Why did Nazgul spare the other hobbits at Weathertop?
Why did the Nazgûl not attack a second time at Weathertop?
Why does Stack Overflow only allow one @User per comment?  This restriction lead me to a non-comment like :     @MiMo - seem my previous comment (Only one @ user per comment)   Which left me wondering why Stack Overflow is blocking me from notifying two users of a response.  I am sure there is a reason for this, can some one share it with me? Connection between rank and matrix product I have a problem understanding the following:     Let $A$ be an $m \times n$ matrix and let t $\in \mathbb{N}$. Prove that   $\operatorname{rank}(A)\leq t$ if and only if there exists an $m \times t$ matrix $B$ and a $t \times n$ matrix $C$ so that $A = BC$.   I know what a rank is but I can't make a connection between the rank and the existence of two matrices such that $A = BC$. Spaces in files names in \input or \includegraphics I'm going crazy...  I have a file I want to include which path is C:\Users\Matthieu\Documents\Folder withOneSpace\myFile.tex. (As you can see with the \, I'm on windows actually)  I tried the following but none works:  \input{C:/Users/Matthieu/Documents/Folder withOneSpace/myFile} \input{C:/Users/Matthieu/Documents/Folder\ withOneSpace/myFile} \input{C:/Users/Matthieu/Documents/Folder withOneSpace/myFile.tex} \input{C:/Users/Matthieu/Documents/Folder\ withOneSpace/myFile.tex}   I also tried to put the file out of the folder with a space and it works (no surprise):  \input{C:/Users/Matthieu/Documents/myFile} --&gt; Works   All I found is to put a \ before the space but it doesn't work at all. --> Please help ! :) What is the random variable $Z$ by the notation $Z =X/Y$? Is it true every ideal of $\mathbb{Z}[x]$ is finitely generated? Let $I$ be any ideal of $\mathbb {Z}[x]$, and $I=\left&lt;{f_1,f_2 , ...}\right&gt;$.  Without loss of generality, $f_1$ has minimal (absolute value) coefficient of constant term (if there exist).  Then, let $g$ be any element of $I$ .  We have for some integer $m$ that $g - mf_1$ is element of $I$ (minimal coefficient and division algorithm).  I guess this process terminate finitely, but I don't know how can I do next step.  ( ex. $g-mf_1, g-mf_1-nf_2, ...$)  Please help me and thanks for in advance. Does seam itself affect baking? Bind application to a specific network interface Where does the extra kinetic energy of the rocket come from? Consider a rocket in deep space with no external forces. Using the formula for linear kinetic energy $$\text{KE} = mv^2/2$$ we find that adding $100\ \text{m/s}$ while initially travelling at $1000\ \text{m/s}$ will add a great deal more energy to the ship than adding $100 \ \text{m/s}$ while initially at rest: $$(1100^2 - 1000^2) \frac{m}{2} \gg (100^2) \frac{m}{2}.$$ In both cases, the $\Delta v$ is the same, and is dependent on the mass of fuel used, hence the same mass and number of molecules is used in the combustion process to obtain this $\Delta v$. So I'd wager the same quantity of chemical energy is converted to kinetic energy, yet I'm left with this seemingly unexplained $200,000\ \text{J/kg}$ more energy, and I'm clueless as to where it could have come from. In the past month, I've been really busy with the university and stuff, and I don't sleep as much as I used to. I usually sleep 4-5 hours a night.  I haven't been doing that extensively ( I always kept a good sleeping routine), but can the lack of sleep affect my health on the long term ? PermGen elimination in JDK 8 I committed and joined  in private beta. As per private beta privilege page casting close and reopen votes requires only 1 reputation. So anyone is able to access close votes and reopen votes review queues. But there is no link of the review page in the top bar or anywhere. So I thought I am not able to access any review queue. I opened the review page by typing the URL manually (to see review statistics for curiosity). But I found that I can access close &amp; reopen votes review queues.        Probably I won't be able to access any review queue if I have less than 350 and the site goes to public beta. But during private beta off topics, unclear, or closable questions are likely to come. So it is very important to close/reopen them during private beta.  So can we please add the "review" link in the top bar if I am able to review some queues? Continuously Differentiable Curves in $\mathbb{R}^{d}$ and their Lebesgue Measure Show that the image of the curve $\Gamma\in\mathscr{C}^{1}\left([a,b]\to\mathbb{R}^{d}\right)$ has d-dimensional Lebesgue measure zero (of course, $d\geq2$).  This can be proved using the absolute continuity of $\Gamma'$ (since $[a,b]$ is compact and $\Gamma'$ is assumed continuous, hence in $L^{1}([a,b])$) together with the fundamental theorem of calculus to obtain an $\epsilon$-small cover of $\Gamma$ by balls.  But I am trying to prove this using more elementary means (i.e. without integration).  Intuitively, since $\Gamma$ is smooth, we ought to (for fine enough partitions) be able to cover $\Gamma$ by boxes which arise from its tangent line.  And by taking the partition of $[a,b]$ to be finer and finer, the "tangent box" cover ought to also get smaller and smaller.  More rigorously, the vector-version of the mean value theorem can be applied: $$|\Gamma(t_{i-1})-\Gamma(t_{i})|\leq(t_{i}-t_{i-1})|\Gamma'(t_{i}^{\star})|\leq M_{i}\Delta t$$ where $t_{i}^{\star}\in(t_{i-1},t_{i})$ and $M_{i}=\sup_{t\in[t_{i-1},t_{i}]}|\Gamma'(t)|$ which exists and is finite since $\Gamma'$ is continuous.  But to me, it's not quite clear how to rigorously construct a cover by boxes from here.  NOTE:  In the proof I mentioned (using integration), essentially you sum the left hand side over all partition intervals of uniform length $\delta$ (which depends on the $\Gamma'$), and "integrate" the right hand side.  Actually, to be more specific, for each $\epsilon&gt;0$ there exists a $\delta&gt;0$ such that $||P||&lt;\delta$ implies $\int_{t_{i-1}}^{t_{i}}|\Gamma'(t)|dt&lt;\epsilon$ (e.g. absolute continuity).  This allows you to define numbers $\epsilon_{i}=\sup_{t,\bar{t}\in[t_{i-1},t_{i}]}|\Gamma(t)-\Gamma(\bar{t})|\leq\epsilon$ so that $\sum_{i=1}^{\#P}\epsilon_{i}\leq||\Gamma'||_{L^{1}([a,b])}$.  Then you can use these $\epsilon_{i}$ to put balls at each point $\Gamma(t_{i})$ of radius (say) $2\epsilon_{i}$, thus giving you an $\epsilon$-small cover.  Again though, this is harder to establish without integration theory. Invisible software updates list in Software Updater Prove $S_4$ has only 1 subgroup of order 12 The subgroup in $S_4$ that I know has order 12 is the subgroup of all even permutations, otherwise known as the alternating group $A_4$. However, I know this from a fact and not because I am able to show a subgroup of order 12 exists in $S_4$ in the first place. If I had not been told there existed a subgroup of order 12 in $S_4$, I would not have known there was one. So I guess this is actually a two-parter for me: 1) How do I go about showing that a subgroup of order 12 does indeed exist in $S_4$? I know by Lagrange that if there exists a subgroup, then the order of that subgroup must divide the order of the group. However, this doesn't say anything about the existence of the subgroup. And Sylow only verifies subgroups of a prime to some power order, which 12 is not. I also know that there might be a cyclic subgroup of order 12, but without manually multiplying every possible permutation in $S_4$, I don't know any other way to check the existence of this or if this subgroup is isomorphic to $A_4$ because I don't know how to write these permutations as functions in order to check if there is a homomorphism (I know that a permutation is bijective by definition). And 2) How do I show that this group of order 12 is the only group of order 12 in $S_4$? I was thinking about something the other day (that is probably stupid), and thought this would be a good place to talk about it.  If someone wanted to draw its family tree over the last 1000 years, he'd write a binary tree with a depth of about 40 (counting 25 years on average between each generations). This tree would have 240 leaves, or about 1012 ancestors on the deeper level. That is obviously much more than the total world population and a stupid figure when you think about it.  The only explanation that I can think of is that a real genealogy tree must have "linked" branches at several levels that change the actual number of ancestors. Anyway, for an average person, how many ancestors would that make? Could one assume that all or nearly all of the persons that lived in its country or world region a 1000 years ago would be part of its ancestors? I have 6 images, and I want to align them in this way  (a)image1 (c)image3 (e)image5  (b)image2 (d)image4 (f)image6  Any one know how to do that?  The problem for me is how to make (a), (b),..., (f) in vertical direction. I am trying to connect my Galaxy Tab to the Internet through an ad-hoc network made via my PC. The PC ad-hoc settings are fine because I can use it from my iPhone and other devices. On which Stack Exchange site should I post subjective questions? For example, is it possible to drink a flying potion and cast greater invisibility spell on myself (both spells would require concentration), so I can fly around invisible? Transit Visas for South Africa Undesirable-Overstay Tourists I received an undesirable status for overstaying a SA visa while hospitalized for injuries sustained in Lesotho. I had to cross into SA from Lesotho. Now I am banned entry into SA until Feb 2016 and though I have submitted a waiver request to remove the undesirable status, I hear from some the waiver request may take 4-5 months or even take longer than the time of banishment.  I have a pre-paid safari planned in October in which I need to fly through Johannesburg to get to Botswana. The Jburg border control requires tourists to pass through SA immigration even when simply changing planes.  1) Will SA grant me "transit" access if I don't leave the airport?  2) Will SA allow me to spend a night in the airport as the layover is 12 hours>  Thanks for passing on any info.