I was conducting some experiments on my sockpuppet, which included handing out a chat suspension to... well, myself, on another account. I discovered, however, that the "your account has been suspended" message that shows up is unreadable:    Highlighting the text, I can see what it says:     your account has been temporarily suspended by a moderator and cannot chat for 59 minutes.   But otherwise, only the two words "your account" are visible, leaving users possibly confused about why they can't send any messages and why there's a link to their chat profile showing up instead of the textbox.
If you visit a chat room attached to a beta site while logged out, the log in message is all but unreadable, being dark grey on a dark blue background. For example, visit  while logged out or in private browsing mode.
EmployeeNumber = string.IsNullOrEmpty(employeeNumberTextBox.Text)     ? null     : Convert.ToInt32(employeeNumberTextBox.Text),  I often find myself wanting to do things like this (EmployeeNumber is a Nullable&lt;int&gt; as it's a property on a LINQ-to-SQL dbml object where the column allows NULL values).  Unfortunately, the compiler feels that  There is no implicit conversion between 'null' and 'int'  even though both types would be valid in an assignment operation to a nullable int on their own. Using the null coalescing operator is not an option as far as I can see because of the inline conversion that needs to happen on the .Text string if it's not null. As far as I know the only way to do this is to use an if statement and/or assign it in two steps.  In this particular case I find that very frustrating because I wanted to use the object initializer syntax and this assignment would be in the initialization block... Does anyone know a more elegant solution? Lost passport with Schengen visa My passport was stolen in Prague on my way to Vienna. I filed a police report and our embassy swiftly acted by giving me a temporary passport for my onward journey to Munich and Rome. Upon exit from Rome, the immigration ask for the Schengen visa, when I told him that I lost my passport, he did not ask for any document to support my claim, he just stamped my new passport with exit. I am not sure if on record, I am officially out of the Schengen area or not. I am scared that without the lost visa, it appears that I am still in the area and might reach the 90 day limit and will be denied when I re-apply as it would appear I broke the rule. What to do? Since I updated Windows to 8.1 my Grub menu disappeared. I ran Ubuntu using Live USB, I used Boot-Repair but something went wrong. If I click on Reccomended repair I get this error:     GPT detected. Please create a BIOS-Boot partition (>1MB, unformatted filesystem, bios_grub flag). This can be performed via tools such as Gparted. Then try again. Alternatively, you can retry after activating the [Separate /boot/efi partition:] option.   So I tried to set my sda9 (I think that it's the Ubuntu system partition, since it's ext4) with the bios_grub flag and then to run Boot-Repair again. It didn't work.  I don't know what to do. My Grub was working fine (using UEFI) before I updated Windows.  I removed the bios_grub flag. Then I used fsck and it didn't find any error. So I tried again with Boot-Repair but it always gives me the "GPT detected" error. My boot is on UEFI mode, Secure Boot is "off".  When I boot my PC with SecureBoot OFF, I can reach the GRUB menu. But it's empty (the only voice is "System Setup").  I mounted my Ubuntu system disk:  sudo mount /dev/sda9 /mnt   and the other things:  sudo mount --bind /dev /mnt/dev sudo mount --bind /proc /mnt/proc sudo mount --bind /sys /mnt/sys   I logged on the system with chroot:  sudo chroot /mnt   I re-installed grub:  grub-install /dev/sda update-grub2   But it still doesn't work. How to install HPLIP on my Ubuntu to support my HP printer and/or scanner? Let's assume that one has some HP printer and/or scanner and checked its support status on the corresponding page of HPLIP site - .  So one have the following information: the device Model name, Min. HPLIP Version and Driver Plug-in need.   What should one install to make Ubuntu fully support the HP printer and/or scanner? Is it possible to understand the One Piece storyline by watching the movies only? I need to prove that     $f_1 \in \Theta(g_1) \land f_2 \in \Theta(g_2) \implies f_1 + f_2 \in \Theta(\max\{g_1, g_2\})$    is relevant, but I have a slightly different case, so I don't know how to translate it into this one, because I need to turn the $f$'s into $g$'s.  Am I allowed to say $\max\{f_1, f_2\} \in \Theta(\max\{g_1, g_2\})$?  Or even $\Theta(\max\{f_1, f_2\}) = \Theta(\max\{g_1, g_2\})$? Spurious @ symbol at start of bash prompt in Debian Jessie Are Living Beings Turned Into Horcruxes Suceptible to Age, Disease, or Injury? From the HP books I know that Horcruxes grant a form of immortality. So, it is not possible for someone who created a Horcrux to die from old age or succumb to diseases. I guess the same is true for living beings turned into a Horcrux. But do living Horcruxes age normally? Is it possible to cripple them or sever their limbs, head or something like that, or do they instantly "repair" themselves like inanimate Horcruxes? If $a$ and $b$ commute and $\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1$, then $\text{ord}(ab)=\text{ord}(a)\text{ord}(b)$. Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. How to get method parameter names? Given the Python function:  def a_method(arg1, arg2):     pass   How can I extract the number and names of the arguments. I.e., given that I have a reference to func, I want the func.[something] to return ("arg1", "arg2").  The usage scenario for this is that I have a decorator, and I wish to use the method arguments in the same order that they appear for the actual function as a key. I.e., how would the decorator look that printed "a,b" when I call a_method("a", "b")? I use Stack Overflow a lot and have noticed a certain trend that I myself got caught up in at one time of using the phrase "I am having a problem" in place of "I have a problem."  I would use this phrase for an event in the future, for example, "I am having a steak dinner tonight," but for a problem that I need solved and am asking for a solution, I'd use "I have a problem..."   What would force us to use this phrase? Is it even valid English for the present tense? Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$) I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a &amp; b &amp; \ldots &amp; b\\ b &amp; a &amp; \ldots &amp; b\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ b &amp; b &amp; \ldots &amp; a\end{bmatrix}$$  where $0 &lt; b &lt; a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated. Why must $a$ and $b$ both be coprime when proving that the square root of two is irrational? Suppose we wish to prove that the square root of two is irrational. We begin by assuming that it is rational.  Namely, where both $a$ and $b$ are integers $$\frac{a}{b} = \sqrt 2  % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GH9aqpdaGcaaqaaiaaikdaaSqabaaaaa!31D3! $$  Why is it so important that both $a$ and $b$ be coprime and the fraction be irreducible?  That is, if $a$ and $b$ are not coprime and the fraction is indeed reducible, then$$\frac{a}{b} \ne \sqrt 2  % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GHGjsUdaGcaaqaaiaaikdaaSqabaaaaa!3294! $$ Do I have to move swap partition to the right side? According to  I just expand my sda4 to include unallocated space, my question is after create 2 new partitions sda7, sda8, do I need to move the swap partition to the most right of the disk? (I really don't know this, I saw  which moving the swap partition to the right side)  Thanks. Is there any value in full 'beginner's kit bundles'? bash: adduser: command not found on centOS I installed CentOS 5.5 on my VMWare 8 recently and I am trying to add a new user on the system. I am unable to add the user unless I use su - option. I believe it has to do something with path not set properly.  I updated the path and here is what it looks like  /usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/bin:/usr/bin:/bin:/usr/X11R6/bin:/home/uone/bin:/sbin   I believe the command is in /sbin dir which is already a part of path. Can anyone suggest me what else I might be missing? What's a concise way to check that environment variables are set in a Unix shell script? Is $\operatorname{lcm}(a,\gcd(b,c))=\gcd(\operatorname{lcm}(a,b),\operatorname{lcm}(a,c))$ Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$ How does the T-1000 see? In Terminator 2: Judgement Day, how does the T-1000 sense things? It's made of mimetic poly alloy; however, does it have a retinal scan camera like the T-X or T-800, or is just molecules that allow it to see what it does? In other words: let's say it is made out of nanobots. Would a group of these nanobots provide the capability of sight? Kind of like humans with cells that  make up the brain? I have always wondered whether James Cameron explained this on a special edition or if the topic was left up for insight from Terminator fans. From what I get from the whole mimetic poly alloy and some of the extra scenes, it seems to feel its way out to get information. Of course, it's very much not blind! Thoughts?