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Is it possible for information (like 1 and 0s) to be transmitted faster than light?   For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end.   The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey.  Would this actually work? How can I change what keys on my keyboard do? (How can I create custom keyboard commands/shortcuts?) Is it possible to assign the same value to multiple keys in a dict object at once? Pass Method as Parameter using C# Don't revert reputation from suggested edits when reviewer's account is deleted I noticed something strange in the reputation of one of my accounts today:        Never saw "reverted" event before, so went to  and realized what happened: the user who approved my suggested edit has been removed/deleted their account.  The edit is still there. The post is still there. The effort spent in editing is still there, untouched. I really can't see any reason to take the reputation away in such a case.  Can this please be changed so that the reputation won't be lost? Personally I don't care that much, and got lots of reputation from other sources, but many people have low reputation and each point matters to them.  P.S.  This might be a bug introduced when implementing this recent change about "", so tagging as a bug too. Proof Verification: Converse of Intermediate Value Theorem A function $f$ is increasing on $A$ if $f(x)\leq f(y)$ for all $x&lt;y$ in $A$. Show that the Intermediate Value Theorem does have a converse if we assume $f$ is increasing on $[a,b]$.  The converse of IVT along with the additional hypothesis would look like this: Lef $f:[a,b] \rightarrow \mathbb{R}$ be an increasing function on $[a,b]$ which satisfies: if $L$ is a real number such that $f(a)&lt;L&lt;f(b)$ (or $f(a)&gt;L&gt;f(b)$), then there exists a point $c\in (a,b)$ where $f(c)=L.$ If the preceding conditions are met, then $f$ is continuous on $[a,b]$.  The proof (if it's correct) I'm going to show is only for the $c \in (a,b)$, but I think for end points ($a$ or $b$), the method should be similar.  We want to show that given $c \in (a,b)$, for all $\epsilon&gt;0$, there exists a $\delta&gt;0$ such that $|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon$.  Since $f$ is increasing, we know that $f(a)\leq f(c)$. If $f(c)-\epsilon&lt;f(a)$, then set $x_1=a$, if $f(c)-\epsilon\geq f(a)$, then we know by hypothesis that a $x_1&lt;c$ exists such that $f(x_1)=f(c)-\epsilon.$ In either case, we have for $x \in (x_1,c]$ $$ f(c)-\epsilon \leq f(x_1) \leq f(x) \leq f(c) .$$  In a similar way, since we know that $f(c)\leq f(b)$, we can deduce that a $x_2&gt;c$ exists such that, for $x \in [c,x_2)$ $$f(c)\leq f(x)\leq f(x_2)\leq f(c)+\epsilon .$$  Pick $\delta = \min \{c-x_1,x_2-c \}$, we then have $$|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon .$$  Is this correct? In the proof that I saw, they originally used $\frac{\epsilon}{2}$, instead of just $\epsilon$, but I think this one works just as well, what do you guys think? Are continuous self-bijections of connected spaces homeomorphisms? I hope this doesn't turn out to be a silly question.  There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.    My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?    For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism. I have a 17 inch LCD, with a maximum resolution of 1280x720. It works fine, but there are some apps that require 1024x768. I know that the maximum resolution is called like that for a reason, but is there no way of increasing the resolution, other than getting a bigger monitor? Alternate property name while deserializing Changing deleted partition back to NTFS without having to recover to another drive Recently, I accidentally deleted a partition off of my external hard disk. I had a copy of Minitool Power Data Recovery left over when the same drive had become RAW about a year ago. When I selected lost partition recovery, it immediately identified my deleted partition and allowed me to access and recover the files that were there.  My question is this. Is there any way to change the deleted partition back to NTFS without having to have a drive that you need to recover the files to and then move them back after formatting, as I don’t have any space left on either the computer or on my other external hard drive, so that I don’t need to buy a new one. The files on the disk were really important.  I am using Windows 7 Home Premium.  I appreciate any help in solving this crisis. Hexadecapole potential using point particles? We can get monopole $1/r$, dipole $1/r^2$, quadrupole $1/r^3$ and octupole $1/r^4$ potential falloff by placing opposite point charges at the corners of a point, line, square and cube, respectively. My book cryptically says "and so on", how does one get a $1/r^5$ and higher potential falloff with a finite number of point-charges in 3 dimensions? The exact phrase from the book of Swan (the context is choosing the words permit or allow) -      The passive structure with it is only possible with permit   Having said that...     It is not allowed to smoke in the kitchen - incorrect     It is not permitted to smoke in the kitchen - correct   I feel absolutely fine for the incorrect sentence.   Your inputs please. Make me understand this. Matrix Ring of a Semisimple Ring I recently read the concept of semi-simplicity of a (not necessarily commutative) ring.  A ring $R$ is said to be semi-simple if $R$ as a left module over itself is a semi-simple module (This in turn means that each submodule (i.e, left ideal) of $R$ is a direct summand for $R$.)  I am grappling with the following question:     If $R$ is semi-simple, then so is the matrix ring $M_n(R)$.   It is a theorem that each left module of a semi-simple ring is a semi-simple module. Thus $M_n(R)$ is a semi-simple module of $R$. But this doesn't help me.  I need to show that each left ideal of $M_n(R)$ is a direct summand for $M_n(R)$. Is there a characterization  of the left ideals of $M_n(R)$ in terms of the left ideals of $R$?  Lastly, does the converse also hold?     If $M_n(R)$ is a semi-simple ring, then is $R$ necessarily a semi-simple ring?   Thanks. R : Format the text file output If $p$ is a prime and $p \mid ab$, then $p \mid a$ or $p \mid b$. Best Practice for Forcing Garbage Collection in C# Bending of light - photon's inertia instead of mass Using classical mechanics, the formula for gravitational attraction is   $$F = G\frac{m_1m_2}{r^2}.$$   This formula does not work for photons, and we need to use Einstein's theory of gravity to account for that, as the photon is massless. What if, instead of using the mass of the photon, we rather used the inertia of it?  The momentum of a photon is     $$p=\frac{h}{\lambda}.$$  Thus, using that, we can calculate its supposed 'mass' (inertia) as momentum is also equal to mass times velocity. So 'mass' of a photon is     $$m=\frac{p}{v}=\frac{h}{\lambda  c}.$$   If we use this mass as the mass of a photon, can we use Newton's equation for gravitational attraction? 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. How to create my own math operator with limits? How can i write my own math operator with limits? I want it to look like:  \sum\limits_{e=1}^{m} but with a capital A (if possible bigger than the normal text) instead of the sum.  Thanks for the help! Maximum power transfer and Thevenin I am trying to find the thevenin resistance value and the Vth for maximum power transfer. I have tried to find Rth by calculating R3 and R2 as parallel resistance and same for R4 and R5 and then adding them with R1 but I am not sure if I have done it the right way. For Vth I have tried using KVL but I think I have calculated wrong current values using current divider rule. I calculated the total current by diving V1 and R1. I hope someone can help. I have attached the picture of the circuit.