My flag as not an answer on the question  has been declined with this message, "declined - flags should not be used to indicate technical inaccuracies, or an altogether wrong answer".   In this situation what should we do? Ignore the answer knowingly, or is there something else  that I do not know?
I've recently flagged the following , which I got in the flag queue.   The answer does not seem to attempt to address the question at all. It looks like a comment on one of the other answers (probably Luca's).  It got rejected with the following reason:  declined - flags should not be used to indicate technical inaccuracies,            or an altogether wrong answer   I did not flag this because I thought it was wrong, I flagged it because I didn't think it was an attempt to answer the question. For what it's worth, I don't get what the guy who provided that answer even tried to say.  Was I wrong to flag it? If so why?
How do I get HTML5 h.264 Video working on Firefox? According the the , h.264 video should now be working in Linux.  It does also say the appropriate gstreamer plug-ins need to be installed.  Does anyone know what these are?    I used this  and the  to verify that h.264 support works in Google Chrome on my Ubuntu 12.04 64-bit, but does not work on Firefox 26. Proving $\mathbb{Z}[\sqrt {10}]$ is not a UFD To Find Large Powers in C++ Alternative of `route -n` of Linux in Mac How can I see the list of routes and its corresponding gateway in a Mac ? The command in linux is route -n .  For a particular IP address, I am able to get the routes for an particular IP address via route get 10.0.0.0. $G$ is the set of all subsets of a set $A$, under the operation of $\triangle\;$: Symmetric Difference of sets.  $A$ has at least two different elements.  I need to check if this is a group, and if it does to show if the group is abelian and/or finite.  Associativity - easy from Symmetric Difference. Identity element - empty group. Inverse element - each element is inverse to itself. Am I right?  abelian? finite? Adding Favorites to the Stack Exchange global inbox We're considering adding favorite notifications to the global inbox, and are looking for feedback on the idea (as we're not completely sold on it).  But first, some history.  Favorites are kind of like bookmarks, in that they're kept in a handy centralized place.  They also follow you across browsers.  Ok, ok... "bookmarks, on the internet" is not the most compelling feature, but Favorites also opted you into notices for those questions as if you were the owner (edits and new answers, at least).  This was probably the most confusing part (though a  feature) of the old envelope behavior judging by erroneous bug reports, so a way to opt out of this .  Now, favorite updates go nowhere.  But we could* put them into the inbox.  Here's how we envision this feature working:   Every current user has their "notify me on favorite changes" preference cleared (new users start with it cleared). The first time someone favorites a question, they're prompted to opt-in to favorite notices. Whenever a question is commented on or answered (or, possibly, edited) a new notice is pushed into the inbox of users who have favorited it, and have opted in.   Opt-in would be per-site, and it would remain possible to opt-out from your profile page.   This is not a final design, but the basic idea is correct  As I mentioned above, we're not 100% sold on this idea.  We've had a policy to date that nothing goes into the inbox that isn't specifically targeted at a user; this would be tweaking that policy.  *We could also put changes into their tab on the user page, but that reduces their visibility. Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$ Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.   I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me. Hello,    This question is about the Command Line Developer Tools that are usually installed with xcode-select --install and updated via a software update from the Mac App Store (at least until macOS 10.13). I use the developer toolchain on a daily basis and it has always worked and updated without issues.    Today I've updated my mac from High Sierra 10.13.6 to Mojave 10.14.1, and I've lost the ability to update the Developer Tools.    After the update I've executed a terminal command that relies on the developer tools being installed. It was a command to update , although I don't think that the specifics matter, as I believe that any task that tried to access the developer tools would have triggered the same error message.    The error was:   xcrun: error: invalid active developer path (/Library/Developer/CommandLineTools), missing xcrun at: /Library/Developer/CommandLineTools/usr/bin/xcrun       The phrasing was new (maybe), but I thought that it was caused by the usual need to re-install the Developer Tools after some macOS updates.    Later I tried to dig a bit deeper into the failure, and found out that:  $ ls -l /Library/Developer/CommandLineTools/usr/total 0drwxr-xr-x  3 root  admin  96  4 Nov 19:32 share$ which xcrun/usr/bin/xcrun$ xcode-select -p/Library/Developer/CommandLineTools    Still, as I normally do, I entered the command to install and update the Developer Tools in the terminal:  xcode-select --install    This started the usual procedure: modal window to confirm, then request to accept the license agreement, followed by a progress bar. Except that it failed very quickly with an unexpected error:     Can&rsquo;t install the software because it is not currently available from the Software Update server.     I've tried several times, to no avail. It gets always stuck.    Sometimes, however, the failure causes the System Preferences to report a pending update:      If I open that preference panel it starts searching for updates, and it always finds nothing except the first time that it happened. The first time, it found this:          It literally suggested me to install the updates for macOS 10.11 and 10.13. I have no idea why.  I closed the preference without installing, and as I said it hasn't shown them again. There is also no update available in new Mac App Store.   Is this a known issue? Is there any way to resolve the problem?     Update  Of course I can download the Dev Tools for macOS 10.14 installer from .  Before installing them manually, though, I'm wondering if there is anything that's broken with the system. A subgroup of a cyclic group is cyclic - Understanding Proof I'm having some trouble understanding the proof of the following theorem  A subgroup of a cyclic group is cyclic  I will list each step of the proof in my textbook and indicate the places that I'm confused and hopefully somewhere out there can clear some things up for me.  Proof Let  G be a cyclic group generated by &quot;$a$&quot; and let H be a subgroup of G. If $H = {\{e\}}$, then $H \text{ = &lt;e&gt;}$ is cyclic. If $H \neq \space {\{e\}}$, then $a^n \in H$ for some $n \in \mathbb{Z}^{+}$. Let $m$ be the smallest integer in $\mathbb{Z}^{+}$ such that $a^m \in H$. We claim that $c = a^m$ generates H; that is, $$H \space = \space &lt;a^m&gt; \space = \space &lt;c&gt;.$$ We must show that every $b \in H$ is a power of c. Since $b \in H$ and $H \leq G$, we have $b = a^n$ for some $n$. Find $q$ and $r$ such that $$ n = mq + r \space \space \space \space for \space \space \space 0 \leq r &lt; m,$$  Alright this is the first part in the proof where I start to get confused. Where does the division algorithm come from and why are we using it? The proof continues as follows:  $$a^n = a^{mq + r} = (a^m)^q \cdot a^r,$$ so $$ a^r = (a^m)^{-q} \cdot a^n.$$ Now since $a^n \in H$, $a^m \in H$, and $H$ is a group, both $(a^m)^{-q}$ and $a^n$ are in $H$. Thus $$(a^m)^{-q} \cdot a^n \in H; \space \space \space \text{that is,} \space \space a^r \in H.$$  This is another point at which I’m a little confused. What exactly about $a^n$ and $a^m$ being elements of $H$ allows us to accept that $(a^m)^{-q}$ and $a^{n}$ are in $H$? The proof continues:  Since $m$ was the smallest positive integer such that $a^m \in H$ and $0$ $\leq r$ $&lt; m$, we must have $r = 0$. Thus $n = qm$ and $$b \space = \space a^n \space = \space (a^m)^q \space = \space c^q,$$ so b is a power of c.  This final step is confusing as well, but I think its just because of the previous parts I was confused about. Any help in understanding this proof would be greatly appreciated Getting the location from an IP address I want to retrieve information like the city, state, and country of a visitor from their IP address, so that I can customize my web page according to their location. Is there a good and reliable way to do this in PHP? I am using JavaScript for client-side scripting, PHP for server-side scripting, and MySQL for the database. fake proof of $\forall a. \forall b. a = b \to 1 = 0$ I saw a less formal version of this fake proof that claimed to prove $2=1$ but because it assumed $a=b$ from the start I knew why it was wrong. It does seem however that the proof can be used to prove an certain implication I find very odd however. Clearly $\forall a. \forall b. a = b \to 0 = 1$ is false because I could pick $a$ and $b$ to be $1$ and then I would have derived 0 = 1 with no other conditions. So there must be something wrong with this proof that I am not seeing.  proof of $\forall a. \forall b. a = b \to 0 = 1$:  $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (assume)\\ a+a=a+b  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x)=a+x)\\ 2a=a+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  (a+a=2a)\\ 2a-2b=a+b-2b \,\,\, (x=y \to f(x) = f(y), f(x)=x-2b)\\ 2(a-b)=a+b-2b \,\, (distributive)\\ 2(a-b)=a-b \,\,\,\,\,\,\,\,\,\,\,\,\,\, (-b=b-2b)\\ 2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x/(a-b))\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x-1) $  now that I have "proven" $\forall a. \forall b. a = b \to 0 = 1$ I can now pick a and b to be 1.  $ 1 = 1 \to 1=0 \,\,\,\,\,\, (a=1, b=1, above)\\ 1 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{reflexivity})\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{modus ponens}) \\ $  So now it seems that I have managed to derive $1=0$. Clearly I messed up given that no one has been able to show $ZFC$ inconsistent and ostensibly this fits well within the most studied aspects of $ZFC$. Either that or a random guy that makes meme's online found the most profound result in mathematics ever.  So where have I missed the mistake? Unicode input retrieved via PrimeFaces input components become corrupted In Fallout 4 the Bobbleheads are back.  Once you pick them up they stay in your inventory with a weight of 0 and a value of 300. Will I need them later ? Do I lose the bonus if I sell them or drop them somewhere ? Same with the magazines.  Is there any disadvantage to selling them ? Why do some Kree have white skin? This seems bizarre to me.  All the Kree we've seen thus far (MCU based) were blue skinned... Ronan the Accuser being the most famous  What do Kree people look like?  Is there anything concrete with their appearance?  It seems odd that Danvers (Captain Marvel) and Jude Law's character (Yon-Rog?) and the "Supreme Intelligence" are all white skinned. Number of solutions of $P(x)=e^{ax}$ if $P$ is a polynomial Equation numbering problems in AMSMath environments with \pause Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:      If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ prove that $\left\{ s_n \right\}$ converges, and that $s_n &lt; 2$ for $n = 1, 2, 3, \ldots$.    My effort:   We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]  Then we can also show that $s_n &lt; s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]  But how to calculate the exact value of the limit? Where does this sequence occur in applications? For f continuous on $[0,1]$, show that there exist points $\alpha_k$ such that $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $ Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and  $f(1)=1$.For every integer $n$ show  that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that    $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $   Is it theorem or i can show it with basic differentiation knowledge .Please give me hint . Suppose $G$ has odd order, show the function $f:G\rightarrow G$ defined by $f(x)=x^2$ is injective. This proposition is easily provable if we assume $G$ is Abelian, but I don't know how to start this without the assumption of being Abelian. View history of commands run in terminal Is there a way to save all my typed terminal commands and view it like history in a log book?