Why we have to do separate packaging to applications for 32-bit and 64-bit Ubuntu Operating systems.I mean why some applications are making 32-bit and 64-bit separately and why some applications are not ?   is there any advantage if we made 64-bit packaging for a 64-bit OS ?
What is the difference between 32-bit and 64-bit Ubuntu? I've heard the 64-bit platform performs better and can detect more than 4GB of RAM. Also, while some apps haven't ported to 64-bit yet, ia32-libs lets a 64-bit machine run them.  If so, why not promote 64-bit over 32-bit?
I am using my notes app on OSX 10.11.5 and iOS 8. For the most part, things are working, but I am noticing that many of my older notes are not searchable. When I type something in the search box, these older notes do not appear in search results even if there is a match. However, if I edit a note or create a new one, it is fully searchable.  My guess is that somehow the search indexes for my notes got corrupted. Is there any way to fix this or rebuild the indexes? Can we balance the whitespace above/below headings? In January, some additional whitespace  below h1, h2 and h3 headings,  more whitespace below the headers than above them.  Could we have the whitespace above the headings be more than the whitespace below them?  I find some posts a bit hard to read, like the current rendering on the left:   &nbsp; &nbsp; &nbsp; Is it OK to crosspost Stack Overflow questions, now that the Portuguese SO is open? How do you overlay an image on a shader in Cycles? Finite groups with exactly one maximal subgroup I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated.     Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.   Now, does this mean exactly one conjugacy class of maximal subgroups? If it really means exactly one maximal subgroup, then what can we say about a finite group that possesses exactly one class of (at least two) maximal subgroups that are all conjugate to each other? I have a SQL script that creates a package with a comment containing an ampersand (&amp;). When I run the script from SQL Plus, I am prompted to enter a substitute value for the string starting with &amp;. How do I disable this feature so that SQL Plus ignores the ampersand? Prevent questions on Hot List from being upvoted by casual visitors (only rep is from association bonus) How to highlight an entire paragraph? I'm trying to find a movie with an underground facility with people wearing white suits. The people have been affected such that they show no curiosity. There is one person that does and asks one of his co-workers where a tube leads to.      At the end there is an uprising where everyone gets to see the surface and sunlight.   What is the name of this movie? Prevent people from downloading / copying / making a copy of data in Google doc Windows batch command(s) to read first line from text file How can I read the first line from a text file using a Windows batch file? Since the file is large I only want to deal with the first line. Prove that the range of the entire function $z^2+\cos(z)$ is all of $\mathbb{C}$. Prove that the range of the entire function $z^2+\cos(z)$ is all of $\mathbb{C}$.   I'm aware this question has been  already, but the explanations were a little shakey and referenced a google sample preview of a text of whose terminology I wasn't very familiar with, so I would like some more input if possible.  Here's my attempt:  Let $f(z)=z^2+\cos(z)$. Through the power series of the cosine function $$ f(z)=z^2+1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots=1+\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots, $$ so $$ g(z):=\frac{f(z)-1}{z} $$ has a removable singularity at $z=0$ and is effectively an entire odd function. Picard's little theorem asserts that the range of $g$ is all of $\mathbb{C} $: if there exists some $z_0$ such that $g(z)\neq z_0$ for all $z\in\mathbb{C}$, then $-z_0$ is also missed by oddness of $g$, which is a contradiction (since $g(0)=0$ we rule the possibility that $z_0=0$). (I'm not sure about the following step) It follows that $zg(z)$ and so $zg(z)+1=f(z)$ is surjective.  Is this method valid? Would I need further justification to show that $z\mapsto z$ and $z\mapsto g(z)$ surjective imply its product -- $z\mapsto zg(z)$ -- is also surjective (if this is even true). Any input is greatly appreciated.  Edit: After thinking about this for a moment longer, I realize this logic would also show that the range of any entire even function is all of $\mathbb{C}$, which seems like quite a strong conclusion. This makes feel like there are holes in my logic. Is there a term for words that when reversed, form other words? I'm aware of what a palindrome is. What do you call words that, when reversed, form other words, such as ton (not) and part (trap)? Incorrect icons on hot network questions As the title, I was asked to show that $2^{44}-1$ is divisible by $89$.  My first attempt is to change the expression into $(2^{22}-1)(2^{22}+1)$.  Then further simplified it into $(2^{11}-1)(2^{11}+1)(2^{22}+1)$, I used my calculator and was able to show that $2^{11}-1$ is divisible by $89$ but then I don't know how to show it with modular arithmetic. I do think that it is quite similar to the form where we can use the Fermat's little theorem. $(\sqrt{2})^{88}-1$. (Though I do understand Flt can only be applied to integer.)  Can someone tell me whether I can take square root in modular arithmetic as well? I am still pretty new to the modular arithmetic. Thank you very much. I'm trying to understand the UMP of the product functor as right adjoint to the diagonal functor but I'm having difficulties understanding the notation in Steve Awodey's . On page 210 he uses $(f_1, f_2)$, $\langle f_1, f_2 \rangle$ and $(f_1 \times f_2)$ all within one paragraph. What are those? How are these defined and how do they differ? "find: missing argument to `-exec'" when using "-exec rm -f {}\" LibreOffice Calc - Disable rounding There are a  on this topic, but no simple and definitive answers.    Can someone please tell me how to permanently disable cell rounding on the .csv file which is created by script?  Basically, '377635637222436751' is being converted to '377635637222437000' automatically and that isn't really acceptable for my needs.  I've tried this: Tools->Options->Calc->Calculate "Precision as shown"   and this: Format > Cells > Numbers > Category > (scroll down for) Text. (but considering my script creates the .csv -- this option isn't very good since it must be done after the file is created)  And neither is working.  The number stays rounded -- even after formatting the column after the file is created.  Does anyone have any suggestions? Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic. How to perform the modular reduce of Rijndael's finite field I am trying to understand how to calculate the modular reduction of Rijndael's finite field.  The example on  page says that {53} • {CA} = {01}, because (11111101111110 mod 100011011) = {3F7E mod 11B} = {01} = 1.  I am able to get to the point where I have 3F7E mod 11B, but cant seem to figure out how the reduction works.        11111101111110 (mod) 100011011      ^100011011             1110000011110       ^100011011             110110101110        ^100011011             10101110110         ^100011011             0100011010            ^100011011 &lt;-- here, right shifted two places?             00000001   From this example, why does it right shift two places at one point?  I also found an example that 0x1b*0xaa equals 0x8c. Trying to solve this myself I get to the point: 0xE0E mod 0x11B   111000001110 ^100011011   11011010110  ^100011011    1010111010   ^100011011     010001100    ^100011011      10010111   This makes it 0x97