When entering a Q&amp;A at the same time (i.e. "Answer your own question" is checked), if you click the preview box for the answer, a black box is displayed  instead of your answer.  I have tested on HTC One V, Android 4.0.3, SE app v0.1.31.  Screenshot
This is a bit of a more unusual bug to report:    The screenshot above is unedited - the black box is rendered by the Stack Exchange application.  Phone:  running Android 4.0.4 (ICS)
What recourse do I have if I believe a moderator has abused their privileges? One or more of the following things happened to me on a Stack Exchange site:   My post(s) were wrongly deleted or locked by a moderator A moderator posted a wrong comment on one of my posts I was suspended, but did not do anything wrong I asked a meta question, but the question was closed and/or deleted A moderator discredited me in public ...   I think a moderator has abused their power, but I feel powerless against them.  What recourse do I have now? What can I do when I feel that a moderator has abused their power, and has been repeatedly unfair in their dealings with me? How to install LINE (messaging app) on Ubuntu I would like to install  chat platform on my Ubuntu 12.04 LTS system. Please, could anyone provide me the steps to successfully install this software?. I've been trying to install this software using wine but I couldn't succeed it. How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? What is the maximum size of a web browser's cookie's key? What is the maximum size of a web browser's cookie's key?  I know the maximum size of a cookie is 4KB, but does the key have a limitation as well? Change the order of integration in:  $$\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x }{\sqrt{4a^2x^2-(x^2+y^2)^2}}dxdy$$ and hence evaluate it.    I changed the order of integration and got the limits $y=0$ to $a$ and $x=a-\sqrt{a^2-y^2}$ to $a+\sqrt{a^2-y^2}$. I then tried changing to polar coordinates, but it is only making the integrand more complex. Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$? Is there a function whose inverse is exactly the reciprocal of the function? That is $f^{-1} = \frac{1}{f}$.  We know that the inverse of a function is not necessarily equal to its reciprocal in general. Can I convert a C# string value to an escaped string literal? In C#, can I convert a string value to a string literal, the way I would see it in code? I would like to replace tabs, newlines, etc. with their escape sequences. If this code: Console.WriteLine(someString);  produces: Hello World!  I want this code: Console.WriteLine(ToLiteral(someString));  to produce: \tHello\r\n\tWorld!\r\n There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal".  The proof I am thinking of is the following.  Suppose the Lie algebra $L$ satisfies $L\neq[L,L]$. Then $\pi\colon L\to L/[L,L]$ is a projection onto an abelian algebra. Choose a codimension 1 subspace $I\subset L/[L,L]$. Then $I$ is an ideal and its preimage $\pi^{-1}(I)$ is necessarily an ideal in $L$, because any subspace containing the derived algebra is an ideal. Moreover, $\pi^{-1}(I)$ has codimension 1. $\square$  The only condition on $L$ is that $L\neq[L,L]$, which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"? I have just started learning Japanese Mahjong and I am a bit lost as to why this hand failed.  I did several pons and chis over the course of a hand, having a total of three open melds (one set and two sequences).  Apart from these I had an incomplete bamboo sequence of II and III as well as a pair of dragons, therefore I was on a single wait for I or IV bamboo stick tile. When the IV tile appeared in one of the discards of other players, I was given a choice to do another chi and declare another open meld, but I couldn't win with a ron. Why was that?  Am I wrong that winning by ron guarantees a single needed yaku to win a hand, just like Closed Tsumo does?  After declaring the chi I was offered with, my hand was left with a single pair of tiles, one of which I had to discard.  Eventually the round finished with a draw.  Why couldn't I win this hand with a ron on the IV bamboo tile, and why was I given only the choice of chi instead? What permissions should my website files/folders have on a Linux webserver? This is a  about File Permissions on a Linux web server.   I have a Linux web server running Apache2 that hosts several websites. Each website has its own folder in /var/www/.  /var/www/contoso.com/ /var/www/contoso.net/ /var/www/fabrikam.com/   The base directory /var/www/ is owned by root:root. Apache is running as www-data:www-data. The Fabrikam website is maintained by two developers, Alice and Bob. Both Contoso websites are maintained by one developer, Eve. All websites allow users to upload images. If a website is compromised, the impact should be as limited as possible.  I want to know the best way to set up permissions so that Apache can serve the content, the website is secure from attacks, and the developers can still make changes. One of the websites is structured like this:  /var/www/fabrikam.com     /cache     /modules     /styles     /uploads     /index.php   How should the permissions be set on these directories and files? I read somewhere that you should never use 777 permissions on a website, but I don't understand what problems that could cause. During busy periods, the website automatically caches some pages and stores the results in the cache folder. All of the content submitted by website visitors is saved to the uploads folder. Is there a difference in disenchanting strong items vs. weak ones? Possible Duplicate:           states that the strength of the enchantment doesn't matter when disenchanting items to learn new enchantments.  Am I reading that right? If I have novice robes with 50% faster regen and disenchant it and my enchantment skill is at 5/5, when I enchant that item will it be the same result as disenchanting master robes with 150% faster regen? Because if you do the math, 5/5 enchanting will give you 100% better enchantment.   Therefore wouldn't it be better to disenchant more powerful items than weaker ones? Let $A \times \emptyset = \{(x,y)| x\in A, y \in \emptyset \}$. We know there is no element in $\emptyset$. But how does it follow that $A \times \emptyset = \emptyset $? What's the simplest SQL statement that will return the duplicate values for a given column and the count of their occurrences in an Oracle database table?  For example: I have a JOBS table with the column JOB_NUMBER.  How can I find out if I have any duplicate JOB_NUMBERs, and how many times they're duplicated? A limit ordinal $\gamma&gt;0$ is said to be indecomposable iff $\nexists\alpha,\beta&lt;\gamma$ such that $\alpha+\beta=\gamma$.  In view of this definition, I’m trying to prove the equivalence of the following statements for limit ordinals $\gamma$:  (1) $\gamma$ is indecomposable.  (2) $\alpha+\gamma=\gamma$ $\forall\alpha&lt;\gamma$.  (3) $\gamma=\omega^\alpha$ for some $\alpha\in\mathrm{ON}$.  I’ve been able to get $(1)\implies(2)$ and $(3)\implies(1)$, but I have nothing more than scratch work for $(2)\implies(3)$.  Any suggestions would be appreciated. I am trying to create a vase that looks similar to this:    If there is another post out there I will happily remove this one or redirect it there.  Specifically, I am trying to create the creased effect in the geometry that looks like it is twisting around the vase.  I have tried using Checker Deselect and extrude->scaling the verts but that really screws up my mesh.  Link to my model if it helps: A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here? Canonical map $R/(I\cap J)\rightarrow R/I\times _{R/(I+J)} R/J$ is an isomorphism Delete all data for a kind in Google App Engine I would like to wipe out all data for a specific kind in Google App Engine. What is the best way to do this? I wrote a delete script (hack), but since there is so much data is timeout's out after a few hundred records. Change pivot or local origin of an object How can I change the local origin of an object, without changing the object location in the world space.  This is the object located at position (0,0,0) in the world space    This is the same object located at position (0,1,0) in the world space, but with its local origin offset so that the object is in the same place    Is there a way to do this in a single step? A scene I made is not rendering, at all. This grey screen stays the same and nothing changes.  Here are my settings:  [