Does any set admit a total order?
Does linear ordering need the Axiom of Choice?
Just don't show UI for actions that can't be actioned Recently I just deleted one of my answer in stackoverflow where there was one comment. After deleting the answer it also show for sometime that we all know but in the comment it still allow us to up to make it useful but as I clicked it it just showed me the message:  If we are not allowed to do then what was the meaning? Why don't it just be removed or make it disable if the answer is deleted. Just don't show UI for actions that can't be actioned Proof of greatest integer theorem: floor function is well-defined I have to prove that  $$\forall x \in \mathbb{R},\exists !\,n \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1\;.$$  where $\exists !\,n $ means there exists a unique (exactly one) $n$. I'm done with proving that there are at least one integers for the solution. I couldn't prove the &quot;uniqueness&quot; of the solution, and so I looked up the internet, and here's what I found:  Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1$ and $m \leq x &lt; m+1$. Since $n \leq x \text{ and } -(m+1) &lt; -x$, by adding both, $n + (-m-1) &lt; (x-x) = 0$. And (some steps here) likewise, $(x-x) &lt; n+m+1$.  Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)? Or should I also prove those addition of inequalities? Thank you :D Every cyclic group with order > 2 has at least 2 distinct generators Every cyclic group with order > 2 has at least 2 distinct generators   Here's what I've got so far:  Either order of our group, $G$, is finite or infinite.  Suppose infinite: then our group is isomorphic to $\mathbb Z$ under addition. This group has two distinct generators, therefore so does $G$.  Suppose finite: Then $G$ is isomorphic to $\mathbb Z_n$ under addition modulo $n$.  We know order of $G$ is $\geqslant 3$. If order of $G$ is odd, both 1 and 2 are generators of $\mathbb Z_n$ under addition modulo $n$, so $G$ has at least two generators.  If order of $G$ is even, then both 1 and some odd number between 1 and $n$ are generators. How do we determine this odd number?  Does this look like I'm on the right track? How do you appear offline to friends on Battle.net? I have a number of friends on my battle.net account. However, sometimes I don't want to appear online to some (or all) of them. Is there a way to be hidden from your friends on battle.net? How do I properly clean up Excel interop objects? What happens when I hover over a link in Chrome? What is meant by a continuous-time white noise process? Decidability of a predicate. I have the following problem:  Let the following predicates be: $P1, P2, Q1, Q1 : R \to \{0,1\} $ .  It is given that $P1 \lor P2$ and $Q1 \lor Q2$ are semidecidable and $P2 = \lnot Q2$  What can it be said about $P1 \lor Q1$?  I have searched all over the internet and I simply cannot understand the concept of decidability of a predicate, let alone a logic operation with predicates. A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form   $$f(x)=ae^{bx}+cx+d = 0$$  where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply  $$x = \frac{\ln(\frac{-d}{a})}{b}$$  but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots? Do comic characters sometimes get drawn like the actor playing them? I noticed in Spider-Man's recent cameo in Captain Marvel, he appeared an awful lot like Andrew Garfield who is playing Spider-Man in the recent movie.    Irrelevant of whether it is just my imagination, do some depictions of comic characters morph to be more like an actor who plays them around the time a corresponding movie is released?  By comparison, here is Spider-Man's depiction in the Spider-Men crossover: Plot draws list of curves in same color when not using Evaluate Fix missing ssl-cert for subdomain www. using redirect or similar I was using the amspset document class for generating exam practice problems. I had wanted to add some referencing to the problems so I started a new counter like this (adapted from some Tex Exchange post)  \newcounter{P} \newcommand{\plb}[1]{\refstepcounter{P}\label{#1} \textbf{P-\ref{#1}.}}  \begin{problem} \plb{p1} A parallel-wire transmission line is constructed of \#6 AWG copper wire (diameter $d=0.162$ in., conductivity $\sigma_c=5.8\times 10^7$ S/m) with a 12-inch separation in air. Assuming no leakage between the two wires, find $R'$, $L'$, $G'$, and $C'$. Assume a working frequency of 1 MHz. \end{problem}  \begin{solution} Using Table.~\ref{tab:tlparam} and Fig.~\ref{fig:tlempl},  \[ Rs = \sqrt{\pi f \mu_c/\sigma_c} = 2.61\times10^{-4}~ \Omega\] \[ R' = \frac{2R_s}{\pi d} = 4.04\times10^{-2} ~\Omega\text{/m} \] \[ L' = \frac{\mu_0}{\pi} \ln{\frac{2D}{d}}=2.0~\mu \text{H/m} \] \[ C' = \frac{\pi \epsilon}{\displaystyle\ln{\frac{2D}{d}}}=5.56 ~\text{pF/m} \] $G'=0$ because the problem states that there is no leakage between the two wires. \end{solution}  \begin{figure}[ht] \begin{centering}     \includegraphics{fig24}     \caption{(Fig.2-4 from FAE) A few examples of transmission lines.} \end{centering} \label{fig:tlempl} \end{figure}  \begin{table}[ht] \caption{(Table 2-1 from FAE)Transmission-line parameters $R'$, $L'$, $G'$, and $C'$ for three types of lines.}     \begin{tabular}{c}         \hspace{3cm}\includegraphics{tb21}     \end{tabular} \label{tab:tlparam} \end{table} ... and more   Now all the figure references stop working, I get "Fig. " instead of "Fig. 1". The table references are alright though. Any idea what the problem could be? MSTest Code Coverage I was asked to prove the following:  If $\{ E_1,E_2,\ldots,E_n\}$ is a fully independent set of events, prove that:  $$ P\left(\bigcup_{i = 1}^n E_i \right) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right]  $$  (Hint: use complements)  I got the following:  Let $\alpha = E_1\cup E_2 \cup \cdots \cup E_n = \{E^C_1 \cap E^C_2 \cap \cdots\cap E_n^C\}^C$. Then, $\alpha^C = \{ E^C_1 \cap E^C_2 \cap \cdots \cap E^C\} $.  Assuming that the set of complements of each ${E_i}$ is also fully independent, we have:  \begin{align} &amp; P(\alpha^C) = \prod_{i = 1}^n P(E_i^C)  = \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ &amp; P(\alpha) = 1 - P(\alpha^C) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ &amp; P(\bigcup_{i = 1}^n E_i) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \end{align}  However, even though I know (by Googling it) that my assumption is true, I'm unable to prove it so I could validly use it as a lemma. Is there another way to prove this equality? Linux: Case-INSENSITIVE Filesystem Why I can't scan the WhatsApp QR? need a regex to parse a csv file with double quotes in php ibus Avro in Ubuntu 14.04 LTS Showing that a finite abelian group has a subgroup of order m for each divisor m of n I have made an attempt to prove that a finite abelian group of order $n$ has a subgroup of order $m$ for every divisor $m$ of $n$.     Specifically, I am asked to use a quotient group-induction argument to show this.  I'd appreciate comments on the validity or lack thereof of my attempted proof below.  Let $G$ be a finite abelian group of order $n$ and let $m$ be a divisor of $n$.  The proposition is true for $n=1$, so we'll proceed by induction and assume $n \ge 2$.  Let $p$ be a prime dividing $m$ and let $x$ be an element of order $p$ in $G$ (which exists by Cauchy's Theorem for Abelian Groups).  By the induction hypothesis, $G/\langle x \rangle$ has a subgroup of order $\dfrac{m}{p}$.  This subgroup is of the form $H/\langle x \rangle$ for some $H \le G$.  Since $|H/\langle x \rangle| = \dfrac{m}{p}$, it follows that $H \le G$ has order $m$.  I chose to use a prime divisor of $m$, but I don't see why it wouldn't work to use any proper divisor of $m$.  Am I correct on this point?  Thanks, I appreciate the help.