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Any group of index 2 is normal I'm trying to understand the proof that any group of index $2$ is normal.   Let $G$ be a group and $H$ a subgroup of index $2$.  I understand that if $x\in H$ then $xH=H=Hx$ since $H$ is a subgroup. However, I don't understand why if $x\in G\setminus H$ then $xH=G\setminus H.$  I know there are other proofs of this on this site but none of them specifically address this question. Alternative version of the Final Value theorem for Laplace Transform Changing the size of the mouse pointer in Ubuntu MATE 18.04 does not work I have Ubuntu 18.04 with the gnome-desktop  and parallel the Ubuntu MATE desktop installed. While I can easily change the size of the mouse pointer (DMZ-White) in the Gnome3 desktop , this does not work on the MATE desktop. I use for this the dconf-editor.  Maybe someone knows a solution?  Thanks a lot. We know that $221 = 17*13$. So we can check if the system has roots to both of those equations separately, which it does:  $x^{5} \equiv 2$ mod $13$ has the solution $6 + 13n$ and $x^{5} \equiv 2$ mod $17$ has the solution $15 + 17n$.   I got these numbers from wolfram, I have no idea how to solve this problem WITHOUT a calculator. And even after finding these numbers. How would one obtain a solution modulo $221$? I was thinking the Chinese Remainder Theorem but I am under the assumption that CRT only applies to problems with powers of $x$ which are $1$.  Thanks. Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?  There are a number of equivalent definitions for the measurability of a function and the most obvious one would be to show that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{R}$.  Thus my strategy has been to consider an arbitrary irrational $y$ and use the density of the rationals in the reals to show that there exists some open set $O$ such that $m^*(O-\{x\in E : f(x)&gt;y\})&lt; \varepsilon$.  I would do this by choosing some rational $q&lt;y$, close enough to $y$, such that the set of values in $E$ which get sent under $f$ to values strictly greater than $q$ and less than or equal to $y$ is small enough.  But that doesn't in general seem possible since I could keep sending intervals of some fixed positive length to smaller and smaller regions just beneath $y$.  However conversely, such a function would not be enough for a counter example, since finding such a function only rules out this strategy, but doesn't necessarily entail non-measurability.  Is there a better definition of the measurability of a set I should use? or does the statement actually not imply measurability?  Thanks. Limit of sequences $V(n)$ and $U(n)$ Why does QgsSpatialIndex.nearestNeighbor(point, 1) sometimes return 2 lines? How to increase swap space? I have, by default, 250MB of swap space in Ubuntu, and I want to expand it to a larger size. I need 800MB, which I think will be enough to open several applications without having to hit the current limit of swap. I hope someone can help me. Lost my old passport for Schengen visa application Windows installer for Ubuntu 14.04 LTS onwards Is there any Windows installer (wubi) for Ubuntu 14.04 LTS?  When I mount the .iso file with Daemon Tools and run wubi.exe, it just asks me if I want to reboot (I think that's useful only when the .iso  is burned on CD), so is there no "real" installation? Repeating an equation including its content and number I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$,  we have $S+T\geq 0$, and $ST\geq 0$.  $T\geq 0$ means $(Tx,x)\geq 0$.    To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.   There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:   $T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.   Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.   Is there a more straightforward way? Delete votes from users who later become moderators shouldn't prevent community undeletion Why is my variable unaltered after I modify it inside of a function? - Asynchronous code reference If all motion is relative, how does light have a finite speed? I've often heard that Einstein shattered the notion of absolute motion (i.e. all things move relative to one another) and that he established the speed of light as being absolute. That sounds paradoxical to me; I cannot understand how the two concepts can be reconciled.  Before going further, I'd like to say: 1) Over the years, I've seen many layman's explanations on these topics (including the nice YouTube video by Vsauce, ). I understand everything that's said (or, at least, I think I do). Just nothing I've found seems to address this apparent contradiction. 2) More recently, I've tried to find the answer on my own. That includes searching the posts on this site. Some come close (), but nothing I've been able to find seems address specifically what I'm asking.  Back to the question: Relativity shows us that there is no universal frame of reference by which to judge motion, so object A might be reckoned as moving at 10&nbsp;m/s relative to object B or as stationary relative to object C. This is fine for me. I can grasp that the universe has no intrinsic coordinate system, that we only think that way on Earth because we have the ground to move over.  Then there's the speed of light (in a vacuum). The speed of light is the ultimate "speed limit," it's often said. But if there is no universal frame of reference, how can there be any such speed? The very idea only make sense if there is a universal frame.   If one object is moving (uniformly) at 60%&nbsp;c and another object is also moving at 60%&nbsp;c, but in the exact opposite direct, then from the perspective of either one (if they could still see each other) the other would appear to violate that speed limit. All these spacetime bending consequences used to explain why nothing can move past this speed only seems to enshrine the concept that there is some ultimate speed standard. If there is only relative speed, then the concept of light having a specific speed in the vacuum should be a nonsensical one since it having speed (x m/s) would only make sense when measured against some other body.   Since I was very young, it has always sounded to me like motion is only mostly relative, that until you get close to the speed of light, the effects of an absolute frame of reference are negligible. Perhaps that there is an actual fabric of space which everything moves relative to, which is why there is something to expand between galaxies (faster than light can propagate) in the metric expansion of space. Growing up, I always thought this would just start makes sense with time. Now I'm up to a first year (college) level in physics, I even know basic calculus, yet I'm still hopelessly confused.  EDIT:  Thank you to whoever suggested this may be a duplicate of . It and others are very much related and at least partially answer my question. Unfortunately, explaining that distances become shorter and time becomes slower as a way to stop you from exceeding the speed of light does not explain how that speed is not an absolute.  By my reckoning, if all speed is relative, then no matter how fast you go light should always race away from you at the same apparent speed. I.e. there should be no speed limit. For there to be a speed which you cannot exceed or you would catch up (and make time irrelevant) requires the very concept of some external speed by which light can travel and nothing else can reach - thus my logical paradox continues unabated. Duality discrete math problem This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm     The dual of a compound proposition that contains only the logical operators $\lor$ , $\land$ , and   $\neg$ is the compound proposition obtained by replacing each $\lor$ by $\land$ , each $\land$ by $\lor$ , each $\def\T{{\rm T}}\def\F{{\rm F}}$   $\T$ by $\F$ , and each $\F$ by $\T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.      a) $p \lor\neg q$     I got $\neg p \land q$       b) $p \land (q \lor (r \land \T))$     My answer was $\neg p \lor (\neg q \land r)$       c) $(p \land \neg q) \lor (q \land \F)$     My answer was $(\neg p \lor q) \land  \neg q$  I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers. I'm trying to find an alternative to the standard tombstone qed symbol for more informal papers. I initially thought of an empty square with Pub written inside, but what would actually be better would be a small beer mug symbol. (much like the one found in this set of icons )  Unfortunately there doesn't seem to such a symbol in the comprehensive latex symbol list (although many funny symbols can be found there). Has anyone thought about this? Is there any quick-and-dirty solution? Set the maximum character length of a UITextField How can I set the maximum amount of characters in a UITextField on the iPhone SDK when I load up a UIView? If I have a $2\times 2$ or $3\times 3$ matrix, how should I go about computing the eigenvalues and eigenvectors of the matrix?  NB: I am making this question to provide a unified answer to questions about eigenvalues of small matrices so that all of the specific examples that come up can be marked as duplicates of this post. See . For the following block of code:  For I = 0 To listOfStrings.Count - 1     If myString.Contains(lstOfStrings.Item(I)) Then         Return True     End If Next Return False   The output is:  Case 1:  myString: C:\Files\myfile.doc listOfString: C:\Files\, C:\Files2\ Result: True   Case 2:  myString: C:\Files3\myfile.doc listOfString: C:\Files\, C:\Files2\ Result: False   The list (listOfStrings) may contain several items (minimum 20) and it has to be checked against a thousands of strings (like myString).  Is there a better (more efficient) way to write this code?