Macos mojave locked up, forcing me to reboot. I was then presented with the prohibitory symbol indicating that the OS can no longer load. It was caused by APFS being corrupted, confirmed by disk utility in recovery mode. Disk utility did nothing to repair the problem.  I reinstalled the OS and restored everything from a backup. One day later macos locked up again, and I once again had to reboot. It now stops booting after filling the progress bar while loading the OS. APFS has once again managed to corrupt itself, evident both when staring in verbose mode and when running disk utility in recovery mode. A checksum for some object is invalid. Disk utility once again makes no attempt to actually fix anything and I have found no evidence of any other APFS tools from apple for this purpose.  I'm not resizing partitions or anything advanced when it happens, just using Firefox and getting a beach ball.  I can't think of any other solution to the situation other than to get rid of APFS:   Reinstalling the OS onto an HFS+ partition. I tried this -- the installer ignores the existing file system and created APFS anyway. Install an older version of the OS on HFS+ Use some hack to force mojave onto HFS+ Use a different OS   The SSD is a 500 GB, less than a year old Samsung 860 EVO that boots Windows 10 with seemingly no problems so the hardware does not seem to be the problem.  Are there any other obvious options that I have missed in my list? (buying new hardware because the file system is useless doesn't qualify)
At WWDC 2018 macOS 10.14 or macOS Mojave was disclosed.  I am currently on macOS Sierra (10.12) and am looking forward to upgrading to macOS Mojave once its released.  I was wondering if it will be possible to update to macOS Mojave without converting to APFS, similar to  trick from OS X Daily on how to upgrade to macOS High Sierra (10.13) without converting to APFS with the command /Applications/Install\ macOS\ High\ Sierra.app/Contents/Resources/startosinstall --converttoapfs NO.
If $ar + bs =1$, then $\gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1$ If you make a search for something that returns more than 10,000 items in result. For example, search for  return 161,338 results on MSE. But if you select a page where 10,000+ result should be shown:        In the edge case  shows zero results:        But this is actual for any page where per_page * page_number is more than 10,000.  Though this does not happen for Stack Overflow in English. It looks like all other sites are affected.  Related bug report on ruSO.meta: A Polygon is inscribed in a circle $\Gamma$ A regular polygon P is inscribed in a circle $\Gamma$. Let A, B, and C, be three consecutive vertices on the polygon P, and let M be a point on the arc AC of $\Gamma$ that does not contain B. Prove that  $MA\cdot MC=MB^2-AB^2$  I inscribed the polygon P in the unit circle and let B=1. Past this, I'm not really sure how to proceed. It might be helpful to have M at 1 as well but given that the polygon would have the n-th roots of unity in it B seemed like a good choice. What would be a good way to proceed from here? Can anyone help me with this limit problem without L'Hopital rule and Taylor series? $$\lim_{x\rightarrow\ -\infty}\left(-2\frac{\arctan(x)}{\pi}\right)^x$$ I keep getting this warning whenever I try to run sudo apt-get update.  W: Duplicate sources.list entry http://archive.ubuntu.com/ubuntu/ precise-updates/main i386 Packages (/var/lib/apt/lists/archive.ubuntu.com_ubuntu_dists_precise-updates_main_binary-i386_Packages) W: You may want to run apt-get update to correct these problems   Below is the output from /etc/apt/sources.list file:  deb http://archive.ubuntu.com/ubuntu precise main restricted deb-src http://archive.ubuntu.com/ubuntu precise main restricted  deb http://archive.ubuntu.com/ubuntu precise-updates main restricted deb-src http://archive.ubuntu.com/ubuntu precise-updates main restricted  deb http://archive.ubuntu.com/ubuntu precise universe deb-src http://archive.ubuntu.com/ubuntu precise universe deb http://archive.ubuntu.com/ubuntu precise-updates universe deb-src http://archive.ubuntu.com/ubuntu precise-updates universe  deb http://archive.ubuntu.com/ubuntu precise multiverse deb-src http://archive.ubuntu.com/ubuntu precise multiverse deb http://archive.ubuntu.com/ubuntu precise-updates multiverse deb-src http://archive.ubuntu.com/ubuntu precise-updates multiverse  deb http://archive.ubuntu.com/ubuntu precise-security main restricted deb-src http://archive.ubuntu.com/ubuntu precise-security main restricted deb http://archive.ubuntu.com/ubuntu precise-security universe deb-src http://archive.ubuntu.com/ubuntu precise-security universe deb http://archive.ubuntu.com/ubuntu precise-security multiverse deb-src http://archive.ubuntu.com/ubuntu precise-security multiverse   How do I fix it? How do I prove that in every commuting family there is a common eigenvector? The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made my own effort into it and tried a different approach. Yet I can't finish it.  Let $\mathscr{F}$ be a commuting family in $M_n(\mathbb{C}^n)$, and $A\in\mathscr{F}$, then $A$ has $n$ eigenvalues. We pick one, say $\lambda$. Let $x$ be one of its eigenvector.  We can easily prove that, if $A$ has no other eigenvector with eigenvalue $\lambda$ that linearly independent with $x$, which means that $\{cx|c\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$, then $x$ is a common eigenvector. Because $\forall B\in\mathscr{F}$ and $\forall y\in \{cx|c\in\mathbb{C}\}$, $$A(Bx)=ABx=BAx=B(Ax)=B(\lambda x)=\lambda (Bx)$$, so that $Bx$ has to be in $\{cx|c\in\mathbb{C}\}$, that is, $Bx=c_0x$ for some $c_0\in\mathbb{C}$, which means $x$ is a eigenvector of $B$ too.  But what if there are vectors satisfying $Ax=\lambda x$ that's not in  $\{cx|c\in\mathbb{C}\}$? Well, then we should have a set of linearly independent eigenvectors $\{x_1,x_2,...,x_k\}$, that $\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$.  Now, I have a reasonable hypothesis that there exists some $x=c_1x_1+c_2x_2+...+c_kx_k$, that can be proven to be a common eigenvector of $\mathscr{F}$. I've tried some approaches to prove it but all failed.  Do you guys believe it's true? And if it is true then how do I prove it? Could you list all possible directories where Android apps may store data, providing description what kind of data are stored in each directory? Permission Subscribe to Dashboards: Add Recipients depends on permission(s): Subscribe to Dashboards Gay rights – does Star Trek ever address this issue? I am aware of the picture of the man wearing the 'dress' from TNG, which promotes gender equality, but are gay rights ever addressed in Star Trek at all? (All seasons accepted.) Why Listen to Music, why not Listening Music Does this exist: software for noise cancellation In the Ubuntu dock on Ubuntu 18.04, multiple windows of the same application are grouped and orange dots are shown to tell you how many windows of the application are opened. When you click on the icon, a small screen capture of each of the windows is shown.  However, these screen captures are shown in an order that's based on which window was last active on the desktop. This is something that clashes with the way I use the dock. I'm used to mentally mapping the windows to their dock positions. As such, I will often be trying to get to "the top one" and then I end up needing several clicks before I actually get to the terminal (or some other application) that I was looking for.  Is there some way to make sure that the order of these screen captures does not change whenever a window becomes active? Show that $\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$ In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$  Where $T$ is the Energy-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.  I know the following relationships:   Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$ Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$ Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$ d'Alembert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$ Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$   So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in , but I still cannot see how to tackle this problem.  Could somebody please give me a hint into the right direction? Is the simulation hypothesis outside of science? On the question of the  (i.e. that reality is a simulation), a friend of mine once remarked he didn't accept it on the grounds of . To me (with my admittedly instrumentalist view of science) this seems a misuse of Ockham's razor in that all my friend is really claiming is that it isn't empirically necessary (and hence outside of science).  Is the simulation hypothesis outside of science? If so, then does it pose a problem for scientific realism? If not, is it falsifiable? Is it an empirical question? If I insert a flash drive and try to drag any files onto desktop - nothing happens.  Or from any other folder.  Drag 'n copy only works if you open desktop in nautilus ("show desktop in files").  I already ran the command:  gsettings set org.gnome.nautilus.preferences use-experimental-views false   It didn't help. How to fix this? Extracting Z data from point layer using ArcGIS Desktop? I have created a point layer with ~150 points. I have a DEM beneath it, which is where I have taken the X and Y values from.   Does anybody know how I can retrieve the Z values, without using the info button?  I am using ArcGIS Desktop 10. I have licences for the spatial analyst etc tools. The Pearson correlation coefficient of x and y is the same, whether you compute pearson(x, y) or pearson(y, x). This suggests that doing a linear regression of y given x or x given y should be the same, but I don't think that's the case.   Can someone shed light on when the relationship is not symmetric, and how that relates to the Pearson correlation coefficient (which I always think of as summarizing the best fit line)? Suppose we have functions $f:A→B$ and $g:B→C$. Prove that if $f$ and $g$ are invertible, then so is $g \circ f$. Suppose we have functions $f:A→B$ and $g:B→C$. Prove that if $f$ and $g$ are invertible, then so is $g \circ f$. Is the converse true? I.e., if $g \circ f$ is invertible, does it follow that $f$ and $g$ are both invertible?  I know that if $f$ and $g$ are invertible, then they are both injective and surjective. Therefore, I assume from this that I have to show $g \circ f$ is injective and surjective as well, thus proving it is invertible. However, I don't know how to show $g \circ f$ is bijective based only on the information I have. There are $2n+1$ people. For each $n$ people there is somebody who is friend with each of them. Prove there is a "know-them-all" person. I have a following problem.     There are $2n+1$ people in the room. For any group of $n$ people there is a person (different from them) who is friends with each person in this group. Prove that there is a person, who knows all $2n$ other people.   One can easily see, that $\min_v \deg v = n$. From this we can get a lower bound for number of edges $N_e$: $$ N_e = \frac{1}{2}\sum_v \deg v \geq \frac{(2n+1) \cdot n}{2} $$ I do not know where to move from this point (and do not think, that this is the right direction) and pretty stuck right now. Can you give a hint? Proving the convergence of $\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$ for $|z| > 1$ $\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?