How to choose the parent category that shows when using multiple custom categories
Determine if Term has Grandparent/Great-Grandparent
For a prime $p \equiv 1$ or $3$ (mod $8$), show that the equation $x^{2} + 2y^{2} = p$ has a solution. For a prime $p \equiv 1$ or $3$  (mod $8$), show that the equation $x^{2} + 2y^{2} = p$ has a solution.   I'd like to see an elementary proof. How to automount NTFS partitions? I have two NTFS partitions, and I don't want to mount them manually every-time I start Ubuntu.   How can I do this?  Is there a tool or a code to use?   If so, is it safe to automount? specially when they are being used by another OS? "Software Updater" & "Software & Updates" not working I havent used my Ubuntu for quite a long time.  Today, when I booted Ubuntu (14.04 LTS), the software updater and Software &amp; Updates (from Settings) are not working.  The software updater's window will show up for around 1 sec, but it will close automatically. Software &amp; Updates is never shown, no matter how many times I click the icon. Note that neither of them show any error message.  Those 2 features were working perfectly the last time I used Ubuntu. I want to do some system upgrade and update.  For now, I use sudo apt-get update &amp;&amp; sudo apt-get upgrade. This command runs very fast, meaning (I think) my Ubuntu is not updated properly. The update from software updater always has a big size hence the download should be slower.  What should I do?   Can I just uninstall and reinstall both of them? If I can, how to do it?  Just to clarify, I don't want to upgrade my Ubuntu version (I want the LTS one), I just want to upgrade the software/packages I used.  Thanks for your time. Derive Variance of regression coefficient in simple linear regression Prove if $(a,p)=1$, then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$. let $p$ be a prime and let $a$ be an integer not divisible by $p$, that is $(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$.  Since $p\nmid a$, $p\nmid\{a,2a,3a,...,(p-1)a\}$ Since $p|p$, $p|\{pa\}$ which is exactly one element in the set and is therefore a complete residue system.   Wondering if this is an acceptable proof to this theorem. Using mask as a garbage matte to hide an object I'm trying to use a mask to hide a part of a render layer, so that when the render layer is shown over a movie, some parts in the movie will appear to be in front of the render layer.  For example, assume I want to add a box to this image, so that it would look like it's behind the person:    I created the following simple object:    If I render the object, it looks great - the object is solid, everything else is transparent (I didn't use any 'sky'):     Now I create a mask to describe the standing person (or at least the part that would otherwise be covered by the box):    What I want now is to create a "hole" in the rendered box, so that when I use Alpha-over, the box look as if it's behind the person, but I can't find an easy way to do that. My first guess was to use 'Set alpha' to apply the mask (inverted) as an alpha for the rendered layer, but this doesn't look good:    The mask did create a "hole" where it should, but now all the other pixels are not transparent. How can I "retain" the transparent pixels while setting more pixels to be transparent based on the mask (use it as a matte)?  Note: I can do this with (I'm ashamed to say) some math work on the alpha channel, but it just doesn't seem right:    Thanks, and sorry for the [oh so] long question. I am trying to understand what mathematics is really built up of. I thought mathematical logic was the foundation of everything. But from reading a book in mathematical logic, they use "="(equals-sign), functions and relations.  Now is the "=" taken as undefined? I have seen it been defined in terms of the identity relation.  But in order to talk about functions and relations you need set theory.  However,  seems to be a part of mathematical logic.  Does this mean that (naive) set theory comes before sentential and predicate logic? Is (naive)set-theory at the absolute bottom, where we can define relations and functions and the eqality relation. And then comes sentential logic, and then predicate logic?  I am a little confused because when I took an introductory course, we had a little logic before set-theory. But now I see in another book on introduction to proofs that set-theory is in a chapter before logic. So what is at the bottom/start of mathematics, logic or set theory?, or is it circular at the bottom?  Can this be how it is at the bottom?  naive set-theory $\rightarrow$ sentential logic $\rightarrow $ predicate logic $\rightarrow$ axiomatic set-theory(ZFC) $\rightarrow$ mathematics  (But the problem with this explanation is that it seems that some naive-set theory proofs use logic...)  (The arrows are of course not "logical" arrows.)  simple explanation of the problem:  a book on logic uses at the start: functions, relations, sets, ordered pairs, "="  a book on set theory uses at the start: logical deductions like this: "$B \subseteq A$", means every element in B is in A, so if $C \subseteq B, B \subseteq A$, a proof can be "since every element in C is in B, and every element in B is in A, every element of C is in A: $C \subseteq A$". But this is first order logic? ($(c \rightarrow b \wedge b \rightarrow a)\rightarrow (c\rightarrow a)$).  Hence, both started from each other? Take $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$ I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.  Proof: Suppse that $|B|&lt;2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $ \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.  What do you think?  Thank you, Shir Who's the random kid standing in the gathering at the end? Why am I not able to shutdown, log-off and restart after an upgrade to Ubuntu 13.10? After the update neither my desktop (Asus) nor my laptop (HP) respond to the shut-down, log off or restart commands in the right top tool bar.  How do I fix the issue? How to remove the "Could not connect to the iTunes store" error message when your network does not allow access to the iTunes Store The speed of light as it approaches a massive body No matter how fast you go, you will aways perceive the speed of light as constant. Taking that as a fact, the special relativity theory was formulated. Now, for what I understand about general relativity, just by standing here on the surface of the planet under the effect of gravity, my frame of reference is a little different from someone in geostationary orbit, and time goes by a little slower for me.  So, considering me and someone in geostationary orbit, does both of us would perceive light at the same speed? If so, does that means that the light is slower down here than up there? Numerical Approximation of Differential Equations with Midpoint Method So, my update to 13.04 didn't go very well. Due to the lack of a decent driver (Yes, i'm unlucky to have a legacy driver which doesn't play nice with the new Xorg.), I had to mess around with stuff to first troubleshoot and afterwards "fix" my system.  I've finally got unity back, but now CtrlAltarrow and CtrlAltShiftarrow don't work anymore. They are bound in "keyboard" > "shortcuts", but nothing happens when I try it. A strange thing is that other keyboard shortcuts, like ctrlaltt to show terminal are still working.  I noticed that CtrlAltarrow doesn't show in the list with "unity shortcuts" (the list that shows when you hold the super key).  Does anyone know how to re-enable this way of workspace switching?  I enabled workspaces in the appearance menu. It doesn't have any effect though. Switching via the workspace-switcher in the unity dash works fine though.  CCSM is installed, but I can't seem to find a way to add keyboard bindings there.     System: ubuntu 13.04 64bit Proving that if $G/Z(G)$ is cyclic, then $G$ is abelian Possible Duplicate:             If $G$ is a group and $Z(G)$ the center of $G$, show that if $G/Z(G)$ is cyclic, then $G$ is abelian.  This is what I have so far:  We know that all cyclic groups are abelian. This means $G/Z(G)$ is abelian.  $Z(G)= \{z \in G \mid zx=xz \text{ for all } x \in G \}$.  So $Z(G)$ is abelian.    Is it sufficient to say that since $G/Z(G)$ and $Z(G)$ are both abelian, $G$ must be abelian? How to print results in SQLcmd window If $E$ is a connected space, then so is $\overline{E}$. I have to prove that if $E$ is a connected space, then so is $\overline{E}$ (the closure of $E$) a connected space. I tried to prove the contrapositive. So suppose that $\overline{E}$ is not connected in a metric space $X$. This implies there are $A\subset X$ and $B\subset X$ such that $A\cap B=\emptyset$, $A\cap \overline{E}\neq \emptyset$, $B\cap \overline{E}\neq \emptyset$ and $\overline{E}\subset A\cup B$. I will try to prove that this $A$ and $B$ will also work for $E$. Suppose that $A\cap E=\emptyset$ and $B\cap E=\emptyset$, this implies that $A\cup B$ is a subset of the limit points of $E$, because $A\cap \overline{E}\neq \emptyset$ and $B\cap \overline{E}\neq \emptyset$. This contradicts the fact that $\overline{E}\subset A\cup B$, so we have to conclude that $A\cap E\neq \emptyset\neq B\cap E$. Since $E\subset\overline{E}\subset A\cup B$, we have that $E$ is also not connected.  My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance!   Edit1: The definition we use for a connected space $E$ in a metric space $X$ is that if $E$ is connected, then there are no open $A,B\subset X$ such that $A\cap B=\emptyset$, $A\cap E\neq \emptyset\neq B\cap E$ and $E\subset A\cup B$.  Edit2: $A$ and $B$ must be open. Why prefer the 18-55mm and 55-250mm lenses vs 18-200mm? Do tenses in a time clause never back-shift in reported speech? Do tenses in a time clause never back-shift in reported speech?  For example, does      "I went to Italy when I finished school."    become      She said she had gone to Spain when she finished school.   -or-   She said she had gone to Spain when she had finished school.   and does      "I will go to Italy when I finish school."    become     She said she would go to Italy when she finishes school.   -or-   She said she would go to Italy when she finished school. I'm a Windows/Mac/Ubuntu and Android user, so I know what I'm talking about, when I say: How do I open binary image files? (.img) They just won't open, on any OS…  I'm an Android dev… I'm currently working on a ROM, (I also program, using Windows) but I need to extract files, from .img files. I've converted them to .ext4.img but they just aren't recognized by Linux (Definitly not by Android), by Mac OS or Windows. In other words, I can't open, extract or mount them.  Can anyone help me? I'm kinda confused…