According to Frobenius theorem,  every associative division algebra over $\mathbb{R}$ is either $\mathbb{R}$, or $\mathbb{C}$, or $\mathbb{H}$ where $\mathbb{H}$ is the quaternion algebra. This clearly implies that $\mathbb{R}^3$ does not have an associative division algebra. But is there an easy way to demonstrate this fact without proving Frobenius theorem?
I am trying to solve the following exercise:     Prove that complex multiplication does not extend to a multiplication   on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division   algebra.   I am aware of Frobenius' theorem that there are only three finite dimensional real associative division algebras. But the exercise would be pointless if the theorem was assumed.   Let $x,y \in \mathbb R^3$. Since $x y$ has to extend the complex multiplication:  $$xy = (x_1y_1-x_2y_2, x_1y_2 + x_2 y_1, ?)$$  I have no idea how I can proceed from here. Could someone tell me how to do this? It should be easy as the other exercises I did so far were also easy.
It seems I put a   Bounty assigned by outside party, two lengthy, reference-citing answers, one "-1" (awarded the bounty), one "-2", two others scored "0" and "-2" respectively, the answers suggesting one or the other is correct, 73 comments and no consensus so far - and me, as the asker, lost without a clue what to think of the answers anymore -  this is no longer "Learners' Level" question, so I thought I'd bring it here and hear what the experts have to say.    Original question:  Plurality of verb depending on plurality of list elements  An edit was suggested to my sentence.     There was were an orange, some grapes, two apples and a small pile of cherries on the plate.   In my native language plurality of the verb always follows plurality of the first element on the list. There were an orange,... sounds awkward to me, no matter what follows up. My simple solution was reordering:     There were some grapes, an orange, two apples and a small pile of cherries on the plate.   But that's not the first time I faced this situation and I'd like to know what the rules of grammar say about that - was my editor overzealous or am I trying to copy rules of my language that don't apply in English?    Someone linked a  for negatives, where the situation is more clear-cut ("There is no..."). Same applies to connection with "or" apparently, per accepted answer there. Still, nothing for "and".  It seems there is a consensus that in that if the verb goes after the list, it will be plural.     an orange, some grapes, two apples and a small pile of cherries were on the plate.   There doesn't seem to be clear consensus for:     On the plate { was | were } an orange, some grapes, two apples and a small pile of cherries   Adding more to the confusion is the abbreviated form: "There's"     One last note. In North America, at least, there is a widespread use of using a singular form like "there is" and "there was", without regard for the subject item or items, and this "there is" is often shortened to "there's":           There's three apples on the table!      Could you please clarify this mess? In my situation, it's for a local build/staging server running ubuntu server. Things I'd like to know are:  Why I should/shouldn't use LVM?  What problems might it cause in terms of either regular server operation or things I might want to do later? (ex. partitioning)  Why I should/shouldn't use encrypted LVM instead of LVM?  What problems might encrypted LVM cause in terms of either regular server operation or things I might want to do later? (ex. partitioning) Delayed eraser experiment (Kim experiment) Will I receive an interference pattern on D0 detector   if D1, D2, D3, D4 exist in the set up of the experiment, but they are thousands of light years far? Simply, to set up the experiment so that I detect all of the signal photons on D0 before I detect any of idler photons! Show that $ a，b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q $ Inspired by , I was wondering if there is a simple logical argument to   Show that $ a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q  \implies  \sqrt{a},\sqrt{b} \in\mathbb Q $  Note that the original link is using a computational method, where as I am looking for a simple logical argument.  I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example $a=1-\sqrt{2},b= 1+\sqrt{2} $) Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$ x^3  + 1 = 2\sqrt[3]{{2x - 1}} $$ I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here? How to correct for small p-value due to very large sample size What is LINQ? I know it's for databases, but what does it do? I have a series of pending jobs in the queue,   $ atq 8   Mon Oct 29 13:21:00 2018 5   Mon Oct 29 10:36:00 2018 15  Mon Oct 29 13:31:00 2018 4   Mon Oct 29 10:35:00 2018 18  Mon Oct 29 15:55:00 2018 3   Mon Oct 29 10:31:00 2018 16  Mon Oct 29 15:54:00 2018 11  Mon Oct 29 13:24:00 2018 12  Mon Oct 29 13:24:00 2018 17  Mon Oct 29 15:53:00 2018 10  Mon Oct 29 13:23:00 2018 19  Tue Oct 30 15:43:00 2018 2   Mon Oct 29 07:41:00 2018 6   Mon Oct 29 10:41:00 2018 7   Mon Oct 29 13:20:00 2018 9   Mon Oct 29 13:22:00 2018   and plan to remove them job on Oct 29. Fortunately, atrm could remove multiple jobs  atrm 8 5 15 4    despite this, a iteration might be a better solution,   $ atq | grep "Oct 29" 18  Mon Oct 29 15:55:00 2018 3   Mon Oct 29 10:31:00 2018 16  Mon Oct 29 15:54:00 2018 11  Mon Oct 29 13:24:00 2018 12  Mon Oct 29 13:24:00 2018 17  Mon Oct 29 15:53:00 2018 10  Mon Oct 29 13:23:00 2018 2   Mon Oct 29 07:41:00 2018 6   Mon Oct 29 10:41:00 2018 7   Mon Oct 29 13:20:00 2018 9   Mon Oct 29 13:22:00 2018   How could I select the first field of the job number so as to iterating them? How can I maintain open ssh connection and use it from shell scripts? Currently i invoke the following:  $ ssh me@host.com my_cmd   This is slow and not easy to automate safely. I would like to establish ssh connection once and have some script that will forward my commands to host.com and print output.  Is that possible ?  Adding my machine to authorized_keys is not an option for me and it wouldn't solve slowness issue. LaTeX packages: automatic versus manual installation, and related issues I often run into problems in LaTeX with packages. I did a reinstall (MiKTeX) and it seemed to fix some issues with knowing where packages coming from or going. In general, I am not sure what should or does happen when using \usepackage command.  Is it standard that these packages are automatically pulled from the internet? Where do they go (by default)? Is there a better place to put them?  If I am troubleshooting, and if we can say packages are usually installed automatically, when, and should I ever, install packages manually? If so, where would I put them?  For instance, I believe the package easylist is installed on my computer. I was able, with some difficulty, to get into the roaming folder, looked around, and really the only package I saw, was several easylist files, such as .sty, and an associated readme file. I did a windows search for easylist, and only found one, with suffix .tpm and in the Program Files (x86)\MiKTeX 2.9\tpm folder.  My problem currently, is that I am trying to use the todonotes package. It seems to require the xcolor package. But this "can not be found". It asks to choose a place to get it. I think default was "choose random", but I just switched it to internet, and some random mirror site in country (USA). That didn't work, and I'm afraid I've now fiddled with the default settings for getting packages, which will create problems later.  Here are some other threads I have been fishing through:             In lieu of picture, I have this:    miktex-pdftex.exe: No connection could be made because the target machine actively refused it.  miktex-pdftex.exe: Data: connect failed in tcp_connect() {}======================================================================  ! LaTeX Error: File `xcolor.sty' not found.  Type X to quit or &lt;RETURN&gt; to proceed, or enter new name. (Default extension: sty)  Enter file name:    and from clipboard:    MiKTeX Problem Report Message: No connection could be made because the target machine actively refused it.  Data: connect failed in tcp_connect() Source: Libraries\MiKTeX\PackageManager\PackageManager.cpp Line: 2461 MiKTeX: 2.9 OS: Microsoft Windows 7 Home Premium Edition, 64-bit Service Pack 1 (build 7601) Invokers: explorer SystemAdmin: no PowerUser: no Root0: C:\Users\aUserName\AppData\Roaming\MiKTeX\2.9 Root1: C:\Users\aUserName\AppData\Local\MiKTeX\2.9 Root2: C:\ProgramData\MiKTeX\2.9 Root3: C:\Program Files (x86)\MiKTeX 2.9 UserInstall: C:\Users\aUserName\AppData\Roaming\MiKTeX\2.9 UserConfig: C:\Users\aUserName\AppData\Roaming\MiKTeX\2.9 UserData: C:\Users\aUserName\AppData\Local\MiKTeX\2.9 CommonInstall: C:\Program Files (x86)\MiKTeX 2.9 CommonConfig: C:\ProgramData\MiKTeX\2.9 CommonData: C:\ProgramData\MiKTeX\2.9    (from "Package Installation" window below)   The required file        tex\latex\xcolor\xcolor.sty  is missing. It is part of the following package:        xcolor  The package will be installed from:  ftp://... [Change...] The package will be installed for... In my question's preview, the attached image looks rotated. So, assuming you are able to see it in the real orientation, can anyone determine what is that dust flake type thing on the top, slightly towards the left?   One looks like a dust flake and there's another dot to the left of it?  Note: I wont have physical access to the camera until the weekend.  Is it just a dust flake on the lens or can it be some problem with the sensor? Its a 3 month old Canon 1500D. I am trying to define a proof for  $A \cup B = B$ if and only if $A \subset B$  I started with:  $$A \cup B$$ $$x \in A \lor x \in B$$   but I am unsure on how to continue. Is there a word for a comment which makes no sense or adds nothing to the current discussion? The title is kind of obscure. What I want to know is if this is possible:  string typeName = &lt;read type name from somwhere&gt;; Type myType = Type.GetType(typeName);  MyGenericClass&lt;myType&gt; myGenericClass = new MyGenericClass&lt;myType&gt;();   Obviously, MyGenericClass is described as:  public class MyGenericClass&lt;T&gt;   Right now, the compiler complains that 'The type or namespace 'myType' could not be found." There has got to be a way to do this. I was wondering how to create a theorem environment similar to the following example:   Specifically, the environment should resemble the box that contains "Claim -- The function g is linear".  This comes from a , who has his style files uploaded  I have tried to mimic what I believed to be the relevant code (as there's a lot of stuff there that isn't related to this), to no avail. I have attached my attempt below, but it may be better to ignore it.  \usepackage{amsmath} \usepackage{amsthm} \usepackage{mdframed} \usepackage{thmtools} \usepackage{tikz} \usepackage{xcolor} \usepackage[framemethod=TikZ]{mdframed}     \mdfdefinestyle{mdgreenbox}{%         linewidth=0.5pt,         skipabove=12pt,         frametitleaboveskip=5pt,         frametitlebelowskip=0pt,         skipbelow=2pt,         frametitlefont=\bfseries,         innertopmargin=4pt,         innerbottommargin=8pt,         nobreak=true,         backgroundcolor=SpringGreen!15,         rightline=false,         leftline=false,         topline=false,         bottomline=false,         linecolor=green,     }     \declaretheoremstyle[         headfont=\bfseries\color{OliveGreen},         mdframed={style=mdgreenbox},         headpunct={\\[3pt]},         postheadspace={0pt},     ]{thmgreenbox} \declaretheorem[style=thmgreenbox]{thrm1}   This returns an error, one for xcolor not recognizing the colors, and the following.  l.17        \mdfdefinestyle                       {mdgreenbox}{% The package mdframed has already been loaded with options:   [] There has now been an attempt to load it with options   [framemethod=TikZ] Adding the global options:   ,framemethod=TikZ to your \documentclass declaration may fix this. Try typing  &lt;return&gt;  to proceed.  Package thmtools Info: Key `mdframed' (with value `style=mdgreenbox') (thmtools)             is not a known style key. (thmtools)             Will pass this to every \declaretheorem (thmtools)             that uses `style=thmgreenbox' on input line 39. Package thmtools Info: Automatically pulling in `thmdef-mdframed' on input line 40. (/usr/local/texlive/2017/texmf-dist/tex/latex/thmtools/thmdef-mdframed.sty Package: thmdef-mdframed 2014/04/21 v66 ) \c@thrm1=\count282 (/compile/output.aux) \openout1 = `output.aux'.   I am looking for a way to mimic the environment more than to fix my methods.  Thank you in advance! Downloading US Census data? export CLASSPATH=.;../somejar.jar;../mysql-connector-java-5.1.6-bin.jar java -Xmx500m folder.subfolder../dit1/some.xml cd ..   is the above statement for setting the classpath to already existing classpath in linux is correct or not System.Exception: Too many SOQL queries: 101 proving of inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$ How can we prove the inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$  $\bf{My \; Try}$::  $1\leq n$  $2\leq n$  $3\leq n$  ....  ....  $n\leq n$  So $1\cdot 2 \cdot 3 ..n \leq n^n$  So $n!\leq n^n\Rightarrow (n!)^2 \leq n^n\cdot n!$  But  did not understand how can we prove $n^n\cdot n! \leq (2n!)$  Help me  Thanks How to find all the used IP addresses on a network In my work there is a lot of computer and I want make a joke. I can close computer over the network but finding IP addresses is hard to me.  How can I easily find all online IP adresses from 192.168.1.aa to 192.168.1.zz?