Background We have a motivational poster in our office that says:  None of us is as smart as all of us.   I think that it's grammatically incorrect, and here is my reasoning:  All of the tigers have spots. All of us are here. None of us are dead yet.  The three examples all sound correct when using the plural &quot;are&quot;, rather than &quot;is&quot;. Question Unfortunately, some of my coworkers disagree with me. They believe that the quote is correct when it uses the word &quot;is&quot;. So my question is this: Is the motivational quote grammatically correct or should it say &quot;all&quot;?
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?
consider the following subsets of complex plane $$\Omega_1=\left\{c\in\Bbb C:\begin{bmatrix}1&amp;c\\\bar c&amp;1\\ \end{bmatrix}\text{ is non-negative definite } \right\} $$  $$\Omega_2= \left\{c\in\Bbb C:         \begin{bmatrix}         1 &amp; c &amp; c \\         \bar c &amp; 1 &amp; c \\         \bar c &amp; \bar c &amp; 1 \\         \end{bmatrix} \text{ is non-negative definite } \right\}$$  Let $$\bar D=\{z\in\Bbb C:|z|\le1\}$$ Then 1) $\Omega_1=\bar D,\Omega_2=\bar D$  2). $\Omega_1\neq\bar D, \Omega_2=\bar D$  3). $\Omega_1=\bar D, \Omega_2\neq\bar D$  4). $\Omega_1\neq\bar D, \Omega_2\neq\bar D$  How to proceed? Thank you. Does $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent? Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot Differentiability of $g:=f(\sqrt{x^2+y^2})$ for a $C^1$ function with $f'(0)=0$ NBHM $2008$ Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.  Define for all $x,y\in \mathbb{R}$,  $$g(x,y)=f(\sqrt{x^2+y^2})$$ Pick out the true statements :   $g$ is differentiable on $\mathbb{R}^2$ $g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$ $g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$   I do not have much idea of differentiation of multivariable calculus.  I just tried out some examples to get some idea..  I have to take some function whose derivative at $0$ has to be zero.  for simplicity I would take $f(x)=x^2$ Then I would get :     $g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$   This would suggest me that second option has more chances as $f(0)=0$ in this case.  I would now take $f(x)=x^2+1$ then :     $g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$   So, I would now eliminate second option and third option also...  So, It is now got confirmed that i have to work out to "Prove" that first option is correct.  Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.  I have to look for partial derivatives first :  $$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$  $$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$  I have used here that $f$ is differentiable on $\mathbb{R}$  As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :  $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.     As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$      As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$   So we have continuous partial derivatives for $g(x,y)$..  So, I am willing to conclude that:     $g(x,y)$ is  differentiable on $\mathbb{R}^2$   I would be thankful if some one can confirm if the answer correct.  I would be much more thankful if some one can let me know if there are any gaps in my argument.  This is just a proof verification Question..  Thank you. In $Sets$, the initial object is $0 = \emptyset$. We have $C \times \emptyset = \emptyset$ for any set $C$. As there are no maps to the empty set, we don't get much of the universal property of the product.  I am now wondering what is special about arbitrary categories $\mathscr{C}$ with $C \times 0 \cong 0$ for all objects $C$ of $\mathscr{C}$, where $0$ is the initial object of $\mathscr{C}$.   From Sheaves in Geometry and Logic by MacLane (p.194), in any topos $\mathscr{C}$, any arrow $k:C \to 0$ is an isomorphism. The universal property of the product would then also not yield very much.   Are there any categories of higher interest? SCJP6 regex issue Let $X=(0,1]$ and $d(x,y)=\left|\frac{1}{x}-\frac{1}{y}\right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that? Prove $\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$ for- What is $48\div2(9+3)$? Why can I not access a file named `-` when quoting? I have a file called -.  I want to display its contents.  One way is to do  cat ./-   since cat - reads from standard input. However, why are cat "-" and cat '-' also interpreted by the shell as cat -? Distance Between Two Points in Matlab How to determine if a type implements a specific generic interface type I encountered The Squirrel at "A tree."              |   |                                                                         |   |                     "   Hello, I'm The                        |             |                        Squirrel. I can                        |             |                      provide you candies,                     |             |                      and lots of things.                                    |                      I know how much you                                    |        _.-'''-,      love candies. But I                                    |      .'        `\     feel alone in this                                    |     /           /          forest.       "                         |        |    /      .--^_^                                                   |        |    |     /  C ,,\                                                           |    |    |   \  _.)                           ___                            |     \   |   /  \                   ____.----'.-'                            |      '-, \./ \)\)               .-' --  __.-'                   |            \        `-/   );/           ___/  - _.-'                        |             \_________''--'-'________.-' --    /                            |              -   --        -  -     -     __.-'                                              -    --    -     --   --  /                                                 _________________________.-'                                 He asks you some questions. What are the correct answers, and what is the reward for getting them correct? What word means what many people think 'ironic' means? Is there a command like head but for line truncation? Where did the convention of using single dashes for letters and doubles dashes for words come from and why is continued to be used?  For example if I type in ls --help, you see:    -a, --all                  do not ignore entries starting with .   -A, --almost-all           do not list implied . and ..       --author               with -l, print the author of each file   -b, --escape               print octal escapes for nongraphic characters       --block-size=SIZE      use SIZE-byte blocks   -B, --ignore-backups       do not list implied entries ending with ~ ...   I tried googling - and -- convention even with quotes with little success. How do I prevent directories mounted with 'bind' from appearing on 'Devices' on nautilus? I have these lines in the fstab   # binds /media/DataNtfs/Music           /home/can/Music                  none    rw,bind  /media/DataNtfs/Pictures        /home/can/Pictures               none    rw,bind  /media/DataNtfs/Downloads       /home/can/Downloads              none    rw,bind  /media/DataNtfs/Documents       /home/can/Documents              none    rw,bind  /media/DataNtfs/Backups         /home/can/Backups                none    rw,bind  /media/DataNtfs/Notes           /home/can/Notes                  none    rw,bind  /media/DataNtfs/Other           /home/can/Other                  none    rw,bind  /media/DataNtfs/Packages        /home/can/Packages               none    rw,bind  /media/DataNtfs/Photos          /home/can/Photos                 none    rw,bind  /media/DataNtfs/Videos          /home/can/Videos                 none    rw,bind  /media/DataNtfs/WorkSpace       /home/can/WorkSpace              none    rw,bind Let $X$ and $Y$ be finite sets. Then the set $Y^X$ is finite and $\#(Y^X) = (\#Y)^{\#X}$.  I can see how to do it combinatorially. Let $y \in Y$. There are $\#(X)$ choices for such y to come from x, therefore we have $\#Y^{\#X}$ choices. I am not sure how to do it bijectively. #include&lt;iostream&gt; using namespace std;  class A  {    int a;    int b;    public:    void eat()    {       cout&lt;&lt;"A::eat()"&lt;&lt;endl;    } };  class B: public A {    public:    void eat()    {        cout&lt;&lt;"B::eat()"&lt;&lt;endl;     }  };  class C: public A {     public:    void eat()     {        cout&lt;&lt;"C::eat()"&lt;&lt;endl;     }  };  class D: public B, C {  };  int foo(A *ptr) {  ptr-&gt;eat();  } main() {  D obj; foo(&amp;(obj.B)); //error. How do i call with D's B part.  }   The above foo call is a compile time error.  I want to call foo with obj's B part without using virtual inheritance. How do i do that.   Also, in case of virtual inheritance, why the offset information need to be stored in the vtable. This can be determined at the compile time itself. In the above case, if we pass foo with D's object, at compile time only we can calculate the offset of D's A part. Is it possible to recover an account deleted at the SE team level? How can I derive effect sizes in lme4 and describe the magnitude of fixed effects?