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BibTeX: Abbreviate first name (aka given name) to 2 or 3 letters (not 1) using gb4e keeping interlinear glosses together on the same page I am using gb4e for several hundred interlinear glosses in my dissertation. Several of the examples span more than one page, that is, the vernacular and gloss will be on one page, while the free translation appears at the top of the next page.   How can I keep all three elements together on the same page? I often have subexamples (a, b, c) that don't need to stay together. That is, 1a can be on one page, and 1b on the next, as long as vernacular-gloss-free translation for each subpart stay together.  In the MWE below, the free translation is at the top of page 2, while the vernacular and gloss are at the bottom of page 1.  \documentclass[11pt,letterpaper]{article} \usepackage{lipsum} \usepackage{gb4e} \noautomath \begin{document} \lipsum[1-4] \begin{exe}     \ex \gll wuz x̃ũch it-ak xohish-em\\              1\textsc{sg.prn} bread eat-\textsc{inf} want.\textsc{prs}-1\textsc{sg}\\         \glt `I want to eat bread.'  \end{exe} \end{document}   A solution was proposed to a similar question (), but it requires me to manually look for what could be dozens of orphans throughout my dissertation (and doesn't work when applied to my example, since it still strands the FT). I would prefer a more elegant solution that takes care of the problem automatically. I want to do something like the following in TeX:  \begin{nobreak}    Text here will not split over pages, it will remain as one continuous chunk. If there isn't enough room for it on the current page a pagebreak will happen before it and the whole chunk will start on the next page.    \end{nobreak}   Is this possible? Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic. Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?    Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.  Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.    Thanks I extracted simple example:  require 'pp' x = 1..3 pp x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end pp x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} }  pp(x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end) pp(x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} })   I am wondering why first pp produces:  [1, 2, 3]   While all the oders are giving:  [{:value=&gt;1, :double=&gt;2}, {:value=&gt;2, :double=&gt;4}, {:value=&gt;3, :double=&gt;6}]   I assume it has something to do with operator precedence. Where can I find good explanation? Does the pilot of a rocket ship experience an asymptotic approach to the speed of light? A question has recently come up that goes beyond my knowledge of special relativity.  Suppose a pilot has his foot on the gas pedal of a rocket ship and keeps it applied to achieve a constant acceleration, and he has a magical engine that can maintain the thrust needed to accomplish such acceleration even as the vehicle acquires mass.  He measures his acceleration at periodic intervals and sums it up as he goes.  Observers perceive the rocket to increase in speed and gradually approach the speed of light (but never reach it). Two possibilities are being debated for what the pilot observes, and my preference is for the first:   Time dilation gives him the impression that the summation of his readings are approaching light speed at a constant rate, but there is an asymptote involved here even in the inertial frame such that he never actually perceives himself to reach, let alone exceed, $c$, because time slows down to a crawl (as seen from the reference frame) before this can happen. Although to observers he is only approaching $c$, it is possible for the summation of the measurements on his accelerometer to indicate that he has surpassed $c$ and is now traveling faster than light.  From the reference frame he is approaching $c$ and his instruments in the inertial frame tell him that he has surpassed it.   I assume (1) is true beyond a doubt?  If so, is there a straightforward explanation for this? What are some uses of template template parameters? I've seen some examples of C++ using template template parameters (that is templates which take templates as parameters) to do policy-based class design. What other uses does this technique have? How do I mail merge from Google Spreadsheet to a Google Document? Prove that $\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$ Problem  Suppose $f$ is a non-negative integrable function on a measure space $(X,\mathcal{A},\mu).$ Prove that $$\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$$  Attempt  Let $E_t=\{x:f(x)\geq t\}$ Note that $f\chi_{E_t}\leq f$. Since $f$ is integrable, the Lebesgue dominated convergence theorem tells us that $$\lim_{t\to\infty} \int f\chi_{E_t}\rightarrow \int \lim_{t\to\infty} f\chi_{E_t}=0,$$ since $\lim_{t\to\infty} f\chi_{E_t}=0$ a.e. $\ ^{(1)}$ The result follows from the fact that $$t\mu(E_t)\leq\int f\chi_{E_t}.$$  Question  I am having trouble proving $(1)$. That is, how can I show that $f\chi_{E_t}\rightarrow 0$ a.e? Passing data between view controllers Where does the term "fun fact" originate?-- namely, not with the compositional meaning but rather with the idiomatic usage to introduce some sort of unusual, esoteric, absurd or otherwise "controversial" fact? Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}...\cot \frac{(m-1)\pi}{2m}=1$ Rotation Path Not Shown in Graph Editor I created a couple of LocRot keyframes but the rotation path and keyframes aren't shown in the graph editor (no, they aren't hidden). I recently traveled from the Basel area to Stuttgart for a weekend.  I had round-trip tickets including (paid) bicycle reservations.  At our first station change, we found that our EC train was canceled.  Information gave us a replacement train schedule, using slow RE trains which don't require reservations for bicycles.  As we continued our trip, every single train we boarded arrived late, eventually causing us to reach Stuttgart roughly 2 hours later than our initial itinerary.  What monetary compensation am I justified in requesting from Deutsche Bahn in this case?  In particular:   It seems clear I should get back the bicycle reservation money, since the train we reserved on was canceled. I know DB offers compensation for late travels.  I'm not sure whether we were 2 hours late or more, or just shy of 2 hours.  But does this apply if we weren't traveling by our original trains? If we had originally scheduled our journey with RE trains and other local transport, we could have bought a Baden-Württemberg ticket for much cheaper than our tickets including an EC cost.  Can I ask them to refund the difference? Am I entitled to any other type of compensation? Can you find all classes in a package using reflection? How do I detect unsigned integer multiply overflow? JavaScript post request like a form submit I'm running Bash scripting, but sometimes I get confused which of these commands I use belong to who. Sometimes man xxx works, sometimes doesn't, so I use --help or info, mostly one of these works to show description of the command. Can anyone tell me how I would know what command belongs to what? Bash builtin, GNU utility, etc. I'm trying to continue the numbering of examples in enumerate environments in beamer. For example, I'd like one slide to feature   Foo Bar   and the next slide   Zip Yadda   My strategy so far has been to define a counter that saves the enumerate counter at the end of the first environment and can then be queried to set the enumerate counter at the start of the second enumerate environment. (This is essentially the proposal also found at )  That works wonderfully so long as I don't try to uncover elements of the slide sequentially. The problem, of course, is that as each of the pdf-pages that make up the slide is created, the counter is advanced. Here's a MWE.  \documentclass{beamer}  \setbeamercovered{highly dynamic}  \newcounter{saveenumi} \newcommand{\seti}{\setcounter{saveenumi}{\value{enumi}}} \newcommand{\conti}{\setcounter{enumi}{\value{saveenumi}}}   \begin{document}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}   \item foo   \item bar%     \seti   \end{enumerate} \end{frame}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}     \conti   \item zip   \item yadda%     \seti   \end{enumerate} \end{frame}  \end{document}   As the second slide is progressively uncovered (pages 3 and 4 of the resulting PDF), the counter at the beginning of the enumerate environment is advanced each time. So page 3 of the PDF has the right numbering, (3) and (4) respectively, while the fully revealed page has the wrong numbering. Pictures follow:   on one of the slides    on the following one Comparison theorem for ODE