When I visit the C:\$RECYCLE.BIN on Windows 10, I can see not only a link to the Reycle Bin but also two other directories, which I cannot access. Upon taking ownership of one of the directories and giving myself full permissions, it no longer appeared as "S-1-5-18", but as another "Recycle Bin". It only reverted back to "S-1-5-18" once I changed all of the settings back and toggled the "read-only" attribute.  What are these additional directories, and why do they display such unusual behaviour?
What does the Recycle Bin in Windows actually do?  Is it just a glorified folder and a holding place for soon to be deleted files, or does it do something specific?  Specifically, are files that are "moved" to the Recycle Bin actually moved on the hard disc, or are just the pointers to the files moved?  I am a fairly experienced user, I just wanted a more depth explanation of the Recycle Bin.
On SO, when I'm scrolling through "questions" some have a yellow background.  Is this intuitive to everyone but me?  What is the yellow background trying to tell me about this particular question? Inverse of a Positive Definite “Are” vs. “is” after “parents and the family” Suppose that I would like to record some video sessions of Angry Birds on my phone.   Are there any apps that I can use? I have root permissions. Automatically generate thumbnails for all subfolders of a directory in Windows 7? Proving by induction that $1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$ Need guidance on this proof by mathematical induction. I am new to this type of math and don't know how exactly to get started.   $$ 1^3  +  2^3  +  3^3  +  \ldots  +  n^3  =  \left[\frac{n(n+1)}{2}\right]^2 $$ implicit operator using interfaces I have a generic class that I'm trying to implement implicit type casting for.  While it mostly works, it won't work for interface casting.  Upon further investigation, I found that there is a compiler error: "User-defined conversion from interface" that applies.  While I understand that this should be enforced in some cases, what I'm trying to do does seem like a legitimate case.  Here's an example:  public class Foo&lt;T&gt; where T : IBar {     private readonly T instance;      public Foo(T instance)     {         this.instance = instance;     }     public T Instance     {         get { return instance; }     }     public static implicit operator Foo&lt;T&gt;(T instance)     {         return new Foo&lt;T&gt;(instance);     } }   Code to use it:  var concreteReferenceToBar = new ConcreteBar(); IBar intefaceReferenceToBar = concreteReferenceToBar; Foo&lt;ConcreteBar&gt; concreteFooFromConcreteBar = concreteReferenceToBar; Foo&lt;IBar&gt; fooFromConcreteBar = concreteReferenceToBar; Foo&lt;IBar&gt; fooFromInterfaceBar = intefaceReferenceToBar; // doesn't work   Does anyone know a workaround, or can anyone explain in a satisfactory way why I shuouldn't be able to cast interfaceReferenceToBar implicitly to Foo&lt;IBar&gt;, since in my case it is not being converted, but only contained within Foo?  EDIT: It looks like covariance might offer salvation.  Let's hope the C# 4.0 specification allows for implicit casting of interface types using covariance. How to get the location of a program in Ubuntu? For example I have Oracle, how to get the folder racine (location) of Oracle? Auto Package download for TeXLive I use MiKTeX on Windows and quite satisfied with it. Recently I started switching all my tasks toward open-source alternatives, and in the course I would love to use Linux. In Linux TeXLive is available as alternative to MiKTeX. The thing I really like about MiKTeX, is its ability to install packages automatically. Can I do the same in TeXLive too? Is there a way I can enable such or install some plugins for it?  I am using Fedora 18, if that's needed. Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.  I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand. Is it bad practice to return from within a try catch finally block? So I came across some code this morning that looked like this:  try {     x = SomeThingDangerous();     return x; } catch (Exception ex) {     throw new DangerousException(ex); } finally {     CleanUpDangerousStuff(); }   Now this code compiles fine and works as it should, but it just doesn't feel right to return from within a try block, especially if there's an associated finally.  My main issue is what happens if the finally throws an exception of it's own? You've got a returned variable but also an exception to deal with... so I'm interested to know what others think about returning from within a try block? How do I change the default file manager back to Nautilus? Given two input layers  The clip layer - an arbitrary layer (though in my case it's a float 1 band geotiff) The mask layer - a paletted raster in which every pixel is either NO DATA or a single &quot;color&quot; value, same for every pixel.  I want to create an output raster whose overall extent matches the extent of the mask layer and each pixel is either NO DATA if the corresponding pixel of the mask layer is NO DATA or the value of the corresponding pixel from the clip layer if the corresponding pixel of the mask layer is not NO DATA. I know I can mask the clip layer with a polygon, but the mask layer is a raster. Polygonizing the raster layer is possible in principle, but would result in a polygon with a lot of very short edges. Is there a direct way to do this in QGIS? I suppose there's probably a path that involves creating a new layer and using the raster calculator, but I'm hoping there's something more direct. Further complicating the matter is that the two rasters might not have the same resolution and origin. I believe this is equivalent to  in ArcMap. Image uploads not working on Chrome When I try to upload an image, I'm presented with the following dialog box,    The dialog box does not allow uploading either from my computer or by posting a link.  This happens only on Chrome browser (running on Windows). Clearing cache and cookies, and restarting computer do not resolve.  Is this just my computer? Or is this a site-wide bug?  Additional info:  Clicking on 'click' to upload does nothing. Dragging an image onto dialog redirects me to the file location on my computer. Clicking 'Add picture' redirects me to http://[SE-domain].com/upload/image?https=true.  UPDATE: A Norton extension for Chrome was the source of the problem. Why does an argument similiar to 0.999...=1 show 999...=-1? I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\ldots = 2.5$ etc. Can anyone point me to resources that would explain what the below argument that shows $999\ldots = -1$ is about? Here is the most usual proof I see that $0.999\ldots = 1$:  $x=0.999\ldots$ $10x=9.999\ldots$ $10x - x = 9$ $x=1$  Using this same argument template I can show $999\ldots=-1$:  $x= \ldots9999.0 $ $0.1x= \ldots9999.9$ $0.1x - x = 0.9$ $x=-1$  What might this mean? Edit from one of the comments: $$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$ Divisibility and the Fibonacci sequence Which attacks are possible against raw/textbook RSA? Uncountable ordinals without power set axiom Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set? I'm currently reading Marcus' "Number Fields," and I'm having difficulty proving the following result:  Corollary 2.2: Let $m$ be a squarefree integer. The set of algebraic integers in the quadratic field $\mathbb{Q}(\sqrt{m})$ is $$ \left\{ a + b \sqrt{m} : a,b \in \mathbb{Z} \right\} $$ if $m \equiv 2,3 \pmod{4}$; and$$ \left\{ \frac{a+b\sqrt{m}}{2} : a,b \in \mathbb{Z} ~,~ a \equiv b \pmod{2} \right\}$$ when $m \equiv 1 \pmod{4}$. $\square$  It is easy enough to see that the claimed sets do, indeed, consist of algebraic integers. The other half of the proof proceeds something like this: suppose that $\alpha = r + s \sqrt{m}$ is an element of $\mathbb{Q}(\sqrt{m})$. Then $\alpha$ is a root of the polynomial $$ x^2 - 2rx + r^2 - s^2m $$ In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers. At this point, the proof stops, and says that this implies the result. And therein lies the problem: I don't see how to make the connection. I do know, for example, that if one of $r,s$ is an integer, then the other one must be, too, i.e., either both are integers, or both are not. However, I cannot get any further than that. Can anyone give me a gentle nudge in the right direction? Thank you very much! Topologist's Sine Curve: A rigorous proof of why $\overline{S} = S \cup V$ Let $S$ and $V$ be the following subsets of $\mathbb{R}^2$.  $$ S = \left\{ \ x \times \sin \frac{1}{x} \ \colon  \ 0 &lt; x \leq 1 \ \right\}, \qquad V = \left\{ \ 0 \times y \ \colon \ -1 \leq y \leq 1 \ \right\} = \{ 0 \} \times [-1, 1]. $$ Then how to show rigorously that the closure $\overline{S}$ of $S$ is given by  $$ \overline{S} = S \cup V? $$  By rigorously, I mean using the smme kind of argument as is used in real analysis or topology, such as in (most of) the proofs in Rudin or Munkres.   Intuitively, it is clear that as $x \to 0+0$, then $\sin \frac{1}{x}$ will fluctuate its full range of $[-1, 1]$, and for any point $0 \times y$, where $y \in [-1, 1]$, there will be points of $S$ arbitrarily close to it.   But I'm unable to rigorously show this.