I wish to know is possible to set in bold a specific word without type every times \textbf{bold}. For example  \documentclass{article} \begin{document} \textbf{dog}, cat,\textbf{dog}, cat,  \textbf{dog}, cat, \textbf{dog}, cat  \end{document}   Is it possible to set \textbf{dog} and just write "dog" which will be in bold after the compilation?
Is it possible and furthermore constructive to make LaTeX automatically format specific words in specific way? Let's say, for example, I'd like to have every occurrence of the word "hello" in a bold cyan monospaced font, whereas the rest of the document remains normal, sans serif and black.   How can I achieve that without having to do a search&amp;replace to place "hello" into a custom command every time before I compile the .tex file?
Being characteristic is transitive Here, I wanted to verify that:     The property of being characteristic is a transitive relation among subgroups of a group $G$.   For subgroups $N,H\leq G$, we have $N$ char $G$ and $H$ char $N$. So $N$ char $G$ implies: $$\forall\psi\in Aut(G); \psi(N)=N$$ $H$ char $N$, so for all elements in $Aut(N)$, $H$ is remained never changing. Especially, when I take $\psi'=\psi|_{N}$ then $\psi':N\to Aut(N)$ would be in $Aut(N)$ and $\psi'(H)=H$. Since the last equality is true for $H$ and the maps, caused by restriction on $N$, then I have $H$ char $G$.  Honestly, I am inly not satisfied form the conclusion here and think I am losing something. Thanks for your hints. I'm from Central Pennsylvania, and apparently, we have a strange language construct in this area. I was recently talking about how "my car needs washed" to a friend from NJ, and she told me that my sentence was grammatically incorrect.   I realize she's right, and I realize my options are "my car needs to be washed" or "my car needs washing". But I'm not clear on the precise rule being invoked. In "Seeing like a state" of James C.Scott there is a sentence at the beginning of a paragraph (Acknowledgements xi):     There are a good many scholars whose writings opened up new   perspectives for me or provided outstanding analyses of issues that I   could not have hoped to study so comprehensively on my own.   What worries me is a good many scholars. Is this some type of inversion or ancient English? Is this grammatically right? Why is a here? I thought that appropriate version would be There are many good scholars. Am I wrong? Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.   $"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$  then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$   stuck at this spot,  $(\sum_{n=0}^{\infty} a_nx^n)(\sum_{n=0}^{\infty} b_nx^n) = \sum_{n=0}^{\infty}(\sum_{k=0}^{n}(a_kb_{n-k}))x^n$  A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n&gt;0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.  EDIT  Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$ I have an alpha version of Ubuntu.  When the stable version is released do I have to upgrade or is it already stable from the installed updates? Let $H$ and $K$ be subgroups of a group $G$, and let $HK=\{hk: h \in H, k \in K\}$, $KH=\{kh: k \in K, h \in H\}$. How can we prove that $HK$ is a subgroup iff $HK=KH$? Links that are italics and bold not showing as links in Mobile Web The following sentence refers to an apocalyptic story where money no longer has any value:     A can of sardines, radio batteries, or a bicycle is/are more precious than money.   Should I use is or are after the series? Calculate Time Remaining How do I move child windows without moving or minimizing parent in GNOME 3? Impose strict time-limit for questions on Hot Network Questions list Changing Custom Post Type Name Order $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ Problem:      Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y  $ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)  $ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)  = g(x) $ for all $x \in  X $.   Attempt at a solution:  Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) &gt; 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y $ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.   I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A $, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.  Also, Would be be possible to prove this without using the Hausdorff condition on $Y$? I had installed VirtualBox but it started giving me problems. So I uninstalled it using:-  sudo apt-get remove virtualbox-\* sudo apt-get purge virtualbox-\*   But when I reinstalled VirtualBox it again resumed to previous settings and started giving same problems. How can i completely remove VirtualBox from my laptop and then reinstall it such that it wont give same problem again? From the :  [1,4,9].map(Math.sqrt)   will yield:  [1,2,3]   Why then does this:  ['1','2','3'].map(parseInt)   yield this:  [1, NaN, NaN]   I have tested in Firefox 3.0.1 and Chrome 0.3 and just as a disclaimer, I know this is not cross-browser functionality (no IE).  I found out that the following will accomplish the desired effect. However, it still doesn’t explain the errant behavior of parseInt.  ['1','2','3'].map(function(i){return +i;}) // returns [1,2,3] android - out of memory exception when creating bitmap Executing cmd.exe commands from Java How to assign same Data to multiple object copies I made many copies of an object, then I thought it would've been better if I made them to share the same Data (being an instance of the same original object), I know how to do it one by one, is there a way to do it for all of them at once? I tried Copy Attribute add-on but didn't find what option can do that, or even if this is possible in the first place. What's the "Bad magic number" ImportError in python, and how do I fix it?  The only thing I can find online suggests this is caused by compiling a .py -> .pyc file and then trying to use it with the wrong version of python. In my case, however, the file seems to import fine some times but not others, and I'm not sure why.  The information python's providing in the traceback isn't particularly helpful (which is why I was asking here...), but here it is in case it helps:  Traceback (most recent call last):   File "run.py", line 7, in &lt;module&gt;     from Normalization import Normalizer Monic $f(x)\in\Bbb Z[x]$ has no rational root if $f(0)\ \&\ f(1)$ odd [Parity Root Test, Modular Root Test] A polynomial problem from my old algebra textbook:     $f(x)\in\Bbb Z[x]$ with leading coefficient $1$, $\deg f(x)\ge 1$, and both $f(0)$ and $f(1)$ are odd numbers, prove:   $f(x)$ has no root within $\Bbb Q$   Eisenstein's criterion seems to be of little help here because we know virtually nothing about the coefficients. So I tried some other approach. Let $$f(x)=a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+x^n\in \Bbb{Z}[x]$$ What we have known so far is $$f(0)=a_0\text{ is an odd number}$$ and $$f(1)=a_0+a_1+a_2+\cdots+a_{n-1}+1 \text{ is an odd number}$$ from which we can also conclude that $$f(1)-f(0)=a_1+a_2+\cdots+a_{n-1}+1\text{ is an even number}$$ To be honest I don't know how to proceed now. It isn't until just now that I realized "irreduicible" is not equivalent to "has no roots".. So all my previous attempt is fundamentally wrong.. Can anybody help or drop a hint? Best regards!