome/SingletonLock: Permission denied (13) [11831:11831:1122/114854.195538:ERROR:process_singleton_posix.cc(433)] readlink failed: Permission denied (13) [11831:11831:1122/114854.195564:ERROR:process_singleton_posix.cc(256)] readlink(/home/basil/.config/google-chrome/SingletonLock) failed: Permission denied (13) [11831:11831:1122/114854.195622:ERROR:chrome_browser_main.cc(1498)] Failed to create a ProcessSingleton for your profile directory. This means that running multiple instances would start multiple browser processes rather than opening a new window in the existing process. Aborting now to avoid profile corruption. [11831:11868:1122/114854.209297:ERROR:browser_gpu_channel_host_factory.cc(108)] Failed to launch GPU process.   Any new approach will be much appreciated Black screen after suspend-after moving to the nvidia official drivers (not priority) Import PLY doesn't display vertex colors I'm trying to import a .PLY mesh in Blender. I've troubles with colors: the mesh appear in a grayscale palette. Other softwares open it correctly (eg. MeshLab).  This is the header of the PLY file:  ply format binary_big_endian 1.0 comment created by Geomagic Sculpt Exporter. element vertex 182488 property float x property float y property float z property uchar red property uchar green property uchar blue element face 364980 property list uchar int vertex_indices end_header   Any tips? Thanks! I'm struggling with the following question;     For every graph $G$ such that $K_3 \not\subseteq G$ (i.e. $G$ does not contain a triangle), prove that $\chi(G) \leq 2\sqrt{n} +1$ (where $\chi(G)$ denotes the chromatic number).   Any hints? I really cannot see the connection between the triangle-free nature of $G$ and a chromatic number being bounded above by $O(\sqrt{n})$. Why can I modify a read-only file? Short Question:  Why can we manipulate a read-only file in Vim using : + w + q + ! even without being an administrator?  Long Question:  I have a text file (myFile.txt) which is read-only for everyone:  navid@navid-ThinkPad-T530:~/ubuntuTest$ ls -l myFile.txt  -r--r--r-- 1 navid navid 26 Aug 22 21:21 myFile.txt   I can open it with Vim without having admin privileges:  navid@navid-ThinkPad-T530:~/ubuntuTest$ vi myFile.txt    I modify it and press: Esc + : + w + q + Enter and I see this error message:  E45: 'readonly' option is set (add ! to override)   So far, everything makes sense.  But when I press: Esc + : + w + q + ! + Enter, Vim saves the changes.  I'm using Ubuntu 16.04 and VIM 7.4. Completeness and Incompleteness Through browsing questions asked in the past about Godel's completeness and imcompleteness theorems, I've come to see that the sense of completeness in both of the theorems are different. However, I can't see how they are different!   Does this distinction boil down to that of truth and proof? Is it the case that the completeness theorem talks about the relationship between truth and proof, while the incompleteness theorems only relate to provability?  Could someone please elaborate on the sense in which both notions of completeness are different? Rigorous proof that $dx dy=r\ dr\ d\theta$ I get the graphic explanation, i.e. that the area $dA$ of the sector's increment can be looked upon as a polar "rectangle" as $dr$ and $d\theta$ are infinitesimal, but how do you prove this rigorously?    1. Geometrically, the exact area would be $$\frac{(r+dr)^2d\theta}{2} - \frac{r^2d\theta}{2}$$$$= (r + \frac{dr}{2}) dr d\theta  $$$$= r dr d\theta + \frac{dr^2 d\theta}{2}.$$  How do we get rid of $\frac{dr^2 d\theta}{2}$? Is it too insignificant in value compared to $r dr d\theta$ so that it vanishes? If so, is it the reason it gets ignored? And what's more, geometrically speaking, why would the two areas - $dx dy$ and $r dr d\theta$, be equal?  2. Approaching it algebraically, setting $$x = r\sin\theta$$$$y = r \cos\theta$$  gives $$\frac{dx}{d\theta} = r\cos\theta,  dx = r\cos{\theta} d\theta$$$$ \frac{dy}{dr} = \cos\theta,  dy = \cos\theta dr. $$  Multiplying both equations, side by side, gives $$dxdy = r\cos^2\theta dr d\theta.$$  Again I get an extra term, which is $\cos^2 \theta$. In both cases I am unable to derive that $dxdy = rdrd\theta$.  What do I do wrong in my reasoning? It all makes me think that I'm getting something essential terribly wrong. Writing my first script so I'm sure this is a basic question, but can someone please explain to me why I can:  cd ~ cd bin cd ~/bin cd 'bin'   But not  cd '~' cd '~/bin'   I need to cd to a directory path with a space in one of the directory names, so I need the quotes (it's Windows Program Files under wine). I can get around it with two cd commands, but why can't I put ~ in quotes?  If I type cd '~' (or cd "~") I get:  bash: cd: ~: No such file or directory Pretty printing XML with javascript Where is the Layers panel in Blender 2.8? I am starting working on Blender 2.8, but I can't find the layer panel. Where is it? In a document I'm writing, I'm using the familiar  option for my floating tables, and I  them. I'd like to avoid repeating that every time I'm creating a table. I assume it's easy to add something in my preamble that makes sure that these two options are used every time I begin a new table environment?  \documentclass{article} \begin{document} \begin{table}[hbt] \centering \begin{tabular}{*3{l}} 1 &amp; 2 &amp; 3\\ a &amp; b &amp; c\\ \end{tabular} \end{table} \end{document} Straighten Vertices Along a Line On a line, I need to line up the bottom vertex points in relation to a top reference vertex point. How could I reset the X Y and Z coordinates of these vertex points to essentially line them up all through use of the vector of the reference point vertex.  I do not have access to Mesh -> Transform -> Scale on Axis for aligning vertices along a line. How to solder with one hand? I am a beginner at electronics.  I don't have many tools and stuff, so I am forced to solder by hand.   I don't have a pcb to solder on, so I usually just solder the wires together, but it's really difficult without a holder.  Is there an easier way to do this? If not, is there a substitute for a wire holder?  Any help would be appreciated. Output from ls has newlines but displays on a single line. Why? I think I may be overlooking a relatively fundamental point regarding shell. Output from the ls command by default separates output with newlines, but the shell displays the output on a single line.  Can anyone explain this to me? I had always presumed that the output was simply separated by spaces, but now that I see the output separated by newlines, I would expect the output to be displaying on separate lines.  Example:  cpoweradm@debian:~/lpi103-4$ ls text* text1  text2  text3   od shows that the output is separated by newlines:  cpoweradm@debian:~/lpi103-4$ ls text* | od -c 0000000   t   e   x   t   1  \n   t   e   x   t   2  \n   t   e   x   t 0000020   3  \n 0000022   If newlines are present, then why doesn't the output display as:  text1  text2 text3 How to escape @ characters in Subversion managed file names? Why are passports checked at the check-in counter, Immigration and even at the gate? I was wondering, why there are passport checks at the check-in counter, Immigration and even at the gate?    For example, I was at Istanbul flying to either UK or US. There's a security check at the check-in counter asking a bunch of questions about my immigration status - even more than the Immigration officer at the border. Also, at the gate, there's also Security officer checking my immigration status and passport asking the same questions.    I thought that would be the responsibility of an Immigration officer rather than some third party company?   The thing I'm curious the most is they don't even know which countries this passport can go to with or without visa. They have to go back to the counter and check with the system which takes a long time. I was wondering why they hire these kind of companies to slow down the process? Each site got a &quot;hot&quot; tab, where its contents are calculated based on . However take a look on the screenshots below and see for yourself that &quot;hot&quot; got nothing to do with many of the questions listed there. :    :    As you can see, there are questions with negative score, no answers, and even closed. I won't be surprised to see spam in there as well. This can't be any worse. What I suggest, instead of fixing the formula or patching it, simply have the hot tab show the same questions as the Hot Network Questions sidebar. (from the specific site only, of course.) Since the hot questions &quot;pool&quot; (as seen ) is limited to 100 for the whole network, either expand it or have the per-site list show questions with less hotness points - it would still be better than what we have now. What is the use case for a Fragment with no UI? Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$? 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.