Rational with minimal denominator between two rationals My question from an easy problem.     $p,q$ are positive integers such that $$ \frac{5}{9}&lt;\frac{p}{q}&lt;\frac{4}{7} $$ find $p,q$ such that $q$ is the smallest number that satisfies this inequality.   Draw the line of $ y&lt;\frac{9}{5}x$ and $y&gt;\frac{7}{4}x$ , we can "observe" that $\frac{9}{16}$ is such number.  However, if the question becomes     $a,b,c,d$ are positive integers such that $$\frac{a}{c}&lt;\frac{b}{d} $$ find $p$,$q$ such that $q$ is the smallest number that satisfies the inequality   $$\frac{a}{c}&lt;\frac{p}{q}&lt;\frac{b}{d}$$   No idea about this.
Minimal $ab$ for Rational Number $a/b$ in an Interval Given rational numbers $L$ and $U$, $0&lt;L&lt;U&lt;1$, find rational number $M=a/b$ such that $L \le M&lt;U$ and $(a\times b)$ is as small as possible---$a$ and $b$ are integers. For example,    If $L=66/101$ and $U=19/29$ then $M=17/26$. If $L=66/101$ and $U=19/28$ then $M=2/3$.   Right now I'm using a naive search that is exponential in complexity. Is there a better method?  If it is of any help, it can be assumed the prime factorization of numerator and denominator of $U$ and $L$ are known.
How do I update Android on my device? This Community Wiki question is designed to be the "canonical question" for questions asking how to upgrade one's device to a new Android version.   I want to get a newer version of the OS for my Android device. How can I do that?  See also: $\LaTeX$ and Mathematica Did I correctly derive this recurrence equation formula I think that the question is contained in the title. Suppose we begin from something that is false for example $1=0$. Is it possible using only $\Rightarrow$ (and of course $\lnot ,\wedge,\lor$)  to prove any possible statement? It has been ages since i studied logic at the university so i would prefer a simple explanation (if possible) Is there a way to export large multi-layered Illustrator file to Photoshop?  When I export with preserve layers and text edit-ability, the complex layers are flattened. I've reorganized my layers to have simple groupings but that's time consuming.  Is there a more effective way of doing this? Or script to automate it? How can I disable the Document Library drag and drop function? Before I start - I'm new to this SharePoint development lark and am slowly familiarising myself with it.  We're using SharePoint online and I have amended the SharePoint out-of-the-box document library to include some extra fields that are required when the user adds a new document.  If the user creates or uploads a new document then they are prompted to enter the required information. However if they use the drag and drop functionality the entry of the necessary fields is bypassed.  During my search I came across  which asks how to enforce metadata entry (which would be another good option for me). It also mentions disabling the drag and drop functionality by editing dragdrop.js and removing content from various methods.  I've had a look at the site collection I'm working on (to which I have full control) and I can't seem to find dragdrop.js (or the layouts folder for that matter).   Is this something that's held at site collection level or is it held at a higher level? I don't have admin rights above my site collection, do I need them? Do I need to use Visual Studio to make the necessary changes? (Currently I'm doing things through the web and having a bit of a play with Napa I don't have Visual Studio installed)  When I get there is it obvious what methods need to be removed? How can I tell SharePoint to use the new version of dragdrop.js? Is it just a case of saving a copy of the existing one to a different name and then editing the existing file?      I suppose the best place for me to have asked this question would probably have been on the existing post, but I am unable to post a comment on there, sorry... I'm not sure if this is possible architecturally, but when judging whether to set a bounty and how much it should be, it would be nice to see the impact based on our previous questions. If you're not visiting the site throughout the day, there is no way to tell how effective that bounty might be--particularly for increasing traffic. There is a similar question with adding and removing tags that increase the visibility in the right places, and the effect on views that may have. I've read  about how some publishers (e.g., Springer) allows authors to submit to arXiv before journal submission, but they make authors wait for a year after acceptance to submit to arXiv.   My question is: Is it okay to submit to arXiv during the time when the paper is under consideration by the journal, for example after it has been "revised and resubmitted" but before final decision? Some of my Apps Icons are changing to a generic app icon, and I have no idea why. Tried to restart the system and nothing. 3 days ago i updated my Skype and then the icon is gone. Show that $a \star b=a \cdot b+a+b$ is binary operation for the group $\Bbb{ Q} \setminus \{-1\}$ The group $\left(\Bbb{ Q}\setminus\{-1\},\star\right)$ has as its underlying set the rational numbers different from $-1$ and the operation $\star$ is defined as $a \star b=a \cdot b+a+b$ where multiplication/addition are the usual operations with rational numbers.Show that this is a binary operation.  What I did (Please help verify):  identity element has to be 0 since $a\star 0 = a+0+a(0)$ so for $a\star b=0$ we have  $ab+a+b=0$ $$ab+a=-b \\ a(b+1)=-b \\ a=\frac{-b}{b+1}$$ so $b$ cant be $-1$ since that solution is not in $\Bbb{Q}$. Similar question might have been answered here. Is this operation true for all groups? I think so.  Thanks for the tips and help. Do I need to tame my twitchy F5 finger? I just got this error message for the first time (I was getting this as the response to any request on the SE network for about the last five minutes, until now):     We're sorry...      There are an unusual number of requests coming from this IP address.   To protect our users, we can't process any more requests from this IP   address right now. We'll restore access as quickly as possible, so try   again soon. If you believe you have reached this page in error,   contact us.   As far as I know, I was not spamming SE or participating in DOS attack, and I find it unlikely that any of my office coworkers who share my IP address were doing so either.  Granted, I was pressing F5 a lot. But I've always pressed F5 a lot on SE, and have never gotten this message before. So my question is: Is it possible to get this error message simply by refreshing too much or too often? Or is that message reserved for a truly massive number of requests that are unlikely to be caused by one person refreshing a lot? (In the former case, I can't explain why I've never gotten it before; in the latter case, I can't explain why I just got it now). Given a DateTime object, how do I get an ISO 8601 date in string format? Given:  DateTime.UtcNow   How do I get a string which represents the same value in an -compliant format?  Note that ISO 8601 defines a number of similar formats. The specific format I am looking for is:  yyyy-MM-ddTHH:mm:ssZ Which of the following is also an ideal? If $U,V$ are ideals of a ring $R$, then which of the following is also an ideal of $R$?   $U+V=\{u+v\mid u\in U,v\in V\}$ $U\cdot V=\{u\cdot v\mid u\in U,v\in V\}$ $U\cap V$   My attempt: I have proved $U+V$ and $U\cap V$ are ideals of $R$.  My problem: I have problem about the second option. I think it is not ideal of $R$. If I am wrong then please prove this otherwise give one counter example to show that $U\cdot V$ is not ideal of $R$. Looking for qt-at-spi for ubuntu 20.04 How to properly secure a wireless home network I'm soon going to be setting up a wireless network for me and some buddies and (hopefully) making it secure. I have some questions and concerns regarding this and figured someone here could give me a hand and steer me in the right direction.   I have a WRT55AG (version 2) loaded with the factory firmware. I've heard that it's possible to load your own firmware onto routers like these. Is it worth it? What can be achieved with it? I was wondering what kind of security I should use on my wireless. Ideally I would want all devices to be 'accepted', but I'm not sure how to accomplish this. I'm guessing it has something to do with MAC addresses, but I don't know the first thing about them. It would be great that ALONG WITH encryption (I'm told WPA2 is the best), every device would need to be added to a list of some sort before even being able to access the AP. Another feature I was thinking of would be completely hiding my wireless from showing up on other people's computer. Is this possible (like an SSID broadcast thing). Is it worth the trouble? One more option I was looking to implement would be bandwidth calculators. Ideally something in the router to measure the amount of bandwidth being used and how much has been used for that month (this has to do with bills and limits, obviously). Is it possible to throttle down speed when a certain limit is reached?   These are alot of points but I was hoping someone who has more knowledge than me in this subject could give me a hand and a few pointers. Hopefully all this is possible and I'm not dreaming :) My PC has a build-in VT1828S chip. The connection between PC and monitor (which has built-in speakers) is through an HDMI cable.  Audio preferences shows 2 devices for audio output: analog stereo and RS880 Audio Device [Radeon HD 4200] digital stereo (HDMI).  When I plug a headphone in the rear jack and I choose the former device, everything works. When I choose the latter device, instead, no sound is heard, and video playback is too fast. It is as if video frames want to synchronise with the audio timing, but the audio timing information is wrong.  HDMI audio worked perfectly with Ubuntu 11.04, but stopped after 11.10 upgrade. Consider a string under tension, for example, a string on a guitar. When a guitar string is plucked, it vibrates at a certain frequency. When the tension on the string is increased by twisting the tuning peg at the end of the neck, the frequency of the string increases. According to , increasing the tension on a string also increases the speed of the wave moving through it. According to  the speed of a wave is higher in a more dense medium.  Does this mean that increasing the tension also increases the density?  This does not seem very logical to me, since density is the "compactness" of a substance, or how close together the individual molecules are. By twisting the tuning peg on a guitar, you rotate the cylinder that the string is fixed to, wrapping it around more tightly. This pulls more of the string around that cylinder, similarly to when you reel in a fishing rod. Since more of the string's molecules are now wrapped around the peg, there should be fewer molecules over the same distance between the bridge and the nut (the two ends of the vibrating portion of the string), therefore bringing the density down. Let $C_k$ be the set obtained in the $k-$th stage of building the Cantor set, where  $$C_1=[0,\frac{1}{3}]\cup[\frac{1}{3},\frac{2}{3}]$$  $$C_2=[0,\frac{1}{9}]\cup[\frac{2}{9},\frac{1}{3}]\cup[\frac{2}{3},\frac{7}{9}]\cup[\frac{8}{9},1]$$  $$\dots$$  So we have   $$C_k=\bigcup_{j=1}^{2^k}I_{k,j}$$  where $I_{k,j}$ are pairwise disjoint closed intervals of length $3^{-k}$ in $[0,1]$, let  $$[0,1]\setminus C_k=\bigcup_{j=1}^{2^k-1}J_{k,j}$$  where $J_{k,j}$ are pairwise disjoint open intervals and each $J_{k,j}$ are located between intervals $I_{k,j},I_{k,j+1}$  define  $$F_k:[0,1]\to[0,1]$$  where  $$F_k(0)=0,F_k(1)=1,F_k(x)=\frac{j}{2^k}\quad x\in J_{k,j}$$  and $F_k$ is linear on each subinterval of $C_k$. We can show that $F_k$ converges to the Cantor function $F:[0,1]\to[0,1]$. Moreover, it can be shown that  $$|F_k(x)-F_k(y)|\leq\frac{3^k}{2^k}|x-y|$$  How can we show that $F$ is Hölder continuous of order $\log_32$ ? In another  regarding a difference between short on and short of, the  says the following:     ... But this is not always the case. A person can be short on/of money,   but lacking a specific amount, someone can only be €10 short or short   €10. This construction never uses a preposition.   but is that really true? I can see sentences like "I am short of 10 dollars" quite often. How can I convert the binary string $x_bin="0001001100101" to its numeric value $x_num=613 in Perl?