Which of the following are quadratic integers in $Q[√−5]$?  EDIT: THANKS TO THE COMMUNITY FOR DIRECTING ME TO THE ORIGINAL QUESTION! I DID NOT KNOW THIS QUESTION ALREADY EXISTED ON MATH.SE  (a) $3/5$   (b) $2 + 3√−5$   (c) $(3 + 8√−5)/2$  (d) $(3 + 8√−5)/5$  (e) $i√−5$  Prove your result in each case.  Here's what I've got so far (I'm not even sure if this is half  right):  $\{1,i\sqrt{5}\}$ forms a basis for the field  $\mathbb{Q}(\sqrt{-5})$ viewed as a vector space over  $\mathbb{Q}$. We note that, $\mathbb{Q}\leq \mathbb{Q}(\sqrt{-5})$ Meaning, that the rationals are contained in this simple extension.  a)Yes, as just explained all rationals are contained in this field.  d)Yes, it is  $\frac{3}{5}+\frac{8}{5}\sqrt{-5}$ so if we select  $a=3/5$, $b=8/5$ in $\mathbb{Q}$ as a linear combination for  $a+b\sqrt{-5}$ then it works!  e)No, unless we are asking for  $i\sqrt{5}$.  I am having trouble on how to explain (b) and (c).
Can someone tell me if $\frac{3}{5}$, $2+3\sqrt{-5}$, $\frac{3+8\sqrt{-5}}{2}$, $\frac{3+8\sqrt{-5}}{5}$, $i\sqrt{-5}$ are all quadratic integers in $\mathbb Q[\sqrt{-5}]$. And if so why are they in $\mathbb Q[\sqrt{-5}]$.
Sitemaps - one per subdomain or one for the base domain? I have more than 5 subdomains on a project, say developer, www, staff and some more, what is the best practice in creating sitemaps?   All subdomains that share a same sitemap at base domain (example.com/sitemap.xml) ? or different sitemaps for different subdomains?(developer.example.com/sitemap.xml and www.example.com/sitemap.xml)   Since the subdomains are several parts of the main application, what are the pros and cons of each, and what should I consider while choosing the one over the other? (one sitemap per subdomain or one for the whole site) Proofs of AM-GM inequality Two diagrams side by side I am trying to put two feynman diagrams side by side, but they get displayed below eachother. I am using feynmf and subfloat. This normally works for pictures.  \begin{figure}[ht!]       \centering     \subfloat{          \begin{fmfgraph*}(40, 25)         \fmfstraight          \fmfbottom{wm,i2,z2}           \fmftop{wp,i1,z1}           \fmf{dashes}{wp,i1,z1}           \fmf{dashes}{z2,i2,wm}           \fmf{dashes,caption=$W^{+}$}{i1,i2}         %  \fmf{dashes,caption=$Z^0$}{i2,z1}          % \fmf{dashes,caption=$Z^0$}{i2,z2}         \fmfv{lab=$Z^0$}{z1}         \fmfv{lab=$Z^0$}{z2}     \fmfv{lab=$W^+$}{wp}     \fmfv{lab=$W^-$}{wm}         \end{fmfgraph*}     }      \subfloat{          \begin{fmfgraph*}(40, 25)         \fmfstraight          \fmfbottom{wm,i2,z2}           \fmftop{wp,i1,z1}           \fmf{dashes}{wp,i1,z1}           \fmf{dashes}{z2,i2,wm}           \fmf{dashes,caption=$W^{+}$}{i1,i2}         %  \fmf{dashes,caption=$Z^0$}{i2,z1}          % \fmf{dashes,caption=$Z^0$}{i2,z2}         \fmfv{lab=$Z^0$}{z1}         \fmfv{lab=$Z^0$}{z2}     \fmfv{lab=$W^+$}{wp}     \fmfv{lab=$W^-$}{wm}         \end{fmfgraph*}     }      \caption{Exchange of W boson}     \end{figure} What is the need of acceleration? As we can usually predict the various forces that a body might be undergoing based on its position and its velocity.Hence what is the need of acceleration? What's the last character in a file? I just read the answers to  and everyone said to delete the last character. My question is,  isn't the eof character the last one? I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.   Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.   Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)? Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.   Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.   What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks. How does Linux know where its swap partition is? I've read that you need to put the swap partition on HDD rather than SSD.   My questions are the following:    When and how is the "checking" done by the distribution (or something else) to find its Swap partition?  Does it happen during boot?  It just checks all available disks and searches for a partition with 'swap' flag?  What happens if there are several partitions like that?  Also, how many swap partitions do I need to have if I run, for example, two different distributions on the same disk, let's say Fedora and Ubuntu? What's the use of a decoupling capacitor near a reservoir capacitor? I've seen some circuits where a decoupling capacitor is used as well as a reservoir capacitor, like this (C4 and C5):    I've read about  and for me it looks as if they are meant to remove small fluctuations in the supply voltage. Then I thought - wasn't that the purpose of a  as well? Why wouldn't the reservoir capacitor be able to filter out the small fluctuations, if it is able to filter out the large fluctuations?  So I feel like I have a basic misunderstanding here. What is the purpose of a decoupling capacitor next to a reservoir capacitor, when we assume we place both equally near to the power consuming part? Or is the only advantage of the decoupling capacitor that it is smaller and can therefore be easily placed more near to the power consuming part? Number of permutations of $n$ elements where no number $i$ is in position $i$ Is there any way to create multiple insert statements in a ms-access query? New passport and entrance Schengen I am from Brazil and travel a lot in Schengen countries. Under the rules of Schengen visa (3 months stay in a 6 months period). In my passport are the entrance and exit stamps, also to chec the days I stay in Schengen.  I need a new passport and will get this in Brazil. So in this new passport are NOT the stamps. Does this mean I start new with the Schengen visa rule? I seems like. The new passport is empty. Or do I miss something? It would be great because I am almost on my 90 days stay in the period of 180 days. Let $N = \{N(t) \}_{t\geq 0 }$ be a Poisson process. I already know that $N(t)- \lambda t$ is a martingale where $\mathbb{E} [ N(t) ] = \lambda t$.   I want to prove that $$ \frac{N(t)}{t} \rightarrow \lambda, \quad t \rightarrow \infty \quad \text{ a.s.}. $$  To prove this I received a hint that I should use the martingale law of large numbers which states that:  Theorem: (Martingale LLN)  Let $S= (S_n = \sum_{k=1}^n X_k)_{n=1,2,\ldots}$ be a martingale with respect to the filtration $(\mathcal{F}_n)_{n=1,2,\ldots}$. Then $\sum_{k=1}^n X_k/k$ converges almost surely on the set $\left\{ \sum_{k=1}^\infty k^{-2} \mathbb{E}\left[ X_k^2 \mid \mathcal{F}_{k-1} \right] &lt; \infty \right\}$. Hence $S_n/n \rightarrow 0$ a.s. on the same set.  Now to use this theorem I think I need to use Doob-Meyer to decompose $N(t)$ as the sum of a martingale and an increasing process. Which would be the martingale $M(t) = N(t) - \lambda t$ and increasing process $A(t) = \lambda t$. But how is $M(t)$ the sum of a martingale? And I do not know how this exactly relates to the thing I have to prove. Thanks for any help. Checking for sublist in list Functions satisfying $f:\mathbb{N}\rightarrow\ \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$ Find all functions $f$ such that $f:\mathbb{N}\rightarrow\  \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$   Let us plug in $n=1$  $f(f(1))+f(2)=3$  Since the function is from $\mathbb{N}$ to $\mathbb{N}$, $f(2)$ can only take the values $1,2$. Now we divide the problem into cases.  Case-1: $f(f(1))=2,f(2)=1$  We can assume that $f(1)=c$ for the time being. Then plugging in $n=3$ and using $f(2)=1$ gives $$f(3)=4-c$$ and again since the range of the function is positive integers,then $4-c$ has to be positive and hence $c$ belongs to  {$1,2,3$}. Now, $$f(1)=c$$$$\implies f(f(1))=f(c)$$$$\implies 2=f(c)$$ by the assumption that $f(f(1))=2$ . Now,since $c$ can only take the values $1,2,3$,we start treating cases. If $c=1$,$$f(c)=2$$$$\implies f(1)=2$$ but we know from the deinition of $c$ that $f(1)=c=1$,a contradiction.If $c=2$,then $2=f(c)=f(2)$ but $f(2)=1$ by assumption. Finally,if $c=3$ $$2=f(c)=f(3)$$ but $$f(3)=4-c=4-3=1$$ which is once again a contradiction. Therefore there are no such functions in this case.  Case-2: $f(f(1))=1,f(2)=2$  Again assuming $f(1)=c$ and using $f(n)\le n$ along with plugging $n=c-1$ will give us that $f(1)=1$ and then it is easy to prove that such a function exists by recursion. I can only give a "sort of recursive" way to define the function. Here it goes $$f:\mathbb{N}\rightarrow \mathbb{N}$$$$f(1)=1$$$$f(n)=n+1-f(f(n-1))$$  But this case is harder to deal with.Some help will be appreciated. Entire function with vanishing derivatives? Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function. And assume that at each point, one of it's derivatives vanishes.  What can you say about $f$?  A hint suggests that $f$ must be a polynomial. How do you teleport a certain player to spawn after they win a 1v1 I managed a KitPvP server in Minecraft and everything works fine except when players wants to duel each other. They get sent to a room where they will be dueling. The main problem is when the winner wins, I have no idea how to teleport them back to spawn. Create ArrayList from array I have an array that is initialized like:  Element[] array = {new Element(1), new Element(2), new Element(3)};   I would like to convert this array into an object of the  class.  ArrayList&lt;Element&gt; arraylist = ???; Can you cast a spell at a higher level, even if it doesn't have an "At higher levels" entry? Does ReLU layer work well for a shallow network? Python and regular expression with Unicode I need to delete some Unicode symbols from the string 'بِسْمِ اللَّهِ الرَّحْمَٰنِ الرَّحِيمِ'  I know they exist here for sure. I tried:  re.sub('([\u064B-\u0652\u06D4\u0670\u0674\u06D5-\u06ED]+)', '', 'بِسْمِ اللَّهِ الرَّحْمَٰنِ الرَّحِيمِ')   but it doesn't work. String stays the same. What am I doing wrong?