I really want to know how can I prevent a program to connect to Internet with Windows 7 firewall?    I have smart security nod32 business edition installed on my computer. should i do this job with nod or there is a way in Windows Firewall?
There are some .exe and .dll files belonging to some programs on my laptop which communicate with the internet without my permission (for self-update generally). I want to prevent those files one by one from accessing the internet. I don't want to install any external firewalls. I'm using Windows 7 Home Pro and if this can be done by Windows Firewall, it's OK. I took a look at Windows Firewall but couldn't find a setting to disable file access.
Send file using POST from a Python script Highlight \verb in inline text How to scale and shift voltage signal? I want to scale and shift my voltage signal by below relation:  $$ V_{out} = \frac {10} {12} \cdot V_{in} - 10 $$  Indeed my primary voltage is from 0v to 24v and I want to put it from -10v to +10v.  I try to use the below circuit but its gain is always greater than 1.    Image from SLOA097 TI.  Does anyone know how I can achieve my goal? I have ±12v and 5v power lines.  EDIT: According to the innovative solution by scorpdaddy, I thought it would be valuable to edit my question:   High input impedance is required to decrease load effects. How can it be achieved? How can I achieve the more and more accuracy? Automorphism group of the quaternion group Let $Q_8$ be the quaternion group. How do we determine the automorphism group $Aut(Q_8)$ of $Q_8$ algebraically? I searched for this problem on internet. I found some geometric proofs that $Aut(Q_8)$ is isomorphic to the rotation group of a cube, hence it is isomorphic to the symmetric group $S_4$. I would like to know an algebraic proof that $Aut(Q_8)$ is isomorphic to $S_4$. Let $D = B(0,1) \subset \mathbb{C} $ a disc, $f$ holomorphic on $D$. I want to demonstrate that if $$ 2|f^{'}(0)| = \sup_{z, w \in D} |f(z)-f(w)|$$  then $f$ is linear.   I know this is a well-known result. Where can I find the proof ? Does the Divergence Theorem Work on a Surface? The divergence theorem in $\mathbb{R}^3$ says that the integral of the divergence of a vector field over a solid $\Omega$ in $\mathbb{R}^3$ equals the flux through the surface of $\Omega$ denoted by $\partial \Omega$.   My question is that can you use the divergence theorem in $\mathbb{R}^3$ where you're integrating the divergence of a vector field over a surface $S$ in $\mathbb{R}^3$ to get the flux through the edges ($\partial S$) of the surface? So if the surface $S$ is closed in $\mathbb{R}^3$ such as a sphere, then clearly $\partial S = 0$ so the integral of the divergence over this particular $S$ is $0$? In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition   Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.  The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."   I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated. I would like to show that  $$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$$  I'm interested in more ways of proofing this.  My method :  suppose that $\sqrt{\frac{n}{n+1}}\in  \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}{n+1}}=\frac{p}{q}$ thus   $$\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2$$ since $p\neq q\implies p^2\neq q^2$ then  $$n=\dfrac{p^2}{(q^2-p^2)}$$  since $n\in \mathbb{N}^*$ then $n\in  \mathbb{Q}$   I'm stuck here and I would like to see different ways to prove $\sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$ $\mathbb Z_{p^n}$ is not the direct product of any family of its proper subgroups I'm trying to solve this question of Hungerford's Algebra book:     $S_3$ is not the direct product of any family of its proper subgroups.   The same is true of $\mathbb Z_{p^n}$.   The first claim is easy: we note that every subgroup of $S_3$ is of order $2$ or $3$ by Lagrange, since its subgroups are of prime order, they are cyclic, then abelian, but $S_3$ is not abelian, contradiction because $S_3$ is not abelian while the direct product of abelian groups has to be abelian.  My problem is with the second claim, I need help. I wanted a simple environment for making lists of equations, so I made one that includes an equation and a split environment. The code won't compile and the error seems unrelated. The most pressing thing I noticed was the &quot;You can't use `\end' in internal vertical mode.&quot; The problem is definitely the split environment. Here is a MWE. \documentclass{article} \usepackage{amsmath}  \newenvironment{env}     {\begin{equation}\begin{split}}     {\end{split}\end{equation}}  \begin{document} \begin{env}     x \end{env} \end{document}  If I remove the split environment from env, everything compiles just fine. Additionally, If the following example compiles just fine. What is the problem then? \documentclass{article} \usepackage{amsmath}  \begin{document} \begin{equation} \begin{split}     x \end{split} \end{equation} \end{document}  Here is the LaTeX error upon running the MWE. math.tex|11 error| LaTeX Error: \begin{split} on input line 9 ended by \end{env}. math.tex|12 error| LaTeX Error: \begin{equation} on input line 9 ended by \end{document}. math.tex|12 error| Missing $ inserted. math.tex|12 error| Missing } inserted. math.tex|12 error| Missing } inserted. math.tex|12 error| Missing } inserted. math.tex|12 error| Missing \cr inserted. math.tex|12 error| Missing { inserted. math.tex|12 error| Missing { inserted. math.tex|12 error| Extra }, or forgotten $. math.tex|12 error| Missing $ inserted. math.tex|12 error| Missing } inserted. math.tex|12 error| You can't use `\end' in internal vertical mode. math.tex|12 error| LaTeX Error: \begin{equation} on input line 9 ended by \end{document}. math.tex|12 error| Missing } inserted. math.tex|12 error| Missing $ inserted. math.tex|12 error| Display math should end with $$. math.tex|| Emergency stop. math.tex|| ==&gt; Fatal error occurred, no output PDF file produced! Why is 8 so special? I have been reading about multi-dimensional numbers, and found out that it's been proven that the Octonions are the composition algebra of the largest dimension. I was wondering why, despite having infinitely many different dimensions of numbers, the only composition algebras are of 1, 2, 4, and 8 dimensions. What's so special about 8? Given fields $K\subseteq L$, why does $f,g$ relatively prime in $K[x]$ imply relatively prime in $L[x]$? $K,L$ are fields, $K\subseteq L$. $f,g \in K[x]$. Suppose that $f,g$ are relatively prime as elements of $K[x]$. Prove they remain relatively prime  in $L[x]$.  I've tried everything I can think of. I feel like working with the contrapositive may be helpful but that's just a feeling. Add ability to flag answers as duplicates Since I upgraded to the latest version of Ubuntu, I cannot install the Java OpenJDK package needed to compile Android custom ROMs. When I run sudo apt-get install openjdk-7-jdk, the following error appears:  Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  openjdk-7-jdk :  Depends: openjdk-7-jre (= 7u51-2.4.6-1ubuntu4) E: Unable to correct  problems, you have held broken packages.   How do I fix this problem? $M$ finitely generated if submodule and quotient are finitely generated. I know this must be easy. I'm new to modules, so perhaps I must be missing  something important.     Let $M$ be an $R$-module. Show that if there exists a submodule $N$ such that $N$ and $M/N$ are finitely generated, then $M$ is finitely generated.   If the exact sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0 $ (where the second and third arrows are inclusion and projection respectively) splits then $M$ would be isomorphic to the direct sum of $M$ and $M/N$ and the conclusion would follow (I think). But I can't show that it splits.  Thank you. Rich Wizards, Poor Muggles How can I prove: if $p $ is prime and $n>1$, then $ p^{\frac1n} $ is irrational? How can I decide between the medium sized intuos tablet or the small one? I am a high school student with an interest for drawing tablets for some hobby. I went to check on a store that sell Intuos, a sales representative (or something) of Wacom told me that most students buy the medium sized tablets. Even the staff said that since it is always bought the supplies of the medium size are a lot more than the small size.  I am on a tight budget (which is why I am buying the Intuos), and the small size seems small to me (when my hands are open, my thumb and my pinky would reach the sides). But then, why would Wacom sell the Manga variant in small? Is the staff just saying that so I would buy the more expensive one?  I have enough savings from my small allowance to buy the small size right now, should I wait for another month so that I can buy the medium size? How to disable javascript modal popup ads For many years I have blocked popup ads with AdBlock Plus.  But it seems more and more these past couple months there is a "new" kind of popup overtaking the web.  They are javascript based "modal" popups which ABP is powerless over.  See attached for an example.  Is there a way to kill these ads without breaking the functionality of anything I need? Inductive proof that there are infinitely many pairs of consecutive powerful integers?