I've been up and down the indexes but I can't seem to find this. There are monsters that are vulnerable to adamantine, silver, and I think cold forged iron, but I can only find rules for the creation of silvered weapons (add 100 gp to the cost).  What is the cost of an adamantine weapon?  What is the cost of a cold forged iron weapon (or did I just make that up)?  I have the DMG, the PHP, the MM, and Princes of the Apocalypse - are the rules for this in any of those books and I have just missed it?
In the Monster Manual, gargoyles, golems, helmed horrors, and xorns all enjoy either immunity or resistance to:     bludgeoning, piercing, and slashing damage from nonmagical weapons that aren't adamantine   Is it actually possible to obtain a nonmagical adamantine weapon?   Although the Dungeon Master's Guide describes adamantine armor (p.150) and Daern's instant fortress (p.160) as being made of this metal, and provides an AC for destructible adamantine objects (p.246), there's no mention of adamantine weapons. The creation spell (PH p.229) can produce an adamantine weapon that lasts for a single minute. The true polymorph spell (PH p.283) can't turn a regular weapon into adamantine, but it might be able to  for an hour. The only way to obtain a (somewhat) permanent adamantine weapon in the game seems to be to loot one from drow elves (MM p.127). A PC (perhaps even a drow) with proficiency in smith's tools could presumably craft a nonmagical adamantine sword during downtime at 5gp per day (PH p.187), except that unlike silvered weapons, there's no pricing for adamantine.   It's seeming like mugging an NPC drow is the only by-the-book way to tool up against gargoyles.
In MK (Morse-Kelley) set theory life is easy: $\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)$. If $X=\left\{\right\}$ then $\bigcap X=U$, where $U$ is the universal class. So the (unary) intersection of the empty set is the class that contains all sets as elements. In ZF (Zermelo-Fraenkel) set theory, instead, proper classes are not allowed. So, how can I define $\bigcap X$ in ZF? I tried with the following definitions:   $\forall X\left(X\not=\left\{\right\}\rightarrow\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)\right)$. This means that $\bigcap\left\{\right\}$ is undefined, which is not that good. $\forall X\forall y\left(y\in\bigcap X\leftrightarrow X\not=\left\{\right\}\land\forall x\left(x\in X\rightarrow y\in x\right)\right)$. This means that $\bigcap\left\{\right\}=\left\{\right\}$, which is the opposite of MK.   I couldn't find any other valuable definition. Any ideas? Thank you. Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?  By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$. My book (Concepts of Physics by H.C. Verma) writes:  It has been reported () that the force between two masses may be better represented by $$F = \frac{G_{\infty} m_{1} m_{2}}{r^2} \left[1 + \left(1 + \frac{r}{\lambda} \right) \alpha e^{-\frac{r}{\lambda}}\right]$$ where $\alpha \approx - 0.007$ and $\lambda \approx 200~\mathrm{m}$.  What is this? Such a horrendous formula! So, what about Newton's? And what's the difference between $G$ &amp; $G_{\infty}$? How to block specific websites without using Parental Controls? Why is there something instead of nothing? A simple but fundamental question.    The "something" means the whole Universe (known and unknown), it could be represented as the reality version of the set of all sets, which is itself . It includes all the  and such.  A better version of the whole Universe could be described with a .  The "nothing" means an absence of the whole Universe, not a particular void or emptiness in the whole Universe but a nonexistent whole Universe. International TLD's vs. duplicate content I currently work at a pretty big website that has visitors from around the globe. My job is to help out on the SEO, and one thing we've been discussing lately is the use of international TLD's. The ones we use range between:   (partly) translated websites like .es and .de that serve most of the content in the country's language non-translated (english) websites for non-english languages (due to a lack of translations) like .ro and .cz english websites for english speaking countries with localized TLD's (.co.nz, .co.uk)   On one hand I really have the feeling this is causing a lot of duplicate content, especially for the last two categories of TLD's.   On the other hand though it seems a lot like country-specific TLD's tend to score a lot better in that country's Google. Would it be advisable to keep on using these domains, or should we canonicalize them all to the .com version? How does one get the MinionPro \qoppa? Since I have switched to the 2014 texlive, one issue is not working: the character \qoppa. The command is defined in polutonikogreek in babel. It is an archaic greek letter, but was used also in classic times for the number 90.  The glyph is available in the cmr font and in MinionPro and in the beautiful MinionMath font of Johannes Küster. I want the latter. In MinionPro/MinionMath, it is unicode character 03d9 (formerly u+03DE, it recently changed, as it turns out), see   How do I get the glyph into latex? I tried  \def\testqoppa{\begingroup\fontfamily{MinionPro}\fontencoding{LGR}\selectfont\char19\endgroup}   and   \def\testqoppab{\begingroup\fontfamily{minionmath}\fontencoding{LGR}\selectfont\char19\endgroup}   But all this gives the cmr glyph, because latex complains  LaTeX Font Warning: Font shape `LGR/MinionPro/m/n' undefined LaTeX Font Warning: Font shape `LGR/minionmath/m/n' undefined (Font)              using `LGR/cmr/m/n' instead on input line 571.   This seems like a beginner error from my side. Can somebody help me to get it right? I need \qoppa only in two places, so a nasty trick is also sufficient. The MinionMath font is installed and is working for all other purposes. Here is an extract of the .cls file:  \usepackage[fullfamily,textlf, opticals, swash, onlytext]{MinionPro} \usepackage[extraops,textcomp,withamsmath,amssymb]{minionmath} \usepackage{amsmath}             \usepackage{minionamsmath} \usepackage[polutonikogreek,greek,latin,french,italian,german,british]{babel} Just started blender, following the donut tutorial. Using Blender 2.8. made another donut to practice, when adding Lights I can't find the 'Use Nodes' in the Light settings. Some help would be appreciated &lt;3 Is it possible to run Snow Leopard on a LATE 2011 MacBook Pro? How do I change Wi-Fi Hotspot settings in Ubuntu with GNOME 3? How do I manage my hotspot connected devices or change the hotspot password? I can only see the Network Name, Security type and the Password but I can't seem to manage users or the network properties. Is it possible to use more than 20 layers? Is it possible to use more than 20 Scene layers or Render layers in Blender? How do I install a .tar.gz (or .tar.bz2) file? I have downloaded tar.gz files. But I don't know how to install it. How do I install this kind of file? Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition. Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition. The man page for GNU find states:    -exec command ;     [...] The  string `{}'  is  replaced  by the current      file name being processed everywhere it occurs in the      arguments to the command, not just in arguments where      it is alone, as in some  versions  of  find.     Both  of  these  constructions might need to be escaped      (with a `\') or quoted to protect them from expansion      by the shell.     That's from the man to find (GNU findutils) 4.4.2.   Now I tested this with bash and dash, and both don't need to have the {} being masked. Here is a simple test:  find /etc -name "hosts" -exec md5sum {} \;    Is there a shell, for which I really need to mask the braces? Note, that it doesn't depend upon whether the file found contains a blank (invoked from bash):   find ~ -maxdepth 1 -type d -name "U*" -exec ls -d {} \;  /home/stefan/Ubuntu One   This changes if the found file is passed to a subshell:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d {}' \;  ls: cannot access /home/stefan/Ubuntu: No such file or directory ls: cannot access One: No such file or directory   which can be solved by:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "$0"' {} \;   in contrast to:   find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "{}"' \;  /home/stefan/Ubuntu One   but that's not what the man page is talking about, is it? So which shell treats {} in a different way? It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.  The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.  This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m&gt;n$? This does not seem obvious.  Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get  $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.   The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$? In propositional logic, the statement "If pigs can fly, then elephants can lay eggs." validates to true because the antecedent validates to false.  In other words, given $a \rightarrow b$, if a is false, the entire statement is true. Why?  Just because the antecendent is false doesn't mean that another fact depends on it, right? Let $a_1,\dots ,a_n \in \mathbb{R}^n$ with $a_i = (a_{i1}, \dots , a_{in})$.  Show that when $\mid a_{jj}\mid \gt \sum_{i \neq j} \mid a_{ij} \mid \; \forall \; j\in \{1,\dots ,n\}$, then $a_1,\dots ,a_n$ are linearly independent.  I want to show that the matrix \begin{matrix} a_{11} &amp; a_{21} &amp; \cdots &amp; a_{n1} \\ a_{12} &amp; a_{22} &amp; \cdots &amp; a_{n2} \\ \vdots &amp; &amp; \ddots &amp; \\ a_{1n} &amp; a_{2n} &amp; \cdots &amp; a_{nn} \end{matrix} can be transformed into an upper triangular matrix with all entries in the main diagonal $\neq 0$.  For the cases $n=1$ and $n=2$ this is easy. However, I fail to show this for the general case; the problem being that the tranformations mess up the coefficients such that I can't use the assumption on the $a_{jj}$'s anymore.  I'm pretty sure that this has already been solved somewhere on here, but I don't know how to search for it. Where do I put my systemd unit file? How do I change my Stack Exchange password? How do I change my password?  I have a Stack Exchange account but can't seem to find the option to change my password.  Help please? Superluminal neutrinos I was quite surprised to read this all over the news today:     Elusive, nearly massive subatomic particles called neutrinos appear to travel just faster than light, a team of physicists in Europe reports. If so, the observation would wreck Einstein's theory of special relativity, which demands that nothing can travel faster than light.   &mdash;  Apparently a CERN/Gran Sasso team measured a faster-than-light speed for neutrinos.   Is this even remotely possible? If so, would it be a real violation of Lorentz invariance or an "" effect?   The ; a webcast is/was planned .  News conference video