For the field $F=\{a+b\sqrt{3}\}$, find $d$ so that $b\cdot d=1$, where $b,d\in F$  Let $b=a_1+b_1\sqrt{3}$  Let $d=a_2+b_2\sqrt{3}$  We need to find $a_2,b_2$ such that $b\cdot d=1$  Doing the math, we get that $b\cdot d=(a_1a_2+3b_1b_2)+(a_1b_2+a_2b_1)\sqrt{3}$  But what from here? How do I decide that I want $(a_1a_2+3b_1b_2)$ to equal $0$, and so I want $(a_1b_2+a_2b_1)$ to be $1$. Or I want the first part of the expression to be $1$, and the next part to be $0$. How do I decide?
Show that the set $\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}$ is a field  I am having a little trouble proving that for each non-zero $b\in\mathbb{Q}(\sqrt{p})$, there is $d\in\mathbb{Q}$ such that $b\cdot d=1$.  My attempt:  Let $b=a_1+b_1\sqrt{p}$  Let $d=a_2+b_2\sqrt{p}$  We want $$b\cdot d=(a_1+b_1\sqrt{p})(a_2+b_2\sqrt{p})$$ $$=a_1a_2+a_1b_2\sqrt{p}+a_2b_1\sqrt{p}+b_1b_2p$$  $$=(a_1a_2+b_1b_2p)+(a_1b_2+a_2b_1)\sqrt{p}$$ $$=1$$  Am I following the right approach? What should be my next step
Expectation values of the position operator is equal to zero in case of even potentials? What are the benefits/detriments to graduate students joining unions? How can you encode a string to Base64 in JavaScript? I have a PHP script that can encode a PNG image to a Base64 string.  I'd like to do the same thing using JavaScript. I know how to open files, but I'm not sure how to do the encoding. I'm not used to working with binary data. How can I make extra money with photography? I have photos in Getty Images, in Shutterstock, Imagekind and I recently added some to RedBubble, but I'm not having the expected results.  Does anyone have some history of success selling photographs or working as a photographer?  Examples of my work are available in .    Also asked by   I am an amateur who sometimes takes pictures of nature or random things. I've been told by several people that my pictures are worth money.  How can I monetize my best shots?  I don't really want to do a lot of work, editing, SEO or whatever. I'm looking for an easy way where I would just upload my pictures somewhere and let it go.  Maybe stock photos? People buying prints? I have no idea how that works...  EDIT: maybe a link to see which type of photography I do can help I can't distinguish the difference in meaning between these  two sentences.   It really doesn't matter.  It doesn't  really matter.    It seems that there is a nuanced difference, but I cannot see what this is. Trim string in JavaScript? How do I trim a string in JavaScript? That is, how do I remove all whitespace from the beginning and the end of the string in JavaScript? I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$ The choice between the gerund and the infinitive in a certain construction I am pretty much sure that for native speakers the issue I am going to bring up might look as an uncalled question as they can easily figure out which form of a verbal part of speech should be used, be it a gerund, an infinitive, or a bare infinitive. However, it can be pretty much misleading for a foreigner because the same construction works differently with different words and I don’t see a logical explanation. For example:  The purpose of this article is to analyze the issue…  My hobby is reading. One may assume, and many foreigners really come to this conclusion, that it’s OK to switch the “gerund” and “infinitive” as the meaning will stay the same, that is:  The purpose of this article is analyzing the issue… (same as 1 above)  My hobby is to read. (same as 2 above)   However, as I was told, this assumption is erroneous. Another example: His task was watching after them. (not good) His task was to watch after them. Is there any principle which governs the choice of the right form in such cases? Additive Roman Numeral 4 I am working on a diplomatic edition of a 15th century book and it numbers its folios using the standard roman numerals of the time.  I've been using the following to get the page/folio in my header:  \newcommand{\rectoverso}[1]{\ifthenelse{\isodd{#1}}{r}{v}} \newcommand{\folio}[1]{\the\numexpr (#1+1)/2\relax} \uppercase\expandafter{\romannumeral \value{\folio{\thepage}}}\rectoverso{\thepage}   This works great, but the problem is that the roman numeral sequence is off of what I need.  Instead of I, II, III, IIII, V, VI, VII, VIII, IX, X, I get the subtractive IV (somewhat understandably, given it's the far more common one in modern usage).  My source text only (and always) uses the additive 4 in the one's position: 24 is XXIIII, 34 is XXXIIII but 44 is XLIIII.  9 is universally rendered subtractively as IX or XC.  Is there a package that allows for some control over which numbers are done additively and which are subtractively, or would my best option be to write my own custom definition to account for the specific style used in my book? I need help to solve following problem from Rudin's Mathematical analysis book:    Convergence of the series $\sum a_n$  implies the convergence of $\sum \dfrac{\sqrt {a_n}}{n}$, if $a_n&gt;0$  I tried to construct a suitable convergence sequence $b_n$ such that $\sum b_n$ converges and $a_n \leq b_n$ but, I am not able to find such sequence $b_n$ .  Thanks for the help and sugestions. I have a default Blender cylinder positioned upright atop a ground plane:    How can I rig the cylinder so that can be animated to wiggle/jitter (perhaps procedurally with noise f-curve modifiers on two of its rotation axis) without penetrating the floor?    A simple parent-child relationship as in the image above is sufficient to rotate (and translate) the cylinder on 1 axis in either a positive OR negative direction.   I'm looking for a setup that would allow me easily to rotate the cylinder on two or even three axis in both a positive and negative direction.  I imagine this would be possible with some drivers applied to the XYZ coordinates using trigonometric equations but the math escapes me. It might also be possible using a particular bone setup with the correct constraints but my attempts to find something that works so far have not succeeded.  Thanks for your assistance. "Interested with the course" or "interested in the course"? Interested in/with the course   Which preposition is correct? in or with, or both. Implementing a Custom Subscription Center in Salesforce Marketing Cloud Has anyone implemented a custom subscription center using landing pages? The example one from the documentation doesn't seem to be working. I would really like to just customized the out of the box one but the amp script is hidden when I view source. Does anyone have and tips/tricks for creating a sub center?  Thanks! You can also contact me to customize your template, or create a new unique one.   Do I need to put a comma between the new and unique adjectives in this sentence?  In my native language (if I am not mistaken) we don't put a comma if the adjectives are of a different type. Not sure about English though, and if it's the case here anyway. Any help would be appreciated! Take most recent (or concatenate) edit comments when editing during grace period What does ‘Sport’ mean when you say ‘the new Apple iPad sports cameras for video conferencing’? What is the difference between a "proof by contradiction" and "proving the contrapositive"? Intuitive, it feels like doing the exact same thing. And when I compare an exercise, one person proves by contradiction, and the other proves the contrapositive, the proofs look almost exactly the same.  For example, say I want to prove: $P \implies Q$ When I want to prove by contradiction, I would say assume this is not true. Assume $Q$ is not true, and $P$ is true. Blabla, but this implies $P$ is not true, which is a contradiction.  When I want to prove the contrapositive, I say. Assume $Q$ is not true. Blabla, this implies $P$ is not true.  The only difference in the proof is that I assume $P$ is true in the beginning, when I want to prove by contradiction. But this feels almost redundant, as in the end I always get that this is not true. The only other way that I could get a contradiction is by proving that $Q$ is true. But this would be the exact same things as a direct proof.   Can somebody enlighten me a little bit here ? For example: Are there proofs that can be proven by contradiction but not proven by proving the contrapositve? The converse of Newton's shell theorem What does the "private" modifier do? Considering "private" is the default access modifier for class Members, why is the keyword even needed? Is 'acronymise/ze' a word? Is it used only colloquially, if at all? I just sent a text to a friend, who didn't understand an acronym I used for a game:      "Ah, I said it in a previous text so thought it was ok to abbreviate   it."   Though, since it was an acronym I had created, could I have used the word 'acronymise/ze' to be more specific?   I'm pretty sure it isn't a (formally used) word, but I wondered, as you often refer to acronyms specifically as acronyms, rather than abbreviations. Can you be more specific with a verb, too?  It sounds too odd to me, so I'm guessing generally no. But I wondered—so perhaps I'm wrong. Or half wrong.