I have only one partition right now. I want to make two partitions. But I am unable to do so. I am using GParted for this. Here is the screenshot of gparted.    When I click on resize. The minimum amount is always 0.
Previously, I have installed Windows 7 on my 320 GB laptop with three partitions 173, 84 and 63 GB each. The 63 GB partition was where the Windows was installed. The rest were for file containers.  Now I changed my OS to Ubuntu 12.04 LTS. I installed Ubuntu by replacing the entire Windows 7 on the 63 GB partition. The rest of the partitions remain as an NTFS Windows partition and I can still access them both (the 173 and 84 GB partitions).  Now I want to change the two partitions of Windows into an Ubuntu format partitions plus most importantly, I want to extend the 63 GB partition to more than a 100 GB because at the moment I am running out of disk space. Whenever I try to install any application, especially using wine, it always complains for a shortage of disk space.  How do I do the extending activity before I entirely format my laptop again and lose all the important files on my partitions?
I like to edit my answers. To clarify, illuminate, add examples, add links, point to new information as it crops up, etc.  It's highly irritating that my effort to improve my answer turns it into a Community Wiki.  Could anything be done to keep me from turning into a collective? I'm having a bit of a trouble with the below question     Given $G$ is an undirected graph, the degree of a vertex $v$, denoted by $\mathrm{deg}(v)$, in graph $G$ is the number of neighbors of $v$.   Prove that the number of vertices of odd degree in any graph $G$ is even. The verbs dare and need do not require auxiliaries when used in the interrogative; for example, “need I?” is as acceptable as “do I need?”  Excluding the auxiliaries themselves (like be, do, have), are there any other such verbs that work that way? Let's say we have this code for a LWC:  &lt;!-- helloExpressions.html --&gt; &lt;template&gt;     &lt;div class="slds-m-around_medium"&gt;         &lt;lightning-input name='firstName' label="First Name" onchange={handleChange}&gt;&lt;/lightning-input&gt;         &lt;lightning-input name='lastName' label="Last Name" onchange={handleChange}&gt;&lt;/lightning-input&gt;         &lt;p class="slds-m-top_medium"&gt;Uppercased Full Name: {uppercasedFullName}&lt;/p&gt;     &lt;/div&gt; &lt;/template&gt;   and this is the JS file:  // helloExpressions.js import { LightningElement, track } from 'lwc';  export default class HelloExpressions extends LightningElement {     @track firstName = '';     @track lastName = '';      handleChange(event) {         const field = event.target.name;         if (field === 'firstName') {             this.firstName = event.target.value;         } else if (field === 'lastName') {             this.lastName = event.target.value;         }     }      get uppercasedFullName() {         return `${this.firstName} ${this.lastName}`.toUpperCase();     } }   firstName and lastName have been decorated with @track. Now as per the documentation, decorating a property with @track makes it private and reactive, while to mark a property public and reactive, use @api.  So if I just change the decorator to @api for both these variables, it should work. I am referring these from the same component only. But it's not working.  UPDATE: 27th Feb -------------------- Somehow it's working today. I just changed the decorator of the variables to api.  import { LightningElement, api } from 'lwc';  export default class HelloExpressions extends LightningElement {     @api firstName = '';     @api lastName = '';      handleChange(event) {         const field = event.target.name;         if (field === 'firstName') {             this.firstName = event.target.value;         } else if (field === 'lastName') {             this.lastName = event.target.value;         }     }      get uppercasedFullName() {         return `${this.firstName} ${this.lastName}`.toUpperCase();     } }   So it's working. Is there a (La)TeX distribution running on iPad? Can a well-timed Twisted Image kill a player using Tree of Redemption? What happens if my opponent activates  and I cast  with the life-exchange ability on the stack?  Looking at the comp rules, these seem most relevant:     701.8a A spell or ability may instruct players to exchange something (for example, life totals or    control of two permanents) as part of   its resolution. When such a spell or ability resolves, if the   entire exchange can’t be completed, no part of the exchange occurs.      Example: If a spell attempts to exchange control of two target creatures but one of those              creatures is destroyed before the spell resolves, the spell does nothing to the other              creature.      112.7a Once activated or triggered, an ability exists on the stack independently of its source.    Destruction or removal of the source   after that time won’t affect the ability. Note that some    abilities   cause a source to do something (for example, “Prodigal Sorcerer deals   1 damage to    target creature or player”) rather than the ability   doing anything directly. In these cases, any    activated or triggered   ability that references information about the source because the   effect    needs to be divided checks that information when the ability   is put onto the stack. Otherwise, it    will check that information   when it resolves. In both instances, if the source is no longer in the   zone it’s expected to be in at that time, its last known information   is used. The source can still    perform the action even though it no   longer exists.   Is the entire exchange ability countered because the Tree is dead by the time it resolves? Or do we use the "last known state" (toughness 0), thereby killing its controller? How does chain fishing work? "FTL" Communication with Quantum Entanglement? Can quantum entanglement make sending a message, whether audio, video, or even Morse code, instantaneous between two points (faster than it could travel normally at the speed of light)?  Let me first be clear, my question is not whether a message can travel faster than light. I simply mean whether a message can be generated here and then a copy of that message heard over there sooner than the message would normally be received.  Say a large number of entangled pairs are generated. One half the particles are given to Jack and the opposite ones to Sally. Jack then hops on a ship and travels 8 months to Mars. Sally then manipulates her particles in the form of a message. Do the corresponding opposite particles get manipulated simultaneously so that Jack receives the message instantly instead of the 16 light-minutes for a classically transmitted message? How do they compute the Gold and Elixir that can be looted? How do they compute the Gold and Elixir that can be looted from a base? Is it 100% of the remaining? or the total (including the uncollected from mine)? Active Web Site as Gnome Background Is there a way to put an active website, such as Gmail, as your Desktop background in Gnome, running Ubuntu 10.10 64-bit? MS Win calls it "Active Desktop". Adding Facebook Online Account - "application you are trying to use doesn't exist or has been disabled" How to execute a Matlab script from C++ Make locking posts more granular Partial vs. Full Volume Wort Boil The starting position for most new extract brewers is to use a partial boil set up.  Defined as using a pot smaller than the intended batch size, all the ingredients are boiled in a smaller volume.  Then the wort is diluted to the final desired batch size.  (Often the dilution step is used as part of a cooling step)  What are the Ups and Downs of both Partial vs. Full boils? $\mathbb Q$ 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. How to view the output of a running process in another bash session? I have left a script running on a remote machine from when I was locally working at it. I can connect over SSH to the machine as the same user and see the script running in ps.   $ ps aux | grep ipcheck myuser  18386  0.0  0.0  18460  3476 pts/0    S+   Dec14   1:11 /bin/bash ./ipchecker.sh   It is simply outputting to stdout on a local session (I ran ./ipchecker.sh form a local terminal window, no redirection, no use of screen etc).  Is there anyway from an SSH session I can view the output of this running command (without stopping it)?  So far the best I have found is to use strace -p 18386 but I get hordes of text flying up the screen, its far too detailed. I can stop strace and then sift through the output and find the text bring printed to stdout but its very long and confusing, and obviously whilst it's stopped I might miss something. I would like to find a way to see the script output live as if I was working locally.  Can anyone improve on this? The obvious answer is to restart the script with redirection or in a screen session etc, this isn't a mission critical script so I could do that. Rather though, I see this as a fun learning exercise. Minecraft LAN "java.net.ConnectException: Connection refused: no further information:" I am trying to play a map on LAN. But every time I try to log on, I get:     java.net.ConnectException: Connection refused: no further information:   This used to work, but for some reason has stopped all of a sudden. I looked it up online, and everything I found said the firewall just wasn't allowing Java through. I followed tutorials to allow Java through the firewall, but even then it won't work. I know for a fact that I'm putting the IP in correctly, and everything seems to be correct, but yet I still get the same error message.  If it helps any, I'm using MC version 1.8.9 (the map requires it), and I don't have any mods. How does the post-2.0 Puzzle Ring work? Patch 2.0.1 introduced a lot of significant changes to the loot system in the game. One of the significant ones was the revamp to the , which now has a new effect:      Summon a treasure goblin who picks up normal-quality items for you. After picking up 12–16 items, he drops a rare item with a chance for a legendary.    What is the cooldown on the goblin item drop? Does the effect stack if you have one on your character and another on your follower (i.e. two goblins)? Is the drop chance for the rare/legendary affected by increased Magic Find? If so, approximately how much? Is there a cap on them max item level of the dropped item? (While &lt; lvl 60, at lvl 60, and when the level cap is raised in the expansion.) Does the item counter reset when you leave the game, change acts, or die? How did the ending happen from Neil's perspective? I am trying to figure out the scene near the finale of Tenet where Neil unlocks the door and gets shot in the face. I can't really see the flow where he opens and holds the door and the protagonist walks in and fights the bad guy. Can you tell the series of events as Neil sees them happen till he gets shot? UPDATE If I am not wrong, I see it like this: Neil was running down towards the door, both the protagonist and his mate are inside with the door locked fighting the bad guy, then, he unlocks the door and walks in and stands behind the door. The protagonist's mate comes out of the room, then the protagonist, then he closes the room, then he look back to get shot in the face. But this doesn't make sense, if he wanted to unlock the door when he was running down, they were already in fighting the bad guy!!!!! DDL Trigger in MySQL?