What's the difference between "" and "!="?
Difference between "not equal" operators  and != in PHP
I am trying to prove the following statement.     If a field $K$ is not algebraically closed, then any $K$-variety $V\subset\mathbb{A}^n$ can be written as the zero set of a single polynomial in $K[X_1,X_2,...,X_n].$     Suppose $V=V(f_1,f_2,...,f_n)$, and I would want to find a polynomial $g\in K[X_1,X_2,...,X_n]$ such that $V=V(g)$.  I know that if I can find a polynomial $\phi\in K[X_1,X_2,...,X_n]$ whose only zero is $(0,0,...,0)$, then it is done, since we could take $g=\phi(f_1,f_2,...,f_n)$.  For $n=1$, then I would just choose $\phi=X$, but I do not even know how to proceed to the case $n=2$. Maybe we could choose an irreducible polynomial of degree $&gt;1$.  Any help will be appreciated. Why isn't the eigenclass equivalent to self.class, when it looks so similar? How do I choose between "while" and "whilst"? When should whilst be used instead of while?  For example, should I use the first or the second sentence?      They don't do this whilst they do that.   They don't do this while they do that.   Which would be correct? To make sure that we are talking about the same, I would like to post the relevant definitions I know first.  Definitions:  A pair $(G, +)$ where $G$ is a set and  $+: G \times G \rightarrow G$  is called a commutative group if it has all of the following three characteristics:   (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$ (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$ (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$ (commutativity): $\forall x,y \in G: x+y=y+x$.     A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and  $+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  $\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  is called a field if it has all of the following three characteristics:   $(\mathbb{K}, +)$ is a commutative group $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$     $\mathbb{N^+} := 1, 2, 3, 4, 5 ...$ (all positive integers)  $\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)  $\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.    My question: Does a field exist that has $\mathbb{N}$ as its set?  I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.   But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.    Related:     The smallest field containing the integers is the field of rational   numbers.   Source: en-Wikipedia:   I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess. Add a UIButton as a subview to a UITabBar ASP.NET MVC custom routing for search Here is my scenario.  For the example lets say that I need to return a list of cars based on a search criteria.  I would like to have a single View to display the results since the output will be the same, but I need several ways of getting there.  For instance, I may have a Form with a textbox to search by year.  I may have another separate page that contains a hyperlink for all red, Toyota cars.  How do I handle these multiple scenarios in the same View and Controller.  My dilemma is that the search could contain several options… year, make, model, etc but I don’t know where to put them.  What is the best approach for this?  Should I define the parameters in the routing or go with query strings, etc? How do you set, clear, and toggle a bit? How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$? I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$ Now, $n(n-1)(n+1)$ is divisible by $6$. Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$. My guess is using Fermat's little theorem but I don't know how. Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real. Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real.  We know $f(0)=0$, and if one assumes differentiability it is quite easy.      But how does monotone make up for differentiability? After removing any part the rest can be split evenly. Consequences? Where do I use delegates? What are some real world places that call for delegates?  I'm curious what situations or patterns are present where this method is the best solution.  No code required. How does Cooper get out of the black hole? In the movie Interstellar,  Cooper (played by ) enters a black hole and sends a message to Murph (his daughter). Later, he is found by rangers, but I don't understand how he gets out of the black hole?  As far as I know, nothing can get out of a black hole, as it has extremely high gravity. Classify all groups with order $pq$, $p$ is not equal to $q$ I am reading the following material:  . At page 5, there is a proof classifying all groups with order $pq$(they are all primes and $p$ is not equal to $q$). I am lost in the last two sentence, which claims the semidirect product are all isomorphic. Could someone help me to understand it? Thanks! This issue came up when I got different records counts for what I thought were identical queries one using a not in where constraint and the other a left join. The table in the not in constraint had one null value (bad data) which caused that query to return a count of 0 records. I sort of understand why but I could use some help fully grasping the concept.  To state it simply, why does query A return a result but B doesn't?  A: select 'true' where 3 in (1, 2, 3, null) B: select 'true' where 3 not in (1, 2, null)   This was on SQL Server 2005. I also found that calling set ansi_nulls off causes B to return a result. I would like to capture the contents of AJAX requests using Greasemonkey.  Does anybody know how to do this? I'm looking to generate a set of 5 random variables and enforce a dependence structure between them and onto a dependent variable $Y$.  I understand how to generate correlated random variables for multivariate normal, but not when mixing different types.  Below is a little more than I need, but I'm hoping someone can give me a general way of solving this problem...   $X_1$ and $X_2$ need to be highly correlated Bernoulli variables.  $X_3$ needs to take one of 5 categorical values, call them "A"..."E".  $X_4$ needs to be normal, and negatively correlated with $X_1$, $X_2$.  $X_5$ needs to approximate test scores from $0$ to $100$ with a high skew, so gamma probably.  $X_5$ needs to be positively correlated with $X_1$, $X_2$, $X_4$.   Each of these variables must impact a "success/occurrence" Bernoulli distributed variable $Y$.  How would I begin?  I would like to enforce correlation both between the values of $X$, and also between each $X$ and $Y$.  (The categorical correlations seem particularly confusing to me.) Ubuntu 18.04 Dock mouse hover tooltips - how to disable? I need to know if it is possible and how to disable the Ubuntu 18.04 Dock tool-tips for mouse hover the pinned application. I was considering playing a Summoner in my next Pathfinder campaign, and I decided that I wanted my Eidolon to be a a four-armed giant (the character being a Gnome with a height complex).  Well, I thought to myself: If it has four arms, then shouldn't it be able to wield two two-handed weapons?  As far as I can tell, it should work. That said, how would that work, exactly? I would assume that it would work like usual two-weapon wielding as far as feats and the like, and it would gain all the bonuses from two-weapon wielding as well. However, as this isn't exactly something that happens often, and thus there are no rules for it (that I can find) I can't be sure of any of this. Does anyone know of anything that could help me with this? What is the maximum convergent $x$ in the power tower $x^{x^{x^{x\cdots}}}$? In the power tower $x^{x^{x^{x\cdots}}}$ where there is an infinite stack of $x$'s, what is the maximum convergent number? I know the answer by playing with the form $x^y=y$ and using Mathematica, but I don't know how to solve this by hand. Can a concentration spell be cast without actually concentrating on it for an "instant" effect?