Will 18.04.5 upgrade to version 20 through GUI or should I force it? I've been waiting for my Ubuntu to give me that nice &quot;Ubuntu 20 is now available&quot; prompt since August the 6th, to no avail. 18.04 just keeps updating itself. Today I'm not in a particular hurry but merely curious: is it even supposed to give me this prompt? Or maybe GUI upgrades have been cancelled long ago and I have to force it through the console? Or is my software updater misconfigured?
Why isn't an upgrade to 20.04 from 18.04 available yet? I want to use do-release-upgrade to upgrade from 18.04 to 20.04, but I can't do that do-release-upgrade Checking for a new Ubuntu release There is no development version of an LTS available.  It seems that  is used, and it still points to 18.04. Do you know when (or how?) can I upgrade to 20.04?
Presentation $\langle x,y,z\mid xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2}\rangle$ of group equal to trivial group Convert .mov to .mp4 I want to convert .mov to .mp4! I saved the video from iMovie but it is in .mov format! How do I convert it to .mp4?  I'm using a MacBook and I want to share it to my phone, but my phone can't open .mov format video. Disk space not freed up after deleting files and emptying Trash Why is using a wild card with a Java import statement bad? It is much more convenient and cleaner to use a single statement like  import java.awt.*;   than to import a bunch of individual classes  import java.awt.Panel; import java.awt.Graphics; import java.awt.Canvas; ...   What is wrong with using a wildcard in the import statement? My current idea is to use the dual graph, and prove that that is a bipartite graph, and so proving the statement. But I can't seem to figure out how even degree for all verticies means that it is bipartite.  Any help is appreciated. Thanks! In the following question  the notation $c\in ]a,b[$ is used. What does this mean? I have never seen it before. I've got a rather large number of files on my Windows drive that have a colon in their name. (These came from unpacking some Unix archives.)   When I try to delete them, Windows (XP) complains that the file does not exist, and refuses to delete it. This happens when trying to delete or rename it from explorer or the command line. Still, chkdsk does not complain about those files nor fix the problem.   Any ideas on how to get rid of them? This is my follow up question. My previous question is here:   Problem:  If I add the image in my table, I get the following output as shown in the image below. Why is the Co-Supervisor going below?   I am using this code:  \begin{tabular}{|l|c|r|} \hline \bf Author: &amp; \hspace{5pt} &amp; \bf Research Supervisor:\\ \hline Mr. X &amp;  \hspace{5pt} &amp; Prof. Dr. Z\\ \hline \includegraphics[width=0.40\textwidth,height=10mm]{Figures/images} &amp;  \hspace{5pt} &amp; \bf Co-Supervisor:\\ \hline &amp;  \hspace{5pt} &amp; Prof. Dr. Y\\ \hline \end{tabular}   What I want is shown below:   PS: Please ignore the table borders, they are for demo only, I won't be using them in my final code.  The final product: that I desire is: How Special Relativity causes magnetism Can't mount an IDE HDD - "can't read superblock" error message I'm trying to mount my HDD,  /dev/sdb, which is connected via an enclosure but every time I try do so I get an error message   mount: /dev/sdb: can't read superblock   How do I solve this problem and mount my HDD? How does ownership in fractional shares work? I've been trying to find out some more information about fractional shares, and how they work. I understand conceptually, you own x% of a share rather than x "full" shares, but I'm curious about how such a system is implemented.  As I understand it, in a "traditional" purchase of shares, the role of a broker is to act as an executor on your behalf, and does not hold any assets themselves. In the case of a fractional share, is this then not the case? For example, the broker holds the complete shares, and then just ensures that they keep an internal record of each partial owner in their system? So there would be no external record of partial ownership?  Thanks! How does a lurker gain reputation to receive privilege for commenting? I have been reading Stack Exchange for years - mostly not logged in, and have never posted (this is my first). I'm simply not very talkative, and very good at googling.  Recently a few occasions have come up when I've wanted to comment on a question.  I wanted either to ask for, or provide, clarification which did not warrant a new answer.  How do I gain enough reputation to comment without:   asking questions which have already been answered or that I can figure out on my own, or posting my comment like as an answer (i.e. being  rude)? Use of .apply() with 'new' operator. Is this possible? In SQL Server, how do I generate a CREATE TABLE statement for a given table? I've spent a good amount of time coming up with solution to this problem, so in the spirit of , I'm posting it here, since I think it might be useful to others.   If anyone has a better script, or anything to add, please post it.  Edit: Yes guys, I know how to do it in Management Studio - but I needed to be able to do it from within another application. What does ':' (colon) do in JavaScript? Does a Grapple or Shove count as an attack for the Mobile feat? The  feat enables you to not get opportunity attacks from creatures you try to attack.  Does it still apply if the attack you made was a Grapple or Shove ? Android app "Unknown error" when uploading photo Spent a half hour on this. Android 5.1.1 latest version SE app (Play Store won't let me update so latest).   Tried to upload photo into an answer. "Something went wrong".  Did tricks to reduce photo size to 275k -- same.    45k rep on the stack.  Here is the photo..... oh wait.  How do I solve? Can I travel to Canada with valid visa on old/expired passport? Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Order of the Rubik's cube group Associated to the Rubik's cube is a group as described in this : $G = \langle F, B, U, L, D, R\rangle$. For example, the element $F$ corresponds to rotating the front face clockwise by $90$ degrees.  According to the article the order of the group is: $2^{27} 3^{14} 5^3 7^2 11$.  I have come across several places (for example ) where this order has been found by just counting various ways to arrange edges and such. Is there a nice way to compute the order by appealing to more "group theoretic" tools?