android screen orientation detect bug?
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If $ \langle G, \star \rangle $ is an abelian group, then for all $a, b \in G$, show that $(a \star b)^{n} = a^{n} \star b^{n}$. If $ \langle G, \star \rangle $ is an abelian group, then for all $a, b \in G$, show that $(a \star b)^{n} = a^{n} \star b^{n}$.  I am stuck at the first step, unable to figure out how to start. I am just beginning on group theory, and know that an abelian group obeys commutativity, associativity, has an identity element, and each element is invertible. I am unable to figure out how to apply this for the proof. Any hints on how I should get started? To give some background, I am attempting to solve the problem from Artin's algebra textbook which was already asked in this link:  Now, I would like to prove the following two claims which are not obvious to me. If I can prove these, I will be able to show that $\zeta_5\notin \mathbb{Q(\zeta_7)}$ as needed. I suspect that both claims will have similar proofs.  Claim 1: $\zeta_5$ has degree $4$ over $\mathbb{Q}$ Claim 2: $\zeta_7$ has degree $6$ over $\mathbb{Q}$  Attempt to prove Claim 1 Let $α=\zeta_5$. I would like to find the monic irreducible polynomial for $α$, but unfortunately I do not know what it is yet. I see $α=cos(\frac {2π}5)+isin(\frac {2π}5)$ and $α^4=cos(\frac {4π}5)+isin(\frac {4π}5)$. Please help me to find the irreducible polynomial, and (ideally) to show that it is irreducible. Why the $(n+1)$th prime number cannot exceed the smallest prime divisor of $p_1\dotsc p_n + 1$ Let $p_n$ be the $n$th prime number in its natural order. We want to show that  $$p_{n+1} \leq p_1\dotsc p_n + 1 \tag{1}$$  I know $p_1\dotsc p_n + 1 &gt; 1$, so it has at least one prime divisor, but the text says that $p_{n+1}$ cannot exceed the smallest of these divisors, but I cannot understand why, especially when I imagine $p_{n+1}$ being very large. Although some examples confirm that this statement is true, I cannot make a connection as to why the $n+1$ prime number cannot exceed the smallest prime divisor. Once this has been shown to be true, $(1)$ immediately follows. How can I get more of this macro photo in focus? Are all n-ary operators simply compositions of binary operators? Take for example $A \times B \cdot C$ = $(A \times B) \cdot C$ where $A, B, C$ are 3-component real vectors.  We can define a 3-nary operator $\times - \cdot$ that is a composition of the two common binary operators $\times$ and $\cdot$.  The same thing happens with most functions (operators) - the way we calculate them is by doing smaller binary problems and adding together.  Every time I try to come up with an $(n &gt; 2)$-ary operator my mind automatically looks for binary operators.  So, the question is, do there exist operators (of some weird kind in some branch of math) that cannot be decomposed into 2-ary and 1-ary operators?  Thanks. 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. List of all active modules used in a site YA dystopia novel of battle school with unusual military control scheme Some sort of battle school (no, not Ender's Game), where teens are being trained as military commanders.   The control system for their army hinges around some sort of system that requires extremely fine motor control -- being able to "push" around digital representations of military units using a special interface that interprets minute finger movements.  The boy protagonist eventually meets a girl in the battle school, and is astonished to learn that most people do not live as well as he does -- for instance, he gets a fresh jumpsuit outfit every day and disposes of the old one; this is a luxury.  One scene I recall is them kissing for a moment in some sort of underground passageway; the boy protagonist notes the exact length of the kiss, as he has been trained to have heightened time perception.  I don't recall much about other commander characters; I think they have different special abilities, and one other major character might have physical deformities, possibly because of postwar-radiation-related birth defects.  They have been training for an upcoming, prearranged final battle between all the existing national governments; the military units are very possibly robotic or entirely digital in nature, so as to avoid wasting materiel / human life.    They ultimately win the battle, just barely; I think the girl protagonist gives them an edge in the competition by hacking the timing / control system in favor of their team, causing a discrepancy which only the boy protagonist notices due to his heightened perceptions.  It was agreed that the winning government gets dominion over one of the few islands of land left that was not irradiated in some previous war; the protagonists get to live there as a reward for their victory.  I believe the boy protagonist later realizes he's gradually losing his ability to exactly quantify the passage of time.  Published no later than 2004, likely somewhere between 1980-2000, probably sometime in the '90s. I'm wondering if it's possible to hide a folder without renaming it with a dot on the Desktop.  Ubuntu 12.04  Thanks "is a field and used like a type" error The Tree of Life in Evangelion I've noticed references to the Kabbalah in a couple of anime I've watched.  One of them is Neon Genesis Evangelion.  Neon Genesis Evangelion has it in the sky in both the opening credits and in the series itself:     I also remember it was on the ceiling at NERV, but the best screencap I could find of that was too tiny to be visible.  So, my question is, Is there any deeper meaning to the use of the Kabbalistic tree of life, or is it just used because it looks cool? How do I write a correct micro-benchmark in Java? Highlight \verb in inline text Suppose that $V$ is a vector space, and $W$ is a subspace of $V$. If   $V$ is finite dimensional, then prove $W$ too must be finite dimensional.   It seems intuitively obvious that the dimension of $W$ must be less than or equal to that of $V$, but I am uncertain of how specifically to show this to be true analytically.  This question has been asked before (). However, there is not a great degree of explanation in the given answers. The answer which labels itself as a hint is most similar to my thoughts towards an approach to proving this particular theorem.  I was thinking that if $\exists n\in\mathbb{R}:n=\text{dim}(V) \implies \exists B = \{\vec{v}_1,...,\vec{v}_n\}$, a basis for $V$, with $n$ vectors $\vec{v}_1, ..., \vec{v}_1 \in V$. Since $W$ is a subspace of $V \implies \beta$, a basis for $W$, which has $m$ vectors in it can be constructed by removing $k\in\mathbb{R}$ vectors from $B$ such that $B\setminus\{\vec{v}_i,...,\vec{v}_j\}=\beta$ where $\vec{v}_i,...,\vec{v}_j$ are those $k$ vectors. This means a basis for $W$ has $m=n-k$ vectors in it, and by the definition of bases $\implies \text{dim}(W)=m$, and since $n-k \in \mathbb{R}$ and $n-k=m \implies m\in\mathbb{R}$. Thus $W$ is finite. $\square$  It seems like there may be some fault with this argument, though. Why is the work done on a charge calculated from infinity? Why is the work done on a charge calculated from infinity to a point? Why not from one particular point to other? Prove that any two nontrivial subgroups of $\mathbb{Q}$ have nontrivial intersection I need to prove that any two nontrivial subgroups of $\mathbb{Q}$ have a nontrivial intersection as part of a larger proof that $\mathbb{Q}$ cannot be represented as a nontrivial direct product.    (Yes, I realize that there are other ways to prove that $\mathbb{Q}$ cannot be represented as a nontrivial direct product, but I am not interested in those, and will not accept answers showing me how to do that).  Based on the density of the rationals, I guess this idea kind of makes sense, but I am doing this proof for an abstract algebra class, and I don't think real analysis proofs would be accepted.   Everywhere I've looked online has said this is "trivial" or "easy to see" without bothering to actually prove it for those of us mere mortals who don't see it as such. As for myself, I have literally no idea where to even get started proving this. Could somebody help me figure this out (without giving me cutesy examples that aren't really all that related with the expectation that I should be able to figure it out from looking at those - that's not really how my brain works)?  Thank you. I am a Saint Lucian and will be traveling to Manchester, UK soon about which I am a bit nervous.  Run Down:   I own a small business in my country (in the Caribbean), I have a certificate mentioning the details of the business from my Government. I pay rent in my country, but I do not have any loans (I hate owing). I pay life insurance in my country. I have a US B-1/B-2 visa (Visitor &amp; Business). I have visited the US once (last year November) on my US B-1/B-2 visa.   What I will be doing in the UK:   I plan to spend three months out of the six months which would be granted if I get leave to enter stamp. I plan to skydive in the first week and spend a few nights at an Airbnb. I plan to visit the Manchester Football Club stadium, I'm not the biggest of football fans but since I'm in Manchester, why not! I plan to do some more sightseeing and enjoy my time there. After the first week I will be staying with my female friend who owns an apartment for the rest of the time showing me around and helping me experience the UK lifestyle until June.   Why I'm leaving in June:   I am a WordPress enthusiast (I'm a Programmer) and WordCamp EU will be happening in France this year so I plan to fly from the UK to Paris where I will spend three days then take another flight from there to my country.   How my flights will be:   I will be taking a one-way Thomas Cook flight from my country to Manchester but will be purchasing my ticket for Manchester to France ahead of time so I could show Thomas Cook as well as Immigration that I don't plan to stay.   My concerns, could someone address them point by point please?:   I have a bank account but I just use it for collecting easy payments from my clients and pay for expenses. I store my savings in a credit union because of higher interest rates paid than the bank. When I applied for my US visa I showed a credit union statement showing my total balance. I'm seeing online that I should provide statements but these statements would look very bad since I my bank account is just for expenses... I mean it sometimes gets as low as $30 then, when my monthly earnings come in from clients, I withdraw savings to the credit union and leave what I need in the account for expenses. I read another post and it said people with bank statements showing poor spending might have difficulty but it's simply because the account is used for expenses not savings. For the later stages of my visit (after the first week) I will be staying with my friend. I will simply be staying with her to save on cost of Airbnbs for the next two months three weeks, even though they could be cheap, free is better right? I will be booking the Airbnb, skydiving and tours prior to traveling to the UK, will that help? Is my stay too long for staying with someone? I won't be depending on her for anything since I have my own funds...it's just convenient that she has a place where I could stay. If it is then I don't have a problem shortening my stay to 1-2 weeks and just head back to my country and take a separate flight to Paris in June.   I am nervous because I really want to visit the UK and see my friend, I am also not sure if the minimum amount of time that could be given to a visitor is six months? What if I get a round trip for two weeks and the Immigration officer gives me exactly two weeks stay, is that possible? Or is the minimum they could give six months?  Your help is very much appreciated, I want soon to try backpacking (nomad life) but I don't think my poor heart could take the stress of worrying about immigration :(  Update  Since people in the comments are asking about my relationship with the friend I plan to stay with even though she's from my country originally but now a UK citizens and we are simply friends. Why do all Winforms programs require the [STAThread] attribute? Java Serialization with non serializable parts I have:  class MyClass extends MyClass2 implements Serializable {   //... }   In MyClass2 is a property that is not serializable. How can I serialize (and de-serialize) this object?  Correction: MyClass2 is, of course, not an interface but a class. How to copy the complete folder structure one location to another location include all files and sub folders I have folder structure like in Shared documents/East using CSOM and C# Main Folder   Folder1   Folder2      File1      File2  I have to copy all folders with files to Shared documents/West ....