I am a novice Linux user and accidentally uninstalled a lot of packages. I was attempting to do a clean install of Wine using the advice on this help page:  and before I noticed a warning by a later user, I opened my terminal and executed:  apt-get remove wine\*   Apparently apt-get interprets characters like * as symbols in a regular expression, not as shell-style wildcards, and I just removed most packages with the string "win" in their names on my computer.  I exited the terminal before everything shut down. Is there a way to reinstall these packages and patch my system without doing a fresh OS install?
I accidentally removed all the packages starting with the name jpeg with:  sudo apt-get remove jpeg*   This screwed up a lot of things, removed Unity, Firefox, Flash player etc.  I somehow managed to recover the desktop and unity but then a lot of things are still buggy. For example .rpm files and .tag.gz just won't open up at all.  What I want is a way to fix everything that's amiss. Something that is the equivalent of a 'repair installation' with a windows disc.  Any tips, anyone?
How to display citations in text grouped as [1-3] in stead of [1,2,3] Prove the Radical of an Ideal is an Ideal I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight. How to generate all permutations of a list? How do you generate all the permutations of a list in Python, independently of the type of elements in that list?  For example:  permutations([]) []  permutations([1]) [1]  permutations([1, 2]) [1, 2] [2, 1]  permutations([1, 2, 3]) [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1] What's the opposite word for "sin"? I would like to know if there is an opposite word for sin in English.  I mean, how could I say the opposite of     I committed a sin   other than using a negation? What oil is best for seasoning a cast-iron skillet I'm having a problem setting up a vhost on Apache 2.4, using CentOS 6.3.  This question seems to have been answered several times on this forum, but none of the solutions work for me.  Basically, I set up a vhost with a particular document root. When I attempt to access the site, I get a permissions error. But I have no problem if I use a different root.  More specifically, if I set the root to be /volume1/web/xxx, I get a 503 because of a permission problem on index.html.  But if I set it to be /volume1/web, the index.html comes up without a problem.  The details:   I set the persmissions on volume1, web, and xxx to 755. I set the permissions on the index.html files to 777. I copy the index.html file to both /volume1/web and /volume1/web/xxx, so that they have identical index files with identical permissions.   In httpd.conf, I set the VirtualHost options so that I can easily switch the document root by changing 2 comment marks  I have the following lines in the conf file:  NameVirtualHost *:80  &lt;VirtualHost *:80&gt;     ServerName localhost     #DocumentRoot /volume1/web     DocumentRoot /volume1/web/xxx      #&lt;Directory /volume1/web&gt;     &lt;Directory /volume1/web/xxx&gt;       Options Indexes FollowSymLinks Includes ExecCGI       AllowOverride All       Order Allow,Deny       Allow from all     &lt;/Directory&gt;      ErrorLog logs/localhost-error_log     CustomLog logs/localhost-access_log common &lt;/VirtualHost&gt;   These lines contain all the suggestions that worked for other people with the same problem.  What I get when I enter localhost in my browser depends on the document root.   If the root is /volume1/web, the index.html page comes up. if the root is /volume1/web/xxx, then I get a 503 error, and the apache error log has this line:  [Thu Jul 19 17:17:52 2012] [error] [client ::1] (13)Permission denied: access to /index.html denied   I also tried adding an index.php file to the directory. I then got permission denied messages for both the php file and the html file.  I also tried setting the user : group to apache : apache on the directories and files. That didn't help.  I've done everything I can think of to fix the permissions, to no avail.  Does anyone have an idea about what's happening here?  Thanks. How to convert a std::string to const char* or char*? How does a lurker gain reputation to receive privilege for commenting? I have a small question regarding using sudo with output redirect &gt;. To enable IP forwrding, someone can use the command:  echo 1 &gt; /proc/sys/net/ipv4/ip_forward   Executing this command will give permission denied as it requires root privileges. However, executing the same command with sudo gives also permission denied error! It seems that output redirect &gt; does not inherit the permissions of the preceding command echo. Is this right?  As a workaround I do:  echo 1 | sudo tee /proc/sys/net/ipv4/ip_forward   Is this the best way to do it? Am I missing something?  Please, note that this is an example and it applies to all commands that use output redirect. Where to place my local website starting with the 2.4.7 version of apache2? I recently installed Ubuntu 14.04, then I installed lamp-server and placed my webpages in /var/www directory, but when I opened localhost in the browser there was nothing. I think that happened because Zend updated Apache.  Whatever the reason may be, I want to know where to keep my php files so that I can access them from my browser. Recommendation for mobile prepaid SIM card for smartphone + home calls in Germany What SIM card would you recommend for a smartphone user for travel in Germany?  It would be probably the only or the most important connection to the Internet, so large amounts of data and relatively fast transfer is an issue.  I need to stay in touch with friends and family, so the calls to other EU countries should be as cheap as possible.   For to the first criteria, I found the T-Mobile offer interesting. The internet costs 10 Euros/month and after reaching the transfer limit it stays relatively fast. According to tests (Galileo from Pro7, AFAIR) it has also the best coverage. However, calls to other countries are very expensive.  For to second criteria, Lebara and Mobilka are interesting. The calls are very cheap, but there are problems with signal strength in many locations, where T-Mobile has no problem. The voice quality, at least in Lebara, is also limited. They both have mobile internet with unlimited transfer, but after reaching the high speed limit, the transfer is reduced to 64 kbps, which is too slow.  I've found out that O2 has foreign calls at moderate prices, but I haven't yet checked for details. There are so many other options, making it is hard to compare them all without spending the whole weekend on it. So, if someone has found a better option than having 2 SIM cards and constantly switching them, this answer would be greatly appreciated. How do you switch from Wayland back to Xorg in Ubuntu 17.10? Some of my applications don't work on Ubuntu 17.10 Wayland. How can I switch back to Xorg? I took some shots recently and I noticed that I'm getting a lot of little black dots in my images.  I cleaned my lens thoroughly with the Nikon camelhair brush and retook the pictures but once again, the dots appear.  I'm shooting with a Nikon D7000 and using a starter lens. I've attached an example (notice the black spots on the chairs, on the ceiling, etc). Plural or singular verb for collective term It is me who am to blame.    It is I who am to blame.      Which one is correct here? I am bewildered about these two sentences. Is $\{0\}$ a field? Consider the set $F$ consisting of the single element $I$.  Define addition and multiplication such that $I+I=I$ and $I \times I=I$ .  This ring satisfies the field axioms:   Closure under addition.  If $x, y \in F$, then $x = y = I$, so $x + y = I + I = I \in F$. Closure under multiplication.  $x \times y = I \times I = I \in F$ Existence of additive identity.  $\forall x \in F$ (i.e., for $x=I$), $x + I = x$, so $I$ is the additive identity. Existence of mulitiplicative identity. $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity. Additive inverse. $\forall x \in F, \exists y = I \in F: x + y = I$ Multiplicative inverse. $\forall x \in F, \exists y = I \in F: x \times y = I$.  However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. Commutativity of addition.  $\forall x, y \in F, x + y = I = y + x$ Commutativity of multiplication.  $\forall x, y \in F, x \times y = I = y \times x$ Associativity of addition.  $\forall x, y, z \in F$, $(x + y) + z = I + I = I$ and $x + (y + z) = I + I = I$, so $(x + y) + z = x + (y + z)$ Associativity of multiplication.  $\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$ Distributivity of multiplication over addition. $I \times (I + I) = I$ and $I \times I + I \times I = I$, so $\forall x,y,z \in F, x(y+z) = xy+xz$   Based on the above, $\{I\}$ seems to qualify as a field.  If $I$ is assumed to be a real number, then the unique solution of $I + I = I$ and $I \times I = I$ is, of course, $I = 0$.  So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it?  Specifically, is it required for the multiplicative identity to be distinct from the additive identity? Why is $\frac{1}{\infty } \approx 0 $ and $ \frac{1}{0} = {\infty}$? First I have checked the search option but found nothing relevant to my problem and also level of math. I just started learning the language of mathematics, on my own and I have trouble understanding why  $  \frac{1}{\infty } \approx 0 $ and  why $\frac{1}{0} = {\infty}$?  I want to know how exactly do I get to these answers. It might sound strange but it's hard for me as a beginner to picture things like this on my own. Can commuting matrices $X,Y$ always be written as polynomials of some matrix $A$? Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification. Get mp3 duration in android Prove that  $ n &lt; 2^{n}  $ for all natural numbers $n$.  I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k&lt;2^{k}$. Considering $k+1 &lt;2^{k}+1$, but where do I go from here?  Any other methods maybe?