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Best way to find all factors of a given number Conditional expectation of $X$ given $\sin(X)$ Undergraduate research field different from PhD field Is it possible to install MySQL 5.5 or 5.6 on Ubuntu 16.04? I recently upgraded to Ubuntu 16.04. Unfortunately I need to work MySQL 5.5 or 5.6. I tried many solutions to have MySQL 5.5 or 5.6 run on Ubuntu 16.04 like this  . But none of them is working. Ubuntu detect the unmet dependencies but says "but it is not going to be installed" for 5.5 or says Package 'mysql-server-5.6' has no installation candidate...  Is it not possible to force an earlier version of MySQL on Ubuntu? Why is that? Will inverse functions, and functions always meet at the line $y=x$? If I have a function, the inverse function, by definition will be a reflection of the original function in the line $y=x$, so if I wanted to find the point of intersection, instead of solving it with equating both of the functions to equaling each other, could I assume that the point of intersection, between the two functions, will always be at the point $y=x$? This would enable me to solve problems much easier, instead of having to solve quartic equations, for example. Let's fix some terminology first. A category $\mathcal{C}$ is preabelian if:  1) $Hom_{\mathcal{C}}(A,B)$ is an abelian group for every $A,B$ such that composition is biadditive,  2) $\mathcal{C}$ has a zero object,  3) $\mathcal{C}$ has binary products,  4) $\mathcal{C}$ has kernels and cokernels.  A category $\mathcal{C}$ is abelian if it is preabelian and satisfies:  5) every monomorphism is a kernel and every epimorphism is a cokernel.  Define the coimage of a map to be the cokernel of its kernel, and the image to be the kernel of its cokernel. We have the following commutative diagram:    where $\overline{f}$ is the only existing map (because of universality of kernel and cokernel).  I'm having trouble proving the following:     A preabelian category $\mathcal{C}$ is abelian iff $\overline{f}$ is an isomorphism.   The converse is easily shown, I'm having trouble proving $\Rightarrow$... There is some conflict or error I don't understand. I can skip it and nothing visible happens. The output is okay and everything seems to be fine. Still I would like to get rid of the error message. If I run the following minimal example:   \documentclass[]{scrbook}  \usepackage[applemac, grafiken]{tahbiblatex} \usepackage[ngerman]{babel}   \begin{document}  Testing äüöaüä.  \end{document}   I get the error message:      ! Missing = inserted for \ifnum. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel} ?  ! Missing number, treated as zero. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel}   Here is my .sty file:  \NeedsTeXFormat{LaTeX2e}\relax \ProvidesPackage{tahbiblatex}[2012, by tah modified by sm]  \RequirePackage{ifthen}  \DeclareOption{latin1}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{applemac}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{cp850}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{utf8}{\PassOptionsToPackage{\CurrentOption}{inputenc}}  \DeclareOption{dvi}{% \newif\ifpdf \ifx\pdfoutput\undefined     \pdffalse \else     \pdfoutput=0     \pdftrue \fi}  \newcounter{Grafiken}\setcounter{Grafiken}{0} \DeclareOption{grafiken}{\setcounter{Grafiken}{1}}  \newcounter{Randausgleich}\setcounter{Randausgleich}{0} \DeclareOption{randausgleich}{\setcounter{Randausgleich}{1}}  \newcounter{Tabellen}\setcounter{Tabellen}{0} \DeclareOption{tabellen}{\setcounter{Tabellen}{1}}  \newcounter{Sync}\setcounter{Sync}{0} \DeclareOption{sync}{\setcounter{Sync}{1}}  \ProcessOptions  \RequirePackage{fixltx2e} \RequirePackage{inputenc} \RequirePackage[T1]{fontenc} \RequirePackage[T1]{url} \RequirePackage{setspace} %\RequirePackage[german]{varioref} \RequirePackage{textcomp}  \newcommand{\urlangabe}[2]{% $\langle$\url{#1}$\rangle$ -- Zugriff am #2}  \def\hierlockern {\begingroup \def\par{\endgraf\endgroup}%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% }  \newcommand{\alleslockern}{%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% } \ifthenelse{\value{Grafiken}=1}   {   \ifx\pdfoutput\undefined \newcount\pdfoutput \fi   \ifcase\pdfoutput       \RequirePackage[dvips]{graphicx}       \DeclareGraphicsExtensions{.eps}       \typeout{Paket tah Meldung: Die Option `grafiken' fuer DVI wurde gewaehlt.}% \else     \RequirePackage[pdftex]{graphicx}     \DeclareGraphicsExtensions{.jpg,.tif,.png,.pdf}     \typeout{Paket tah Meldung: Die Option `grafiken' fuer PDF wurde gewaehlt.}%   \fi}   {}  \ifthenelse{\value{Sync}=1}   {     \ifx\pdfoutput\undefined \newcount\pdfoutput \fi     \ifcase\pdfoutput     \RequirePackage{srcltx}     \typeout{Paket tah Meldung: Die Option `sync' fuer DVI wurde gewaehlt.}%     \else     \RequirePackage{pdfsync}     \typeout{Paket tah Meldung: Die Option `sync' fuer PDF wurde gewaehlt.}%     \fi }   {}   \ifthenelse{\value{Randausgleich}=1}   {\typeout{Paket tah Meldung: Die Option `randausgleich' wurde gewaehlt.}%   \RequirePackage[activate]{pdfcprot}}   {}  \ifthenelse{\value{Tabellen}=1}   {\typeout{Paket tah Meldung: Die Option `tabellen' wurde gewaehlt.}%   \RequirePackage{tabularx}   \RequirePackage{ragged2e}   \RequirePackage{dcolumn}   \RequirePackage{booktabs}   \newcolumntype{L}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}l}   \newcolumntype{Y}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}X}   \newcolumntype{d}[1]{&gt;{\DC@{.}{,}{#1}}l&lt;{\DC@end}}   }   {}    \endinput   I also don't really need all the features. So it would be nice to reduce it to minimal features, which are needed.. Prove that if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$. Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $ where $I$ denotes the identity matrix of order $n$. How to prove this?   what I did:   let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u  \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A  $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$ Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 &amp; i=j \\ u_iv_j &amp; i\neq j \end{cases}$, and $A_{1j}$ is the  submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.  Ps. , in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys I mistakenly typed cd // instead of cd /. To my surprise, current directory showed up as //.   What is that directory? Why does it exist?  apple@snipped $ pwd /home/apple apple@snipped $ cd / apple@snipped $ pwd / apple@snipped $ cd // apple@snipped $ pwd // apple@snipped $ cd /// apple@snipped $ pwd / glob() - sort array of files by last modified datetime stamp I'm trying to display an array of files in order of date (last modified).  I have done this buy looping through the array and sorting it into another array, but is there an easier (more efficient) way to do this? Do accepted answers still serve a useful purpose? In the Stack Exchange voting system, we have two classes of votes: standard up / down votes that anyone can cast, and special  that only the asker can.   Although accepting an answer does not always pin it at the top of the answer list, that green check mark does  about the correctness of the answer. This gives a single person much greater weight than the rest of the community. That green check mark can seem confusing to visitors, who see  or  answers marked as being correct. I've heard this from several friends recently who were surprised when they came across factually wrong answers that had been downvoted, yet still had that check mark next to them.  Moderators regularly field flags from even more experienced users who demand that the accepted answer be changed from one that is incorrect or even harmful to another. We decline these, because we're not supposed to be making technical judgments like that and because we have no way of accepting answers on behalf of the asker. However, these flags do show that some people have problems with the presentation of accepted answers vs. the community voting on all the others.  There's also all the fun that came with arguments over accept rate, and people badgering askers to accept their answers.   Questions on the Stack Exchange network are intended to be broadly useful, to help out many others in the future beyond the person asking the question. If that's our goal, why should the person asking the question have any more say than the rest of the world? If it's a broadly applicable question, why should one person be able to highlight something as correct simply because they asked it first?  I guess what I'm asking is: if we did away with accepted answers, converting them to upvotes where appropriate, and only let community voting sort out the best answers, would we lose something important? I thought I'd ask after reading  to , because I wasn't sure how I felt about accepted answers anymore. Display error message if same record exist with combination of some fields? Which notation and why: $\text{P}()$, $\Pr()$, $\text{Prob}()$, or $\mathbb{P}()$ Using competitors domain with different ending to boost traffic I cannot update my Kali Linux, when trying to execute apt-get update I get this error message:  # apt-get update  Get:1 http://kali.mirror.garr.it/mirrors/kali kali-rolling InRelease [30.5 kB] Err:1 http://kali.mirror.garr.it/mirrors/kali kali-rolling InRelease   The following signatures were invalid: EXPKEYSIG ED444FF07D8D0BF6 Kali Linux Repository &lt;devel@kali.org&gt; Reading package lists... Done W: GPG error: http://kali.mirror.garr.it/mirrors/kali kali-rolling InRelease: The following signatures were invalid: EXPKEYSIG ED444FF07D8D0BF6 Kali Linux Repository &lt;devel@kali.org&gt; E: The repository 'http://kali.mirror.garr.it/mirrors/kali kali-rolling InRelease' is not signed. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details.   If you need my kernel version:  # uname -a 4.13.0-kali1-amd64 #1 SMP Debian 4.13.10-1kali2 (2017-11-08) x86_64 GNU/Linux   How can I fix this? Why is the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group realized as the vector space of Hermitian $2\times 2$ matrices? Prove Uncountable set minus a countable set is uncountable Let B be a countable set, and let A be an uncountable set. Prove that A-B is uncountable Some people (me) would prefer the straight GNOME experience as opposed to the Unity style desktop offered in Ubuntu 17.10.  How can I get the standard GNOME desktop and interface? Generating ordered combination of numbers I can form numbers with only 0,2,4,6,8.  The sequence is as follows  0,2,4,6,8,20,22,24,26,28,40,42,.....  How to generate an ordered sequence of numbers from combinations of 0,2,4,6,8.