Which passport to use for airline check-in for dual citizenship I am a Canadian living in Canada with another citizenship from a country (Country A) that does not recognize dual citizenship.  Citizens of Country A needs visa to enter Canada.  Will be travelling to Country A from Canada soon.  Very confused as to which passport to use for airline checkin in Canada and Canada immigration.  Which passport for entering Country A.  For returning to Canada, which passport to use for airline checkin in Country A and Country A immigration.
I have two passports/nationalities. How do I use them when I travel? I am a citizen of two different countries, and have two passports.  How should I use my passports when traveling?
Sole minimal element: Why not also the minimum? A minimal element (any number thereof) of a partially ordered set $S$ is an element that is not greater than any other element in $S$.  The minimum (at most one) of a partially ordered set $S$ is an element that is less than or equal to any other element of $S$.    Let's consider the power set $\mathcal P (\{x,y,z\})$ together with the binary relation $\subseteq$.  The Hasse diagram shows what element(s) we're looking for:    It's easy to see that:   $\emptyset$ is a minimal element $\emptyset$ is the minimum   Now if we remove $\emptyset$ and consider $\mathcal P (\{x,y,z\})\setminus \emptyset$ instead, we get the following:   $\{x\}$, $\{y\}$ and $\{z\}$ are minimal elements there is no minimum   (1) We know that a minimum is unique and it is always the only minimal element.  (2) And from the example above, it seems that, if a sole minimal element exists, it is always the minimum.  But I read that (2) is false. Why? VBA array sort function? I'm looking for a decent sort implementation for arrays in VBA. A Quicksort would be preferred. Or any other  other than bubble or merge would suffice.  Please note that this is to work with MS Project 2003, so should avoid any of the Excel native functions and anything .net related. 'Wouldn't wake up' vs. 'didn't wake up' I'm reading an article about two pets rescuing their owners. After the cat managed to wake up the mother, she tried to wake up other members but they wouldn't wake up.     Cathy tried to tell her husband and son to leave the house, but they wouldn't wake up.   Why wouldn't and not didn't?   (I checked possible duplicates but they weren't convincing at all. Besides, none of the answers were confirmed. One said would is used for permission and another said it's for emphasis!) As discussed here:  offering actual money as a bounty sets the wrong incentives. However I would suggest giving users the ability to pay with money for offering a bounty regardless of their reputation. The bounty would still count as reputation for the receiving user but would not be discounted from the paying user's reputation. Find $\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$ (a question asked at trivia) My friend's trivia league had this math question:  $$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$$  After computing a few values, one could guess the answer is $e$ = 2.718...But how can we prove that is the limit?  Someone offered up a hand-wavy proof like this:  \begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] &amp; = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &amp;= \lim_{n \to \infty} e \\ &amp;= e \end{align}  The part about substituting $e$ is hand-wavy since technically this is an indeterminate form of $\infty - \infty$. And using $e$ as as upper bound did not lead me to an easy proof either.  Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove $e$ was a lower bound, but I could not find a suitable upper bound converging to $e$, (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated.  So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally?  Sources  Trivia question:   Thread on trivia:   Hand-wavy proof:   Idea for mean value theorem: How do we know Dark Matter isn't simply Neutrinos? For example block cipher AES-128, Key size is 128bit and it is used to make a 10 round key which is total 320bit.  Question 1. If i use another Key schedule algorithm in AES, then security decreased or remain same? In other words, if the key schedule algorithm of AES has some relationship with AES encryption itself, then it will be decreased, or it will be same if the new key schedule algorithm is safe enough.  Question 2. Why we expand key? brute force will take O(2^128) time regardless of the key schedule is strong or not. what is the advantage of making key expend? I was curious about this when I was learned about 3DES. Which developed because the former DES has weak key length 56bit so vulnerable to brute force attack, i was wondering why doesn't they just remove key schedule process and use the whole round key as a key. Compression API on the iPhone Can you combine multiple lists with LINQ? Say I have two lists:  var list1 = new int[] {1, 2, 3}; var list2 = new string[] {"a", "b", "c"};   Is it possible to write a LINQ statement that will generate the following list:  var result = new []{      new {i = 1, s = "a"},     new {i = 1, s = "b"},     new {i = 1, s = "c"},     new {i = 2, s = "a"},     new {i = 2, s = "b"},     new {i = 2, s = "c"},     new {i = 3, s = "a"},     new {i = 3, s = "b"},     new {i = 3, s = "c"} };   ?  Edit: I forgot to mention I didn't want it in query syntax. Anyway, based on preetsangha's answer I've got the following:  var result = list1.SelectMany(i =&gt;  list2.Select(s =&gt; new {i = i, s = s})); Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? Where can I find the location of downloaded .deb files?  I want to copy this folder before I reinstall Ubuntu so I don't have to redownload the packages.  I am running Ubuntu 12.10 32-bit.  How can I do this? I am looking for the combinatorial argument for the identity:  \begin{equation} k\binom{n}{k} = n\binom{n-1}{k-1} \end{equation}  This is easy to show algebraically as:  \begin{equation} \binom{n}{k} = \dfrac{n(n-1)(n-2)(n-k+1)}{k(k-1)!} \end{equation}   What is the combinatorial argument? What are some general ideas to get started?   Here is a clarification of 2. From what I have seen so far, proving (combinatorially) an identity with an addition sign usually implies that we need to partition a set (this makes sense because of the addition rule and provides a nice visual). On the contrary, the previous observation leads me to believe that multiplication in identities can be resolved with the multiplication principle, but what is the "visual/interpretation" for this? Could someone provide such an interpretation for the example identity given above? I use dwm as a window manager on Ubuntu, I like tiled windows and I like them as few as possible. I sometimes need to run an application (let's say it's the pdf viewer mupdf in x11) from the terminal, the application will be launched in a new window but the terminal from which I launched becomes this prompt-less console for that application to log whatever.  I use dmenu to launch applications, but sometimes an application needs arguments (file names) to start. I use uxrvt within x11 managed by dwm. Database design for products with bundles of products I am building a database system for my retail business. I have set some tables which are:   Product Purchase Sales Balance   All are connected one another and are able to show my inventory level.  The problem I am having is I also sell bundles of products - that have different prices than their individual prices. Example: I sell an orange for $1, an apple for $1.2; I sell fruit package 1 (2 oranges and 2 apples) for $3.8, package 2 (4 oranges and 4 apples) for $7.  Is there a right way how to create relationship for these product bundles?  PS: I am using FileMaker Pro creating this. I have put his on wolfram and obtained answer as follows:  $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$  And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$  However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper? GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light How to reinstall yum? I made a big mistake: I removed yum. So now when I for example run the following it gives me an error:  $ yum clean all yum command not found.   I can download the package succesfully:  wget mirror.centos.org/centos/6/os/i386/Packages/yum-3.2.29-40.el6.centos.noarch.rpm --2014-01-15 15:42:59--  http://mirror.centos.org/centos/6/os/i386/Packages/yum-3.2.29-40.el6.centos.noarch.rpm Resolving mirror.centos.org... 192.133.139.3 Connecting to mirror.centos.org|192.133.139.3|:80... connected. HTTP request sent, awaiting response... 200 OK Length: 1019540 (996K) [application/x-rpm] Saving to: “yum-3.2.29-40.el6.centos.noarch.rpm.1”  100%[====================================================================================================================&gt;] 1,019,540    649K/s   in 1.5s  2014-01-15 15:43:01 (649 KB/s) - “yum-3.2.29-40.el6.centos.noarch.rpm.1” saved [1019540/1019540]   Trying to install the downloaded package gives the following error:  # rpm -ivh yum-3.2.29-40.el6.centos.noarch.rpm Preparing...                ########################################### [100%]     file /usr/lib/python2.6/site-packages/yum/__init__.py from install of yum-3.2.29-40.el6.centos.noarch conflicts with file from package yum-3.2.29-30.el6.centos.noarch [...] file /usr/share/yum-cli/yummain.pyo from install of yum-3.2.29-40.el6.centos.noarch conflicts with file from package yum-3.2.29-30.el6.centos.noarch   There is no directory "yum" in /usr/bin.  When I use the command "rpm -qa | grep yum".  yum-metadata-parser-1.1.2-16.el6.x86_64 yum-plugin-fastestmirror-1.1.30-14.el6.noarch yum-3.2.29-30.el6.centos.noarch   The distro is CentOS 6.3. How can I reinstall yum? If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum. Wrong Java resource bundle loaded Given a recurrence relation:   $$ x_{n+2} = px_{n+1} + qx_n + r $$   Find the general term $x_n$ given initial conditions $x_1 = a$ and $x_2 = b$, where $a,b,p,q,r$ are some given numbers, $n \in \mathbb N$.   Since this is a non-homogeneous linear recurrence relation i've started with solving for $x_n^h$ which is a solution for homogeneous relation:  $$ x_{n+2} - px_{n+1} - qx_n = 0 \\ \lambda^2-p\lambda - q = 0 $$  I'm assuming here that two distinct roots $\lambda_{1,2}$ exist for this equation (single root case is handled similarly). So the general term is in the form:  $$ x_n = C_1\lambda_1^{n-1}+C_2\lambda_2^{n-1} $$  With the initial condition it's possible to find the values for $C_1$ and $C_2$:  $$ C_1 + C_2 = a \\ C_1\lambda_1 + C_2\lambda_2 = b $$  After solving the system of equation with respect to $C_1$ and $C_2$:  $$ C_1 = \frac{b-a\lambda_2}{\lambda_1 - \lambda_2} \\ C_2 = \frac{b-a\lambda_1}{\lambda_2 - \lambda_1} \\ $$  Which eventually results in:  $$ x_n^h = \frac{(a\lambda_2 - b)\lambda_1^{n-1} - (a\lambda_1-b)\lambda_2^{n-1}}{\lambda_2 - \lambda_1} $$  Final solution consists of a sum of particular and homogeneous solutions, but that is where I got stuck. What is the way to find the particular solution for the given problem? I'm pretty new to such kind of problems so any details are very much appreciated.  upd:  Based on the technique in @rtybase answer here are my findings:  $$ \begin{align} x_1 &amp;= a \\ x_2 &amp;= b \\ x_3 &amp;= pb - qa + r \end{align} $$  Writing the system for $c_1, c_2, c_3$:  $$ \begin{align} a &amp;= c_1 + c_2 + c_3 \\ b &amp;= c_1\lambda_1 + c_2\lambda_2 + c_3 \\ pb+qa+r &amp;= c_1\lambda_1^2 + c_2\lambda_2^2 + c_3 \end{align} $$  Now solving for the coefficients gives a huge formula. The answer to this problem suggests that:  $$ x_n = \frac{(\lambda_2(a+\gamma) - b -\gamma)\lambda_1^{n-1} - (\lambda_1(a + \gamma) - (b-\gamma))\lambda_2^{n-1}}{\lambda_2 - \lambda_1} - \gamma $$  Where:  $$ \gamma = \frac{r}{p+q-1} $$  Unfortunately I was not able to arrive at exactly the same result, but testing my solution (not posting since it's monstrously big) using SameQ in Mathematica with the given answer shows that they are equivalent. Those transformations are not very obvious to me, so i skipped them since the answer matches the given one (yet the symbolic expressions are different)