I am taking a flight from Amsterdam to Delhi via London Heathrow Airport. I have a British Airways flight to London and a Virgin Atlantic flight from London to Delhi. The customer care on British Airways has confirmed that I can do a through check-in, meaning I don't have to collect my baggage at London and check it in again. I have a layover of about 6 hours in London, and the next flight is on the same calendar day. I am an Indian national.   Can anyone who has been through the same situation clarify if I would need a transit visa for London? Also, I don't hold any other visas right now (USA, Australia etc). I am traveling to Europe on a tourist (type C) visa, and this is my flight back to India. Drop a comment if you'd need additional information!  According to the UK website, I would need a transit visa, but is there any possibility that people have traveled along the same route and it wouldn't be required if I'm not leaving the airport? My flights are non-refundable and Virgin Atlantic has declined to refund any partial amount as well.
Is there a website or some other way to find out if I need a transit visa for a short stop in the UK on my way somewhere else? Ideally this should give me the answer for all nationalities, and take into account any other visas I may have.
How to prove $C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$? I want to prove the following theorem (no idea whether it has a name):  Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $\|\cdot\|$ be a norm on $V$. Then, there exist $C_1, C_2 &gt; 0$ such that for all $x \in V$:  $$C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$$  I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $\|x\|_\infty$ and normalized the vector $x$ such that $\|x\|_\infty = 1$. That left me with  $$C_1 \leq \|x\| \leq C_2$$  but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers. How to make child process die after parent exits? Installed Teamviewer using a 64-bit system, but I get a dependency error I was using , implemented in scikit-learn, for my research work. But I have no good understanding of this.   Can anyone please give a simple, good explanation of one class SVM? How do I know whether my bike can use disk brakes? Initialize class fields in constructor or at declaration? I've been programming in C# and Java recently and I am curious where the best place is to initialize my class fields.  Should I do it at declaration?:  public class Dice {     private int topFace = 1;     private Random myRand = new Random();      public void Roll()     {        // ......     } }   or in a constructor?:  public class Dice {     private int topFace;     private Random myRand;      public Dice()     {         topFace = 1;         myRand = new Random();     }      public void Roll()     {         // .....     } }   I'm really curious what some of you veterans think is the best practice. I want to be consistent and stick to one approach. Can I use virtualbox with a 64 Bit image in a 32 bit host I have Virtualbox 4.1.20 in Ubuntu 12.04 32 Bit (With PAE) and I want to run Ubuntu Server 64 Bit on the Guest. Is this possible?. I ask since I get the following warning:    I want to make sure that there is no option to "emulate" this. I have a 64 bit CPU that supports virtualization. The Intel DZ68DB motherboard with an Intel I7 2600 CPU. What are the rules for splitting words at the end of a line? What are the rules in English language to split words at the end of a line?   Where exactly must the hyphen split the word? How to sort a Google Sheets Query by a custom ordering I have a table of choir members with a “name” and a “voice type” column. How can I write a query that orders them by decreasing pitch level of the voice type (i.e. soprano > contralto > tenor > bass)?  Is there in the  something similar to ?  I also tried the , but it does not seem to have an option for custom ordering, either. Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. I define a map $$\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n$$ by $R \mapsto (0,r^{\frac{n+1}{2}})$ and $M \mapsto(1,m)$. Then I am stuck at showing the map is bijective. By the way, do we need to show that $(0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)$ ? If we can prove this, then what can we conclude? I know that there exist some attack when using same modulus.      But with a little modification, $m$ is the plain-text $N$ is the RSA modulus $r_1, r_2$ is the random padding $e, s$ is the public exponent $C_e, C_s$ is the cipher-text  encrypt as follow $$С_e = (m + r_1)^e   \bmod N$$ $$С_s = (m + r_2)^s   \bmod N$$  If attacker only knows $C_e, C_s, r_1, r_2, e, s, N$. Is it possible to know $m$?  It seems the  does not work? R stem and leaf values Viewer-independent copyable spaces at the beginning of a line? Prove that $f_n$ converges uniformly on $\mathbb{R}$ How to see the full date on macOS Big Sur? How can I upgrade Ubuntu LTS to the next one offline? I'm using Ubuntu 12.04 on my computer and I want to upgrade to Ubuntu 14.04.  I do not have internet at home, so I need to do the upgrade offline.  On openSUSE there a way to make the upgrade offline simple and effective way is by using  the iso DVD.  My question is: Is there any way to make Ubuntu upgrade offline as in openSUSE ? What is the most preferred and easiest way to do pagination in ASP.NET MVC?  I.e. what is the easiest way to break up a list into several browsable pages.  As an example lets say I get a list of elements from a database/gateway/repository like this:  public ActionResult ListMyItems() {     List&lt;Item&gt; list = ItemDB.GetListOfItems();     ViewData["ItemList"] = list;      return View(); }   For simplicity's sake I'd like to specify just a page number for my action as parameter. Like this:  public ActionResult ListMyItems(int page) {    //... } Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain. Windows Managers vs Login Managers Vs Display Managers Vs Desktop Environment I posted a  and noticed people weren't distinguishing correctly between many of these things: Windows Managers vs Login Managers Vs Display Managers Vs Desktop Environment.  Can someone please clear this up, i.e. tell us the difference between them and how they are related perhaps?  What category does Xorg fall under?  What about Gdm/Kdm/Xdm?  People also talk about X.  What is X? Why can you see virtual images?