I must prove that the ring of parts of set $X$, with the binary operations symmetric difference (as the sum) and the Intersection (as the product) is isomorphic to the ring formed by the functions of the form $f: X \rightarrow \mathbb{Z}_{2}$ under the sum and the usual product in $\mathbb{Z}_{2}$. I have been looking for functions that constitute a homomorphism and have not been able to find it, I appreciate your help.
I'm studying for an abstract algebra exam and one of the review questions was this:     Let $X$ be a set, and $\mathcal P(X)$ be the power set of $X$. Consider the operations $\Delta$ = symmetric difference (a.k.a. "XOR"), and $\bigcap$ = intersection.   a) Does $\Delta$ and $\bigcap$ make $P(X)$ into a ring?   b) If so, is it a ring with unity?   c) Is the ring commutative?   d) Is it a field?   For parts a) and b), I think it does form a ring with unity, but I'm not quite sure how to get started on proving it.   For part c), it is a commutative ring since $\mathcal P(X)$ is closed under symmetric difference and intersection, right?  Not even sure how to get started on d).  I'd really like to understand this question fully, so any kind of input would be tremendously helpful. Thank you!
Use Results of Find Command to Call vi Is there a way that I can pipe the output of this command (assuming there is only one file found) to vi.  I want to find this command, and then call it up in vi.  Thanks.  find . -name 'id.properties' What counts as "observation" in Schrödinger's Cat, and why are superpositions possible? So if I understood correctly, Schrödinger's Cat is a thought experiment that puts a cat inside a box, and there's a mechanism that kills the cat with 50% probability based on a quantum process. The argument is that the cat now must be in a superposition of dead and alive.  This makes sense at first, but the state of the cat inside the box will affect the outside world in an observable way, right? For example if the cat dies, it might meow loudly which would be audible. If it didn't meow, it would produce a thud on the ground when it dies. And even if the ground was very solid, the redistribution of mass inside the box will affects its gravity field which means the whole universe theoretically immediately observe's the cat's death.  So extending this argument to all superpositions, the different states would cause different effects on the rest of the universe, usually a slight change in the gravity field is the minimum. This gravity perturbation would propagate throughout the universe, and even all the experimenters go to sleep with thick, thick earplugs, somebody or something in the universe is going to inadvertently observe the event and the superposition immediately collapses. Thus superpositions cannot exist beyond an extremely short amount of time.  What's wrong with my reasoning? I recently bought one of these ESR meters, a 'MESR-100':       It claims to be able to test capacitor ESR with the caps left in-circuit.  It appears to work well from my tests so far.  However, how much can I trust the in-circuit readings?  Are they going to be less accurate?  Are there certain circuit configurations where it wouldn't be possible to measure ESR in-circuit accurately?  [EDIT: Added the model as requested] : Eeeek! Code blocks are misbehaving! So code blocks in questions and answers on the android app (inline included) seem to be scared of me putting my finger in them. When I poke them, they only move over two columns!  (For clarity, these are the code blocks that overlap the TextArea view because they've been eating too much)  The intended behaviour here is that when I press my finger into a code block, the touch event should register a drag; at current, I try to scroll inside a code block and I end up switching to question or answer view instead, over and over.  EDIT:  Something tells me this is status-bydesign because the text is deliberately cut off (probably inviting you to click on the code block and expand it; UX Decision?): Understanding a probability paradox Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 1 3 to 1 2 because he would then be one of two prisoners. What do you think of the jailer’s reasoning?  If the jailer refuses to say anything, then the probability that prisoner $A$ is excecuted is $\frac{1}{3}$. If the jailer says to prisoner $A$ that prisoner $B$ will walk free, then $2$ prisoners remain to be considered, $A$ and $C$. One dies, one does not. Heads or tails essentially. $\frac{1}{2}$ ought to be the conditional probability that $A$ dies given that $B$ walks free no? Apparently not though, allegedly the correct answer is still $\frac{1}{3}$. Even my attempt to calculate the correct answer yielded the result $\frac{1}{2}$.   Let $A_D$ and $C_D$ respectively denote the event of $A$ and $C$ dying. Let $B_F$ denote the event that $B$ walks free. Assume that the jailer tells prisoner $A$ that prisoner $B$ will walk free.   Here's my attempt.  $$P(A_D\mid B_F)=\frac{P(A_D\cap B_F)}{P(B_F)}=\frac{P(A_D\cap B_F)}{P((B_F\cap A_D)\cup (B_F\cap C_D))}=\frac{P(A_D)P(B_F\mid A_D)}{P(A_D)P(B_F\mid A_D)+P(C_D)P(B_F\mid C_D)}=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times 1}=\frac{1}{2}$$  What am I doing wrong?  Edit 1: Intuitively I am still troubled but I understand now that $B_F$ may occur even though the jailer does not necessarally say $B$.   Edit 2: I suppose that it makes some sense if the faiths of the prisoners had already been decided before prisoner $A$ asked the jailer the question. If the jailer decides to reveal one of the others who will walk free then that must've been the case. I would like to make cross references to own defined counter on environments. So far I have tried the following,   \documentclass{book} \newcounter{entryc}[subsection] \newcommand{\thetheorem}{\thesubsection.\arabic{entryc}} \newenvironment{qd}[1]{\refstepcounter{entryc}\noindent\textbf{Quasi Definition \thetheorem.}\begin{itshape}} {\end{itshape}} % \begin{document}  \chapter{First chapter} This is the first chapter.  \section{First section} This is the first section.  \subsection{First subsection} This is the first subsection.  \begin{qd}{}\label{qd:emp} This is my environment. \end{qd}  \noindent In quasi definition \ref{qd:emp} it \dots  \end{document}   What happens is that     But I would prefer "...In quasi definition 1.1.1.1". Do anyone have some suggestions of how solve this? and How to avoid the          \noindent  thing before the continuation of the text?   Additional information: if possible I would try to not use ntheorem. Find count of all combination of numbers whose sum is x What's the difference between using > and >> in shell? Notepad doesn't open files created by Gedit? Does Ubuntu follow the Linux Standard Base (LSB)? This is a continued discussion from () by return count only.     Problem      Given four lists A, B, C, D of integer values, compute how many tuples   (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.      To make problem a bit easier, all A, B, C, D have same length of N   where \$0 \le N \le 500\$. All integers are in the range of \$-2^{28}\$ to \$2^{28} - 1\$   and the result is guaranteed to be at most \$2^{31} - 1\$.      Example:      Input:  A = [ 1, 2] B = [-2,-1] C = [-1, 2] D = [ 0, 2]       Output:  2       Explanation:      The two tuples are:         (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0   (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0      I'm wondering if there are any ideas to have a solution less than \$O(n^2)\$ time complexity.  Source code in Python 2.7,  from collections import defaultdict def four_sum(A, B, C, D):     sum_map = defaultdict(int)     result = 0     for i in A:         for j in B:             sum_map[i+j] += 1     for i in C:         for j in D:             if -(i+j) in sum_map:                 result += sum_map[-(i+j)]      return result  if __name__ == "__main__":     A = [1, 2]     B = [-2, -1]     C = [-1, 2]     D = [0, 2]     print four_sum(A,B,C,D) Is the 9-speed Sora crankset compatible with 105 11-speed chain? Google Drive search for part of a file name Is there any way I can search based on the last three digits of a file name?  I have a file with many invoices that are easy to reference with the last three numbers. Is there any way I can search something like *765 to search for ends with 765  Another example.. sometimes files are prefixed with a 0, sometimes not.. 093241251 vs 93241251 and if I just search 93241251 it will not find the 0 prefix version Measuring accuracy of latitude and longitude? Notation for infinite product in reverse order Remotely uninstall apps? I have tried through the Google Play store and it does not seem possible from there, but does anyone know if it is possible to remove an application remotely?   In other words, if I do not have direct access to my phone, can I make use of some website, app or service to uninstall an app without being able to interact with the phone directly. Splitting words into specific fragments Summary table with a field that shows components of each summarized record I have a table with 2 fields the first is for community and the second for the main community that it belongs to, as in the image  I want to summarize the second field (Belongs_to) but with an extra field (multiple value field) that shows components of each main community, as in the image  Could anyone help me please? line 56 syntax error: unexpected end of file I have typed out the code for this mychoice.sh bash example and I keep getting the syntax error: unexpected end of file.  Am I missing something?  mychoice.sh  #!/bin/bash  E='echo -e';e='echo -en';trap ";exit" 2 ESC=$( $e "\e") TPUT(){ $e "\e[${1};${2}H";} CLEAR(){ $e "\ec";} CIVIS(){ $e "\e[25l"} DRAW(){ $e "\e%\e(0";} WRITE(){ $e "\e(B";} MARK(){ $e "\e[7m";} UNMARK(){ $e "\e[27m";} R(){ CLEAR ;stty sane;$e "\ec\e[37;44m[J";}; HEAD(){ DRAW     for each in $(seq 1 13);do     $E "   x                    x"     done     WRITE;MARK;TPUT; 1 5     $E "BASH SELECTION MENU         ";UNMARK;}     i=0; CLEAR; CIVIS;NULL=/dev/null FOOT(){ MARK;TPUT 13 5     printf "ENTER - SELECT,NEXT         ";UNMARK;} ARROW(){ read -s -n3 key 2&gt;/dev/null &gt;&amp;2      if [[$key = $ESC[A ]];then echo up;fi      if [[$key = $ESC[B ]];then echo dn;fi;} M0(){ TPUT  4 20; $e "Login info";} M1(){ TPUT  5 20; $e "Network";} M2(){ TPUT  6 20; $e "Disk";} M3(){ TPUT  7 20; $e "Routing";} M4(){ TPUT  8 20; $e "Time";} M5(){ TPUT  9 20; $e "About  ";} M6(){ TPUT 10 20; $e "Exit   ";} LM=6 MENU(){ for each in $(seq 0 $LM);do M${each};done;} POS(){ if [[ $cur == up ]];then ((i--));fi     if [[ $cur == dn ]];then ((i++));fi     if [[ $i -lt 0 ]];then i=$LM;fi     if [[ $i -gt $LM ]];then i=0;fi;} REFRESH(){ after=$((i+1)); before=$((i-!))     if [[ $before -lt 0 ]];then before=$LM;fi     if [[ $after -gt 0 ]];then after=0;fi     if [[ $j -lt $i ]];then UNMARK;M$before;else UNMARK;M$after;fi     if [[ $after -eq 0 ]] || [$before -eq $LM ];then          UNMARK; M$before; M$after;fi;j=$i;UNMARK;M$before;M$after;} INIT(){ R;HEAD;FOOT;MENU;} SC(){ REFRESH;MARK;$S;$b;cur='ARROW';} ES(){ MARK;$e "ENTER = main menu ";$b;read;INIT;};INIT while [[ "$0" != " " ]]; do case $i in     0) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(w     )\n";ES;fi;;     1) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(ifconfig  )\n";ES;fi;;     2) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(df -h )\n";ES;fi;;     3) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(route -n  )\n";ES;fi;;     4) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(date  )\n";ES;fi;;     5) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS($e by oTo )\n";ES;fi;;     6) S=M2;SC;if [[ $cur == "" ]];then R;exit 0;fi;;  esac;POS;done; JavaScript + Unicode regexes How can I use Unicode-aware regular expressions in JavaScript?  For example, there should be something akin to \w that can match any code-point in Letters or Marks category (not just the ASCII ones), and hopefully have filters like [[P*]] for punctuation, etc.