Excuse my limited acquaintance on English usage; which sentence is grammatically correct, and if any, which meaning does each convey?   I saw them play chess. I saw then playing chess.
I saw him crossing the road.   I saw him cross the road.   Which one is correct and why?
Precedence between implication and bi-implication I came across this question:  Let p, q, and r be the propositions  p : Grizzly bears have been seen in the area. q : Hiking is safe on the trail. r : Berries are ripe along the trail. Write these propositions using p, q, and r and logical connectives (including negations).     If berries are ripe along the trail, hiking is safe if and   only if grizzly bears have not been seen in the area.   I see an If r,q iff ~p where If r,q is the part "If berries are ripe along the trail, hiking is safe" and ~p is "grizzly bears have not been seen in the area"  and I came up with (r -&gt; q) &lt;-&gt; ~p  But the answer given is: r -&gt; (q &lt;-&gt; ~p)  Can someone please explain the difference in precedence? Can bosons that are composed of several fermions occupy the same state? It is generally assumed that there is no limit on how many bosons are allowed to occupy the same quantum mechanical state. However, almost every boson encountered in every-day physics is not a fundamental particle (with the photon being the most prominent exception). They are instead composed of a number of fermions, which can not occupy the same state.  Is it possible for more than one of these composite bosons to be in the same state even though their constituents are not allowed to be in the same state? If the answer is "yes", how does this not contradict the more fundamental viewpoint considering fermions? How do I use F3 to get my coordinates when my laptop makes F3 a special key? Let $f(z)$ be an entire function such that $$|f(z)|&lt;\frac{1}{|\text{Im}(z)|},\qquad z\in\Bbb C-\Bbb R.$$ The question asked me to prove that $f(z)=0$. At least looking at it, it really seems to have an application of Liouville's theorem lurking around somewhere, but I haven't found it.   My thoughts first led me to think about doing this by contradiction and using Picard's little theorem. So, I've considered a strip containing the real axis (say of width $2$ for simplicity). This will imply the complement maps into the unit disk, which implies that this strip has to map almost everywhere else on the complex plane. Now, on the imaginary axis, I know that $f(z)$ will vanish as $z\to\infty$, and this seems like it might be useful. I'm sort of under the impression $f(z)$ might have an essential singularity at $\infty$, which would mean $f(1/z)$ has an essential singularity at $0$. Either way, it of course can't have a pole at $\infty$ because of $f(z)$ vanishing on the imaginary axis, and if it is a removable singularity, it must be $0$, which still gives the solution. Therefore, I think it must have something to do with $f(z)$ having an essential singularity at $\infty$.   I was hoping to combine this with the above inequality and deduce a contradiction, but I haven't thought of one yet.   Any hints or suggestions? I'm studying for a prelim. exam, so this isn't homework.  Update: So, in the spirit of searching methods using the essential singularity at $z=0$, we have $$\left|f\left(\frac{1}{z}\right)\right|\leq\frac{1}{\left|\text{Im}\left(\frac{1}{z}\right)\right|}=\frac{|z|^2}{|y|}.$$ Thus, for $|y|&gt;1$, we have $$\left|f\left(\frac{1}{z}\right)\right|&lt;|y|\left|f\left(\frac{1}{z}\right)\right|&lt;|z|^2.$$ Thus, if I can show that $|f(1/z)|$ is a polynomial, we'll know it can't have an essential singularity at the origin, concluding my proof. Apple has introduced , also known as "rootless", with OS X 10.11, El Capitan. I understand this is a step for general protection against malware but as a developer I need write access to some of the files it locks away.  How do I disable this protection? Fastest Square Root Algorithm What is the fastest algorithm for finding the square root of a number?  I created one that can find the square root of "$987654321$" to $16$ decimal places in just $20$ iterations   I've now tried Newton's method as well as my own method (Newtons code as seen below)  What is the fastest known algorithm for taking the second root of a number?  My code for Newton's Method (*Edit: there was an error in my code, it is fixed in the comments below):        a=2   //2nd root     b=97654321   //base     n=1   //initial guess     c=0   //current iteration (this is a changing variable)     r=500000   //total number of iterations to run     while (c&lt;r)      {         m = n-(((n^a)-b)/(a*b))  //Newton's algorithm         n=m         c++;         trace(m + "  &lt;--guess   ...   iteration--&gt;  " + c)     } If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian? I know that any group satisfying $x^2=1$ for all $x$ is abelian. Is the same true if $x^3=1$? I don't think it is, but I can't find a basic counterexample. I have come across these three definitions of the limit superior (or upper limit) and the limit inferior (or lower limit) of a sequence of real numbers and I wonder how to establish the equivalence of these.    Walter Rudin: PRINCIPLES OF MATHEMATICAL ANALYSIS, 3rd edition:  Definition 3.16: Given a sequence $\{s_n\}$ of real numbers, let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \to x$ for some subsequence $\{s_{n_k}\}$. This set contains all subsequential limits ..., plus possibly the numbers $+\infty$, $-\infty$. We now ... put  $$s^{*} = \sup E,$$ $$s_{*} = \inf E.$$ The numbers $s^{*}$, $s_{*}$ are called the upper and lower limits of $\{s_n\}$; we use the notation  $$ \lim_{n \to \infty} \sup s_n = s^{*}, \, \, \, \lim_{n \to \infty} \inf s_n = s_{*}.$$ Tom M. Apostol: MATHEMATICAL ANALYSIS, 2nd edition:  Sec. 8.3: Definition 8.2: Let $\{a_n\}$ be a sequence of real numbers. Suppose there is a real number $U$ satisfying the following conditions:    i} For every $\epsilon &gt; 0$ there exists an integer $N$ such that $n &gt; N$ implies $$a_n &lt; U + \epsilon.$$ ii) Given $\epsilon &gt; 0$ and given $m &gt; 0$, there exists an integer $n&gt; m$ such that $$a_n &gt; U - \epsilon.$$ Then $U$ is called the limit superior (or upper limit) of $\{a_n\}$, and we write $$U = \lim_{n \to \infty} \sup a_n.$$ Statement (i) implies that the set $\{a_1, a_2, a_3, \ldots \}$ is bounded above. If this set is not bounded above, we define $$\lim_{n\to\infty}\sup a_n = +\infty.$$ If the set is bounded above but not bounded below and if $\{a_n\}$ has no finite limit superior, then we say $\lim \sup_{n\to\infty} a_n = -\infty$. The limit inferior (or lower limit) of $\{a_n\}$ is defined as follows: $$\lim_{n\to\infty}\inf a_n = -\lim_{n\to\infty}\sup b_n,$$ where $b_n = -a_n$ for $n= 1, 2, 3, \ldots$.    Robert G. Bartle and Donald R. Sherbert: INTRODUCTION TO REAL ANALYSIS, 3rd edition: Exercises for Section 3.3: Problem 10: Let $(x_n)$ be a bounded sequence of real numbers, and for each $n\in \mathbb{N}$ let $s_n \colon= \sup \{x_k \colon k \geq n\}$ and let $t_n \colon= \inf \{x_k \colon k \geq n\}$. Prove that $(s_n)$ and $(t_n)$ are monotone and convergent. Also prove that if $\lim (s_n) = \lim (t_n)$, then $(x_n)$ is convergent. [One calls $\lim (s_n)$ the limit superior of $(x_n)$ and $\lim (t_n)$ the limit inferior of $(x_n)$.]   Now how can one show that the above three definitions are equivalent (i.e. these three definitions are of the same pair of numbers)? Installing Wi-Fi driver for Realtek Semiconductor RTL8723DE Device [10ec:d723] How do I install Wi-Fi drivers for Realtek RTL8723DE device in Ubuntu 16.04?  lspci -v | grep -i network 02:00.0 Network controller: Realtek Semiconductor Co., Ltd. Device d723 I would like to draw a squiggly line with an arrow somewhere in the middle of it. TikZ has a pathmorphing decoration that draws a squiggly line and a markings decoration to place an arrow tip. So a solution would be to use them together:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{decorations,decorations.pathmorphing,decorations.markings} \begin{document}  \begin{tikzpicture}  \draw[          decoration={snake},          decorate,          postaction={decoration={markings,mark=at position 0.6 with {\arrow{>}}},decorate}      ]         (0,0) -- (2,0); \end{tikzpicture}  \end{document}   Unfortunately this does not work and results in  ! Dimension too large. &lt;to be read again&gt;                     \relax  l.11         (0,0) -- (2,0);   I suspect there is an infinite recursion going on somewhere.  Is there a workaround? If not, is there a different simple solution to the problem?  (Note: This is probably related to my previous question ) Prove that two distinct numbers of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if $a$ is even and have $\gcd=2$ if $a$ is odd.   My attempt: If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$ Then, $$a^{2^{n}} + 1 = 2^{2^{n}s}\cdot k^{2^n} + 1$$ and $$a^{2^{m}}  + 1 = 2^{2^{m}s}\cdot k^{2^m} + 1$$ To prove that they're relatively prime, we need to show that their gcd = 1. And I was stuck here, how could I prove that gcd of two numbers is $1$?     A hint would be sufficient. Thanks. How do I check if a string is a number (float)? What are some good books for learning the concepts of Kinematics, Newton laws, 2D Motion of Object etc.? block certain dangerous commands? In my  High school there are few computer having Ubuntu 14.04 installed and many students know the Root  password, how  can I block/disable  certain dangerous  commands like :- rm -rf / rm -rf * :(){:|:&amp;};: to avoid Data loss ? Scenario:   Create SymLink: C:\Data\workspaces --&gt; C:\Users\****\Documents\workspaces Navigate to C:\Data\Workspaces in explorer Click and drag the the path from the address bar onto "Quick Access" in the navigation pane.   Expected Result:   Clicking on the shortcut takes you to C:\Data\workspaces   Actual Result:   Clicking on the shortcut takes you to C:\Users\****\Documents\workspaces   How do you make it so that the shortcut points to the correct path. (the SymLink)  Command Used to create the SymLink:  mklink /D C:\Data\workspaces C:\Users\****\Documents\workspaces   Updated with pictures:  Please note the workspaces symlink inside C:\Data:    When I double-click on the symlink it works correctly. (note the address bar.    When i pin it to the Quick access bar, however, it uses the address that the symlink points to and NOT the symlink address. I have a Queue&lt;T&gt; object that I have initialised to a capacity of 2, but obviously that is just the capacity and it keeps expanding as I add items.  Is there already an object that automatically dequeues an item when the limit is reached, or is the best solution to create my own inherited class? How Can I View All The Reputation Gained/Lost For One Specific Question If I would like to view all of the reputation I have gained or lost due to one specific question of my own, how can this be done?  Is it possible for someone with, say, 421 rep on the site in question? Can the milky way actually be photographed ? I know the image is manipulated, and is probably a composite, but how do you capture such contrast in the milky way? Is an equitorially mounted scope required? Any advice or pointers are welcome, thanks.  I'm heading out towards much less light polluted areas middle of next week and I want to try some nighttime sky captures with my 7D.  Additional info: I just found  on wikipedia that says: 54s, tripod mounted 5D, 16mm lens, f/2.8, ISO800. What else do I need to be aware of? My 7D is only an APS-C and the widest lens I have is 19mm. I'm thinking that might not be wide enough to capture an interesting shot? How can I find the XY coordinates of an HTML element (DIV) from JavaScript if they were not explicitly set? I'm stuck in a teleporting loop. What can I do? I set a command block to /tp. It was supposed to activate only if you had 0 time left, and to activate the command block with a hopper clock to keep giving it power.   But it is not checking if you have 0 time left, so it keeps teleporting me. I can't get close enough to the command block to break it, so I'm stuck in a neverending teleport loop.  Any ideas?