I have seen a few questions which have been marked as duplicate although they the questions have been asked before the question they are marked duplicate of.  For eg,   (I will call this as Q1) has been marked duplicate of  (I will call it Q2). But as you can see that Q1 is asked a year before Q2, then how can it be flagged as a duplicate?  Another example -    and  and the Q1 has been asked 2 years before Q2.  Isn't it necessary that the time of asking should be considered before marking a question as duplicate.  But also what I can see is that the newer question has better answers than the older ones. But this is unfair to the asker.  How should StackOverflow deal to such problems?   One solution of which I could think of is merging questions. I know that moderators have merging question permission but it's hardly used and according to me, it should be done on every question marked as duplicate.  So instead a marking a question as duplicate and closing it, merge the questions together. Older reputations earned by the question should be retained by the user while newer ones earned after merging the question should be retained by the person who asked the question earlier. This might probably work on all the questions marked as duplicate, so when a user clicks on link of question which is marked as duplicate, he should be redirected to the original question (message of redirection could be shown). Answers should be merged too.  This will prevent users from earning free reputation by rephrasing a question.
I have seen this many times, a question is being asked because a previous, duplicate question was asked and answered a few years ago and has no longer relevant answers.  The  is to edit the old question, or add new answers to it, but it is not always happening.  I was wondering, should I still mark it as duplicate and vote to close it? Or just leave it as is?
Is there a cross-browser onload event when clicking the back button? Reflexive and relation Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes. Context: I was looking at . I didn't know the result referred to in Brian M. Scott's answer is hard enough to require a separate proof   ― .  ― .  (1)  ― What's the intuition? Is Bezout Identity supposed to be intuitive?  (2) — Why doesn't this try work? Let $c$ be any common divisor of $a, b$. Therefore $c\mid a, c\mid b \iff cj_1 = a, cj_2 = b$ for some integers $j_1, j_2$.     Call $g = \gcd$. By definition $\gcd(a,b)\mid a$ and $\gcd(a,b)\mid b$ and $c \le \gcd(a,b)$. Therefore $g\mid a, g\mid b \iff gk_1 = a, gk_2 = b$ for some integers $k_1, k_2$.   The two paragraphs together precipitate $cj_1 = a = gk_1,  cj_2 = b = gk_2$ but this only proves $c\mid gk_1, c\mid gk_2$?    Origin - Elementary Number Theory, Jones,  p9, Exercise 1.8 Countable axiom of choice: why you can't prove it from just ZF This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ Problem:      Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y  $ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)  $ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)  = g(x) $ for all $x \in  X $.   Attempt at a solution:  Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) &gt; 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y $ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.   I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A $, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.  Also, Would be be possible to prove this without using the Hausdorff condition on $Y$? This is a  about  the use of a *AMPP's stack.    I recently had a talk with some experienced people and they suggested to me not to use a WAMP stack, and instead install apache, mysql and php separately.  I don't understand why they have suggested this, though, so can anyone tell me?  Is there a particular disadvantage of WAMP, or a particular advantage to installing all of them separately?  Since a WAMP stack itself is composed of apache, mysql and php, then what's the difference between using the WAMP stack and installing them all separately? Velocity Maxwell-Boltzmann distribution for dummies I have a volume with $N$ molecules; I need to assign to each particle a velocity vector: $$|\mathbf{v}_{i}|=[v_{x}, v_{y}, v_{z}]^{T}$$ for the $i$-th molecule; the velocities must follow the Maxwell-Boltzmann distribution; here is the question: I know the definition of the M.B. distribution, how do I extract the velocities that I have to assign to each molecule? Basically what I want to know is how to calculate $$v_{x}=?$$ $$v_{y}=?$$ $$v_{z}=?$$ for each molecule. Several lines of formulas to the left of one curly brace pointing to equation number How to make several lines of equations under one curly brace from the left side, such that it point to a reference number.  Now I do something like this:  \begin{align} \left. \begin{array}{ll} a = b + c \\ &amp; \dots  \\ a = b + c \\ &amp; \dots  \\ a = b + c  \end{array} \right\} \label{formulas}\\ \end{align}   But equation number appears in the bottom of curly brace. How do CSS triangles work? I'm having problems with a WFS GetFeature request. I'm using OpenLayers with an OSM map as base layers which use the Google projection. Overlaying the layer with WMS works perfect. When I do a request to return all features within the bounds of a polygon, no features are found.  My layer uses EPSG:28992. the base layer uses: EPSG:900913  My layer configuration in GeoServer:    When I set the declared SRS to EPSG:900913 and set "SRS handling" to "Reproject native to declared" it works fine, but I need OpenLayers to reproject it, not GeoServer.  Below is an example request I would do. I think the problem is that GeoServer does not convert the spatial filter to the native SRS.  http://mydomain.com/wfs?request=GetFeature&amp;version=1.1.0&amp; srsName=epsg:28992&amp; typeName=Namespace:Layer1&amp; propertyName=(*)&amp; CQL_FILTER=year = 2011 and INTERSECTS(geometry,POLYGON((565585.61613069 6816272.786200799,565585.61613069 6817032.3791692,564362.62367835 6817032.3791692,564362.62367835 6816272.786200799,565585.61613069 6816272.786200799)))&amp; outputFormat=GMl2&amp;   Update  I just tried re-projecting some points from OpenLayers (EPSG:900913) via proj4js to EPSG:28992 (PostgreSQL and GeoServer). I took 3 points and  compared them. They all showed an offset.   Converted: POINT(233492,24349424947 582191,0681508556)  In Database: POINT(233530,852147064 582315,680920578)  Difference x: 38,60865281453 Difference y: 124,6127697224 Converted: POINT(133459,1980658749 455770,7481867363)  In Database: POINT(133486,356985081 455879,67223342)  Difference x: 27,1589192061 Difference y: 108,9240466837 Converted: POINT(176920,43262818802 317930,36182729155)  In Database: POINT(176954,154083767 318023,515324756)  Difference x: 33,72145557898 Difference y: 93,15349746445   If the offset was constant I could correct it. Now I'm not sure if the offset is caused by GeSserver or by proj4js. I'm guessing proj4js since the WMS layer does not show any offset. How do I handle the window close event in Tkinter? How do I handle the window close event (user clicking the 'X' button) in a Python Tkinter program? Why do Tcler suggest to brace your `expr`essions? I can't destroy blocks in Minecraft Creative Mode (singleplayer) Interference between strokeopacity and fillopacity I have a few circles in a drawing, and want to highlight some of them.  \begin{tikzpicture}  \coordinate (A1) at (0,0); \coordinate (A2) at (1,0); \coordinate (A3) at (1,1); \coordinate (B)  at (1,0);  \draw (A1) -- (A2); \draw (A1) -- (A3); \draw (B)  -- (A3);  \foreach \point in {A1,A2,A3,B} {    \fill [black,draw=white](\point) circle (3pt); }  \pgfsetstrokeopacity{.3} \pgfsetfillopacity{.3} \fill [red,draw,rounded corners,line width=9pt] (A1) -- (A2) -- (A3) -- cycle; \pgfsetstrokeopacity{1} \pgfsetfillopacity{1} \end{tikzpicture}   I would like the whole area surrounding the A points to be of the same color, but the strokeopaticy and the fillopacity interfere.  Is there some ay I can prevent this?  Note: in my actuall use scenario, there are too many coordinates for manual adjusting to be an option (otherwise I would just do with a fill with rounded corners, with points calculated manually). I'm talking about this nonsense:   Now, I'm aware that there are problems with the practicalities (or possibilities) of methods involving exotic matter with negative mass, and that kind of thing has been addressed here:   But in this case my question is different and more specific. Sweeping all the other problems of constructing such a beast under the rug, wouldn't this violate causality?  SR says that the simultaneity and relative ordering of events is dependent on the relative motion of the observer, but that no matter how it works out, if event A causes event B, no observer will see B first. One of the many problems with FTL travel (or signals) is that if any information is sent faster than light, there will exist an inertial frame in which the signal arrives before it is sent, a blatant violation of causality. (I think we can all agree, can we not, that the threshold should be extremely high before breaking assumptions that we're in a causal universe?)  So how is any scheme to travel faster than light, even if it may locally satisfy the equations of GR, not a blatant violation causality according to SR?  Is my understanding of GR incomplete and there's something in it that obviates the causality implications of SR? Or am I correct that FTL claims should be dismissed with extreme prejudice on this basis alone? (Pending a serious replicable experiment showing causality to be breakable.) Easy way to write contents of a Java InputStream to an OutputStream I am interested in publishing a GML file as a WFS. I don't want to use a database like PostgreSQL. What would be the rough steps I need to do for that? I like to use open source software (geoserver). I don't have much experience. I am just looking for a rough guide what kind of software I should look into and which step follows which step.. Deriving Chinese Remainder Theorem from gcd Bezout identity We want to find the solution $x$ to the congruence system  $$\begin{align} x &amp;\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &amp;\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &amp;= r_1 + m_1 j\\ x &amp;= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$? I accidentally removed all the packages starting with the name jpeg with:  sudo apt-get remove jpeg*   This screwed up a lot of things, removed Unity, Firefox, Flash player etc.  I somehow managed to recover the desktop and unity but then a lot of things are still buggy. For example .rpm files and .tag.gz just won't open up at all.  What I want is a way to fix everything that's amiss. Something that is the equivalent of a 'repair installation' with a windows disc.  Any tips, anyone? How to pronounce fractions with denominators larger than 20 where the last digit of the denominator is 1 or 2? eg 4/31 Disclaimer: I speak British English.  I've noticed a lot of differences between the way Americans and Brits pronounce numbers.1  Since the question concerns this, I thought it might be appropriate to draw attention to it case we inadvertently confuse each other.  My question is not about these differences, I just wish to highlight them in case they cause confusion.  1. You seem to happily call a quarter "a fourth" sometimes, we always call 131 "one hundred and thirty one", we pronounce double digits in phone numbers like 12449 as "one two double four nine", and I think we are much more likely to use expressions like "thirteen hundred" to mean 1300.    Question  When talking about fractions, I have frequently heard  1/2    a half 1/3    a third 1/4    a quarter 1/8    an eighth 1/64   a sixty fourth 1/56   a fifty sixth   etc.  Essentially the rule seems to be that, except for "a whole", "a half", and "a quarter", the word matches the ordinal number; that is to say:      Numeral       Ordinal         Fractional     one           first           whole     two           second          half     three         third           third     four          fourth          quarter     five          fifth           fifth     six           sixth           sixth     fifty-seven   fifty-seventh   fifty-seventh  Even though 1/4 is a quarter, 1/64 is a sixty-fourth.  So what’s 1/62?  A  sixty-twoth?  A sixty-second? Surely not a sixty-half!  I know that simply saying one over sixty-two can usually work, but I'm asking specifically for the word itself, i.e.  if I divide a huge pizza into 21 pieces, what are the pieces?  Other than baker’s twentieths.  Summary: Can anyone point me to any sources (whether style guides or common usage studies or anything else) that discuss the pronunciation of fractions; specifically one that discusses this separately from ordinal numbers, rather than how to form ordinal numbers in the first place.