Android, add a condition to my button (SMS app)
What is a NullPointerException, and how do I fix it?
Showing $F[x,y]/(ax^2+by^2-1)$ is a Dedekind domain How to fix broken taskbar icons in Windows 7 Why did Dumbledore tell Harry there were 6 Horcruxes when he already knew Harry was a horcrux? Harry: "He made seven Horcruxes?"   Dumbledore: "...Not seven Horcruxes: six. The seventh part of his soul, however maimed, resides inside his regenerated body..." (HP6, US Hardback, page 503).    I would think that Dumbledore had already known that there were 6 horcruxes:    the diary  Gaunt's ring  Salazar's locket Helga Hufflepuff's cup Nagini (the whole essence divided thing)   Now those are the five that he was sure of. He knew that when Riddle came to Hogwarts to ask for the DADA position, he was really there so he could hide a horcrux. (But he never got a chance to tell Harry that, right?) Did Dumbledore know of the diadem? Or did he just think Riddle wanted to hide the cup?   Did Dumbledore say there were 6 and not 7 because he didn't know about the diadem but did know about Harry being a horcrux? Or did he say there were 6 to ensure Harry didn't guess he was a horcrux? Sorry if this makes no sense.  I know that in the end there were 7 horcruxes and an 8th part of Voldemort's soul right? How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.  This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated. When I visit a website it sometimes puts cookies on my browser. Usually cookies keep track of browsing info from that particular site. If I don't delete those cookies and browse other websites, will the cookies be able to know which site I am browsing? Also, can any website read browsing history which is saved by browsers? Can any website know about my browsing history by help of cookies, browsing histories saved by browsers, or by any scripting language e.g. javascript? Can ubuntu be installed in intel kabylake 7th gen processor? I am using a 7th generation i5 processor(7200,2.50 ghz).The processor also have intel hd graphics 620 and nvidia geforce 940mx 2gb graphics card. I read previously that only windows 10 can be installed to this specific processor.  Is it true no version of ubuntu can be used with this processor?  If it's possible to install ubuntu using this processor, what additional steps other than the normal installation process should I undertake to get Ubuntu working? How do you address unknown peers in email? This is one bit of academic etiquette I'm not sure about. When I write to another researcher, a peer (we're both professors). I've never exchanged with before and never met him but I know of him (and he probably knows of me).  Erring on the side of formality, I write to him “Dear professor Mortimer”. I sign with my first name, because that's how I've always signed my emails (there's a formal signature block below, with full name and contact details) and I think it indicates a willingness to be called by my first name (which I prefer, but I don't want to force people one way or the other).  Is that going to be well received? Will people take the hint? Or should I just start using first names from the start? (it does sound weird to write “dear Philipp” to someone you don't know) Is the "Roman alphabet" what we use for English? I understand that the alphabet for the English language is not strictly English as languages such as French, Dutch and many more use the same alphabet, with few additions in other languages. Is Roman alphabet the common term to refer to this set of symbols for writing texts? Why is $c+q$ still a primitive root modulo $q$? Question: Let $p$ and $q$ be distinct odd prime numbers.  By considering primitive roots, we need to show $\exists c\in\mathbb{Z}$ with the property that:  $\bullet$ $c^n\equiv 1\pmod{p}$ whenever $n$ is a multiple of $p-1$, and   $\bullet$ whenever $n$ is not divisible by $q-1$, then $c^n\not\equiv 1\pmod{q}$  I am only interested with the beginning of this answer.  Answer: We let $c$ be a primitive root modulo $q$.  We are going to use Fermat's little theorem so we need $(c,p)=1$  If $(c,p)\neq 1$, then we have $p\mid c$, but then we have $(c+q, p)=1$  and since $c+q$ is a primitive root modulo $q$, we can replace $c$ with $c+q$ for $(c,p)=1$ for using Fermat's little theorem.  I understand the general method and argument, but...  Contention: Why is $c+q$ still a primitive root modulo $q$?  When is this generally true? Is there an identity for when things remain a primitive root?  Why is $(c+q)^{\phi(q)}\equiv 1\pmod{q}$? I want to install Windows XP on my desktop computer (only XP), but my CD drive is not working.  I've only found installation tutorials on how to install it on the pen drive rather than from the pen drive.  How can I install Windows XP from a USB flash drive? Multiplicative nature of the separability degree In what follows, let $E / F$ be an algebraic extension, $h(x),f(x)\in F[x]$ polynomials, $h(x)$ irreducible.   Definitions.   We say $h(x)$ is separable if it has not repeated factors. We say $f(x)$ is separable if all its irreducible factors are separable. We say $\alpha\in E$ is separable over $F$, if its minimal polynomial over $F$ is. The extension $E/F$ is separable if all the elements of $E$ are separable over $F$. We define the field $$\newcommand\sclo{\operatorname{Scl}}\sclo_F E= \{\alpha\in E : \text{ $\alpha$ is separable over $F$ }\}$$ as the separable closure of $F$ in $E$. The separability degree of the extension $E/F$ is defined as $$[E:F]_s = [\sclo_F E:F].$$   Now, suppose the extension $E/F$ is finite.  My question is: how can one proof that with these definitions, if $B$ is an intermediate field of the extension $E / F$, then $$[E:F]_s = [E:B]_s[B:F]_s?$$  I know that the purpose of the separability degree is to count in how many ways we can extend certain injective homomorphism to an automorphism. If one starts defining separability degree this way, as done in the book by Lang, the result follows without much difficulty.  I wonder how to achieve it in this way. "He walks as if he is drunk." Grammatically correct ? Any difference in meaning from "...as if he were drunk."? Is it grammatically correct to use an indicative verb after "as if", or "as though" for that matter?        I've heard someone say: "He walks as if he is drunk."  Would there have been any difference if he had said "He walks as if he were drunk."?   A second example might be:        It looks as if it is going to rain.   It looks as if it were going to rain. When casting a spell as a bonus action, you can't cast another spell on the same turn except a 1-action cantrip - can the order be "reversed"? Can you cast a cantrip first as a bonus action, then any other spell? If not, would it be broken to allow it?  I'm thinking a Sorcerer casting a Quickened true strike, then casting a leveled spell requiring a touch attack. Seems like a great use of the cantrip + Metamagic feature, but it seems like it is not allowed. Prove that the power set of an $n$-element set contains $2^n$ elements Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.      Proof. The proof is by induction on the numbers of elements of $X$.      For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.      Now the induction step. Suppose $|X|=n$; by the induction hypotesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$      From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.  The construction of a new set $S$ is the union of the old set with cardinality $n$ and an new element $\{a\}$, therefore the set that do not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element...  Can someone please elucidate? Proving an inclusion related to algebraic sets and interpreting it Proof that all abelian simple groups are cyclic groups of prime order Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order. Simple extension fields If I am correct simple extension fields are extensions generated by one element. I have learned that this means that elements of a simple extension can be written as powers of that element as long as the power is greater than the degree of that element. Also this means that all elements of the simple ext. Can be written in terms of decending powers of the generating element where the highest power is the degree of the generator minus 1. This concept seems to work for generating elements whose irreductable in a certain field is a descending degree polynomial. Looking at the construction of the complex numbers by starting with $\Bbb R$ (and doing $\Bbb R [x]/\langle x^2+1\rangle$), it is said that all elements are of the form $a+b\cdot \text{generator}$. But when the generator is evaluated for the irreductable, $x^2+1$, we see... $(\text{generator})^2=-1$. Thus there is no first power of gen. Can someone explain where all elements of the extension just explained are of the form $a+b\cdot \text{generator}$ comes from or explain this situation for anytime where the irreduc ( gen, F)  is not a descending degree polynomial. I hope my question is clear.  Thank you I know how to detect when resistors are arranged in parallel or series arrangement and I can also find their equivalent resistance in simple circuits or when resistances are connected in form of triangle but what happens when the arrangement is complex like this :   Which resistors are parallel and which are in series ? How can I find the equivalent resistance in such cases ? Is there rule or method for figuring this out ? It would be nice to see what users have recently been active. This is somewhat standard with many forum-style websites, showing how active the site has been in the last n-minutes or so. As a functional advantage, this also serves to let you know when is a good time to get a question in. Using Gimp to batch convert images to another format in Windows As implied by the title, does anybody know how to use the gimp-console-[version].exe program to batch convert images between formats (with default settings) in Windows ?