Jacob Rees-Mogg MP, speaking by the invitation of the Speaker of the House of Commons in a , quoted a Financial Times article and, in passing, 'corrected' its grammar.  Quoting the article, Rees-Mogg said of entries on a Wikipedia list (of entries in a table) that 'none are' from Europe and then corrected the article saying that it should be 'none is'.  'None' may refer to 'not one' in which case I would say it is singular. But 'none' may also refer to 'not any' in which case I would suggest it is plural.  Thus I think that Mr Rees-Mogg is not correct in correcting the FT article.  'None of the entries' (that is, not any of the entries) are from Europe' seems quite correct to me.  'Not one of the entries is from Europe' is also correct, I would say.  Is Mr Rees-Mogg correct ?     entry for 'none' :     a. Not any (one) of a number of people or things. Also: neither of two persons or things (now regional).   ======================================  EDIT : I have read the suggested duplicates and they are inconclusive. I was hoping for a definitive (and thus an academic), decisive answer to the question.  =======================================
In my grammar book (English Grammar, HarperCollins Publishers), I read that none is occasionally treated as plural, but it is usually regarded as singular. Can you give me an example of sentence where none is used as plural pronoun?
Fermi in his lecture asserts:     The second law of thermodynamics rules out the possibility of constructing a perpetuum mobile of the second kind.   So, this means there can be no machine which just transforms all the heat energy gained by cooling surrounding bodies into mechanical work.  But how does actually the Second Law prohibit this?   The hot reservoir provides heat energy to the system. Does it cause a decrease in entropy of the universe(system + hot reservoir)? How? In order to receive heat wouldn't the system have to be cooler than the reservoir? If so, then entropy increases as the heat energy gets expelled from the reservoir at a higher temperature than the temperature the system receives the heat energy. Is it so?  I'm not getting how the Second Law nullifies the existence of this machine. Could anyone please explain how the Law prohibits the perpetual machine of the second kind? Is there anything special with complex fraction $\left|\frac{z-a}{1-\bar{a}{z}}\right|$? Is there anything special with the form:  $$\left|\frac{z-a}{1-\bar{a}{z}}\right|$$ ? With $a$ and $z$ are complex numbers.  In fact, I saw it in a problem:   If $|z| = 1$, prove that $|\frac{z-a}{1-\bar{a}{z}}| = 1$ If $|z| &lt; 1$ and $|a| &lt; 1$, prove that $|\frac{z-a}{1-\bar{a}{z}}| &lt; 1$   I can easily prove the first one with expansion:  $$z=\cos\theta + i\sin \theta \\ a = m +in$$  But it will be terrible to use in the second one. What's more, I found this form a little special so maybe there is some clever trick without using expansion?  Thank you a lot! Are PPAs safe to add to my system and what are some "red flags" to watch out for? Why is a T distribution used for hypothesis testing a linear regression coefficient? In practice, using a standard T-test to check the significance of a linear regression coefficient is common practice. The mechanics of the calculation make sense to me.  Why is it that the T-distribution can be used to model the standard test statistic used in linear regression hypothesis testing? Standard test statistic I am referring to here:   $$ T_{0} = \frac{\widehat{\beta} - \beta_{0}}{SE(\widehat{\beta})} $$ There are various articles on Hearthstone opening strategy which suggest it's a bad idea to use early-game spells and summons to damage your opponent directly, and that you should instead save them for minion removal.  An example: your starting hand contains an Elven Archer, a 1-cost 1/1 minion that does a point of damage when summoned. I've seen it advised that if going first, with that card in your hand, you should not play it, but save it for minion removal.  Another example:  Going second against a Rogue with a 2-cost 2/1 Loot Hoarder, Deathrattle: draw a card in your hand. The advice was not to play it and save the coin, because on her second turn, the Rogue can generate a dagger and kill the Hoarder.   I understand why, in most circumstances, it's better to remove a minion than damage the enemy. What's bugging me about these particular opening examples is why they're a better idea than doing nothing and ending the turn.  In both cases, you've wasted a point of mana, and the chance to take the initiatve. By having a minion on the board first, you can use the following turn to attack with that minion, possibly removing an enemy summon and leaving you clear to summon onto an empty board.   The Loot Hoarder one I find particularly puzzling. If the Rogue gets a dagger and kills it, you still have the initative, you've forced the Rogue to make what's possibly a suboptimal turn-2 play, you've done 2 damage to the Rogue and you got a card into the bargain. Sounds like a pretty good use of the coin to me.  What am I missing here? Present continuous and simple future In my preamble, you'll find:  \renewcommand{\baselinestretch}{1} \renewcommand{\familydefault}{\sfdefault} \usepackage[sfmath]{kpfonts} \usepackage[mathrmOrig]{sfmath} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{animate} \usepackage{anysize} \usepackage{arcs} \usepackage{array} \usepackage[scale=0]{background} \usepackage{blindtext} \usepackage{calc} \usepackage{cancel} \usepackage{chngcntr} \usepackage{CJK} \usepackage{CJKspace} \usepackage{CJKpunct} \usepackage{colortbl} \usepackage{enumitem} \usepackage{epstopdf} \usepackage{etoolbox} \usepackage{fancybox} \usepackage{fancyhdr} \usepackage{float} \usepackage{framed} \usepackage{graphicx} \usepackage{hyphenat} \usepackage{ifthen} \usepackage{latexsym} \usepackage{lmodern} \usepackage{multicol} \usepackage{multirow} \usepackage{mwe} \usepackage{pgffor} \usepackage{rotating} \usepackage{tabularx} \usepackage{textcomp} \usepackage{tgheros,tgtermes,tgcursor} \usepackage{tikz} \usepackage{times} \usepackage{tipa} \usepackage{varwidth} \usepackage{wasysym} \usepackage{wboiboites} \usepackage[abs]{overpic} \usepackage[breaklinks,colorlinks=true,linkcolor=red,citecolor=red, urlcolor=blue]{hyperref} \usepackage[framemethod=TikZ]{mdframed} \usepackage[breakable, theorems, skins]{tcolorbox,empheq} \usepackage[default]{gillius} \usepackage[document]{ragged2e} \usepackage[latin1]{inputenc} \usepackage[spanish]{babel} \usepackage[T1]{fontenc} \usepackage[total={18.59cm,26.5cm},left=1.5cm,top=2.5cm]{geometry} \usetikzlibrary{matrix,arrows, positioning,shadows,shadings,backgrounds,calc, shapes, tikzmark} \usetikzlibrary{shadows.blur} \usetikzlibrary{shapes.symbols} \usetikzlibrary{patterns,fadings} \tcbuselibrary{skins,breakable,listings,theorems}    I'm trying to draw an arrow for X axis and Y axis.   \begin{tikzpicture} \draw[yellow!10, fill=blue!5](-1, -1) rectangle (7, 5); \draw[dotted] (-0.75,-0.75) grid (6.75,4.75); \node[below] at (-0.25,0){$O$}; \draw[-&gt;,black, thick] (-0.75,0)--(6.75,0); \end{tikzpicture}   But I get this error:    Additionally, if I remove -> from the options of \draw the tikz picture compiles normally. Language specific settings for field formatters I've created a custom field formatter widget for numeric fields. The formatter has A LOT of settings to customize the visual output of that number, to whatever designers/marketing wants (adding a lot of HTML markup, with[out] leading/trailing zeros, zeros as dash, one or two fractions, rounding, prefixes/infixes/suffixes,... ).  Some of those settings are language-dependant though, so I'd like to enter different formatter settings in  /en/admin/structure/types/manage/MY_CONTENT_TYPE/display /de/admin/structure/types/manage/MY_CONTENT_TYPE/display /fr/admin/structure/types/manage/MY_CONTENT_TYPE/display  I've noticed that there is a $langcode parameter in   but I don't know how to make  language-aware.  Is it possible at all to have language-specific settings in a field formatter? If yes, how to I store and retrieve those settings in the correct language? Update iOS 5.1.1 on iPad 1 to at least iOS 6 Similar to this question posted in August 2014:       Has the fact that  (as  in The Fires of Pompeii, and  in Torchwood’s Children of Earth season) yet been addressed/explained either in, or out of universe? I own a Nikon D40X and just the kit lens and I'm considering buying either a macro lens or some extension tubes.  I don't necessarily want to do true 1:1 macro, just some close ups of bugs, fungi and flowers. I thought that the quotation mark (") was simply a type of grouping marker but I'm debugging some NHibernate code and notice that while  SELECT * FROM site WHERE site_id = 3;   Works fine  SELECT * FROM "site" WHERE site_id = 3;   fails with a table or view does not exist error.  What gives? Reducing size of KML file (after converting from Shapefile) Can not switch Node.js to a default version of my choice I've heard Google's algorithm is based on PageRank, and that it has something to do with the links to your site. Can someone explain it, and more importantly, how I can increase my site's PageRank? I have a project where I have a few layers (State of Texas, cities, roads, etc.). They are all resaved and projected to NAD83 / UTM zone 14N EPSG: 26914. The project CRS is also NAD83 / UTM zone 14N EPSG: 26914 and on the fly projection is enabled. I added a text delimited file and cannot get them to line up. I checked the attribute table of the delimited file and the coordinates are correct. I cross-checked this with a point close to where a few of the points from the delimited file are and they are very close so in theory they should be lining up. I tried to save the delimited file and save it with the same projection and add the shp file back in and that did not work. Anyone have any ideas? I can't think of anything else to try. Automorphism group of the quaternion group Let $Q_8$ be the quaternion group. How do we determine the automorphism group $Aut(Q_8)$ of $Q_8$ algebraically? I searched for this problem on internet. I found some geometric proofs that $Aut(Q_8)$ is isomorphic to the rotation group of a cube, hence it is isomorphic to the symmetric group $S_4$. I would like to know an algebraic proof that $Aut(Q_8)$ is isomorphic to $S_4$. Intermediate ring between a field and an algebraic extension. Is it possible to install MySQL 5.5 or 5.6 on Ubuntu 16.04? I recently upgraded to Ubuntu 16.04. Unfortunately I need to work MySQL 5.5 or 5.6. I tried many solutions to have MySQL 5.5 or 5.6 run on Ubuntu 16.04 like this  . But none of them is working. Ubuntu detect the unmet dependencies but says "but it is not going to be installed" for 5.5 or says Package 'mysql-server-5.6' has no installation candidate...  Is it not possible to force an earlier version of MySQL on Ubuntu? Why is that? I think I have heard that the following is true before, but I don't know how to prove it:     Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.   Is this true? Would anyone be willing to provide a proof?    Attempt at a proof:  Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.   Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.  Does this look to be correct?    More generally, one might prove the following     Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$.