Is she not a girl? Isn't she a girl? Am I not here? Am not I here?   Are tey all correct or right structured?
(1) Does he not know? (2) Doesn't he know?  I don't usually see and hear questions formed in the first style. I was even surprised to know that it is grammatically correct. And actually it is the most &quot;proper&quot; way to form an interrogative sentence.  shows that the first style was much more popular before than now. What are your suggestions on choosing between these styles? And I will probably add the third one.  (3) He does not know?  EDIT: The answers have explained how to form a question, but have not actually provided an account on the usage of the first two forms. Especially I'm interested in the first form: does it sound formal, is it used in speech or only in writing, et cetera.
I created a bunch of filters in Gmail for a "master account" that receives forwarded mail from all of my accounts.  The filter list and web of labels/folders has grown to be rather long, and therefore a bit burdensome to manage. I would like to edit / delete / consolidate a lot of these filter rules (and folders) but it seems the only way I can interface with them is through the laundry list at Setting/Filters. It'd be cool there was an option to "View/Edit Filter" on an individual email's page, but I'm not sure if something like that currently exists.  What's the best way to view a filter rule currently being applied to an email (in Gmail)? What is the probability that all three tosses result in the same outcome? A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used? I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance. warning message at command line `sudo` Typo in First Posts queue If you go to the First Posts review queue, and if there's an answer in the queue and if you hover with your mouse over "No Action Needed", then you get a tooltip "this answers seems to be valid":  But it should be "this answer seems to be valid". This is my first question in /photography. I'm a very amateur photographer, doing things wrong and by eye with lots of trial-and-error all the time.  I currently own a reflex Nikon D70 my uncle gave us after he bought a better one. It has some nice settings, and he bought a a Nikkor 28-80mm F/3.5-5.6 zoom lens for it. It behaves nicely, but it's getting old (it has those old cartridges before SD cards existed) and every new charge takes longer than the last.  I'm planning to upgrade to a better camera, and there are some neat offers on a nearby superstore (MediaMarkt), this one catching my eye:  Sony DSC-H400 63x, Sensor Super HAD 1/2/3 CCD 20.1MP -> for less than 300€ taxes included  I don't have any type of extra information aside from that, all I can get from the official site is clinical data from other cameras that I suspect it's not 100% accurate, and there's no photography expert on the store to ask.  The common things I use my D70 for are: taking pictures of scenery (castles, towns, forests, waves hitting rocks, etc...), nearby (&lt;5m) persons and objects, and well, things where I want to save a memory in digital format, not for professional purposes at all.  Being that my objective is to have a decent, good-zooming (superzoom looks awesome), amateur (but with manual settings) camera, is this Sony a good replacement? Will I miss any features? Is there any better, similarly-priced alternative you know of?  Basically, I want to know whether I will get better quality photos than I currently do, given the examples, with any averagely-priced bridge camera similar to that Sony, and if it's worth it.  examples of current camera results  Full zoom-out of town at night, bad quality I know; notice the pink blur at the left.    Maximum zoom on that pink blur from almost the same spot:    Daylight, cloudy, modified with Adobe Lightroom, scenery:    Same day, same beach, different settings:    Same day, same beach, rocks: Polynomial in $n$ variables Let $f$ be polynomial in $n$ variables. If we have $f(a_1,\ldots,a_n) = 0, a_i \in S_i$ and $S_1,\ldots,S_n$ are infinite subsets of a field $k$, how can I show that $f=0$? If posed with the subject question, given that I will not do the action in question, then what is the correct answer     No, I will not do it.   or      Yes, I will not do it.   #1 sounds better to me, but #2 is more logical. #1 sounds somewhat paradoxical in context of the question. So this is what I'm trying to do in Drupal 6:  Node type A is video, node type B is a person Nodetype A has a reference field where you can enter a reference to any Node from nodetype B  how can I create a block view which will be displayed on nodes from node type b, which will show all the nodes from node type A where the current node from node type B is referenced? What are the regulations for transporting medication abroad? This is in reference to medications, both prescription and non-prescription, not illicit drugs. Since the legality may vary on country, let's say a round trip from the US to Singapore, as an example.  When travelling with prescription medication, I am aware that you are supposed to bring your prescription with you. What about over-the-counter medication, such as Tylenol/aspirin? How about medication which is restricted or prescription-only in one country, yet available OTC in another (think Viagra)? Is it possible to get the data off of the hard drive of a PlayStation 3 that has crashed? I have already voided the warranty so I am not worried about that and I have access to quite a few tools.   I have a new PlayStation and a few saves that were not backed up. I am not worried about them but if it is possible I would love to get it moved over.   The old PlayStation will not turn on at all and I think I am going to have to pull the drive. My main concern is if I am able to transfer the files without running into any issues. I have heard that some of the stuff is specific to each PlayStation, is that true? In which situations is it OK to omit articles in short sentences? Recently, in many new online software I often see that developers omit articles in short sentences, especially on action buttons or tooltips.     For instance: "Add new task" (not "Add a new task"), "Create project" (not "Create a project"), but often in the same software I can see "Make a copy". What are situations it can be justified? How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? Mapping JSF .xhtml files to no extension I'm new to Linux (currently running Ubuntu 10.04) and I have just finished setting up SSH access to my Linux Machine.  Currently, I have to use ssh root@x.x.x.x to connect but would much rather be able to swap the IP for the computer name (like in Windows with \\name) or a domain name (like computername.example.com). I don't really know where to start so any help would be most appreciated.  Please go slowly, as mentioned - I am still new to this.  EDIT 1 Completely forgot to mention that I am trying to connect from Windows 7 (via PuTTY) - sorry. \overline outside of math mode If I want to draw an overline over some text, not an equation, how can I do it? \overline seems to only work in math mode, and when used outside of math mode, it automatically says "Missing $ inserted". If the imaginary part of an entire function is never zero, the function is constant Let $f : \mathbb{C} \to \{z\in\mathbb{C}:\Im(z)\neq0\} $ entire . Show that $f$ is constant.  I took $g(z)=\frac{1}{f(z)}$ and I think that g is bounded, therefore it is constant (due to Louville's theorem) but how do I prove it? Is $|g(z)|=\frac{1}{|f(z)|}=\frac{1}{u^2+v^2}\le M$ valid? \pagestyle{empty} doesn't seem to work for page with \maketitle Pick an urn and pick two balls from the urn There are 5 urns, the nth urn contains n − 1 red balls and 5 − n green balls (in total 4 ineach urn). We pick an urn at random and take 2 balls out of it without replacement.  (a) What is the probability the second ball is green?  (b) What is the probability the second ball is green given the ﬁrst is green?  (c) What is the probability we picked either the ﬁrst urn or second urn given both balls are green?  for (a), I did:   P(2nd ball is green) = 5*[{P(i th urn picked) * P(first ball drawn red/ ith urn picked) P(2nd ball drwan green/first ball red)}+ {P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)}]  = 5*[[{1/5 * (n-1)/4 * (5-n)/3 } + {1/5 * (5-n)/4 * (4-n)/3}] = 5* [{(n-1)(5-n)/60} + {(5-n)(4-n)/ 60}]  =[(n-1)(5-n) + (5-n)(4-n)]/12 = (5-n)(n-1+4-n)/12 = (5-n)*3/12 = (5-n)/4  And for b): 5*{P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)} = 5*{1/5 * (5-n)/4 * (4-n)/3} =(4-n)(5-n)/12  c), P(1st urn or 2nd urn/ both are green) = [ [P(1st or 2nd]* (4-n)(5-n)/6 ] / [{P(1st or 2nd]* (4-n)(5-n)/6} + {{P(3rd or 4th]* (4-n)(5-n)/6}] = {(4-n)(5-n)/15} /  2*{(4-n)(5-n)/15} = 1/2  But I am really uncertain about my solutions. If anyone has different solution or easier way to get the answer please let me know! Thanks a lot! Show that if $p$ is prime then $\Bbb Z_p$ is a field Check my proof please.  Let $\Bbb Z_p:=\Bbb Z/p\Bbb Z$ be the quotient ring modulo $p$. I want to prove that if $p$ is prime then $\Bbb Z_p$ is a field.  Known facts about the quotient rings of the kind $\Bbb Z_n$ (I dont list all the characteristics of a ring, just some ones):   They are commutative rings with unity. $[0]=[n]$ is the identity of addition or zero. Then $[n]+[a]=[a]$ and $[n]\cdot[a]=[n]$ for all $[a]\in\Bbb Z_n$. $[1]$ is the multiplicative identity or unity. Then $[1]\cdot[a]=[a]$ for all $[a]\in\Bbb Z_n$. The addition is defined as $[a]+[b]=[a+b]$, and the multiplication is defined as $[a]\cdot[b]=[ab]$. $[a]=[b]$ means that exists $z\in\Bbb Z$ such that $a=b+nz$. $|\Bbb Z_p|=p$.   A ring $\Bbb Z_p$ dont have zero divisors, i.e. doesnt exist $[a],[b]\in\Bbb Z_p$ distinct of $[p]$ such that $[a]\cdot[b]=[p]$. Proof by contradiction: suppose that exists such $[a],[b]$ distinct of $[p]$ that are zero divisors. Then exist some $z_j\in\Bbb Z$ such that:  $$(a+pz_1)(b+pz_2)=pz_3\iff ab+pz_4=pz_3\iff ab=pz_5$$  but this is a contradiction with the assumption that $a,b\neq p$ and $p$ prime, so dont exist divisors of zero on any $\Bbb Z_p$.$\Box$  For any $[a]\in\Bbb Z_p$ distinct of $[p]$ exist $[b]$ such that $[a][b]=[1]$. Because $\Bbb Z_p$ dont have zero divisors and is finite for some $[a]\neq[p]$ we have that if $[b]\neq[c]$ then $[a][b]\neq[a][c]$. Proof by contradiction: if $[a][b]=[a][c]$ in the above conditions then  $$[a][c]=[ac]=[ab]=[a][b]\iff ac+pz_1=ab+pz_2\iff a(c-b)=pz_3$$  for some $z_j\in\Bbb Z$. But $pz_3\in[p]$, then $[a][c-b]=[p]$ but because there are no zero divisors in $\Bbb Z_p$ then or $[a]=[p]$ or $[c-b]=[p]$ what contradicts the above conditions.$\Box$  Because $[a][b]\neq[a][c]$ for $[a]\neq[p]$ and $[b]\neq[c]$ then for any $[a]\neq[p]$ we have that  $$[a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1]$$  In other words: every element of $\Bbb Z_p$ but $[p]$ have a multiplicative inverse as stated above.$\Box$  Because $\Bbb Z_p$ is a commutative ring with $[1]\neq[0]$ and have multiplicative inverse for all their elements but $[p]=[0]$ then $\Bbb Z_p$ is a field.$\Box$