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$\mathbb R$ has the same cardinality of any interval I want to see all URLs that for packages that would be downloaded for apt install &lt;pkg&gt; without actually installing the package.  Normally, the URLs can be extracted from the screen output after installing the package (as above).  An example of a URL I want to list:    Is it only possible to get such list of URLs beforehand by looking through all the apt's package source lists? Installing python module within code Is there a general formula for finding all subgroups of dihedral groups? Summation of series $\sum_{n=1}^\infty \frac{n^a}{b^n}$? Is Tactical Down-Voting ever valid? Has anyone noticed any tactical downvoting of their, or other people, answers?  For example, in a recent answer to a question, my answer described a possible solution.  Shortly afterwards, another answer appeared following the solution I had mentioned.    I know that this is completely in the spirit of the site, however my answer had been down voted even though it was still as valid as it ever was.  Another poster also had his equally valid solution down voted.  There is no way of knowing the reason behind the down voting, as no comments were left, but I am suspicious as to the ethics behind down voting someone else's valid answer in the hope of making their own answer more prominent.  Are people really that desperate to gain reputation?  I have seen comments in other posts that suggest that tactical downvoting has occurred.  So, does anyone think that it is acceptable to downvote another post, even though it is still correct but may not provide a complete answer, to make their own appear more favourably? Voting to Close after Accidentally Voting to Leave Open When reviewing on SO, I accidentally pushed the "Leave Open" button on a post that clearly should've been closed. I pushed the back button and saw that in the review panel my vote to keep open had been locked and it said:     PhpMyCoder voted to leave open 1 minute ago   So I clicked on the question heading to go to its page and I clicked the close link on the question page and voted to close for the relevant reason. However, when I returned to the review page (and did a hard refresh) it still said:     PhpMyCoder voted to leave open 2 minutes ago   Is it possible to vote to leave open and vote to close at the same time? I assume not. In which case, I'd say this is a bug because it should've updated to say:     OtherUser and PhpMyCoder voted to close 2 minutes ago Does app review rejection list all issues? Our app was rejected recently with reason Google+ signin, which we understood is not according to Apple policy and need to fix.  While discussing about this issue we came across an interesting question with Apple review process, does Apple review team list all issues (if more than one exists) while rejecting app (or) would they stop reviewing other areas as soon as they found an issue (in our case Google+ signin)? I remember (vaguely) reading somewhere that it is wrong to use the present perfect when the subject of a sentence is someone who is long dead:     *Einstein has visited the Philippines.   My question is, What if the dead person is not the subject but an object:     Leo is the best president our country has/had ever had. (The ones who   succeeded him were inferior, even our current president.)   Which tense is proper here Prove that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ converges absolutely I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise:  Exercise 2.7.5 (a) Show that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ also converges absolutely. Does this proposition hold without absolute convergence?  I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here.  Anyway, here's how I prove the absolute convergence of $\sum{a_n^2}$, and note that I never use the fact that $\sum{a_n}$ is absolutely convergent:  Proof: Let $s_n = \sum_{i=1}^n{a_n^2}$. I want to show that $(s_n)$ is a Cauchy sequence. So, let $\epsilon &gt; 0$ and $n &gt; m$, and consider, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\                 &amp;= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\                 &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\   \end{aligned} \end{equation} Now, since $\sum{a_m}$ converges, the sequence $(a_m)$ has limit 0. Therefore I can choose an $N$ such that $|a_m| &lt; \sqrt{\epsilon/n}$ for all $m &gt; N$. So for all $n &gt; m&gt; N$,  we have, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\                 &amp;= \epsilon   \end{aligned} \end{equation} Therefore $(s_n)$ is Cauchy and $\sum{a_n^2}$ converges. And since $\sum{a_n^2} = \sum{|a_n^2|}$, the series $\sum{a_n^2}$ is absolutely convergent. &#x220E;  Am I doing something wrong here? Am I right in thinking that $\sum{a_n}$ need not be absolutely convergent for $\sum{a_n^2}$ to be absolutely convergent? How to prove this decomposition of sum of squares? I am reading a paper   It has a mathematical step shown as  $$\sum_{i=1}^{n}(x_i-\mu)^2=n(\mu-\overline x)+\sum_{i=1}^n(x_i-\overline x)^2$$  Here the $\mu$ is unknown parameter and $\overline x$ is the known mean of $\mathbf x=(x_1, x_2, x_3, \dots,x_n)$  How do I demonstrate this identity? How to run/use popcorn-time? I got popcorn time installed using this link:    But I have no clue how to use or run Popcorn time.  Do I open my torrent client and download a movie and then it gets going? or do I have to open it? If I have to open it, how do I do that, I opened the dash and tried clicking on it but nothing happened. Please help me!! :D Whenever I boot Ubuntu, I get a message that it cannot mount my windows partition, and I can choose to either wait, skip or manually mount.  When I try to enter my Windows partition through Nautilus I get a message saying that this partition is hibernated and that I need to enter the file system and properly close it, something I have done with no problem so I don't know why this happens.    Here's my partition table, if any more data is needed please let me know.     Device Boot      Start         End      Blocks   Id  System /dev/sda1            2048    20000767     9999360   83  Linux /dev/sda2        20002814   478001151   228999169    5  Extended /dev/sda3   *   478001152   622532607    72265728    7  HPFS/NTFS/exFAT /dev/sda4       622532608   625141759     1304576   82  Linux swap / Solaris /dev/sda5        20002816   478001151   228999168   83  Linux Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.   Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing.  My work so far below:  Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$  Inductive Step:  $7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$   So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here... When I'm sneaking around, I can move more slowly and quietly by holding shift.   But walking this way is really slow. I would like to move a little bit faster, but not full speed.    On a console I imagine you have much finer control over how fast you move with the joysticks. Is there any way to accomplish this on the PC (without an external joystick)?  Perhaps there is a console command or config file setting that will change the % speed modifier that holding down shift uses? How to pass parameters to an alias? Disk full, du tells different. How to further investigate? I have a SCSI disk in a server (hardware Raid 1), 32G, ext3 filesytem. df tells me that the disk is 100% full. If I delete 1G this is correctly shown.  However, if I run a du -h -x / then du tells me that only 12G are used (I use -x because of some Samba mounts).  So my question is not about subtle differences between the du and df commands but about how I can find out what causes this huge difference?  I rebooted the machine for a fsck that went w/out errors. Should I run badblocks? lsof shows me no open deleted files, lost+found is empty and there is no obvious warn/err/fail statement in the messages file.  Feel free to ask for further details of the setup. How useful is that rule and how many times we can use that in one game?  Under what conditions can the en passant rule be used?  How often does it come up and is it a good idea to use it? Since , how do I uninstall it from my desktop system?  (Please note: This question is not about Unity 7, which is still supported at the time of this writing and should remain installed if present.) Prove that the normalisation of $A=k[X,Y]/(Y^2-X^2-X^3)$ is $k[t]$ where $t=Y/X$ (Reid, Exercise 4.5) This is a problem about finding the normalisation of a quotient polynomial ring. So I have to find the integral closure of the ring in its field of fractions. The problem statement is as follows:     Let $A=k[X,Y]/(Y^2-X^2-X^3)$. Prove that the normalisation of $A$ is $k[t]$ where $t=Y/X$.   Can I do this by showing that the field of fractions $\text{Frac} A$ of $A$ is equal to $k[t]$, and subsequently showing that the field of fractions is normal? (This could be done by showing that $k[t]$ is a UFD?)  I am lost at calculating/determining $\text{Frac}A$, or similarly proving that $k[t]=\text{Frac}A$. Also, how do I show that it is normal?  I hope you can help!