Mathematics in power sets of Reals
Are there number systems corresponding to higher cardinalities than the real numbers?
Can you cast a spell learned from the Magic Initiate feat using spell slots? Say I play a wizard who learns cure wounds by taking the Magic Initiate feat. Could I cast that spell using the Wizard's spell slots? How can I install more recent versions of software than what Debian provides? The other day I was looking at Stack Overflow on my Android phone. All is made very nice and all, but I could not find the Review button/link. Is it being implemented, did I miss it, or it is not available? Sorry for the vague title--I don't want the title to be a spoiler.  I've seen the movie three times, and the last two times I looked specifically for clues to this question, but found none.  The final scene in  shows       Cobb's top spinning, as he goes off to be reunited with his children.  The top continues to spin, then the movie ends, without telling us if the top topples--and therefore revealing whether he's in the "real world" or still in a dream. Do we know if he was in a dream at the end of the movie?   I'm assuming the answer is that we simply cannot know.  But did I miss anything?  Were there clues that I missed? Could someone advise me on how to determine all entire functions $f$ such that $|f'(z)| \leq |f(z)|, \forall z\ ?$  Hints will suffice, thank you. Array versus List: When to use which? MyClass[] array; List&lt;MyClass&gt; list;   What are the scenarios when one is preferable over the other? And why? Apply for Schengen visa in a third country Can a Kenyan citizen studying in Russia and temporarily visiting family in Kenya apply for a Schengen visa in Nairobi and not in Russia, where he has his residence? I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The  contains a delta function in the formula, but  gives only the formula without the delta function,  $$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$  Is it intentionally left out?  My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge. Characterizing all ring homomorphisms $C[0,1]\to\mathbb{R}$. This is something I've been trying to work out this evening.      Let $R$ be the ring of continuous real-valued functions on $[0,1]$ with pointwise addition and multiplication. For $t\in [0,1]$, the map $\phi_t\colon f\to f(t)$ is a ring homomorphism of $R$ to $\mathbb{R}$. I'm trying to show that every ring homomorphism of $R\to\mathbb{R}$ has this form.   Suppose otherwise, that there is some $\phi\neq\phi_t$, and thus there is some $f_t\in R$ such that $\phi(f_t)\neq \phi_t(f_t)=f_t(t)$. Define $g_t=f_t-\phi(f_t)1\in R$. Here $\phi(f_t)1$ is the constant function sending $[0,1]$ to $\phi(f_t)$. Then $g_t(t)\neq 0$. My first small question is why does $\phi(g_t)=0$? It seems only that $\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)$.  I would like to conclude that there are only finitely many $t_i$ such that $g(x)=\sum g_{t_i}^2(x)\neq 0$ for all $x$. Then $g^{-1}=1/g(x)\in R$, but $\phi(g)=0$, contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many $g_{t_i}$ such that the sum of their squares is never $0$? Thanks. Simulating the Windows key on an old keyboard Someone told me that the English language is special (compared to German, at least) in the way that every noun could be used as a verb. I think this phenomenon is called supine. Is this correct?  Examples:   house → to house well → to well phone → to phone table → to table lamp → to lamp   At first it seemed to me that this would work with all nouns. But then I thought of the following nouns, and I think they can't be used as a verb, am I right?   person →(?) to person human →(?) to human baker →(?) to baker computer →(?) to computer example →(?) to example   It seems that nouns describing people (person, human, baker) are exceptions? But it works for "author" ("to author"). Hmm. "Computer" seems smiliar to a person's job to me (could be a human instead of a machine that computes).  But what about "example"? I think "to example (something/someone)" is not possible, is it? What would a native speaker understand/think if someone says:      He examples.      I want to example it/him.   Which are the nouns that can't be used as a verb? I am using ArcMap 10.1. The background: I have an .mxd that I intend to roll out across the company as our default template for all future maps/figure production. This .mxd contains two Data Frames; the first (&quot;Main&quot;) is larger (occupying almost the entire page) and is designed to be used to hold the majority of mapped data whilst the second (&quot;Inset&quot;) is much smaller and is to be used as an overview map. The problem: When choosing to import this new .mxd to an existing .mxd (using the Change Layout button on the Layout toolbar) the wizard automatically selects the smaller (&quot;Inset&quot;) Data Frame to occupy the larger part of the page. The screenshot below shows this interface with a red box indicating the two possible locations of each Data Frame:  Whilst it is possible to manually alter the order of the Data Frames using the Move Up or Move Down buttons, how do I change the default option for this setting (i.e. so that the &quot;Main&quot; Data Frame always appears in the larger space - &quot;2&quot;)? The reasoning is that I would like the process of using this new .mxd template to be as simple as possible, this is especially important as we have a wide range of users with varying abilities. Does anyone know how to choose/change this default setting? It does not appear to be in the Data Frame Properties &gt; Data Frame menu where I would expect it to be. Why does my query run fast in Environment A, but slow in Environment B? I have a piece of SQL that seems to run really fast in Environment A, but the exact same query runs really slow in Environment B!  Environments are supposed to be the same, so what should I do and/or where should I look in order to see why the query doesn't perform the same? Run TTY script every time I open a GUI terminal When I log in to a TTY I get the following nice text:  Ubuntu 16.04.03 LTS Log-in: username Password: Last log-in date/time Welcome to Ubuntu 16.04.03 LTS   * Documentation:  * Management:  * Support:  9 packages to update 4 updates are security updates   I would like the last 2 lines repeated every time I open a GUI terminal for all users in the adm group (but will settle for having the entire thing displayed each and every time I open a GUI Terminal for all users)  I tried the obvious:  fab-root@fab-ux-predator:~ $ cd /etc/update-motd.d/ fab-root@fab-ux-predator:/etc/update-motd.d $ ./90-updates-available  fab-root@fab-ux-predator:/etc/update-motd.d $ cat /var/lib/update-notifier/updates-available cat: /var/lib/update-notifier/updates-available: Permission denied fab-root@fab-ux-predator:/etc/update-motd.d $   What am I missing?  P.S. Obviously I can chmod o+r the file, but how future-proof would that be? Exercise from Atiyah-Macdonald, Chapter 1, 2.iv) Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an indeterminate $x,$ with coefficients in $A.$ Let $f=a_0 + a_1x+\cdots+a_nx^n \in A[x].$ $f$ is said to be primitive if $(a_0,a_1,\ldots,a_n)=(1).$     Prove that if $f,g\in A[x],$ then $f$ and $g$ are primitive $\Rightarrow$ $fg$ is primitive.   I know Gauss lemma holds true for UFDs, but in this case it isn't required $A$ to be UFD: is it possible to drop the assumption, or is some hypothesis missing here? Solvability of $a \equiv x^2 \mod b$ I want to burn a Windows ISO to a USB device in Ubuntu. How do I do this?   I know how to burn a Ubuntu ISO into a USB device, but with a Windows ISO it's not the same. I saw a thing called VPN connections under the Internet button at the top of my screen.   So I would like more info on this, and possibly instructions on how to set one up. Thanks! Biblatex - capital names in references, regular in cites I am trying to achieve a biblatex style that would do the following:  in the case of \parencite{} I would get ...like that (Smith 2010).   in the case of \textcite{} I would get Smith (2010) says that...  but in References I would get SMITH, J. Bookname.  How can I achieve this?  I use \usepackage[style=authoryear, firstinits=true]{biblatex}  I tried \renewcommand*{\mkbibnamelast}[1]{\MakeUppercase{#1}} but that makes the name uppercase even in the text, which is undesirable especially with \textcite{} Group of order $8p$ is solvable, for any prime $p$ Consider the following question:     Show that a group $G$ of order $8p$ is solvable, for any prime $p$.   I am kind of stuck, but here are my first attempts:  I chose the series of subroups $G&gt;H_8&gt;H_4&gt;H_2&gt;H_1=\{e\}$ where $H_k$ has order $k$. All of these subgroups exist due to Sylow's Theorems. We have the quotients $H_8/H_4\cong H_4/H_2 \cong H_2/H_1 \cong \mathbb Z/2\mathbb Z$, so $G&gt;H_8\rhd H_4\rhd H_2\rhd H_1=\{e\}$. Only the factor $G/H_8 \cong \mathbb Z/p\mathbb Z$ is causing me a headache, because although its of prime order (and hence abelian) I don't know whether $H_8$ is normal in $G$ (unless $p=2$).  If $p\ne 2$, then the number $k$ of Sylow $2$-subgroups is $1$ mod $2$. Since $k$ divides $|G|$ we might have $k=1$ or $k=p$. If we had $k=1$ then $H_8$ would be the only Sylow $2$-subgroup (and hence normal in $G$). But how could we show this?  Or is there even an easier way to approach this problem?  $\ $  edit: I am going to try a different approach: The case $p=2$ is clear. When $p=3$ or $p=7$ the group has order 24 or 56 and since the smallest simple non-abelian group has order 60, the group must also be solvable in these exceptional cases (as Mariano pointed out).  In any other case we get a Sylow $p$-subgroup $H_p$ of order $p$, which is is normal in $G$. The quotient $G/H_p$ has order $8=2^3$; and prime power order implies solvable. So $G/H_p$ is solvable and $H_p$ is also solvable. And since the composition factors of $G$ are those of $H_p$ together with those of $G/H_p$, we conclude that $G$ is solvable iff $H_p$ and $G/H_p$ are solvable, which is the case. Hence $G$ is solvable if $|G|=8p$.