Ruby: function (appears to) return correct value but fails rspec
expected true to respond to true?
Does $G\oplus G \cong H\oplus H$ imply $G\cong H$ in general? How do I transfer a file to my server using PuTTY? $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $n \mid \frac{a^{n}-b^{n}}{a-b}$ How does one prove that if $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $ \displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}$ where $a,b, n \in \mathbb{N}$.  What i thought of is to consider $$(a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)$$ and if we suppose that $n$ is odd then we have, $$(a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)$$ and since $n \mid (a^{n} - b^{n})$ we have $$(a-b)^{n} \equiv 0 \ (\text{mod} \ n) $$  I think i am far away from the conclusion of the problem, but this is what i could work on regarding the problem. The $n$ complex $n$th roots of a complex number $z$ Suppose $z$ is a nonzero complex number, so $z=re^{i\theta}$. Show that there are only $n$ distinct complex $n$-th roots, given by $r^{1/n}e^{i(2\pi k+\theta)/n}$ for $0\leq k\leq n-1$.   My proof: Let $z=re^{i\theta}\in\mathbb{C}$ and $a\in\mathbb{C}$ s.t. $a^{n}=z$. Let $a=\rho e^{i\varphi}$. Then, $a^{n}=(\rho e^{i\varphi})^{n}$. Setting $\rho^{n}e^{i\varphi n}=re^{i\theta}$ and rewriting in polar form, we get $$\rho^{n}(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(θ)+i\sin(\theta)).$$ Thus $\rho^{n}=r$ $\rightarrow$ $\rho=r^{1/n}$ and $n\varphi=\theta\rightarrow\varphi=\theta/n$. Thus the roots of $a\in\mathbb{C}$ have the form $$r^{1/n}(\cos(\tfrac{\theta}{n})+i\sin(\tfrac{\theta}{n}))=r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))$$ for some $j\in\mathbb{Z}$ due to the periodicity of trig functions. Then, $$r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))=r^{1/n}e^{i\left(2\pi j+\tfrac{\theta}{n}\right)}=r^{1/n}e^{i(\theta+2\pi k)/n}$$ for $0\leq k\leq n-1 ,j=nk\in\mathbb{Z} $.  I feel like there could be a more concise way to put this. How can I remap a keyboard key? One of my laptop's keys has fallen off. Is there any way I can remap another key to serve as that key?  I'll 'sacrifice' the other key because I never use it. Is this proof of $\liminf E_k \subset \limsup E_k $ correct? I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. I have 24 accounts, how can I delete them all in one go? Is there a way to delete all stack exchange accounts at the same time?  I also want to remove all traces of my posts on the site. Is that possible?   Will I have to delete all of my posts manually?  Can I use the stack exchange API for this? Usually I use the package mhchem for writing chemical formulas, but until now I was not able to find a solution for writing crystals such as Nd^{3+}:YAG properly with this package. When simply writing   \ce{Nd^{3+}:YAG}   it results in an error. My current alternative solution is to write it as   \ce{Nd^{3+}}:\ce{YAG}   Is there a more elegant way to do that? Android design convention for saying a user needs to complete an action first? I've read the Android design conventions, but can't see a good solution.  I have a 3 tab mobile application. Tabs 2 and 3 need a selection made on screen 1 before they will work. The user enters at Screen 1, and the tab options are along the top. If they attempt to open screen 2 or 3, I would like to advise them to make a selection first. 'notifications' are the closest answer I can see from the Android design guidelines. Anybody have any other solutions?  Thanks  Andy How can I double the newlines in an output stream How could Argus Filch clean up the whole castle by himself without magic? Unable to uninstall Universal Apps through PowerShell Use Yaml for MongoMapper Config got me wondering about the differences between these three ways of measuring size: a kibibyte, a kilobit, and the conventional kilobyte.  I understand that these measurements have different uses (data transfer rate is measured in bits/sec), but I'm not quite sure if I can tell the difference between Mb and MB and MiB.  Here is a comment, reproduced below, taken from  (emphasis mine).     The C64 has 65536 bytes of RAM. By   convention, memory size is specified   in kibiBytes, data transfer rates in   kilobits, and mass storage in   whatever-the-manufacturers-think-of-now-Bytes. Harddrives use T, G, M and k on the   label, Windows reports the size in Ti,   Gi, Mi and ki. And those 1.44MB   floppys? Those are neither 1.44MB nor   1.44MiB, they are 1.44 kilokibibytes. That's 1440kiB or 1'474'560 bytes. –   Third How to configure lightdm to allow manual logins in Ubuntu 14.04 I used to modify /etc/lightdm/lightdm.conf, adding the line greeter-show-manual-login=true, in order to allow login for hidden users and root.  Now I've just installed the new Ubuntu 14.04 and the file lightdm.conf is no longer located in /etc/lightdm. It's in /etc/init and has been totally modified.  How can I force lightdm login screen to show a field to input a login name? How to create quizzes with macqtex? keyboard hard remap keys? Understanding the -exec option of `find` If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication) How to extend summations to fractional orders?