Given an arbitrary nonzero vector space $V$, is there a nonzero linear functional on $V$, without assuming axiom of choice? I know that by assuming existence of a basis for $V$, we can consider the dual basis for a subspace of $V^*$, which justifies the existence of nonzero linear functional, but this argument fails without axiom of choice. I guess there may not always be a nonzero linear functional, but my knowledge on axiom of choice and infinite-dimensional vector spaces is lacking. A quick search on Google fails to give an answer.  Note that I am not talking about normed spaces or continuous linear functionals, just plain vector spaces with no additional structure.
For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of choice ?
a, b are integers. p is prime. I want to prove: $(a+b)^{p} \equiv a^p + b^p \pmod p$  I know about Fermat's little theorem, but I still can't get it I know this is valid: $(a+b)^{p} \equiv a+b \pmod p$ but from there I don't know what to do.  Also I thought about $(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1} \binom{p}{k} a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$ Any ideas?  Thanks! Story identification: Old man has hacker neighbor who mysteriously dies Does anyone know this story? It was a novelette published with another one (I don't remember what it was) in . The premise was that the narrator is an old man whose neighbor, a hacker, dies under mysterious circumstances. A younger woman who is an expert in computer security is hired by the police to investigate what happened, and she has a fling with the narrator. After a while she dies as well, and the narrator realizes that it is an evil force that comes through electric lines is responsible for the killings. The title has interesting formatting in that it ends with a square (much like a command prompt cursor: [_] ) I'm creating a color object using the following code.  curView.backgroundColor = [[UIColor alloc] initWithHue:229 saturation:40 brightness:75 alpha:1];   How can I retrieve RGB values from the created color object? Why is variance squared? I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does. What to do with cross-site duplicates? What's the right course of action for cross-site duplicates?  There are two flavors:   A question is posted which belongs on another site and ends up already having been asked there. I can't find an example right now, but it's bound to happen soon if it hasn't already. The same question is posted in two places at the same time. Take  for example, which is a duplicate of , posted within a half hour of each other by the same user.   Possible actions:   Move it to the proper site and close it once it gets there as a duplicate. Provide a way to close it right away as a cross-site duplicate, without moving it first. Flag it and let a moderator merge them. Close it as "no longer relevant" or something else.   Additionally, flavor 2 is a bit of a special case; perhaps the question really spans both sites in terms of content and intended audience. Should multiple-postings in this case be allowed? If so, should they somehow be linked to each other? If not, how do we avoid a migration war? How can I know if a branch has been already merged into master? Word for a task which is flawed or doomed to failure but which you have to do anyway? I am looking at the 4-vector treatment of special relativity, but I have had no formal training in Tensor algebra and thus am having difficulty understanding some of the concepts which appear.  One such concept is the notion of contra- and co-variant vectors.  describes a contra-variant vector as varying with the co-ordinates, i.e. the inverse of the reference axes. It gives the example of a change of basis matrix $\mathbf{M}$, and a contra-variant vector $\vec{v}$ and states that in the new set of co-ordinates, we have:  $$\vec{v}'=\mathbf{M}\vec{v}$$  It then states that a covariant vector varies with the co-ordinate axes.     Does this mean that for a covariant vector $\vec{u}$ and change of basis matrix $\mathbf{M}$ we have: $$\vec{u}'=\mathbf{M}^{-1}\vec{u}?$$   I am also having problems understanding the concept of raising and lowering indices. For instance, Wikipedia states that for two four-vectors $\vec{A},\vec{B}$ and using Einstein summation we have:  $$\left\langle \vec{A} \middle|\vec{B} \right\rangle=A^{\mu}B_{\mu}=A_{\nu}B^{\nu}$$  And I am having difficulty understanding how this relates to the other definition of the inner product:  $$\left\langle \vec{A} \middle|\vec{B} \right\rangle =\mathbf{A}^{T}\mathbf{\eta}\mathbf{B}$$  If anyone could help give me a more intuitive understanding of these concepts I would be grateful. Why does $x$ divided by zero not equal $x$? What do you call the process of combining two words to create a new one? Please anyone tell me when we have to use "has been, have been , had been" and something  started in past and still continuing is it possible to use has been ? I know that, whenever I want to see how some control sequence, say \mycontrolsequence, is defined, I can use \show\mycontrolsequence  then compile my code and look up the .log file. However, robust commands seem impervious to \show. For instance, if I compile the following code, \documentclass{article}  \newcommand\mycommand{foo} \show\mycommand  \DeclareRobustCommand{\myrobustcommand}{bar} \show\myrobustcommand  \begin{document} hello \end{document}  my log file contains  \mycommand=\long macro: -&gt;foo. l.4 \show\mycommand  \myrobustcommand=macro: -&gt;\protect \myrobustcommand  . l.7 \show\myrobustcommand   Note that, whereas the definition of \mycommand is exposed (foo), nothing transpires of \myrobustcommand's definition. Of course, in practice, I could always identify in which file a given robust command is defined and look up its definition there. However, finding the file in question may not be obvious. Hence my question: Is there an alternative way of peering into the definition of a robust command? Full List of inventory items After beating the game, I still have several empty inventory spots.  I most likely missed several items in my addicted frenzy to finish the game.  What are all the items that can be found in the game? And where can they found? SSH Server stops working after reboot, caused by missing /var/run/sshd A word for a worldly wise person who pretends to be naïve? What is a word for someone who is experienced and wise, but who deliberately acts naïve?    I don’t intend it for sarcastic use; I’m trying to describe someone succeeding in making people think that that person truly is naïve.   Is there a word for this? I am trying to prove the following statement.     If a field $K$ is not algebraically closed, then any $K$-variety $V\subset\mathbb{A}^n$ can be written as the zero set of a single polynomial in $K[X_1,X_2,...,X_n].$     Suppose $V=V(f_1,f_2,...,f_n)$, and I would want to find a polynomial $g\in K[X_1,X_2,...,X_n]$ such that $V=V(g)$.  I know that if I can find a polynomial $\phi\in K[X_1,X_2,...,X_n]$ whose only zero is $(0,0,...,0)$, then it is done, since we could take $g=\phi(f_1,f_2,...,f_n)$.  For $n=1$, then I would just choose $\phi=X$, but I do not even know how to proceed to the case $n=2$. Maybe we could choose an irreducible polynomial of degree $&gt;1$.  Any help will be appreciated. Deleting Objects in JavaScript Every finite commutative ring with no zero divisors contains a multiplicative identity? Short brandable domain vs. long keyword domain I need a counsel about my websites' domain.  Let's say I have just bought two domains: xyz.com and xyzphotography.com.  As we can see, the first one is shorter but way less meaningful than the second one. The second one contains a photography keyword but it's longer.  I have only one website which is my photography portfolio. Can you advise me how to set those domains to make it all as SEO-friendly as possible? Which one should be the main one, does the rest should work on their own or have some kind of SEO-friendly redirect to the main domain?