I'm not a Linux superuser. However I decided to install official JDK-9 so I done those steps:  wget --no-cookies --no-check-certificate --header "Cookie: oraclelicense=accept-securebackup-cookie" \ http://download.oracle.com/otn-pub/java/jdk/9+181/jdk-9_linux-x64_bin.tar.gz \ -O jdk-9_linux-x64_bin.tar.gz   Move to /opt/  sudo mv jdk-9_linux-x64_bin.tar.gz /opt/   unzip:  cd /opt/ sudo tar -xzf jdk-9_linux-x64_bin.tar.gz   update-alternatives:  sudo update-alternatives --install /usr/bin/java java /opt/jdk-9/bin/java 1000 sudo update-alternatives --install /usr/bin/javac javac /opt/jdk-9/bin/javac 1000 sudo update-alternatives --install /usr/bin/javadoc javadoc /opt/jdk-9/bin/javadoc 1000 sudo update-alternatives --install /usr/bin/javap javap /opt/jdk-9/bin/javap 1000  sudo update-alternatives  --config java   I choose [1]  0 /usr/lib/jvm/java-8-openjdk-amd64/jre/bin/java    [Auto] 1 /opt/jdk-9/bin/java                               [manual mode] 2 /usr/lib/jvm/java-8-openjdk-amd64/jre/bin/java    [manual mode]   After that I got this:  update-alternatives: using /opt/jdk-9/bin/java to provide /usr/bin/java (java) in manual mode   After all steps I still getting jdk 1.8.0 after using  java -version   So I edited /etc/environment/Now it's look like this:  PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games" export JAVA_HOME=/opt/jdk-9 export JAVA_HOME=/usr/lib/jvm/java-8-openjdk-amd64/jre/bin/java export PATH="$PATH:$JAVA_HOME/bin"   And I also edit ~/.bashrc (when I install JDK-8 I added to end of file)  export JAVA_HOME='/usr/lib/jvm/java-8-openjdk-amd64/jre/bin/java' PATH=$JAVA_HOME/bin:$PATH_HOME/jre/bin:$PATH   After logout I can't login. And of course I getting infinity loop login. I can fix something with ctrl+alt+F1 console but I don't know what... I think I spoiled something in /etc/environment or ~/.bashrc  Can someone help me fix this?
My Ubuntu is stuck in a login loop when trying to enter my desktop. When I login, the screen gets black and soon after that the login screen comes back.   I've read that the problem might be caused by an error depending on the graphics, here's my graphics card: ATI Radeon 7670M
What young adult SF book is about a boy saved by aliens and domed colonies? How to get first N elements of a list in C#? I would like to use Linq to query a bus schedule in my project, so that at any time I can get the next 5 bus arrival times. How can I limit my query to the first 5 results?  More generally, how can I take a slice of a list in C#? (In Python I would use mylist[:5] to get the first 5 elements.) How should I apply brick to an interior wall? How to make password-less login work I generated a key and copied it to my target:  ssh-keygen ssh-copy-id username@hostname   I left the the location of the key as default and did not set any passphrase.  For a small while after that, I was able to ssh username@hostname without being prompted for a password, and then it stopped working for some reason. javascript document.getElementsByClassName compatibility with IE I'm trying to make a command block that spawns an item in front of you if you are holding an item with a specific name. I've got these two working scripts;  /testfor @p[r=10] {Inventory:[{tag:{display:{Name:"Item Name"}}}]}  /testfor @p[r=10] {SelectedItemSlot:0,Inventory:[{Slot:0b,­id:"minecraft:diamond_sword"}]­}   but I can't get them together. The block should only look for an item in the active slot. I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers.  First, take the numbers between zero and one, exclusive.  The following sequence of real numbers is suggested so that we have bijection.    0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.  Obviously, this includes repeats, but this set is countable.  Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable.  Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable.    Please help me with this.  I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.  Thanks. I know that the Naruto movies are outside the timeline in the sense that they don't affect whatever happens after them. However, they must fall somewhere in the Naruto timeline — based on who are in the movies and how the characters look like, etc. For example, if Sasuke isn't in the movie, it should take place after he left. For each of the movies, approximately what timeframe (between episodes/arcs) are they set?  The movies:   Ninja Clash in the Land of Snow Legend of the Stone of Gelel Guardians of the Crescent Moon Kingdom Naruto Shippuden the Movie Bonds The Will of Fire The Lost Tower Blood Prison Road to Ninja The Last - Naruto the Movie Boruto: Naruto The Movie Height of an HTML select box (dropdown) Can someone confirm that its not possible to change the height of a dropdown that is shown when you click on a select box.   The size attribute of the select makes it look like a list, the height attribute in the CSS doesnt do much good either. I wish to familiarize myself with the capabilities of QGIS up to and beyond QGIS version 3.0.   Can you recommend some good web resources and tutorials that would show me QGIS' capabilities?   I am interested in both using QGIS through its GUI and in developing applications using QGIS. Enlarged partition, filesystem still reports old size I resized the root partition with GParted, from 20 GB to 40 GB, but the filesystem still reports a size of 20 GB.  How can I use all 40 GB?    Update  The resized partition was an LVM Physical Volume (/dev/sda2), and the boot partition is ext4 (/dev/sda1). Metric is continuous, on the right track? Let $X$ be a metric space with metric $d$. Show that $d:X\times X\rightarrow \mathbb{R}$ is continuous.   The problem is taken from Munkres Topology second edition, Section 20.  I know that if $d$ is a metric on $X$ then $d:X\times X\rightarrow \mathbb{R}$. My thinking is that the topology that is on $X$ is the topology induced by the metric $d$, and that the topology on $X\times X$ is the product topology on that space where we take the basis to be $$\mathcal{B}=\{ U \times V \mid \text{ $U,V$ both open in $X$}\}.$$ Am I on the right track to say that we define some new metric on $X\times X$ and show that this metric induces the same topology as the product topology and then work with the function $d$ as a function between metric spaces to show continuity? The question doesn't mention anything about defining some new metric and I've tried to solve the problem by looking at $X\times X$ as having the product topology, but in picking some point $(x,y) \in X\times X$ and some neighborhood around $d(x,y)$ in $\mathbb{R}$, I haven't yet found the way to make a neighborhood around $(x,y)$ which maps into the neighborhood around $d(x,y)$. Related to the question :  It is well known that   $$\zeta(2n) = f(2n) \pi^{2n}$$  where $f(n)$ is an function in rationals: (the denominator = OEIS : 6, 90, 945, 9450).  Apery proved that $\zeta(3)$ is irrational, and presumably $f(2n+1)$ is so for all (but the latter is open).  But what else is known about $f(2n+1)$? What sort of pattern does it share (if any) with $f(2n)$? Numerically does it fit well? Is there any clue/progress since Apery as to its form? Of course, $f(2n+1)$ may in effect divide out the $\pi$, meaning that somehow $\pi$ is not involved in $\zeta(2n+1)$ in any essential way. 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? How to convince airport security that I am not dangerous when having metallic parts in my body? Is there any Android XML documentation? Bootable USB of Ubuntu Minimal CD I want to burn Ubuntu Minimal CD ISO to an old, 1 GB USB thumbdrive. Regular ISO doesn't fit. I already tried Startup Disk Creator and Unetbootin but both failed. I tried installing GRUB and pointing it to boot like this:  insmod part_msdos linux /linux initrd /initrd.gz   but it doesn't work. I also tried dd but I ended up with not bootable drive.  Rufus worked on Windows with the same thumbdrive but I no longer use Windows.  What else can I do? $f^2$ and $f^3$ are holomorphic implies $f$ is holomorphic. If $|G| = pq$,  $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems? In FHS-2.3, we have /media that holds mount points for removable media such as CD-ROMs and we have /mnt that holds temporarily mounted filesystems.  On the other hand, we have /run/media and /run/mount. For me, the CDs and USBs are mounted on /run/media.  I don't see any clear distinction between them(/media, /mnt, /run/mount) . What are their differences?     I have seen similar trend (mount on /run/media) in fedora 20 - GNOME 3.10.4 and ubuntu 14.04.1 (installed on virtual box) with GNOME 3.10.4. But when I plugged in a USB flash (with auto-mounter script) on a system with Centos 6 and GNOME 2.28.2 it was mounted on /media