I know a good proof, but i want to proof it in my way, im asking if its right. Say $1 = g^2$ is a Group, so $p\cdot p = 1$ with $p$ element of $P,$ prove that that the group is commutative?   $1 = e,$ the neutral element. $1.$ so we know for every group exist associativity. $e = g\cdot g,$ from that we can conclude that every is the inverse element of itself.   i proof that $(g\cdot a) = (a\cdot g)$  $g\cdot a$ is also inverse to itself, so $(e\cdot g)\cdot(e\cdot g) = e.$ now the rest.  $$(g\cdot a)^2 \cdot (a\cdot g)^2 = e $$ $$(g\cdot a) \cdot(g\cdot a)\cdot (a\cdot g)\cdot(a\cdot g)=e$$  because of associativity $$(g\cdot a)\cdot g \cdot(a\cdot a)\cdot g)\cdot(a\cdot g)=e$$ $$g\cdot(a\cdot g)\cdot (g\cdot a)\cdot g=e$$ because of associativity i can also start with $(a\cdot g)$ and $(g\cdot a)$ and for that its only true if they are the inverse off another thus the same.
If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?    Proof: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
The plot is simply that aliens visit Earth and their interest in collecting cow pies (and just like other collectibles, special attributes make them more valuable) creates interest in humans in obtaining them to sell. Then the aliens leave and the market plummets. I do not recall the name of the story and also at the end I think some new object is of interest which maybe someone remembers. As silly as the premise sounds, if you look at what people really do collect and the amazing prices, the story kind of resonates. Digital crypto-kitties or whatever and more recent things selling for tens of millions or cryptocurrencies themselves seem not so different than cow pie collecting. Moreover, the fundamental theorem of collectibles is that to make a huge amount, paradoxically the collectible had to not only start out as more or less worthless but also something people tended to use/read and then destroy or at least throw away. Old Nat Geo magazines are worth almost nothing because everyone kept them; Action Comics #1 on the other hand, well moms could not wait to toss those when their son went away to college or got married. The stories I have in just my own family. I remember after Stargate: Universe was cancelled, that I read, somewhere on the web, either part of an interview or an article by one of the producers about what they were planning if there had been a third season.  (He said they weren't sure whether they'd set it soon after the ending of the second season, or later, or were going to do something else.)  He also wrote that, while he felt okay discussing that, he did not feel like he could reveal how the story about the superstructure the Destiny had found would turn out.  He said that was something the other two producers had worked out and he didn't want to reveal it and would leave it up to them to reveal it if they wanted to.  Was this ever revealed?  Did the other producers ever discuss or reveal anything about what was going to happen with that storyline?    Addition: I'm referring to the superstructure that is the message found in the background radiation throughout the universe, not anything from the planet builders or other easily seen physical structures. Which one should be used?      To me, it makes no difference, but I'm not really sure why.   vs     For me, it makes no difference, but I'm not really sure why. Why would a partition be in use by the system if it isn't mounted? I've been migrating my system from btrfs to ext4 after running into performance issues with VMs. I have two hard drives in my laptop to work with. I've successfully moved my home partition, but the same steps I used aren't working for root.  Progress so far:  I've dd'd my root partition from /dev/sda3 into /dev/sdb3. I modified /etc/fstab to the following:  $ cat /etc/fstab #  # /etc/fstab: static file system information # # &lt;file system&gt; &lt;dir&gt;   &lt;type&gt;  &lt;options&gt;       &lt;dump&gt;  &lt;pass&gt; # UUID=95f13c34-96ca-49e3-bcb2-ff594df31506 /dev/sdb3               /               btrfs           rw,noatime,ssd,space_cache,discard      0 0  # UUID=0fe04f59-599f-41e2-ac30-2ad0f17a9727 /dev/sda2               /boot           ext2            rw,relatime     0 2  # UUID=44741e0f-924a-4841-80ef-2132bef84182 /dev/sda4               /home           ext4            rw,noatime,discard      0 0   and run sudo mkinitcpio -p linux. It seems to work. I'm able to boot by mounting the partition on the second disk. df shows:  $ df Filesystem      Size  Used Avail Use% Mounted on /dev/sdb3        28G   18G  9.8G  65% /   So, clearly, sdb3 is mounted, not sda3. Here's the problematic step: When I try to format sda3, which is supposedly unused, I get the following:  $ sudo mkfs.ext4 /dev/sda3 [sudo] password for stew:  mke2fs 1.42.11 (09-Jul-2014) /dev/sda3 contains a btrfs file system Proceed anyway? (y,n) y /dev/sda3 is apparently in use by the system; will not make a filesystem here!   sda3 is in use. How and why could it possibly be in use?  As per casey's comment, the output of mount:  mount | grep sd /dev/sdb3 on / type btrfs (rw,noatime,ssd,discard,space_cache) /dev/sda4 on /home type ext4 (rw,noatime,discard,data=ordered) /dev/sda2 on /boot type ext2 (rw,relatime)   As per Warwick's comment, unmounting:  $ sudo umount /dev/sda3 umount: /dev/sda3: not mounted   Mounting and umounting sda3 elsewhere works successfully, but changes nothing.  Update: More fishy behavior:  $ mount | grep sd /dev/sdb3 on / type btrfs (rw,noatime,ssd,discard,space_cache) /dev/sda4 on /home type ext4 (rw,noatime,discard,data=ordered) /dev/sda2 on /boot type ext2 (rw,relatime) $ sudo mount /dev/sda3 mnt [sudo] password for stew:  $ mount | grep sd /dev/sda3 on / type btrfs (rw,noatime,ssd,discard,space_cache) /dev/sda4 on /home type ext4 (rw,noatime,discard,data=ordered) /dev/sda2 on /boot type ext2 (rw,relatime) /dev/sda3 on /home/stew/mnt type btrfs (rw,relatime,ssd,discard,space_cache)   After mounting sda3, sdb3 is no longer mounter. Weird, huh?  As per mikeserv:  $ rmmod btrfs rmmod: ERROR: Module btrfs is in use   This is very much expected, since sdb3 is btrfs and supposed to be mounted to root. From my mkinitcpio.conf file:  MODULES="" HOOKS="base udev autodetect modconf block filesystems keyboard fsck" Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits. Where was this picture taken (screensaver on LG Oled TV)? Can someone please tell me where this picture was taken? Why is the sky not purple? I realise the question of why this sky is blue is considered reasonably often here, one way or another.  You can take that knowledge as given.  What I'm wondering is, given that the spectrum of Rayleigh scattering goes like $\omega^4$, why is the sky not purple, rather than blue?    I think this is a reasonable question because  (or, strictly, violet or indigo) in rainbows, so why not across the whole sky if that's the strongest part of the spectrum?  There are two possible lines of argument I've  and I'm not sure which (if not both) is correct.  Firstly, the Sun's thermal emission peaks in the visible range, so we do actually receive less purple than blue. Secondly, the receptor's in our eye are balanced so that we are most sensitive to (roughly) .  Our eyes are simply less sensitive to the purple light than to the blue. Where should the comma be placed in the salutation of a letter? Sometimes I see a comma after the proper name:     Hello Mr. Black,      In order to give you....   But my native language is not English and I think that the comma in this phrase should be placed before the proper name:      Hello, Mr. Black.      In order to give you....   What is the correct punctuation in English? I want to create a standalone python script using QGIS python API (PYQGIS) for education. First I create a model and export that model to python script.  original python script from the export :  ##=name ##raster=raster ##stat=output html outputs_QGISRASTERLAYERSTATISTICS_1=processing.runalg('qgis:rasterlayerstatistics', raster,stat)   Next I read this  and I tried to follow this tutorial to create my script.  Final script :  import sys from qgis.core import QgsApplication from PyQt4.QtGui import QApplication app = QApplication([]) QgsApplication.setPrefixPath("/usr", True) QgsApplication.initQgis()  # Prepare processing framework  sys.path.append('/home/username/.qgis2/python') from processing.core.Processing import Processing dem='aster.tif' slope='slope.tif' aspect='aspect.tif' outputs_SAGASLOPEASPECTCURVATURE_1=processing.runalg('saga:slopeaspectcurvature', dem,6,1,1,slope,aspect,None,None,None,None,None,None,None,None,None,None) Processing.initialize()  # Exit applications QgsApplication.exitQgis() QApplication.exit()   If I run this script it works fine without error but I don't take some export like aspect and slope (these can't be found?). Any idea why I can't take exports ?  I use Ubuntu and python 2.7 and QGIS 2.18.1.    i try to this script :  import sys from qgis.core import QgsApplication sys.path.append('/usr/share/qgis/python/plugins') import processing a=processing.alglist() print a processing.alghelp("saga:slopeaspectcurvature")   and i take that print any idea why ?  Algorithm not found   but if i use this module in QGIS work fine. 17.10 Software installed in software center, but not accessible In Ubuntu 17.10 I installed applications from the software center. At first everything worked fine, but after a few days they kind of disappeared. They are neither found when searching in the dash, nor when the executable name directly in the terminal. However, the software center shows them still as installed and they can be launched from there. Any help to fix this strange behavior is appreciated. Loading OpenStreetMap data in QGIS? Can I find out which tags I have created? I have heard that .NET is not free and that I have to pay Microsoft if I develop a .NET application.  Is this true?  If so:   What do I have to pay? I'm a subscriber to the MSDN via the Academic Alliance (we download a variety of Microsoft Software products for free, like VS2008 and Windows Server 2008). Do I have to pay for the software that I make with these tools? What about web apps with ASP.NET? Is there a payment due when I host my application or do I just pay for the technology?   EDIT:  So what you are saying is that it's the same to develop a web application using ASP.NET or PHP (ignoring technical issues, just payment side), all what I have to pay is the hosting fees?  EDIT 2:  .NET rules :) What is the VTable Layout and VTable Pointer Location in C++ Objects in GCC 3.x and 4.x? Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others? How to add spaces at every X digit in a binary or hexadecimal number? I would like to format binary or hexadecimal numbers with a space separator at each 2 or 4 digit (or any other value like 8 digit or more).  A small picture is better than a long explanation, here is what I would like to do:    As you can see in this mwe, I use the fmtcount package with padzeroes:      \documentclass{article}     \usepackage[utf8]{inputenc}         \usepackage[T1]{fontenc}               \usepackage{fmtcount}     \begin{document}             Binary value : {\padzeroes[16]\binarynum{129}}$_{(2)}$ %             should display 0000 0000 1000 0001$_{(2)}$              Hexadecimal value: {\padzeroes[8]\Hexadecimalnum{65000}}$_{(16)}$ %             should display  00 00 FD E8$_{(2)}$     \end{document}   What I want to do is displaying numbers with various radix (2, 3, 7, 8, 10, 16) given a decimal number and format these numbers correctly with spaces and leading 0 (without having to perform the conversion by myself).  I already found some posts with similar questions, but none of them achieve this (they display number with thousand separators, like the numprint package, every 3 digits; or they do not make it automatic). Here are these posts :         EDIT two other linked questions:       I also prefer to use pdflatex. What is the limit of For a distribution function $F(x)$ and constant $a$, integral of $F(x + a) - F(x)$ is $a$. For any distribution function and any $a \geq 0$, $\int_{-\infty}^{\infty} (F(x+a)-F(x))dx = a$.   In this case, "distribution function" means a right continuous function F with $F(-\infty) = 0$, $F(\infty)= 1$.   This problem comes after a section in the book on mathematical expectation, so I suspect I should use some property of expectation to prove it. Expectation appears to be something that applies to random variables, so I think I should be somehow transforming the problem into something in terms of expectations of random variables, but I'm not sure how. I know that given a distribution function, I can get a probability measure, but I'm not sure how to go from there to a random variable and actually get a relation between the expectation of the random variable and this integral. I'm sure whether this is the right way to go about this problem, but if it is, I'm missing a connection somewhere. It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.  The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.  This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m&gt;n$? This does not seem obvious.  Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get  $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.   The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$? jQuery select and wrap textNode