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Agreement in "[Singular Noun] Is/Are [Plural Noun]"? My fish's native habitat is rice fields. My fish's native habitat are rice fields.   Which one is correct? I'm pretty sure it's the first, since 'is' modifies 'habitat,' but it still sounds weird... This question is about finding reasons for compile errors when using tikz.  Consider the files compiles.tex, compiles-not1.tex and compiles-not2.tex. Then it is clear that compiles-not1.tex does not compile, because the name C has not been assigned. The file compiles-not2.tex does not compile, because \draw (A) -- (A -| B ); contains a space after the name B. However, this is difficult to see from the error message.   Is it possible to tweak the error messages generated by tikz to perhaps mark tikz-code and names in some way? What I mean is that in the example, an error message ! Package pgf Error: No shape named 'B ' is known. might be more helpful, as the space in the name B\␣ (B with a trailing space) is more clearly delimited from the message.   File compiles.tex compiles without errors.  \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture}     \node (A) {A};     \node[above right of=A] (B) {B};     \draw (A) -- (A -| B); \end{tikzpicture} \end{document}   On compilation compiles-not1.tex generates ! Package pgf Error: No shape named C is known.  \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture}     \node (A) {A};     \node[above right of=A] (B) {B};     \draw (A) -- (A -| C); \end{tikzpicture} \end{document}   compiles-not2.tex generates ! Package pgf Error: No shape named B  is known.  \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture}     \node (A) {A};     \node[above right of=A] (B) {B};     \draw (A) -- (A -| B ); \end{tikzpicture} \end{document} I've been investigating the Cartesian graph of $x^y=y^x$. Obviously, part of the graph is comprised of the line $y=x$ but there is also a curve that is symmetrical about the line $y=x$. (We can prove this symmetry by noting that the function $x^y=y^x$ is self-inverse; all self-inverse functions are symmetrical about the line $y=x$.) An image of the graph is shown below:  I decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $(e,e)$. The following is my method: If the gradient of the line $y=x$ is $1$, the gradient of the curve at the intersection point must be $-1$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $\frac{dy}{dx}=-1$. Now to find $\frac{dy}{dx}$. We have $x^y=y^x$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\ln{x^y}}=e^{y\ln{x}}$$ and $$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $x^y=y^x$ and using implicit differentiation and the product rule gives us: $$(\frac{dy}{dx}\times \ln x +\frac{y}{x})x^y=(\ln y +\frac{dy}{dx}\times \frac{x}{y})y^x$$ So $$\frac{dy}{dx}x^y\ln x +yx^{y-1}=\frac{dy}{dx}xy^{x-1}+y^x \ln y$$ Extensively rearranging gives: $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ Let $\frac{dy}{dx}=-1$: $$y^x \ln y -yx^{y-1}=xy^{x-1}-x^y\ln x$$ But we know $x=y$ since we are at the intersection point with the line $y=x$: $$x^x \ln x -x^x=x^x-x^x\ln x$$ So $2x^x \ln x -2x^x=0$ $$2x^x(\ln x -1)=0$$ This means either $2x^x=0$ or $\ln x -1=0$ but we know $x^x$ is always greater than $0$ so $\ln x =1$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome. Thanks in advance Get three highest values from the array How do I create bookmarks with pdfLaTeX? I'm using MiKTeX and using pdfLaTeX. I can generate the PDF, but there are no bookmarks. In the .tex-file it is using:  \chapter \section \subsection   (with labels)  I originally used make with sphinx to output LaTeX. It must work somewhat or I wouldn't get the PDF. When I say bookmarks, I mean when you are in Acrobat and the left side has the Bookmarks panel.  I am reading   But I am not sure where to apply this.  EDIT:  I applied this, \usepackage{bookmark}, and got bookmarks, but I never get the subsection bookmarks. How can I double the newlines in an output stream Adding grouped icons to favorite bar in Ubuntu 18.04 Is there anyway to add icons groups to the favorites bar aka Ubuntu dock (something like drawers or unity-launcher-folders) in Ubuntu 18.04 ? In a table of contents that has been generated with LaTeX's \tableofcontents command, I would like to add a vertical space before a Chapter item in the table of contents. One way that I have in my notes is to use  \addtocontents{toc}{\protect\vspace{10pt}}   at a desired location in the document so that this gets processed by \tableofcontents to include an appropriate vertical space in the table of contents. I have two questions:   Is the above the correct way to get a vertical space in a table of contents? Why is it necessary to use the \protect command here?   Thank you... Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals. I've got a simple Leaflet map with a basemap and one map-layer coming from my own hosted GeoServer.  I made the layers show up nicely in a map.  Now I want to add popups on the point layer showing information stored in MapService.  I couldn't find too much information about that topic.   There's one link to a working example but the link seems to be broken. I found another example which I addopted. You can see it on . I'd like to display the content of the name field in the attribute table but nothing happens when the user clicks the map.  Any idea? I can't get into my field research prize How can I concatenate regex literals in JavaScript? Is it possible to do something like this?  var pattern = /some regex segment/ + /* comment here */     /another segment/;   Or do I have to use new RegExp() syntax and concatenate a string? I'd prefer to use the literal as the code is both more self-evident and concise. Find the Resistance between Two Terminals Let's say R = 49 ohms, and we want to find Req between a and d, how do we do that? Do we have to find Req for the whole circuit? What is the meaning of "this" in Java? While browsing on Safari and listening to iTunes I come across a video and then press the "play" button on the keyboard (to stop the music while I watch the video) it starts/stops the video instead of the music (I have no control over the music).  How can I quickly play/pause my music without affecting the video? Can I mix monk traditions? I'm wanting to mix archetypes (traditions) for my monk. I'm wanting specifically the 3rd and 17th level feats from Way of the Open Fist and the 6th and 11th level feats from the Way of the Shadows. I know we can't multiclass into the same class to gain both but in the PHB page 79 2nd column "Monastic Traditions" reads "Three traditions of monastic pursuit are in common in the monasteries scattered across the multiverse." I'm fine with that but where I question is "Most monasteries practice one tradition exclusively, but a few honor the three traditions and instruct each monk according to his or her aptitude and interest."   So my question would be can I build a monk that is split like that without homebrewing to much? How do I check OS with a preprocessor directive? Why can't I use switch statement on a String? 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). Will stationary charge radiate E.M waves if observer is accelerating?