Square root property of congruences
How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$?
Intutitive meaning behind the formal definition of sufficient statistic? According to the definition of sufficiency, a statistic is sufficient for a parameter if the conditional distribution of $X$ given a value of statistic does not depend upon the parameter.   What I am trying to understand is how does conditional distribution of $X$ not being a function of the parameter fit into the intuitive meaning of sufficiency i.e statistic value holding the same amount of information as that of the respective sample. There appears to be a poltergeist in my speakers (no, really) I've no idea if there's a suitable Stack Exchange for poltergeists. It appears to be software, hardware or virus related, so I thought I'd try here, seeing as Ghostbusters is just a film, so I can't call them.  I'm not going mad, and this isn't a joke.  There's what appears to be a radio stream coming out of my speakers, except I can only hear it when the volume knob is turned down to the min setting whilst the speakers are on, which makes no sense. Sometimes it's French, which also makes little to no sense.  The speakers are Logitech 5.1s, I forget the model number, and I'm running XP 64bit. If that makes a difference.  I'd suspect it could be some sort of crazy virus, but I've been randomly killing processes whilst the poltergeist is active and it hasn't killed it, so I'm not sure about that.  Once again, I'm perfectly sane and this really isn't a joke. Perhaps other people have had this problem but have been too afraid of the reaction to ask anyone (please don't have me committed to an insane asylum).  Googling for strange noises coming from speakers didn't really get me anywhere. Has anyone heard of a virus that can do this and affect the volume controls on the hardware in that way? I'm trying to "add" an 3D object onto a photo. I have added the photo to the scene in three ways so far, but none of them render (eevee):   Camera > Background Images Empty a. Add > Empty > Image (Empty gets added to current Collection) b. Add > Image > Backround (Empty gets added to Scene Collection)   These all show up in the 3D View and I was able to align the camera and the object so its perspective matches the scene in the photograph. But when I Render the scene, I only see the object, not the background image.  For the latter, it's simple: Empties are not rendered at all. There are few questions about this, since 2.8 has changed how it handles background images. The most popular seems this one (it includes references to others):      But I seem not to be able to find a method to have the background image render.  Another method might be to use the Compositor (Nodes) to Mix or Alpha Over an Image with the Render Layers. But this does not give the same alignment as achieved with the background image (also this requires that the World background does not get rendered (Alpha channel); for Eevee, this can be achieved by checking Film>Transparent).  What other options are there?  Docs: A result of Stieltjes Integral While studying the Riemann - Stieltjes integral, I read this exercise in my book:  "Let $f:[a, b]\to \mathbb{R}$ be a bounded function. Denoted $L(P, f)$ the lower Darboux sum and $|P|$ the mesh of the partition $P$. Prove that for every $\epsilon&gt;0$, there exists $\delta&gt;0$ such that for every partition $P$ with $|P|&lt;\delta$, we have: $\underline{\int_{a}^{b}}f(x)dx-L(P, f)&lt;\epsilon$."  My question is: Can we generalize this problem for Stieltjes integral, with $\underline{\int_{a}^{b}}f(x)d\alpha$ instead of $\underline{\int_{a}^{b}}f(x)dx$?  Particularly, can we have this theorem?  Let $f:[a, b]\to \mathbb{R}$ be a bounded function and $\alpha$ is a monotonic function on $[a, b]$ . Denoted $L(P, f, \alpha)$ the lower Darboux sum and $|P|$ the mesh of the partition $P$. Prove that for every $\epsilon&gt;0$, there exists $\delta&gt;0$ such that $\underline{\int_{a}^{b}}f(x)d\alpha-L(P, f, \alpha)&lt;\epsilon$ for every partition $P$ with $|P|&lt;\delta$?  I don't know how the mesh of the partition relate to the function $\alpha$. Other than cost, is there a reason to not use higher rated components than called for by design? Display solution/file path in the Visual Studio IDE I frequently work with multiple instances of Visual Studio, often working on different branches of the same solution.   used to display the full path of the current source file in its title bar, but Visual&nbsp;Studio 2005 doesn't appear to do this. This makes it slightly more awkward than it should be to work out which branch of the solution I'm currently looking at (the quickest way I know of is to hover over a tab so you get the source file's path as a tooltip).  Is there a way to get the full solution or file path into the title bar, or at least somewhere that's always visible, so I can quickly tell which branch is loaded into each instance? How many passes do you need to wipe/shred your files to make them not undeletable? Why doesn't the Doctor go back to Gallifrey before the Time War? According to , it’s just the Time War that is time locked. So why does Eccelston's Doctor tell Rose:     My whole planet is gone. My people, my family, all gone. Don't you think I want to go back and save them?   Why hasn't he ever gone back to before the war?   All the other Doctors basically see Gallifrey as gone from them, and when you say the Doctor cannot go back on his own timeline, there will have been plenty of time periods he could go to; maybe to a time before they could regenerate, or maybe when Time Lords were using Tardises — as in "The Doctor's Wife" she says to Matt Smith:     I was already a museum piece when I stole you Appendix in LaTeX 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. I am planning to visit the UK on a USA travel document but there’s nowhere I am able to find information about visas: I’m a Zimbabwean but using a USA refugee travel document do I need to apply for a UK visa before I travel or I can they just give them on arrival. How much willpower do gravely wounded soldiers lose? The text says they suffer a permanent reduction? MySQL Conditional Insert Strange black shading. Cannot fully light mesh. (Cycles) Transpose/Unzip Function (inverse of zip)? I have a list of 2-item tuples and I'd like to convert them to 2 lists where the first contains the first item in each tuple and the second list holds the second item.  For example:  original = [('a', 1), ('b', 2), ('c', 3), ('d', 4)] # and I want to become... result = (['a', 'b', 'c', 'd'], [1, 2, 3, 4])   Is there a builtin function that does that? Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous? No secondary button (right click) on touchpad I'm experiencing a serious problem with the new Ubuntu 18.04:  System Dell XPS 13 with a freshly installed Ubuntu 18.04 LTS.  Problem description The touchpad provides only primary button (single click and double click), but no secondary button. In particular, when using the "test Your Settings" option in Settings -> Devices -> Mouse &amp; Touchpad, any click on the touchpad is signaled as primary button, no matter where I click (left or right side of the touchpad). This remains the case even after I change the Primary Button from Left to Right (in Settings -> Devices -> Mouse &amp; Touchpad -> General).  In comparison, an external mouse connected to my laptop (via usb) works fine: it provides both primary and secondary buttons with sides corresponding to the choice of Primary Button (in Settings -> Devices -> Mouse &amp; Touchpad -> General).  This may be related to   Any ideas?  Update: Meanwhile, I found out that in addition to the solution proposed by EntropyWins, one can also control this touchpad behavior via the Tweaks tool (a.k.a. gnome-tweaks) under Keyboard &amp; Mouse -> Touchpad -> Mouse Click Emulation. In any case, as EntropyWins explains in his answer, this is not a bug. Wrong place for hyperref link in scaled tikz picture I have a TikZ picture with a node containing a minipage with an itemize environment; each item is a hyperref to other sections of the document.   When scaling the image (for example with \begin{tikzpicture}[scale=.7, transform shape] I've a strange behavior: links in itemize are not in the right place: they present a somewhat vertical offset, as if they don't scale as the rest of image.   Here the MWE:  \documentclass[a4paper]{article} \usepackage{tikz} \usepackage[pdftex]{hyperref} \begin{document}  Here something interesting\label{text:interesting-one} \newpage  Here something interesting, too\label{text:interesting-two} \newpage  and here there are very very interesting stuff\label{text:interesting-three} \newpage  \newpage Now the problem: links rectangles have some offset, making links unusable in case of long itemize  \begin{figure}[htb] \begin{tikzpicture}[scale=.7, transform shape] \node at (0,0) (mynode) { \begin{minipage}{4cm} \begin{itemize} \item \hyperref[text:interesting-one]{First link} \item \hyperref[text:interesting-two]{Second link} \item \hyperref[text:interesting-three]{Third link} \item \hyperref[text:interesting-one]{First link again} \item \hyperref[text:interesting-two]{Second link again} \item \hyperref[text:interesting-three]{Third link again} \end{itemize} \end{minipage} }; \end{tikzpicture} \end{figure} \end{document}   And here a snapshot of the problem    Note: This is evident using the default settings of hyperref package, which draws the red rectangle around links, but it took me some time to found it when only the text is coloured, i.e. when using \usepackage[linkcolor=blue,colorlinks=true]{hyperref} I have a doubt on something. I have a theorem that says that if $R$ is a ring, then $R/I$ is a domain iff $I$ is prime. Since $X^2-1$ is note prime, $\mathbb Z[X]/(X^2-1)$ is not a domain. But, $X^2-1=(X-1)(X+1)$, and since $X-1$ and $X+1$ are prime in $\mathbb Z[X]$, there are coprime, and thus, by the chinese remainder $$\mathbb Z[X]/(X^2-1)\cong \mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1),$$ since $\mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1)$ is a domain, then so is $\mathbb Z[X/(X^2-1)$. What is the mistakes here ? Before I met her I had a poor opinion of her.   As person had a poor opinion before a particular time in past (before He met her) so I think there should be had had.   There are some more example like this:   Before I went to the university, I worked/had worked as a carpenter for 5 years. Before I opened the door, I put/had put the key in the lock. Before I opened the door, I looked/had looked carefully into the barn for snipers.   In all these sentence, for me, it should be perfect tense.  I asked that question on another platform where I was told that if sequence of past event is obvious and sentences uses "before"and "after", we often use past tense. Which means we should use past tense in all these sentences. But I have another sentence "The train had left before I reached the station." If I follow above mentioned rule there should be "left" instead of "had left". But I am damn sure this sentence is correct with "had left".   Am I correct?  Thank you