I tried installing hadoop and I no longer need stuff from it. When launching shell , it gives the following and many commands are unrecognized now , how do I reset it to default ?  bash: export: `/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin: /bin:/usr/games:/usr/local/games:/usr/local/hadoop/bin': not a valid identifier astitva@astitva-Vostro-3446:~$
I accidentally unset all the directories of $PATH while trying to add a new one in ~/.bashrc. I opened a new terminal window as I was editing and now $PATH is empty. I'm worried if I boot from another drive to find the $PATH I won't be able to boot into this drive again.  Basically, what is the default result of echo $PATH?
I'd like a couple of Atmega328s to run the same program, but they're going to interface with each other and need unique IDs. Is there any such ID or serial number hard-coded on the chip? Why don't my female dwellers give birth to children? I am playing Fallout Shelter on my IPad. There were some women who have already gave birth, but two female dwellers are carrying their babies for about 18(!) hours. I know that this process takes three hours, so this is not normal. What do I do? $\mathbb Q$ is not the direct product of two non-trivial groups. I know this has been asked before, in a bunch of places, so I expect it to be closed as a duplicate at some point, but I'd like to know if my own proof works. If the proof is incorrect, please limit answers to pointing out the errors. If the proof is correct, then thoughts on improving it would be appreciated.  Prove that $\Bbb Q$ is not the direct product of two non-trivial groups.  —Aluffi exercise II.3.5. My proof Suppose that $\Bbb Q = G \times H$, neither $G$ nor $H$ trivial. First note that $0_G$ is the only element of $G$ with finite order: If $m\ne 0$ and $m\cdot g=0_G$, then $m\cdot(g,0_H)=0_{G\times H}$, contradicting the fact that $\Bbb Q$ has no non-zero elements of finite order. Suppose that for some $m\ne 0$ and some $n\ne 0$, $\pi_G(m/n)=0_G$. Then $0_G=n\cdot\pi_G\left(\frac m n\right)=\pi_G\left(n\cdot\frac m n\right)=\pi_G(m)=m\cdot\pi_G(1)$. Since $\pi_G(1)$ has finite order, we conclude that $\pi_G(1)=0$. Since $\Bbb Z=\langle 1\rangle$, $\pi_G$ maps every integer to $0_G$. Now let $a$ and $b$ be non-zero integers. Then $b\cdot\pi_G\left(\frac a b\right)=\pi_G\left(b\cdot \frac a b\right)=\pi_G(a)=0_G$. As above, this shows that $\pi_G\left(\frac a b\right)=0_G$. Thus if $\pi_G$ is not an injective function, then it must map everything to $0_G$, so $G$ will be a trivial group. The same result holds for $\pi_H$. Since by assumption, $G$ and $H$ are non-trivial, $\pi_G$ and $\pi_H$ are both injective functions, which is of course impossible. According to the original paper of , there is no key validation needed when using Curve25519 for Diffie-Hellman Key Exchange. However, where does this property come from? Is there any proof or argumentation available in order to understand this really practical property of Curve25519? Does this mean that each $x$ coordinate could belong to a point $P=(x,y)$ from Curve25519? Is that even possible? Mass Lead Conversion - Programmatically or Using Data Loader Can I insert elements to the beginning of an element using .appendChild()? If the user selects an option in a dropdown box, there must be added a label and a textbox. Using appendChild, these elements get added to the end of the container.  var newFreeformLabel = document.createElement('label'); newFreeformLabel.innerHTML = 'Omschrijving:';  var newFreeformField = document.createElement('input'); newFreeformField.className = 'textfield'; newFreeformField.name = 'factuur_orderregel[]';  var newFreeformSpacer = document.createElement('div'); newFreeformSpacer.className = 'spacer';  container.appendChild(newFreeformLabel); container.appendChild(newFreeformField); container.appendChild(newFreeformSpacer);   The issue is that these elements should be inserted at the beginning of container, not at the end.  Is there a solution to this in PrototypeJS? Method to create a live USB disk (with persistence) which actually works? I've tried so many combinations - all I want is a fast Ubuntu Live USB disk with persistence that works. What I've tried so far:   Using unetbootin on my Mac, choosing some "Space used to preserve files across reboots (Ubuntu only)".   Result: Doesn't boot at all.  Using unetbootin on my Mac, choosing no such space.  Result: Boots fine and very fast (doesn't even show the annoying dialog "Try ubuntu/install ubuntu" at startup!), but has no persistence. (So I have to choose my keyboard layout, install programs I need, etc. every time)  Using Startup Disk Creator from my Ubuntu 11.04 installation.  Result: I cannot choose "reserved extra space" at all. It's greyed out. I know I know, there was bug , and many people say that the workaround from  works for them. But it doesn't work with Ubuntu 11.04 anymore (see also ).  Using Startup Disk Creator from my Kubuntu 10.10 installation, gnome version (usb-creator-gtk).   Result: I can select the "reserved extra space" option, but installation doesn't work (fails when installing the bootloader).  Using Startup Disk Creator from my Kubuntu 10.10 installation, kde version (usb-creator-kde).   Result: Actually works: Reserved space can be selected, installation works. But after booting, I get the screen "Try ubuntu/install ubuntu". Clicking on "Try ubuntu" takes me to the desktop - but only after waiting 5 minutes (every time!). It also doesn't seem to keep my keyboard settings.    I know, that instead of creating a startup disk, I could install a full ubuntu system on the USB stick. But as far as I know, this isn't very good, because /tmp and similar directories will then also be on the USB disk, which is a) slow, and b) not very good for the USB disk due to frequent overwriting.  I'm frustrated. So has anybody managed to create a working Ubuntu 11.04 Live USB disk with persistence? How do you select a particular option in a SELECT element in jQuery? Java inner class and static nested class sudo -k vs. sudo -K I know what the man page has to say about the -K and -k :     ‑K       The ‑K (sure kill) option is like ‑k except that it removes the user's   cached credentials entirely and may not be used in conjunction with a   command or other option. This option does not require a password. Not   all security policies support credential caching.      ‑k[command]       When used alone, the ‑k (kill) option to sudo invalidates the user's   cached credentials. The next time sudo is run a password will be   required. This option does not require a password and was added to   allow a user to revoke sudo permissions from a .logout file. Not all   security policies support credential caching. When used in conjunction   with a command or an option that may require a password, the ‑k option   will cause sudo to ignore the user's cached credentials. As a result,   sudo will prompt for a password (if one is required by the security   policy) and will not update the user's cached credentials.   So if there is no need to use a command in conjunction with these options, am i correct in assuming K is always the better option to use rather that k? Where to find the Mac OS X Lion security configuration guide? How to prove $\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}$ is measurable, if $X$ and $Y$ are measurable? Given a probability space $(\Omega ,\mathcal{F} ,\mu)$. Let $X$ and $Y$ be $\mathcal{F}$-measurable real valued random variables.  How would one proove that $\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}$ is measurable.  My thoughts: Since $X$ and $Y$ are measurable, it is true, that for each $x\in\mathbb{R}:$ $\left\{ \omega|X(\omega)&lt;x\right\} \in\mathcal{F}$ and $\left\{ \omega|Y(\omega)&lt;x\right\} \in\mathcal{F}$.  It follows that $\left\{ \omega|X(\omega)-Y(\omega)\leq x\right\} \in\mathcal{F}$  Therefore $\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$, for $n\in\mathbb{N}$.  Therefore $0=\bigcap_{n\in\mathbb{N}}\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$.  Am working towards the correct direction? I appreciate any constructive answer! How to fix "sudo: unable to open ... Read-only file system"? The title might not be as descriptive as I would like it to be but couldn't come up with a better one.  My server's file system went into Read-only. And I don't understand why it does so and how to solve it.  I can SSH into the server and when trying to start apache2 for example I get the following :  username@srv1:~$ sudo service apache2 start [sudo] password for username: sudo: unable to open /var/lib/sudo/username/1: Read-only file system  * Starting web server apache2                                                                                                                                                                                                               (30)Read-only file system: apache2: could not open error log file /var/log/apache2/error.log. Unable to open logs Action 'start' failed. The Apache error log may have more information.   When I try restarting the server I get :  username@srv1:~$ sudo shutdown -r now [sudo] password for username: sudo: unable to open /var/lib/sudo/username/1: Read-only file system   Once I restart it manually it just start up without any warning or message saying something is wrong.  I hope somebody could point me into the right direction to resolve this issue. EDIT: As of Java 8, static methods are now allowed in interfaces.  Here's the example:  public interface IXMLizable&lt;T&gt; {   static T newInstanceFromXML(Element e);   Element toXMLElement(); }   Of course this won't work.  But why not?  One of the possible issues would be, what happens when you call:  IXMLizable.newInstanceFromXML(e);   In this case, I think it should just call an empty method (i.e. {}).  All subclasses would be forced to implement the static method, so they'd all be fine when calling the static method.  So why isn't this possible?  EDIT: I guess I'm looking for answer that's deeper than "because that's the way Java is".   Is there a particular technological reason why static methods can't be overwritten? That is, why did the designers of Java decide to make instance methods overrideable but not static methods?  EDIT: The problem with my design is I'm trying to use interfaces to enforce a coding convention.  That is, the goal of the interface is twofold:   I want the IXMLizable interface to allow me to convert classes that implement it to XML elements (using polymorphism, works fine). If someone wants to make a new instance of a class that implements the IXMLizable interface, they will always know that there will be a newInstanceFromXML(Element e) static constructor.   Is there any other way to ensure this, other than just putting a comment in the interface? PHP Echo Line Breaks Define functions in structs English equivalent for the Persian idiom "send someone out in pursuit of black chickpeas" "To send somebody out in pursuit of some black " is a Persian idiom that implies 'to make or ask someone to run an errand so that you be able to have/ buy some time in order to deal with your private affairs in their absence', like having or continuing a (secret) conversation with someone else.   According to its etymology, since 'black garbanzos/chickpeas' were rarely available in the markets, when someone wanted to buy them, they had to search many stores for a long time, and most of time they couldn't find them at all.   Today, its meaning has changed somehow, and it doesn't necessarily mean that we ask someone to run an impossible errand, but it means we use that errand as an excuse to send them away for a while just to have time to do something in their absence.   Example scenario:   Suppose your classmate has come to your house and you are talking about some girly issues in the sitting room. Now you want to tell her a big secret about one of your classmates, but not in the presence of your younger sister who is sitting there too. So you ask her to go to the market and buy some snacks for you. In this way, you will have some time to continue your conversation until she is back. So you might say to your friend:     Well, I sent my sister out in pursuit of some black chickpeas.  I'm sure that she won't be back for another 15 minutes, so we can talk more freely now."   And your sister, before leaving the house, might sarcastically say to you:     Okay, but I know this is just an excuse, and you are just sending me in pursuit of some black chickpeas!!! so don't think that you can fool me!   Is there any idiom, expression or proverb that conveys the same meaning or connotation in English (i.e. to make or ask someone to do an errand just in order to buy time so that you can deal with some of your private affairs)?   I have found to be/go on  but I'm not sure if I can use it or not. How to show $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ Every once in a while, someone comes to Meta to report a bug about something not working and it turns out that person is visiting one of the network sites over HTTPS/SSL. The problem has been growing since browsers started to outright block mixed content instead of showing a warning.  The usual answer is that HTTPS is not yet fully supported, and the person should try again using HTTP. Sometimes things get fixed, improving the situation a bit.  Is there a roadmap for full support? Partial support has been there for  now.  Is this actively being worked on? If not, what are the reasons to partially support it instead of disabling it completely to avoid these problems? Transferring through US airport. Do I need traveling visa?