if $a \equiv b \bmod m$ and $c \equiv d \bmod m$ and $gcd(c,m) = 1$ prove that $a c^{-1} \equiv b d^{-1} \bmod m$ .
Suppose I have this:   $\frac{6^{666}}{2^{6}}$ (mod $125$)  I saw it is possible to reduce only the numerator's power modulo Euler's phi function. Can someone explain why is that possible?  It is essentially this:   $\frac{6^{666 \space (mod \phi(125))}}{2^{6}}$ (mod $125$)
How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$ indicates that the following relation is &quot;easily shown&quot;: $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it.  I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$).  Can one of you walk me through it or link me to site that will?  Thanks.  Edit: Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet): $$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$ so $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$ where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$ and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$ Thus: $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$=  x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$ From here it doesn't look like those higher order terms are all going to cancel out. Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.  Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:  $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$  WLOG one can assume that $0\le a\le b\le c$.  It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$. Let $p &gt; 3$ be a prime such that $p \equiv 2 \pmod 3$. Define the elliptic curve $E$ over $\mathbb{F}_p$ by $y^2 = x^3 + 1$. Prove that $E(\mathbb{F}_p)$ consists of $p+1$ points.  Using Fermat's little theorem you can prove that $x^3 + 1$ is a bijection on $\mathbb{F}_p$. Hence, $$\#E(\mathbb{F}_p) = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x^3 + 1\} + \#\{\infty\} = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x\} + 1.$$ But then I am stuck trying to prove that $y^2 = x$ has $p$ solutions $(x,y) \in \mathbb{F}_p^2$. Mouse pointer and keyboard Freeze when pluggin in external hard drive so as I say in the title I have this very big, annoying problem. On my new PC (HP Omen 2020, i7-10750h, Nvidia RTX 2070, 16GB RAM) when I connect an external hard drive (precisely an WD My Passport Ultra, 3TB) I get occasionally mouse and keyboard freezes. Practically, the system becomes unusable. The mouse and keyboards works, they even show in device manager, but from time to time (too often) they &quot;freezes&quot;: mouse pointer freeze, and it won't respond to my input, and keyboard too, if I type something it will not respond. First I thought this was a behaviour due to bad drivers, or something related to my new PC, the fact that it has USB 3.1 ports. But then I've noticed that this fact happens only when I connect my WD Hard Disk (which is USB 3.0) to my pc. I've tried some stuff online, suggesting to update drivers, change some windows settings, but with no success. My guess is that the hard disk and the mouse-keyboard does interfere with each other in some ways. I guess that it's a software problem, at driver levels maybe, not a physical one. When the mouse doesn't respond, I've tried to use the touchpad on the notebook, and it does work. So it's not a windows freeze, it's just the device (mouse-keyboard-receiver) that doesn't respond. The mouse and keyboard are from Logitech, bot connected with the Unifiying receiver (a wireless receiver they provide for the mouse and the keyboard). Model: MX Anywhere 2S for the mouse, K360 for the keyboard. Please, help me solve this problem, monday I start taking my engineering class lessons from MS Teams, and I need my pc up and running and well responsive (I need to access data from the hard disk while I take classes). PS: I've seen that there are some other questions like this in the forum, but none of them seems to provide a solution to the problem. I would also change the external hard drive in the future, but I need to know that the problem with this one won't show with another hard drive, or other devices. I've already sent this very same post to Microsoft community, I really need help with this thing otherwise it will be a problem following my online classes. Thank you if you will try to help me EDIT: On the notebook in question I have 4 USB 3.1 Ports, with two ports on each side, and a single Type-C Thunderbolt port (I'm not counting that one cause both the hard disk and the logitech receiver are USB 3.0 type A connectors). I've already tried switching USB port, connecting the WD Hard Disk to another port, doing the same on the logitech receiver, and I've even tried putting them one on the left side and the other to the right side, to minimize &quot;interference&quot; as someone suggested. But nothing, I still got this problem, which seems driver-related. Hope this will give more info Proof: in $\mathbb{R}$, $((0,1),|\cdot|)$ is not compact. Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.  What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$? Why does my iPhone power off when the battery is down to 30%? Today I was walking using my iPhone 6S running iOS 9.3 as a pedometer and it shutdown.   When I got home and plugged it in, it said 30% capacity.   That seems strange. I would expect it for 2% or 5% to preserve RAM, but 30%? Maybe there is a setting I don't know about.   My question is: Why does my iPhone power off when the battery is down to 30%?   Edit: this is not a duplicate of the low battery crash as I witnessed the low power shutdown and it was not a crash. Permanently removing apache2 Word for the satisfaction of curiosity There is a good feeling you get when your curiosity is satisfied, especially if is about a question you found interesting or important, or have spent a lot of time and energy trying to answer. Depending on how much I wanted to know, this feeling can be quite pleasing. Is there a word specifically for either the satisfaction of a curiosity, or the feeling that results? "No such file or directory" when trying to remove a file, but the file exists? Indentation in tcolorbox package I'd like to have the same indentation within a tcolorbox as I have in the rest of the document.  How can I get it?  Thanks.  MWE:  \documentclass[a4paper]{book}  \usepackage{lipsum} \usepackage[most]{tcolorbox}  \begin{document}  \chapter{Example with \texttt{tcolorbox}} \lipsum[1-8]  \begin{tcolorbox}[breakable,enhanced] \lipsum[9-14] \end{tcolorbox}  \end{document}   The first block of text has the typical indentation, but that in the box doesn't. Does non-mass-energy generate a gravitational field? At a very basic level I know that gravity isn't generated by mass but rather the  and when I wave my hands a lot it seems like that implies that energy in $E^2 = (pc)^2 + (mc^2)^2$ is the source of gravity.  If the total energy of a particle contributes to the generation of a gravitational field, that seems to imply some strange things.  For example, you could accelerate a particle like an electron to such a high velocity that it has enough momentum / kinetic energy to collapse into a black hole.  An observer traveling with the electron would see an electron and stationary observers would see an arbitrarily massive black hole flying by.  These observations seem to be contradictory.  The same would be true of an extremely high energy photon.  Another consequence is that a stationary particle could appear to have so much mass that it could attract / deflect the trajectory of a rapidly moving observer.  It seems odd that at low velocities you wouldn't even notice a stationary electron as you pass it by but at high velocities the gravitational field you observe could be significant enough to alter your path.  Am I correct that non-mass-energy generates a gravitational field?  If it does, how are these strange observations like a particle looking like a black hole in one reference frame but not another reconciled? How can I force SQL Server Management Studio to update the IntelliSense cache? IntelliSense in SSMS isn't updating unless I shut it down and restart. I've tried disconnecting from the server and reconnecting while the SSMS session is active, but that isn't working.  Steps to recreate:    Create a table  Select something from that table SSMS displays squiggly red lines under it with the tip "invalid object name" Execute the query   How can I force IntelliSense to update without restarting SSMS? Why don't you have to press CTRL+ALT+DEL to logon anymore in Windows 8? In previous versions of Windows, you had to press Ctrl + Alt + Delete to logon, or to unlock a locked workstation.  This was because this key sequence was recognized only by the OS and thus other software couldn't intercept it and display a spoofed logon screen to capture passwords.  Starting with Windows 8, you now just have to press Enter to get to the logon screen.  What is to stop someone from writing a fake logon screen?  Did Windows 8 add some sort of new security mechanism to mitigate this security issue? What's wrong with XKCD's Frequentists vs. Bayesians comic? makes fun of a frequentist statistician who derives an obviously wrong result.  However it seems to me that his reasoning is actually correct in the sense that it follows  the standard frequentist methodology.   So my question is "does he correctly apply the frequentist methodology?"    If no: what would be a correct frequentist inference in this scenario? How to integrate "prior knowledge" about the sun stability in the frequentist methodology? If yes: wtf? ;-) Determine group G as cyclic If $G$ has no proper subgroup, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.     I know that since $G $is a group with no proper subgroups, $g \in G$ is not just the identity. I don't know where to go from there. Creating a dictionary from a CSV file Change style properties of several layers simultaneously? Prove that $(a+1)(a+2)...(a+b)$ is divisible by $b!$ The problem is following, prove that:  $$(a+1)(a+2)...(a+b)\text{ is divisible by } b!\text{ for every positive integer a,b}$$    I've tried solving this problem using mathematical induction, but I don't think that i did it correctly. Here's what i've done  $1.\ b=1\ (Basis)$  $$b!| (a+1)(a+2)...(a+b)$$ $$1 | a+1 \text{, which is true}$$  $2.\ b=k\ (Induction\ Hypothesis)$  $$k! | (a+1)(a+2)...(a+k)\text{, we assume it's true}$$ $$k!*n = (a+1)(a+2)...(a+k)\text{, n is some positive integer}$$  $3.\ b=k+1\ (Inductive\ Step)$  In order to prove I should get:  $$(k+1)!*m = (a+1)(a+2)...(a+k)(a+k+1)\text{, where m is some positive integer}$$ $$(a+1)(a+2)...(a+k)(a+k+1) = k!*n*(a+k+1)$$ $$(a+1)(a+2)...(a+k)(a+k+1) = (k+1)!*n + k!*n*a$$ $$(a+1)(a+2)...(a+k)(a+k+1) - a(a+1)(a+2)...(a+k) = (k+1)!*n$$ $$(a+1)(a+2)...(a+k)(k+1) = (k+1)!*n$$  And as you can see I'm returning at the beginning.  Also I've tried to use the fact that in b consecutive numbers, there must be at least one that divides b, but since i eliminate that number (because I couldn't be sure that the quotient is divisible with (b-1) or (b-2) or ... or 2. And after this i can't continue with (b-1) using this method, because it's not necessary for the other (b-1) to be consecutive. How to compile self-referencing functions (to perform recursive a task) I know that I can use  to make cross-references and hyperlinks clickable.  That makes the clickable areas outlined in fluorescent green, however.  How can I make the green boxes go away?