Could there be 3 particles (instead of only 2 particles) that could be connected by same quantum entanglement (like minus, plus and "some third position")?
So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.)
I have IIS7 on my computer, and I can access it through other computers in my house on the same network through my computer's IP address - 192.168.1.10x  Instead of typing in http : //192.168.1.10x/ I want to be able to go to http : //somename/  I know you can change the hosts file to add it but I'd have to do that on each computer and I'd still have to type in the IP to access it from my iPod/Wii... etc. Prove that $L(V,W)$ forms a vector space Let $V$ and $W$ be vector spaces over a field $F$. Let $L(V,W) = \{T:V\to W : T \text{ is linear} \}$, that is, $L(V,W)$ is the collection of all linear functions from $V$ to $W$. For $S,T \in L(V,W)$ and $a \in F$, define new functions $(S+T):V \to W$ and $ aT :V \to W$ by $(S+T)(x) = S(x)+T(x)$ ; $(aT)(x) = a(T(x))$.   Prove that $S+T$ belongs to $L(V,W)$ and $aT$ belongs to $L(V,W)$. My laptop has 500gb hard disk in which for 250gb I installed Windows xp on C drive(primary partition). I want to install Ubuntu 12.04 in the remaing hard drive with dual boot option. When I am going to install Ubuntu it is showing something like  /boot /boot/   Can anybody help me in this regard that where and how to install Ubuntu? mod Distributive Law, factoring $\!\!\bmod\!\!:$ $\ ab\bmod ac = a(b\bmod c)$ I stumbled across this problem  Find $\,10^{\large 5^{102}}$ modulo $35$, i.e. the remainder left after it is divided by $35$  Beginning, we try to find a simplification for $10$ to get: $$10 \equiv 3 \text{ mod } 7\\ 10^2 \equiv 2 \text{ mod } 7 \\ 10^3 \equiv 6 \text{ mod } 7$$ As these problems are meant to be done without a calculator, calculating this further is cumbersome. The solution, however, states that since $35 = 5 \cdot 7$, then we only need to find $10^{5^{102}} \text{ mod } 7$. I can see (not immediately) the logic behind this. Basically, since $10^k$ is always divisible by $5$ for any sensical $k$, then: $$10^k - r = 5(7)k$$ But then it's not immediately obvious how/why the fact that $5$ divides $10^k$ helps in this case. My question is, is in general, if we have some mod system with $a^k \equiv r \text{ mod } m$ where $m$ can be decomposed into a product of numbers $a \times b \times c \ \times ...$, we only need to find the mod of those numbers where $a, b, c.....$ doesn't divides $a$? (And if this is the case why?) If this is not the case, then why/how is the solution justified in this specific instance? I want to select the value from the dropdown I am testing www.paytm.com  In that I have filled in the mobile number but I am not able to focus and select the mobile operator which is present in the dropdown just after the phone number section.  I want to select 'vodafone' as the value from the dropdown.  How to do that using webdriver  Code I have written till now  `  public class MMT {   public static void main (String[] args)  {  System.setProperty("webdriver.chrome.driver","E:/som/chromedriver/chromedriver.exe");  WebDriver driver = new ChromeDriver();  driver.get("http://paytm.com");  driver.findElement(By.id("mobile-number")).sendKeys("9876543212");`   What should be my next steps? I'm learning about private/public IPs, port forwarding and NAT, and still fail to see the answer to a simple question:  Suppose two users on the same network are communicating with a single server outside their network (say, me &amp; my wife send http requests to yahoo.com).  The server sees both users as a single public ip, and often - as in this case - communicating on the same port.  How does the router route the packets according to the two different connections? Is there any routing info inside the packet beyond ip/port?  Is there some 'session table' maintained somewhere? By whom and what exactly does it include? How to remove the whitespace BEFORE itemize/enumerate? I'm somewhat of a LaTeX noob, and this is driving me completely nuts.  When I type  \documentclass{article} \usepackage{enumitem}  \begin{document}     Some text     \begin{itemize}[noitemsep]         \item Item 1         \item Item 2         \item Item 3     \end{itemize} \end{document}   I don't want the space which appears in between the text "Some text" and the itemization to be there:    When I search around on Google all I see is tons of long, complicated advice on this site, StackOverflow, forums, etc. on how to not fix the problem ("parskip is evil", etc.), or tons of vague/confusing answers on what package to look at, with no actual example for me to learn from.  So while this question is already asked many times, I have never found a single concrete answer to this question.    The question couldn't be simpler: How do I "properly" get rid of this extra space? Why does "under the sea" mean "in the sea"? I have been there first time.   The reason why I think this might not be natural is that the expression "first time" seems to describe "past". You do not use the present perfect with an adverb indicating "past", in my understanding.  However, I feel the phrase below is not that funny.     This is the first time that I have been there.   I am wondering about the relationship between "the first time" and "I have been there". "That" could behave like a relative adverb. If that is the case, the antecedent of "that"—"the first time"—is shared with the following clause: "I have been there". Then, is it OK to say "I have been there first time"?  By the way, Google searches for "" and " in the UK domain show just 33 hits for the first phrase and 10 hits for the second. "They are quite full up this year." Well I thought this is obvious. since $\gcd(a,b)=1$,  then we have that $a\not\mid b$ AND $a\mid bc$. This implies that $a$ divides $c$. But apparently this is wrong. Help explain why this way is wrong please.   The question tells you give me two relatively prime numbers $a$ and $b$ such that $a$ divides the product $bc$, prove that $a$ divides $c$. how is this not obvious? Explain to me how my "proof" is not correct. Out of Bag Error makes CV unnecessary in Random Forests? We can prove that there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:     Is it true that for every pair of primes $p&lt;q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p&lt;q$ and an integer $n&gt;p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n&gt;6$ then $2^n-1\nmid 5^n-1$?   Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &amp;3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &amp;7^{4}-1\mid 151^{4}-1. \end{array}  Now I prove there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1.$  Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.  If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again. Sync folder with Android device I would like to have a folder with all my music which is automatically synced on my Android device (Wifi only, plugged in only, like in SugarSync). Is there any hope for this feature to be implemented in the near future.  This would actually make me pay for U1. What is the easiest way to resolve apt-get BADSIG GPG errors? I frequently cross this issue, and always have to google for an answer. Does anyone have a permanent fix for BADSIG errors from apt-get?     W: GPG error:    lucid   Release: The following signatures were   invalid: BADSIG 54422A4B98AB5139   Oracle Corporation (VirtualBox archive   signing key) Determining direction of longest segment of polygon using ArcGIS Desktop? I have a shapefile with 9000 agriculture fields and what I like to do is to determine for each field (=each polygon) the direction of the longest segment.  For instance, the selected polygon in the screenshot has its longest segment from west to east so I would like to mark give the polygon the attribute W (for west) or either E (for east).   Is there any solution to solve my problem in a fast but generalized way?   I do not need very detailed results. It's just for the interpretation of a question.   I'm seeking an answer using ArcGIS 10.2 for Desktop.  To be more precise on what I'm looking for I try to explain more practical: I'm doing winderosion analysis for a specific area with 9000 fields. As result I have a potential susceptibility for each field. I like to check, if the fields with high susceptibility are correlated to a certain orientation of the fields, e.g. check if most of the fields with high risk are oriented in the main winddirection or not.   The search for the longest segment is not really important for me. It was just an idea I had to determine the direction. So forget about the longest segment if there is a way to determine the direction with another method ;) Combining all the address fields into one? How do I restore the default repositories? Notify-send ignores timeout? Maybe I'm doing something wrong, but shouldn't the commands (run separately) notify-send -t 1 &quot;test&quot; notify-send -t 1000 &quot;test&quot; notify-send -t 10000 &quot;test&quot;  have different timeouts? The first being nearly instantaneous, the second taking 1 sec and the third 10 seconds? In all cases it seems to take about six seconds for me. Is there a way around this behaviour? As the developers label this as a &quot;feature&quot; instead of a bug, I would like some alternatives. System.ListException: Before Insert or Upsert list must not have two identically equal elements - Deep Dive I had the following exception thrown at me when trying to process more than one selected records via a custom button in the list view.    Class Code  global class MerchandiseInventoryClass {      webservice static void ConverttoInventory(String[] merchandise_id)    {        //Convert Merchandise collection into Inventory Collection...         Merchandise__c[] merchandise_collection = [SELECT ID,Author__c,Book_Title__c,Edition__c,Publisher__c,Total_Merchandise_QtyTotal_Qty__c,Status__c FROM Merchandise__c WHERE ID = :merchandise_id];          Inventory__c i = new Inventory__c(); //Line 1        List&lt;Inventory__c&gt; newInventory = new List&lt;Inventory__c&gt;();        for(Merchandise__c m : merchandise_collection )        {            if(m.Status__c !='Fully Converted')            {            i.Author__c = m.Author__c ;            i.Book_Title__c = m.Book_Title__c;            i.Edition__c = m.Edition__c;            i.Publisher__c = m.Publisher__c;            i.Qty_Available__c = m.Total_Merchandise_QtyTotal_Qty__c;            i.Merchandise__c = m.Id;            m.Uninventoried_Qty__c = 0;            m.Status__c = 'Fully Converted';            newInventory.add(i);            }        }        insert newInventory;        update merchandise_collection;    }  }   By moving the Line 1 to within the for loop solved the issue for me.  Thanks to this post (refer the Best Answer in that post)    What puzzles me is that WHY i have to instantiate "i" for every iteration within the loop.  Why the list collection ("newInventory") is complaining that are duplicate elements and why should it even confuse in the first place when I am performing an "insert" even ?