Android app for decorating destination numbers
Prepend phone numbers with a calling card dial in number
$n &gt; 2$ and $n$ is a prime number then $n$ is odd.  Prove by contradiction assume $n$ is even then there is some $k\in\Bbb N, n = 2k$ then $n &gt;2, 2k &gt; 2 , k &gt; 1$  Is this  a sufficient proof? Nice looking progress bar in java How to reverse a user input without using array or any library function(any function for reversing)? Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost. I need to use an alias in the WHERE clause, but It keeps telling me that its an unknown column. Is there any way to get around this issue? I need to select records that have a rating higher than x. Rating is calculated as the following alias:  sum(reviews.rev_rating)/count(reviews.rev_id) as avg_rating How can I break on an unhandled exception in ASP.NET? Is there a general theorem stating why the restricted Lorentz group's exponential map is surjective? The exponential map for the restricted Lorentz group is surjective. An outline of why is shown on the wiki page .  Is there a more general theorem that states that for some class of Lie groups or Riemannian manifolds (which includes the restricted Lorentz group), the exponential map is surjective?  There is a theorem stating that compact, connected Lie groups have surjective exponential maps. But as the restricted Lorentz group is not compact, this isn't applicable. Why is the [I'm Done] button unavailable after only flagging comments in the First Posts review queue? I just flagged two comments during a review of a question in the First Posts queue (on another SE site). Then I wanted to select the button I'm Done, but it's unavailable (I can only select No action needed). Even if I refresh my page (while still on that review page), it remains the same.  If however I also downvote the question (only temporary, as a test, since I don't want to downvote the question being reviewed), then it seems to work as I'd expect it to: the button I'm Done becomes available, and No action needed becomes unavailable.  Why is the I'm Done button unavailable after only flagging comments in the First Posts review queue?  Notes:   I'm not convinced that you're supposed to review the actual post, and not (only) the comments underneath it (as mentioned in 's comment). Because "if" I would post a new comment, or just upvote an existing comment, I can actually select I'm Done. While in this case I flagged existing comments which IMO had become obsolete. Flagging comments is "acting on comments". And if my flag gets accepted (in this case: "no longer needed"), and the comment is actually deleted (by the mod), then it makes the question a bit more readable (= don't waste your time digesting a comment which is obsolete). Therefor "IMO", in the end I also "acted on the post". Think about all those posts with a zillion of comments, of which many are obsolete (e.g because the post got updated already). I am not sure (yet) if a similar issue exist in any other review queue (i.e. the Late Answers queue). Switching languages in bibliography with biblatex I would like to have a bibliography with both Cyrillic and Latin and the main document language being English. How can I make biblatex to automatically switch between languages? (without using XeTeX as suggested by some) The minimal example returns an error if the main language is English. However, with the main language being Russian the document works.  \documentclass{article}  \usepackage[utf8]{inputenc} \usepackage[russian,english]{babel}  \usepackage[backend = biber,babel=other]{biblatex} \usepackage{csquotes} \addbibresource{library.bib}  \begin{document} %\selectlanguage{russian} %if the main document language is Russian the bibliography is compiled without any problem. This is a test. \parencite{Bara2006,Baranov2001}  \printbibliography  \end{document}   The .bib entries are  @book{Bara2006, address = {New York}, author = {Bara, Judith}, publisher = {Routledge}, title = {{English Citation entry}}, year = {2006} }  @book{Baranov2001, address = {Санкт-Петербугр}, author = {Баранов, Николай А}, title = {{Эволюция взглядов}}, year = {2001} } I am confused. I don't understand the difference a ARMA and a GARCH process.. to me there are the same no ?  Here is the (G)ARCH(p, q) process  $$\sigma_t^2 =      \underbrace{     \underbrace{      \alpha_0       + \sum_{i=1}^q \alpha_ir_{t-i}^2}     _{ARCH}        + \sum_{i=1}^p\beta_i\sigma_{t-i}^2}     _{GARCH}$$  And here is the ARMA($p, q$):  $$ X_t = c + \varepsilon_t +  \sum_{i=1}^p \varphi_i X_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i}.\,$$  Is the ARMA simply an extension of the GARCH, GARCH being used only for returns and with the assumption $r = \sigma\varepsilon$ where $\varepsilon$ follows a strong white process? How do you solve a recurrence with a summation function inside: $t(n) = 1 + \sum\limits_{j=0}^{n-1} t(j)$ Show that  $$t(n) = 1 + \sum_{ j=0}^{n-1} t(j)$$  is the same as  $$t(n) = 2^n$$  Initial condition $t(0) = 1$ This is a question from the free Harvard online abstract algebra .  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see  post.  This problem is from assignment $5$.     Prove directly that distinct cosets do not overlap.   Let $H$ be a subgroup of $G$ and let $a$, $b$, and $c$ be elements of $G$ such that $b\not\in aH$ and $c$ is in both $aH$ and $bH$.  Then there are elements $h$ and $h^\prime$ in $H$ such that $c=ah$ and $c=bh^\prime$.  So $ah=bh^\prime$ and $b=ahh^{\prime -1}$.  But $hh^{\prime -1}\in H$ so $b\in aH$.  This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset.  Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.  Thanks. Curve Bevel function, scale? Is there a way to scale up the bevel object on one end of a curve, a little bit over each point?  I notice the vertex "weight" slider will tweak how much influence each vertex has on the others, is there something like this that can scale the bevel object up a little bit?  Preferably with the option to constrain X,Y,Z independently Column vs Field: have I been using these terms incorrectly? I feel kind of embarrassed here, I've always used the terms "column" and "field" completely interchangeably, which recently caused some confusion in a technical discussion.   I was told, though, that this wasn't correct, that it should be (translating each term into spreadsheet terminology, ignoring data types and all the other stuff that make databases useful):   Database Column: like a spreadsheet column  Database Record: like a spreadsheet row Database Field: like a spreadsheet "cell" (a specific column of a specific row)   Is this right? I could have sworn that column and field are used more interchangeably than that. I certainly have been.  So we don't add fields to a table, we add columns to a table, and fields are only relevant when talking about data within a record?  Other thoughts on column vs field?  Edit: to clarify, the current context is MS SQL Server. My background before SQL server was MS Access, which might influence my use of these terms. Java Generics: List, List, List Regex for quoted string with escaping quotes Is there a way to render an object with a subdivision surface modifier with a wireframe which is the original geometry as well as the wires that make the subdivision surface?  Example:   This wireframe here is curved. What if there was displayed the straight wireframe. The wires that make are made by the subdivision surface. Thats what I want to achieve. How to do it?  What I mean is - How to render the curved wireframe (which is the original geometry with the straight wireframe (which is the wireframe the is made by the subdivision surface)? System program problem detected? Windows Server 2016 scheduled task schedule must be in future We have a 2012 server that has about 20 scheduled tasks that are used for monitoring.  We have built a 2016 server to replace it and as I move tasks to the new server I have run into an issue.  Most of our tasks run every few minutes.      In Server 2012 we would schedule the task to start at 1200AM of the current day (so a time in the past) and reoccur every X minutes.  The task would start at its next scheduled time.  Perfect.  In Server 2016 schedules that start in the past don't run.  So we had to schedule it to start in the future.  This is fine until you reboot.  Now the start time is in the past and it won't run.  It doesn't even register a failed run.  The Next Run Time column shows that it should run but doesn't.    Besides creating two triggers, one that starts in the future and another that starts on reboot (which we don't want to do since some tasks should only run at specified times), is there a better way?  This seems to be a bug.  The 2012 version worked great.  UPDATE  There is a workaround for this.  Create a one time schedule that runs indefinitely.  This works after restarts just like you would expect.  I know this isn't great for schedules that are "weird" but works fine for things that we want to run every 3 minutes or whatever. Why is the automatic login disabled for users with encrypted home? I've tried to setup an autologin, but that isn't possible, because my home is encrypted. What are the reasons for this?  PS. Here I have described my problem: .