Okay here's the question:  Ubuntu 12.04 Precise will not shut-down it just Reboots again!  Its as simple as that. I have tried every command out there and not a single one of these commands works in grub or Ctrl + Alt + T or any other method suggested.   sudo shutdown -p now &lt;-This does not work Type in terminal:   sudo -i (to get a root shell, sudo gedit is not recommended) gedit /etc/default/grub   Find the line: GRUB_CMDLINE_LINUX_DEFAULT="quiet splash"   Change this to: GRUB_CMDLINE_LINUX_DEFAULT="quiet splash acpi=force" Save the file and close the file. Finally, in terminal: update-grub exit (to end the root shell) &lt;-This does not work  sudo poweroff &lt;-This does not work alt + F2 and type gksudo poweroff &lt;-This does not work Code:   sudo gedit /etc/default/grub   --> remove quiet splash, then :  sudo update-grub   before trying to shut down again, run:  sudo apt-get install --reinstall ubuntu-desktop sudo dpkg-reconfigure -phigh -a  &lt;-This does not work  halt or shutdown -h now &lt;-This does not work sudo init 0 &lt;-This does not work acpi=norq added to grub &lt;-This does not work Typing shutdown or restart in Dash Home &lt;-This does not work sudo shutdown -h 0 &lt;-This does not work sudo halt &lt;-This does not work sudo shutdown -h 0 &lt;-This does not work   So lets try to figure it out together, lets solve it since no updates are coming in to fix it!
After two days of installation, strangely my Ubuntu machine is restarting instead of shutdown from both desktop GUI and command line.  sudo shutdown -h now sudo halt  # (doesn't shutdown, instead freezes on boot screen Plymouth)   I had force shutdown by powering off the machine.  This is not happening all the times.  I have been using Ubuntu for 6 months and never experienced this kind of problem.
Switch from single to double column in the same page Is there something like Linux console in Mac OS X? Like "terminal", but without running graphics environment. I want to do this using the Math.Round function I have a collection that contains linked assets. I would like to override their materials and give them a unique material. Is it possible to assign a material to a whole collection, and how ?  Thanks ! Use of Named Credentials seem to be tied to External Data Source Prove that $\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$ Theorem  Let $\{a_n\}$ and $\{b_n\}$ be convergent real sequences. Assume that there exists a $N\in\mathbb{N}$ so  $a_n\le b_n$ (eq. 1)  for all $n\ge N$. Then  $\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$.  My attempt at proving  Let $a=\lim_{n\to\infty}a_n$ and $b=\lim_{n\to\infty}b_n$. $a$ and $b$ are real numbers because the limit of a sequence of real numbers is a real number (which is a corollary in my textbook). For proof by contradiction, assume  $b&lt;a$.  Let $\epsilon=\frac{a-b}{2}$. Then we can find $N_a,N_b\in\mathbb{N}$ so  $|a-a_n|\le\frac{\epsilon}{2}$ when $n\ge N_a$, (eq. 2)  $|b-b_n|\le\frac{\epsilon}{2}$ when $n\ge N_b$. (eq. 3)  In a textbook I found that I should do the following:  Choose a $n$ so $n\ge\max(N,N_a,N_b$), then eqs. (1-3) are satisfied. Then  $a-\epsilon\le a_n+\frac{\epsilon}{2}-\epsilon$  How do I get that last thing? Is it necessary to do a proof by contradiction or can one do a direct proof? Is there any way to find out what happened to a post card sent from other country? How to zoom in & zoom out While using 10.04 and the previous versions I used to put Ctrl+F6 to zoom in and subsequently Ctrl+F7 to zoom out. Now (using 11.10, Unity) I can't even find the options to zoom in and zoom out in the "keyboard shortcuts". I tried "the Magnifier" in the Compiz but really can't understand what is going on right there. There is simple question I would like to ask:   What to do so as to be able to zoom in with Ctrl+F6 and zoom out with Ctrl+F7? How do I make windows save their position on screen after re-attaching a display? When I attach a second monitor to my MacBook, I drag many windows onto it.  When I detach the monitor, all the windows move back onto my MacBook's display.  How can I make it so that when I re-attach the monitor, all the windows automatically move back to their position on the second monitor again?  I use my personal MacBook at work where I have the extra monitor.  But this means I need to detach and re-attach the monitor every day.  I have many windows I leave open from the previous day's work, so this is a real pain.  I know about an app called , which is almost exactly what I want.  But I need something that works on OS X 10.5. If $\Gamma$ is satisfiable then $\Gamma$ is consistent We say that $\Gamma$ is consistent if for no $\phi$, then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$. We say that $\Gamma$ is satisfiable if there is a structure $M$ s.t. $\models_M\gamma[s]$ for alle $\gamma\in\Gamma$.  I want to show that if $\Gamma$ is satisfiable, then $\Gamma$ is consistent.  I'm thinking we assume that $\Gamma$ is not consistent, and then show that $\Gamma$ is not satisfiable. So if $\Gamma$ is not consistent then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$ for some $\phi$. But by soundness theorem then $\Gamma\models\neg\phi$ and $\Gamma\models\phi$, meaning for structure $M$, we have $\models_M\neg\phi[s]$ (whereas $\not\models_M\phi[s]$) and $\models_M\phi[s]$. So would this mean that $\Gamma$ is not satisfiable? I'm a little confused, because wouldn't this instead be a contradiction? Prove that any complex matrix can be expressed in term of Hermitian and anti-Hermitian matrices How can we prove that any complex matrix can be expressed in term of hermitian and anti-hermitian matrices of the following form $A = H_1 + iH_2$ Where, $H_1$ is hermitian and $H_2$ is an anti-hermitian matrix ToList()-- does it create a new list? How do you get a submit button in the Windows Phone keyboard? How to grep for text in a file and display the paragraph that has the text? Make a div fill the height of the remaining screen space Overfull hbox using biblatex Overdispersion and Underdispersion in Negative Binomial/Poisson Regression I was performing a Poisson regression in SAS and found that the Pearson chi-squared value divided by the degrees of freedom was around 5, indicating significant overdispersion. So, I fit a negative binomial model with proc genmod and found the Pearson chi-squared value divided by the degrees of freedom is 0.80. Is this now considered to be underdispersed? If so, how does one go about handling this? I have read a lot about overdispersion and believe I know how to handle this but information on how to handle or determine if there is underdispersion is scant. Can anyone assist?  Thanks. $\mathbb{R}$ \ $\mathbb{Q}$ and $\mathbb{R}^2\setminus\mathbb{Q}^2$ disconnected? If I want to prove that $\mathbb{R} \setminus \mathbb{Q}$ is disconnected, does it suffice to say that there are two open disjoint sets that cover $\mathbb{R}\setminus\mathbb{Q}$, namely:  $$(- \infty, 0), (0, \infty)\text{ ?}$$  Along the same lines, I want to prove or disprove that $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$ (points $(x, y) \in S$ that have at least one irrationais connected. I feel that it is also disconnected; does it suffice to say that there are two open disjoint sets that cover $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$, namely $((- \infty, - \infty), (0, 0))$ and $((0, 0), (\infty, \infty))$? (Very iffy on my assertion and my notation, sorry.)  Thanks! The  was deleted last night by the Community user.    Normally, I wouldn't see an issue with this, but for this particular deletion, I can see three issues:   We try not to delete good content There are multiple upvoted answers We often use previous answers on Meta (as well as badly asked questions) to form the precedent (and to use as duplicates) when users ask questions like this in the future   On a personal note, I find it easier to link to one of my previous answers whenever possible.  This seems like a bug with the deletion algorithm, , the deletion algorithm should never delete questions with upvoted answers.  I'm going to go ahead and undelete it, but I'm bringing this up in case there was either:    A policy change A bug   So is it either of the above or something else? How can Ohm's law be correct if superconductors have 0 resistivity?