I'm trying to install software on Ubuntu but whenever I write the command at terminal (e.g. sudo apt-get install scalpel) it shows following error:  user@OCAC:~$ sudo apt-get install scalpel [sudo] password for user:  Reading package lists... Done Building dependency tree Reading state information... Done N: Ignoring file 'trusty-stella-datong.volatile' in directory '/etc/apt/sources.list.d/' as it has an invalid filename extension N: Ignoring file 'trusty-stella-datong.volatile' in directory '/etc/apt/sources.list.d/' as it has an invalid filename extension  E: Unable to locate package scalpel   How can I solve this problem?
How do I get to the Universe Repository in supported versions of Ubuntu?
Why isn't the product rule used in the definition of mechanical work? This is a question from the free Harvard online abstract algebra .  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see  post.  This problem is from assignment $5$.     Prove directly that distinct cosets do not overlap.   Let $H$ be a subgroup of $G$ and let $a$, $b$, and $c$ be elements of $G$ such that $b\not\in aH$ and $c$ is in both $aH$ and $bH$.  Then there are elements $h$ and $h^\prime$ in $H$ such that $c=ah$ and $c=bh^\prime$.  So $ah=bh^\prime$ and $b=ahh^{\prime -1}$.  But $hh^{\prime -1}\in H$ so $b\in aH$.  This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset.  Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.  Thanks. I'm running Apache2 in a docker container, and want to write nothing to the disk, writing logs to stdout and stderr. I've seen a few different ways to do this (,  ) but these seem like hacks. Is there no way to do this by default?   To be clear, I do not want to tail the log, since that will result in things being written to the disk in the container. Area51 proposals are being advertised now all over the network? jQuery animate() How to rename objects with python? How would I prove there is no smallest positive rational number? what is the best method to prove this statement? In java SSL over https without certificate Hyphen or no hyphen when modifying an adjective with a quantity? I have a sentence which has an object that is described with an adjective:     We need to inform our interested patrons of this change.   If I modify "interested" with "more" or "less", do I connect the words with a hyphen or not?  Example sentence:     We need to inform our less-interested patrons of this change.     Also, please let me know if I have not used the correct terms for sentence parts in this question. Suppose $s$ is a complex number with $\Re(s) \in (0,1]$ and $\{a_n\}$ is a complex sequence converging to $a \neq 0$. Must the Dirichlet series $$\sum_{n=1}^\infty\frac{a_n}{n^s}$$ diverge? I have georeferenced some images (ArcGIS -> Update Georeferencing) which creates a *.tfwx file with the new parameters for georeferencing.   I renamed it to *.tfw but QGIS is not able to read the georeferencing information.   Does anybody has any useful tips how to fix this? Set "this" variable easily? I have a pretty good understanding of Javascript, except that I can't figure out a nice way to set the "this" variable.  Consider:  var myFunction = function(){     alert(this.foo_variable); }  var someObj = document.body; //using body as example object someObj.foo_variable = "hi"; //set foo_variable so it alerts  var old_fn = someObj.fn;   //store old value someObj.fn = myFunction;   //bind to someObj so "this" keyword works someObj.fn();               someObj.fn = old_fn;       //restore old value   Is there a way to do this without the last 4 lines?  It's rather annoying...  I've tried   binding an anonymous function, which I thought was beautiful and clever, but to no avail:  var myFunction = function(){     alert(this.foo_variable); }  var someObj = document.body;        //using body as example object someObj.foo_variable = "hi";        //set foo_variable so it alerts someObj.(function(){ fn(); })();    //fail.   Obviously, passing the variable into myFunction is an option... but that's not the point of this question.  Thanks. Is there a simple way of proving that $\text{GL}_n(R) \not\cong \text{GL}_m(R)$? Letting $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and letting $n \geq 3$ and $m$ be natural numbers, it is known that $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m=n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$ (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967").  So, given a field $\mathbb{F}$, if $n \neq m$, then $\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})$, and thus $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$.  Letting $R$ be a commutative ring (with unity), and letting $n, m \in \mathbb{N}$ be such that $n \neq m$, is there a simple "algebraic" way of proving that $\text{GL}_{m}(R)$ and $\text{GL}_{n}(R)$ are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$ for a field $\mathbb{F}$?  Certain special cases of this problem trivially hold, for example in the case whereby $\mathbb{F}$ is finite, in which case $|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|$. Cardinality of the set of all (real) discontinuous functions This is a question from the book Introduction to Set Theory (Hrbacek and Jech), chapter 5, question 2.6.     (Show that) The cardinality of the set of all discontinuous functions is $2^{2^{\aleph_0}}$. [Hint: Using exercise 2.5, show that $|\mathbb{R}^\mathbb{R}-C|=2^{2^{\aleph_0}}$ whenever $|C|\leq 2^{\aleph_0}$.]   Exercise 2.5, as referred to in the hint, provides the following result:     For $n&gt;0$, $n \cdot 2^{2^{\aleph_0}} = \aleph_0 \cdot 2^{2^{\aleph_0}} = 2^{\aleph_0} \cdot 2^{2^{\aleph_0}} = 2^{2^{\aleph_0}} \cdot 2^{2^{\aleph_0}} = (2^{2^{\aleph_0}})^n = (2^{2^{\aleph_0}})^{\aleph_0} = (2^{2^{\aleph_0}})^{2^{\aleph_0}}=2^{2^{\aleph_0}}$.   The only way I could answer this question was by making use of two theorems stated later  in the same textbook. First, it is easy to prove that $|\mathbb{R}^\mathbb{R}-C|$ is an infinite set, since if it was finite we would have \begin{equation}2^{2^{\aleph_0}}=|\mathbb{R}^\mathbb{R}|=|\mathbb{R}^\mathbb{R}-C|+|C| \leq n+2^{\aleph_0}=2^{\aleph_0}&lt; 2^{2^{\aleph_0}},\end{equation}which is a contradiction. Now I make use of a theorem requiring the axiom of choice: For every infinite set $S$ there exists a unique aleph $\aleph_\alpha$ such that $|S|=\aleph_\alpha$. So I let $|\mathbb{R}^\mathbb{R}-C|=\aleph_\alpha$ for some ordinal $\alpha$. Now assume $\aleph_\alpha &lt; 2^{2^{\aleph_0}}$. Here I make use of another theorem, which i am not sure I am allowed to do: For every $\alpha$ and $\beta$ such that $\alpha \leq \beta$, we have $\aleph_\alpha+\aleph_\beta=\aleph_\beta$. So I basically then say either $|\mathbb{R}^\mathbb{R}-C|+|C|=\aleph_\alpha$ or $|\mathbb{R}^\mathbb{R}-C|+|C|=2^{\aleph_0}$, depending on whether $\aleph_\alpha \leq 2^{\aleph_0}$ or vice-versa. The assumption I am making is that $2^{\aleph_0}=\aleph_1$, which as I understand it is basically equivalent to the continuum hypothesis, as we are saying that the least uncountable number is $\aleph_1$ (but by the previous theorem requiring the axiom of choice this seems reasonable?). So anyway assuming I can do that, we again have a contradiction since then we are saying $|\mathbb{R}^\mathbb{R}|=2^{\aleph_0}&lt;2^{2^{\aleph_0}}$ or $|\mathbb{R}^\mathbb{R}|=\aleph_\alpha &lt; 2^{2^{\aleph_0}}$.  My main problem with my proof is that I am not making use of the result given in exercise 2.5 as suggested in the hint. Can anyone help with a proof making use of this result (stated in the second block quote above). Forward local port or socket file to remote socket file Quick question - I run two linux boxes, one my own desktop and the other my VPS. For security reasons on the VPS end I opted for socket connections to MySQL (/var/run/mysqld/mysql.sock). I know I can tunnel like this: ssh -L 3307:127.0.0.1:3306 user@site.com if I set up the remote sql server to listen on some port, but what I want to know is can I do something like: ssh -L /path/to/myremotesqlserver.sock:/var/run/mysqld/mysql.sock thereby tunnelling two sockets, as opposed to two ports?  A perfectly acceptable solution would also be to forward a local port to the remote socket file, but where possible I'm trying not to have tcp servers running on the remote box.  (and yes, I know tcp would be easier). Every partial order can be extended to a linear ordering How do I show that every partial order can be extended to a linear ordering?      I think that I manage to prove that claim for finite set, how can I prove it for infinite set?  Thank you. Can't install php5 on Ubuntu 16.04 so I've recently installed my first Ubuntu, to be precise, a daily build of 16.04, on my new laptop since the hardware was to new for 15.10. Now I'm in the process of setting it up and I want to install php5 and some extensions.  When I execute sudo apt-get install libapache2-mod-php5 I get the error message that there is no installation package available for php5 and libapache2.  The error message is in german, so I doubt that this would help, but anyways here it is. :-) E: For package »libapache2-mod-php5« existiert kein Installationskandidat.  Do I need to add a apt-get repository? Or what am I doing wrong? Within a custom command, I'm doing the following to write to a file:  \newcommand\mycommand[1]{   \newwrite\file   \immediate\openout\file=foobar.tex   \immediate\write\file{     foo     bar     #1     blarg   }   \immediate\closeout\file }   Of course when I invoke it as \mycommand{baz}, I get the following contents in foobar.tex:  foo bar baz blarg   with all new lines turned into spaces by LaTeX. My question is, is there a way to output these lines into the file "as expected", i.e. with real new lines at the end of each line instead of a space? I'm thinking some \catcode wizardry should do the trick, but I'm not sure which character I need to redefine and to what category code.  Update  The solution in  works perfectly. The only problem is when #1 contains embedded newlines. I can of course send it to mycommand with its own slew of ^^J, I was wondering if there is a way to avoid this. The search parameter isanswered ignores acceptances The SE definition of "answered" question is: has an answer that is either accepted or has positive score. But the search parameter isanswered appears to ignore the "accepted" clause: when set to "no" it still returns questions with an accepted answer, as long as it has 0 score.    to reproduce the bug (unless meta effect kicks in and the answers get upvotes).     By the way, it seems that isanswered works only when is:question is present, but I'm not complaining about that... yet.   Also, this parameter is not listed on the , but I'm not complaining about that either... yet. Can I use Beetel 450TC1 as wireless router?