How to center my divs with bootstrap
How to horizontally center an element
\leqslant causes "Undefined control sequence" I have the following line: $0 \leqslant b \leqslant 1$  I want to display 0 &lt;= b &lt;= 1, but I get Undefined control sequence  thanks Java Server Socket not connecting Error: the question could not be loaded When I was trying to open  I just added using the iOS app I received this error:     Error The question could not be loaded   Reproduction scenario:   add a question; in my case it didn't open it directly; it doesn't show up in the list yet; I opened my profile and clicked the question; bang! Evolutionarily speaking, why do humans have 46 chromosomes override Home key in android ICS Capacitor on USB cable? I was messing around with a USB camera. I wanted to change the cable because is trimmed but as I was cutting it I saw a small thing (image below). As I was trying to cut the casing and see what it has inside I saw a spark. I tried to connect it with another cable but I don't know if I'm gonna damage the camera or the USB port.    What is it and can I connect the camera without it? Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n&gt;0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$  This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$  I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$  It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$ or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$ So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here). How can I install software or packages without Internet (offline)? I have a friend who has got a computer that is not connected to the Internet. Is there any way to install software offline easily? Show that the set of functions $\mathbb{N}\to\{0,1\}$ is not countable Recall that a countable set $S$ implies that there exists a bijection $\mathbb{N}\to S$. Now, I consider (0,1). I want to prove by contradiction that $(0,1)$ is not countable.  First, I assume the contrary that there exists a bijection $f$, and I can find an element in $S$, but not in the range of $f$. But I can't find such element. How can you construct such $f$? Including package to create MATLAB labels using LaTeX Configure Postfix to use external MX servers for delivery of local mail if user is unknown Movie/TV series, which bullets carried DNA of the shooter How do I remove an alias? I want to remove gs alias from my PC. When I type gs it will open GhostScript. But I checked everywhere in the home directory .alias  .bash_aliases .bashrc  I also overwrite the gs with my custom alias.   I can't remove it. And I also type alias in terminal, in the list I couldn't find it.   Please I want to remove it... Database schema documentation for the public data dump and SEDE Stack Exchange releases "data dumps" of all its publicly available content roughly every three months , and also makes that information queryable over the Internet at the  (SEDE) (updated weekly, on Sunday). Over time, as new features and other bits of data are introduced to Stack Exchange sites, the "anatomy" of the data dump is likely to change. I have some programs running on start-up like . UAC prompts every time the Windows starts. How can I set a program to make sure that no intervening prompt would appear during start-up? The difficulty is that it should be cross platform. Windows 2000, XP, Vista, OSX, Linux, other unix variants. I am looking for a snippet of code that can accomplish this for all platforms, and a way to detect the platform.  Now, you should be aware of  that user.home does not work correctly, so please do not provide me of textbook answers, I can find these myself in the manuals. When lecturing about conservation of energy in the 1960s, Richard Feynman remarked:     It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives “28” —always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.    50 some years later, have we made any progress in terms of understanding the fundamental nature of what energy is? Accessing UI thread handler from a service ref to subsection number only is there a way to reference the subsection part only? Example:  \documentclass{article} \begin{document}  \section{This is my section} For further information see subsection \ref{subsecthree} within this section. \subsection{One} \subsection{Two} \subsection{Three}\label{subsecthree}  \end{document}   Usually gives (without the stars):  1 This is my section For further information see subsection **1.3** within this section.   But I want:  1 This is my section For further information see subsection **3** within this section.   Is that possible?  [Edit] To make it more clear: I am looking for a command that works like \ref, but prints (the arabic number of) the value of the subsection counter at the position of the \label.  ======================= == Solution in LaTeX ==  There is another solution below. This is a different solution I came up with  (I cannot post an answer to my own question, so I put it here):  \documentclass{article}  \newcommand{\labelsubseccounter}[1]{     \renewcommand\thesubsection{\arabic{subsection}}     \addtocounter{subsection}{-1}     \refstepcounter{subsection}     \label{#1}     \renewcommand\thesubsection{\thesection.\arabic{subsection}} }  \begin{document}  \section{This is my section} For further information see subsection \ref{subsecthree} within this section,  which is called \ref{onepointthree}, not \ref{onepointfour}. \subsection{One} \subsection{Two} \subsection{Three}\label{onepointthree}\labelsubseccounter{subsecthree} \subsection{Four}\label{onepointfour}  \end{document}   (I would like to insert the compiled code, but don't know how, sorry) FireFox 66 Quantum disabled all my extensions at midnight!