I just would like to have $$\frac{643,2-759,9}{340,09} = -0,34$$  in italic. Is there an easy way? It doesn't work with \textit{} here... I am looking forward for some help. Thank you!
I often find myself adding a lot of \textsl{} around inside \emph{} or \textit{} blocks because I find italicized digits to look too "showoffy" and "baroque" and to hinder legibility (I'm using $$ when the number has a mathematical sense, but this situation occurs in less technical texts like CVs, and such.).  Could someone help me write a renewcommand to automate this, or better, a single low level command that makes \emph, \itshape, \textit and friends choose slanted over italic for numbers? I'm not too familiar with the low level latex font selection system nor with a way to detect digits from regular letters.  Edit: MWE of what I need and what I want to avoid  \documentclass{article} \begin{document} He thought: '\emph{Why 1234567890 apples?}' then he realized '\emph{ah yes, he has \textsl{1234567890} childs}' \end{document}   gives:   I find the italicized version (first one) of 2,3,4 and 7 to look especially bad.
I know how to compute modular multiplicative inverses for co-prime variables $a$ and $b$, but is there an efficient method for computing variable $x$ where $x &lt; b$ and $a$ and $b$ are not co-prime, given variables $a$, $b$ and $c$, as described by the equation below?   $ a x \equiv c \mod b $   For example, given  $ 154x \equiv 14 \mod 182 $, is there an efficient method for computing all the possibilities of $x$, without pure bruteforce?  Please note that I'm not necessarily asking for a direct solution, just a more optimized one.    I do not believe that the Extended Euclidean Algorithm will work here, because $a$ and $b$ are not co-prime.   Edit:  Follow up question, since the first one had a shortcut:  Could the be computed efficiently as well?  $12260x \equiv 24560 \mod 24755$.  $107$ needs to be one of the computed answers. Sometimes chicken breast can be thick and I never know how to tell when it's cooked! How can I be certain it is cooked all the way through? Why does sudo change the PATH? Let $\vec{k}$, $\vec{v}$, and $\vec{u}$ be vectors, where $\vec{u}$ is unknown and $\vec{k}$ and $\vec{v}$ are known vectors. Given:     $\vec{u}\cdot\vec{k}=c$  $\vec{u} \times \vec{k}= \vec{v}$   From this relations, how can i determine the vector $u$? I tried to construct orthogonal coordinate system from $(\vec{k}, \vec{v},\vec{k}\times \vec{v})$ but i couldn't proceed from there. Any idea? Can't find the reason why root directory usage is so large Gradient and Swiftest Ascent I want to understand intuitively why it is that the gradient gives the direction of steepest ascent. (I will consider the case of $f:\mathbb{R}^2\to\mathbb{R}$)  The standard proof is to note that the directional derivative is $$D_vf=v\cdot \nabla f=|\nabla f|\,\cos\theta$$ which is maximized at $\theta=0$. This is a good verification, but it doesn't really help me understand the result. Removing OS from Dual Boot computer, and recovering HD space. I'm participating in the  and wanted to refer to the on- and off-topic questions on the  proposal.  However, I couldn't find a link to it anywhere in the SE site.  If I'm not missing the obvious, could we have one? What is a "callback" in C and how are they implemented? From the reading that I have done, Core Audio relies heavily on callbacks (and C++, but that's another story).    I understand the concept (sort of) of setting up a function that is called by another function repeatedly to accomplish a task.  I just don't understand how they get set up and how they actually work.  Any examples would be appreciated. On the proof that the product of diagonal matrices is a diagonal matrix? Let $A [a_{i,j}]$  and $B =[b_{ij}]$   be two diagonal matrices in   $M_n(F)$. Let $i \ne j \in \left\{1, \ldots, n \right\}$. Using the   product rule, we obtain      $$(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$$      We claim that $a_{ik}b_{kj} =0$ for all $k \in \left\{1, \ldots, n \right\}$ thus $(AB)_{ij}=0$ for all $i \ne j$ and $AB$ is diagonal. To    prove the claim, note that since $i \ne j$ , we have $i \ne k$ or $    k\ne j$. Thus either $a_{ik} = 0$ (since $A$ is diagonal) or $b_{kj} =  0$ (since $B$ is diagonal), thus in all cases $a_{ik}b_{kj}=0$ and the   claim is proved.   Where does the bit "we have $i \ne k$ or $k\ne j$" come from? I understand that since $i \ne j$, and $k$ and $j$ run over the same set of natural numbers, we also have $i \ne k$. But the "or $k \ne j$" baffles me. I understand that $k \ne j$ implies that we can have $i=k$. But that's from the other direction ($\Longleftarrow).$  Also, when it says $a_{ik}=0$ or $b_{kj}=0$ it's talking about when $i \ne k$ or $k \ne j$ right? Many thanks. Change default shell for Terminator I want to change the default shell of Terminator to zsh. In /etc/passwd/ I've changed the default shell to /bin/zsh. This settings changes the default shell in LXTerminal (the default terminal on Lubuntu), but not in Terminator. So I tried chsh -s /bin/zsh, but that doesn't work either. The Terminator  says:  How do I change my default shell? Terminator will always start the shell specified by the SHELL environment variable. (It will fall back to /bin/sh if the SHELL environment variable isn't set.) You could simply influence Terminator's environment, but the best solution is to tell your operating system which shell you'd like, and it will ensure that SHELL is set correctly for all applications, not just Terminator. See the documentation for your operating system for details on how to change your default shell.  So I expected my attempts to work, but they don't. How can I change the default shell in Terminator. Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$ Somewhere on Stack Exchange I saw the equation     $$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$   I had never seen this before, so I started trying to prove it. Without success...  Can anyone explain me (so actually prove) why this equation is true?  And can we say the same when replacing the '$2$' by any integer number '$a$'? Equivalence Relation problem Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas? I am replacing a door with a pre-hung door including the jamb. The old door was 29 11/16" wide. The new pre-hung door is 28".  Is this too big of a gap?  Should I make it plumb on one side and make up the difference on the other side? Other threads suggest making up space on the handle side, but this is more space than than what other people were asking about. The trim is 2 1/2" so the gap would be covered. Printing labels along with equation numbers Working with documents with huge amount of equations it becomes cumbersome to go back and remind yourself what you labeled a certain equation to (I try to name them in a smart way, but there are too many).  Is there a convenient way to have equations print the label along with the tag/number, so that I can easily check the label in the compiled document?  Something like this would be perfect:  x = y               (14.1) eq:my_label Is there a Stack Exchange site for architecture? Operators can be cyclically interchanged inside a trace: $${\rm Tr} (AB)~=~{\rm Tr} (BA).$$ This means the trace of a commutator of any two operators is zero: $${\rm Tr} ([A,B])~=~0.$$ But what about the commutator of the position and momentum operators for a quantum particle? On the one hand: $${\rm Tr}([x,p])~=~0,$$ while on the other hand: $$[x,p]~=~i\hbar.$$ How does this work out? How to change envelope from address using PHP mail? Does seam itself affect baking? Possible Duplicate:           My name is Greg — this is Greg 's post.  If my name ended with an 's', I am aware of the proper apostrophe usage (James&nbsp;→ James'), but how should this be pronounced?  Phonetically, am I Jameses best friend or James best friend?