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How long typically are paper reviews? Is there such a thing as “too long”? When you review papers submitted for publication, is there an “optimal” length for reviews? In my experience as an author and referee, I have seen a large range of review lengths (for reference, a paper in my field is typically between 3 and 8 printed pages):   Zero length: for some of the papers I have authored, the reviewer just clicked the “accept as is” or “reject” checkmark on the review sheet, without adding any comment (at least, not any comment visible to me). It's not so common, but it has happened. Short length: a lot of time, I received reviews who consisted of a single paragraph. Medium: one full page, maybe two pages.   Although I have never received any such long reviews, I have myself written on a few occasions reviews that exceed two pages, including once or twice a four or five-page review.  These were cases where the paper was good, but could be (in my view) much improved and some of the aspects/consequences had escaped the authors' consideration.  I am wondering how useful reviews of various length are to editors. I often consider that “unmotivated” reviews are useless, as they do not give any real insight about the paper to the editor. For example, if the editor gets two conflicting non motivated reviews, how is she to decide?  On the other hand, although I write some from time to time, I have never received long reviews, so… is this something frowned upon? How to avoid pagenumering with text in scrlttr2 Have a Letter in scrlttr2. I cannot avoid/exclude the text before the page number 'Seite' (see print-screen). The  recommends to add \pagenumbering{arabic}. But for my document, it has no effect.    %--------------------------------------------------------------------------- \documentclass%% %---------------------------------------------------------------------------   [fontsize=12pt,%%          Schriftgroesse %--------------------------------------------------------------------------- % Satzspiegel    paper=a4,%%               Papierformat    enlargefirstpage=on,%%    Erste Seite anders    pagenumber=footcenter,%%   Seitenzahl oben mittig %--------------------------------------------------------------------------- % Layout    headsepline=on,%%         Linie unter der Seitenzahl    parskip=half,%%           Abstand zwischen Absaetzen %--------------------------------------------------------------------------- % Formatierung    draft=on%%                Entwurfsmodus ]{scrlttr2} %--------------------------------------------------------------------------- \usepackage{german} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc}  %--------------------------------------------------------------------------- \begin{document} %--------------------------------------------------------------------------- % Briefstil und Position des Briefkopfs \LoadLetterOption{DIN} %% oder: DINmtext, SN, SNleft, KOMAold. \makeatletter \pagenumbering{arabic}  %--------------------------------------------------------------------------- % Seitenstil \pagestyle{plain}%% keine Header in der Kopfzeile %--------------------------------------------------------------------------- \begin{letter}{Herr\\                 Mustermann\\                Straße 1\\                12345 Stadt} %--------------------------------------------------------------------------- \setkomavar{subject}{Beispielschreiben} %--------------------------------------------------------------------------- \opening{Liebe Herr Mustermann,}  here is some Text.  \pagebreak  here is some Text. %--------------------------------------------------------------------------- \end{letter} %--------------------------------------------------------------------------- \end{document} %--------------------------------------------------------------------------- Proof of an elementary property of Projection Operators I'm asked to show the following:  Let $X$ be a linear space, and let $P : X \rightarrow X$ be a projection operator. Restricted to the linear space $range(P)$, the projection $P$ is the identity operator, that is, $Px = x$ for all $x \in range(P)$.  If anyone could offer a hint for how to show this. I'm rusty with linear algebra so don't really know what I'm supposed to bring in outside of what's specified in the lemma/definition of projection. What is so special about the range of P that would make this so? What is the difference between public, protected, package-private and private in Java? In Java, are there clear rules on when to use each of access modifiers, namely the default (package private), public, protected and private, while making class and interface and dealing with inheritance? How can I show that "A center of a free group that is non-cyclic is trivial" ? MySQL cartesian product between two SELECT statements Error : Warning: strpos() [function.strpos]: Offset not contained in string - can not find solution "Results found containing ___" too close to header Predicate Calculus Practice I've been doing some practice questions in the textbook for an upcoming predicate calculus lecture and I think I've managed to get A and B (possibly C) correct but I am clueless on how I can manage D and E.   For A I got: ∃x(R(x)∧¬B)  For B I got: ∀x(¬R(x)∧B)  For C I got: ∃x(R(x)∧(x=y))   - Not sure if this is even remotely correct    R(X) : ”X is rich”  X = Y : ”X equals Y”  B : ”Bill Gates”  a) Someone besides Bill Gates is rich.  b) Bill Gates alone is rich.  c) At least two people are rich.  d) Exactly one person is rich.  e) Exactly two persons are rich.  The universe of discourse is the set of all people. What would I do if I want to have a generic method that only accepts types that have overloaded an operator, for instance the subtraction operator. I tried using an interface as a constraint but interfaces can't have operator overloading.   What is the best way to achieve this? What is the difference between old style and new style classes in Python? The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? If every pair of congruence equations admits solutions, then the entire system admits solutions Let a system of three linear congruence equations in integers be given;      \begin{cases}x\equiv b_1\mod c_1\\ x\equiv b_2\mod c_2\\ x\equiv b_3\mod c_3\\ \end{cases}   with $c_1,c_2,c_3\in\mathbb Z_{+}$. Show that, if every pair of congruence equations admits solutions, then the entire system admits solutions.    Now, if every pairs admits solutions, then for example $b_1\equiv b_2\equiv a\mod \gcd(c_1,c_2)$ for some $a$ then;   $x-a\equiv b_1-a\mod c_1$  $x-a\equiv b_2-a\mod c_2$  replacing $x$ by $x':=\frac{x-y}{\gcd(c_1,c_2)}$, we obtain:  $x'\equiv b_1'\mod c_1'$  $x'\equiv b_2'\mod c_2'$  and since $\gcd(c_1,c_2)=1$ by Chinese Remainder Theorem we have a unique solution $\mod{c_1'c_2'}$, so $x\equiv (\gcd(c_1,c_2)x'+a)\mod{\operatorname{lcm(c_1',c_2')}}\quad(1)$  We can do it with another pair (take $2$nd and $3$rd, since we know that such a pair admits solutions;  $x\equiv(\gcd(c_2,c_3)x''+a')\mod{\operatorname{lcm(c_2',c_3')}}\quad(2)$  Now how can I show that for the pair $(1)$ and $(2)$ also solutions exist ? Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.  So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$  and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$  So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that How to check file types of uploaded files in PHP? On the PHP website, the only real checking they suggest is using is_uploaded_file() or move_uploaded_file(), . Of course you usually don't want user's uploading any type of file, for a variety of reasons.  Because of this, I have often used some "strict" mime type checking. Of course this is very flawed because often mime types are wrong and users can't upload their file. It is also very easy to fake and/or change. And along with all of that, each browser and OS deals with them differently.  Another method is to check the extension, which of course is even easier to change than mime type.  If you only want images, using something like getimagesize() will work.  What about other types of files? PDFs, Word documents or Excel files? Or even text only files?  Edit: If you don't have  or  and system("file -bi $uploadedfile") gives you the wrong file type, what other options are there? I'm going to install 64 bit 14.04 on a new machine with something like 1 to maybe 128GB RAM and even more RAM later. How much space should I allocate to the swap partition?   And especially: WHY those recommendations for these numbers?    (None of the existing answers  and  explain any of the why and are a bit dated if we look at 1GB or RAM and up) Why can I not access a file named `-` when quoting? I have a file called -.  I want to display its contents.  One way is to do  cat ./-   since cat - reads from standard input. However, why are cat "-" and cat '-' also interpreted by the shell as cat -? How to convert coordinates from vertex to world-space? How do I convert from object-space to world-space?  I suspect it is:  vert_os = obj.vertices[foo] vert_ws = vert_os.getPositionFromMatrix(obj.matrixWorld)   but does the matrix also handle translations, or do I need to handle that separately?  Can someone link to the relevant documentation? Where do AirDrop files go when I transfer them between an iPad, iPhone and iPod? I AirDropped some Google Docs from my iPad to my iPhone but nothing happened; where did my files go? The file was a list of tools. Probably not with Trello, but maybe with some integration.  What I need is a solution for archiving cards because I'm not going to work with them now, but later they would come back, so I would remember doing what should be done.  Something in the lines of some email "follow up" services out there that help you remind things and at the same time keep your inbox clean.