Euclidean Domain and Divison Algorithm In ED definition we need for all $a, b$ not zero then there exist $q$ and $r$ such that $a = bq+r$ with some condition with respect to norm.   Actually I am interested whether $q$ and $r$ are unique in ED or not  Further Questions: 1. Can we define some norm in ED with respect to which q and r are unique    And in how many ways we can define such norms   If it is not possible what algebraic structure like this i.e. ED with uniqueness of q and  r, has some special properties.
Euclidean domain in which the quotient and remainder are always unique Let $R$ be a Euclidean domain in which the quotient and remainder are always unique. Does it follow that the ring $R$ is either a field or a polynomial ring $F[X]$ for some field $F$?
I am a graduate student of math right now but I was not able to get a topology subject in my undergrad... I just would like to know if you guys know the best one.. I want to install a portable Tex Live in windows 10. But after searching online I just get this from .  The -portable installer option (or V command in the text installer or corresponding GUI option) creates a completely self-contained TEX Live installation under a common root and forgoes system integration. You can create such an installation directly on a USB stick, or copy it to a USB stick afterwards. To run TEX using this portable installation, you need to add the appropriate binary directory to the search path during your terminal session, as usual. On Windows, you can double-click tl-tray-menu at the root of the installation and create a temporary ‘tray menu’ offering a choice of a few common tasks, as shown in this . The ‘More…’ entry explains how you can customize this menu.  Any more instructions in detail? What should I do? Disable emojicons introduced with WP 4.2 So WP 4.2 introduced emojis (smileys) that basically adds JS and other junk all over your pages. Something some people may find shocking. How does one completely erase all instances of this? How should you deal with (deliberate) selective deafness? Should you gain rep for asking a duplicate question? The Feeds bot’s translate command is no longer working for moderators I've got a question about how to drive the car in Europe on Highways. In the US, I can overtake another car on either side, and I am fine.  In Europe I was told that I have to drive on the right side (Germany), and to overtake on the left side of the other driver.  Now I had following situation, and I did not know how to handle this properly: I was driving along on the right lane, the Autobahn was 3 lanes wide on my side, and was going around 110 km/h. Someone overtook me, but he stayed on the middle lane, and suddenly got slower. He wasn't really in front of me, so I could not change to the left lane, but I was told I may not overtake on the right side.  Would I have to brake in such a situation, or should I just ignore it, and overtake him again on the right side?  (I overtook him, and he was on the phone, so I guess that was why he got distracted and slowed down that much). I understand the difference between LET and LET* (parallel versus sequential binding), and as a theoretical matter it makes perfect sense.  But is there any case where you've ever actually needed LET?  In all of my Lisp code that I've looked at recently, you could replace every LET with LET* with no change.  Edit: OK, I understand why some guy invented LET*, presumably as a macro, way back when.  My question is, given that LET* exists, is there a reason for LET to stay around?  Have you written any actual Lisp code where a LET* would not work as well as a plain LET?  I don't buy the efficiency argument.  First, recognizing cases where LET* can be compiled into something as efficient as LET just doesn't seem that hard.  Second, there are lots of things in the CL spec that simply don't seem like they were designed around efficiency at all.  (When's the last time you saw a LOOP with type declarations?  Those are so hard to figure out I've never seen them used.)  Before Dick Gabriel's benchmarks of the late 1980's, CL was downright slow.  It looks like this is another case of backwards compatibility: wisely, nobody wanted to risk breaking something as fundamental as LET.  Which was my hunch, but it's comforting to hear that nobody has a stupidly-simple case I was missing where LET made a bunch of things ridiculously easier than LET*. Positioning the arrows on a PCA biplot I am looking to implement a biplot for principal component analysis (PCA) in JavaScript. My question is, how do I determine the coordinates of the arrows from the $U,V,D$ output of the singular vector decomposition (SVD) of the data matrix?  Here is an example biplot produced by R:  biplot(prcomp(iris[,1:4]))     I tried looking it up in the  but it's not very useful. Or correct. Not sure which. How to encrypt folders (not in Home)? a year on: what does it mean? PLease, dear native English speakers, help me to understand the meaning of the expression "a year on" within the context "The reality is that, a year on, a true 4K standard is still very much a work in progress..." Thank you a lot! Digging in deeper to Javascript Remoting and I'm a bit confused.  The  starts off by showing how to set up everything and uses code like the following:  function getRemoteAccount() {     var accountName = document.getElementById('acctSearch').value;      // This remoting call will use the page's timeout value     Visualforce.remoting.Manager.invokeAction(         '{!$RemoteAction.AccountRemoter.getAccount}',         accountName,          handleResult     ); }   However, I've also seen examples where Visualforce.remoting.Manager.invokeAction is never called.  They end up looking something like the following:  function getText() {   var text = "";   JSRemotingBasicsController.doGetText(     function(result, event) {       // do something here     }   ); }   I'm wondering what is the point to using Visualforce.remoting.Manager.invokeAction when you can just reference the RemoteAction directly like the second example?  Is there something I'm missing?  Is there a time where the second example doesn't work? A doubt from Milnor's "Topology from a Differentiable Viewpoint". This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise. How do I pipe each command given to the shell? I'd like to edit my .bashrc so that every command executed on the shell is piped to something, for example:   $ sudo apt update   _________________ &lt; sudo apt update &gt;  -----------------     \   ^__^      \  (oo)\_______         (__)\       )\/\             ||----w |             ||     ||   I have managed something rather similar, but not entirely:  $ bash $ exec &gt; &gt;(cowsay) $ echo "Hello AU!" $ exit  _______ &lt; Hello AU! &gt;  -------     \   ^__^      \  (oo)\_______         (__)\       )\/\             ||----w |             ||     ||   It is not the desired result, as it only happens after exiting the current shell.  It's mainly for fun/learning purposes. Let $S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}$, is  $S$ a dense set on $\mathbb{R}$? Deseasonalizing count data I used stl() in R to decompose count data into trend, seasonal &amp; irregular components. The resulting trend values are not integers anymore. I have the following questions:   Is stl() an appropriate way to deseasonalize count data? Since the resulting trend is not interger valued anymore, can I use lm() to model the trend components? Does moving leave the object in a usable state? Finding the Dimension of a Matrix Polynomial: $W$ = { $p(B)$ : $p$ is a polynomial with real coefficients} Let $W = \{ p(B) : p \text{ is a polynomial with real coefficients}\}$, where $$B= \begin{pmatrix} 0 &amp; 1 &amp; 0\\  0 &amp; 0 &amp; 1\\  1 &amp; 0 &amp; 0 \end{pmatrix}$$  Which of the following possibilities presents the tightest bounds on the dimension $d$ of the vector space $W$?     $4 ≤ d ≤ 6$    $6 ≤ d ≤ 9$    $3 ≤ d ≤ 8$    $3 ≤ d ≤ 4$ Should it be /about_us or /about-us?  From usability point of view, I personally think /about-us is much better for end-user yet Google and most other websites (and javascript frameworks) use underscore naming pattern. Is it just matter of style? Are there any compatibility issues with dashes? $M$ connected $\iff$ $M$ and $\emptyset$ are the only subsets of $M$ open and closed at the same time I'm trying to understand this proof that:  $M$ connected $\iff$ $M$ and $\emptyset$ are the only subsets of $M$ open and closed at the same time  Which is:  If $M=A\cup B$ is a separation, then $A$ and $B$ are open and closed. Recriprocally, if $A\subset M$ is open and closed, then $M = A\cup(M-A)$. What? I know that if $M=A\cup B$ is a separation, $A$ and $B$ are both open. But why closed? Also, the 'recriprocally' part is totally nonsense to me. Anybody could help?  Also, there's another proof, which states: $M$ and $\emptyset$ are the only subsets of $M$ at the same time closed and open $\iff$ if $X\subset M$ has empty boundary, theb $X=M$ or $X=\emptyset$  which is proved as the following:  given $X\subset M$, we know the condition $X\cap \partial X = \emptyset$ implies $X$ is open, while the condition $\partial X \subset X$ implies $X$ is closed. Then, $X$ is open and closed $\iff$ $\partial X = X\cap \partial X = \emptyset$, this show $\iff$ for the theorem above.  First of all, I think that the condition $X$ has boundary empty implies that $X\cap \partial X$ is empty, but who said anything about $\partial X\subset X$? Also, where's the $\rightarrow$ of this proof? I can only see $\leftarrow$