I'm Unable to Boot into Mac OS Partition, partition type FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
OS volume shows as type 'FFFFFFFF-FFFF-FFFF-FFFF-FFFFFFFFFFFF'
Say $R$ is a commutative ring and $I\in R$ is an ideal. Let us consider the quotient $R/I$. It is created by taking every element $a\in R$, and adding all the elements of $I$ to it. The elements of $R/I$ are of the form $a+I$ and $b+I$, $\forall a,b\in R$.   Now $$(a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I$$  $$(a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I$$  We know that $a.i_{2}+b.i_{1}+i_{1}i_{2}\in I$.   If $(a+I).(b+I)=ab+I$, then taking suitable $(i_{1},i_{2})\in I\times I$, we should be able to prove $(a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I$. However, can every element in $I$ be generated by taking suitable $i_{1},i_{2}\in I$? And if not, is that the reason why $(a+I).(b+I)$ has to be defined as equal to $ab+I$ in violation of the distributive property of ring elements? This has confused me for a long time.   EDIT: I figured addition does not have to be defined as such because $(a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}$, where $i_{1}+i_{2}\in I$. In fact, every element $i\in I$ can be constructed by taking $i_{1}=0$ and $i_{2}=i$. Hence, $(a+I) + (b+I)=a+b+I$ naturally. This is the logic I followed to determine that $(a+I).(b+I)$ doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody.   Thanks for your help in advance! In Python, is there a concise way of comparing whether the contents of two text files are the same? I don't care what the differences are.  I just want to know whether the contents are different. "Speak to" vs. "Speak with" What are the differences between these two phrasal verbs and what are the best situations to use each? Do I have to leave the Schengen area if I have a short term (C) visa that will expire right before my National Visa (D) officially starts? dpkg: error: dpkg status database is locked by another process I just wondered if somebody could tell me what this means:  E: dpkg was interrupted, you must manually run 'sudo dpkg --configure -a' to correct the problem. reidsr@ubuntu:~$ sudo dpkg --configure -a dpkg: error: dpkg status database is locked by another process reidsr@ubuntu:~$    How do I run sudo dpkg --configure -a  manually? MVC DateTime binding with incorrect date format Why are my PHP sessions dying? And why can't I restore them? I have an app using PHP and the PayPal API. The basic way it works to get a payment is that you do a web service call to PayPal to get a token and then do a browser redirect to PayPal with that token for the user to pay. After the payment details have been confirmed, PayPal redirects back to the URL you originally set in the service call.  This all works, millions of people use it every day, et cetera.  Strange thing is, when PayPal redirects back, the PHP session is gone. It's a .  First question: why is this happening? Both pages are on the same domain, both use HTTPS. The session works for all requests up until the PayPal redirect back.  The linked forum thread suggests a workaround, to persist the session ID in the PayPal request and then to retrieve it later and restore the session. Great, except it doesn't seem to work.  I can add some log statements:  log(session_id());   before and after the various redirects. 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I am trying to create a MATLAB class with a member variable that's being updated as a result of a method invocation, but when I try to change the property within the class it (apperently, from what I understood from MATLAB's memory management) creates a copy of the object and then modifies it, leaving the original object's property untouched.  classdef testprop       properties           numRequests=0;       end       methods           function Request(this, val)               disp(val);               this.numRequests=this.numRequests+1;           end       end   end     .  &gt;&gt; a=testprop; &gt;&gt; a.Request(9); &gt;&gt; a.Request(5);   &gt;&gt; a.numRequests    ans = 0 False proof that $ρe^{iθ} = ρ$ and so complex numbers do not exist? My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form $\rho e^{i\theta}$, so the proof begins there) $$\large\rho e^{i\theta} = \rho e^{\frac{2 \pi i \theta}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho$$ $$Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1$$ Since we started with the general form of a complex number and simplified it to a real number (namely, $\rho$), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step $4$ to $5$ , but I am not sure if that really is the case. QM and Renormalization (layman) I was reading Michio Kaku's Beyond Einstein. In it, I think, he explains that when physicsts treat a particle as a geometric point they end up with infinity when calculating the strength of the particle's field as you approach the particle.  First, did I get that part right?  Second, how does (in very simple terms) renormalization try to fix this in Quantum Mechanics? If you can include some type of word picture that would be great. Accepted answer will go to the clearest explanation.  Update  I got a little further in the book (lol) and Kaku talks about using symmetry to remove the divergences in the math. I'd appreciate an answer that incorporates this also. Thanks guys! If $\mathop{\mathrm{Spec}}A$ is not connected then there is a nontrivial idempotent How to resume baking fluid simulation? It happens sometimes that after finishing bake (e.g., for fluid simulation) we wish we could continue the simulation for two more seconds :(    How to use the cached bake to continue the simulation?  We played with the settings, but with no success. Getting image dimensions without reading the entire file Prove that $\sum_{n=2}^\infty (\ln n)^{- \ln n}$ converges As the title suggests, I'd like to prove that the sum $$ \sum_{n=2}^\infty (\ln n)^{- \ln n} $$ is finite.  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