I have realtek rtl8723be network adapter. I have installed the diver from . I use following commands every time I turn on the computer:  sudo modprobe -rv rtl8723be sudo modprobe -v rtl8723be ant_sel=2   This worked well. But today I have problem with my wifi. It keeps disconnecting and the signal strength is really low even if the router is next to the computer. I don't see the other networks at all. I tried with ant_sel=1 too but no luck. My Ubuntu version is 16.04. Is there any way to troubleshoot and fix the issue?
I have a HP 14-AN001NA laptop. It came with Windows 10 installed on which the Wi-Fi worked fine anywhere in the house.   However I removed Windows 10 and installed Ubuntu 16.04 instead and since I have had terrible Wi-Fi anywhere aside from about three foot away from the router. I don't know if it's relevant but my Wi-Fi is not called wlan0 but wlo1 for some reason (I didn't change it).  When running lspci -knn | grep Net -A3     I get the following output:  05:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8723BE PCIe Wireless Network Adapter [10ec:b723] DeviceName:   Subsystem: Hewlett-Packard Company RTL8723BE PCIe Wireless Network Adapter [103c:81c1] Kernel driver in use: rtl8723be Kernel modules: wl, rtl8723be
How to show that $\mathbb{Q}$ is not $G_\delta$ I read a section of a book and it made mention of the set of rationals not being a $G_\delta$. However, it gave no proof. I read on  about using contradiction, but it made use of the Baire category theorem, which is unfamiliar to me. I was wondering if anyone could offer me a different proof; perhaps using the fact that the complement of $G_\delta$ is $F_\sigma$.  Thanks. Is my solution correct for solving $\cos(z)=2$, $z\in \mathbb{C}$ Is password confirmation necessary on a Change Password dialog? Finding Stack Overflow/Exchange users based on geographic area? Get date for monday and friday for the current week (PHP) Is there a known well ordering of the reals? So, from what I understand, the axiom of choice is equivalent to the claim that every set can be . A set is well ordered by a relation, $R$ , if every subset has a least element. My question is: Has anyone constructed a well ordering on the reals?  First, I was going to ask this question about the rationals, but then I realised that if you pick your favourite bijection between rationals and the integers, this determines a well ordering on the rationals through the natural well order on $\mathbb{Z}$ . So it's not the denseness of the reals that makes it hard to well order them. So is it just the size of $\mathbb{R}$ that makes it difficult to find a well order for it? Why should that be?  To reiterate:   Is there a known well order on the Reals? If there is, does a similar construction work for larger cardinalities? Is there a largest cardinality for which the construction works? I have done major updates to my website, and renamed all my URLs.  How can I tell Google that I have renamed these URLs and let Google refresh its index?  My website URL: http://www.pndmasr.com/  My sitemap: http://www.pndmasr.com/sitemap.xml  I have uploaded my sitemap via Google Webmaster Tools many times.  However, every time I search Google for "pndmasr", I still get results with old pages. I have waited more than 3 days but the problem remains. Any suggestions?  Is there a problem with my sitemap? Overstayed my US visa for over 9 years, never got deported, can I return as a tourist? What is the best way to restore the state of ViewController What is the square root of $i^4$? What is the $\sqrt{i^4}$?   $i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?  Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?  When I plug it into my TI-89 Titanium, I get $1$.  Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$ How do you disable browser autocomplete on web form field / input tags? Checking incoming call in iphone No Surname in indian passport and given name is considered as last name in E ticket for an international flight I have booked tickets for my parents to visit Mauritius for a week and my mom's passport (given name is xxxx) doesn't have surname, so in the e-ticket form her name has been printed as "Mr xxxx". I have spoken to the travel agency and Air Mauritius, but didn't get any information regarding the issue.  Can anyone please tell me if there might be a problem boarding the flight?  I am very anxious right now since its their first international flight. Best practices for making a WordPress site "movable"? I'm planning to buy a domain and install a WordPress site on it at a web-hosting company.  But first I'll need to iron-out some bugs...  I got a working LAMP-stack at my home-computer, so I'm thinking about first installing my site locally - off the Internet.  I'll be mostly tweaking and making "test-posts" to check that everything works, but I'll probably also make some posts I'd like to keep when I move the site and take it online.  When I'm done chasing bugs; I guess I'll just package the public_html|www directory-tree and upload it to the web-hosting company, and also export the database from my home-computer to the web-hosting.    What I'm wondering if there are some "best practices", some do-s and don't-s, some major pits to take note of?    For example, how can ensure the smooth transition of domain-name in posts and links as such, from localhost to whatever-i-end-up-buying.com (I know from previous attempt of running WordPress at my home computer, that this didn't always work quite well...). I've never really got how chmod worked up until today. I followed a tutorial that explained a big deal to me.   For example, I've read that you've got three different permission groups:   owner (u) group (g) everyone (o)   Based on these three groups, I now know that:   If the file is owned by the user, the user permissions determine the access. If the group of the file is the same as the user's group, the group permission determine the access. If the user is not the file owner, and is not in the group, then the other permission is used.   I've also learned that you've got the following permissions:   read (r) write (w) execute (x)   I created a directory to test my newly acquired knowledge:  mkdir test   Then I did some tests:  chmod u+rwx test/ # drwx------ chmod g+rx test/ # drwxr-x--- chmod u-x test/ # drw-r-x---   After fooling around for some time I think I finally got the hang of chmod and the way you set permission using this command.    But...  I still have a few questions:   What does the d at the start stand for? What's the name and use of the containing slot and what other values can it hold? How can I set and unset it? What is the value for this d? (As you only have 7=4+2+1 7=4+2+1 7=4+2+1) Why do people sometimes use 0777 instead of 777 to set their permissions?   But as I shouldn't be asking multiple questions, I'll try to ask it in one question.  In UNIX based system such as all Linux distributions, concerning the permissions, what does the first part (d) stand for and what's the use for this part of the permissions? In the addon ArchiMesh there is a Japaneses curtain feature, but, it looks straight and does not have the fold that I want in it. Is there an addon that can do this for me? If not, how can I make a curtain (like in the images below)? Proving $\lim_{n\to\infty} a^{\frac{1}{n}}=1$ by definition of limit given a sequence $a_n=a^{\frac{1}{n}}$ for $n\in\mathbb{N}^*$, $a\in\mathbb{R},a&gt;1$ then proof that $\lim\limits_{n\to+\infty}a_n=1$ by definition.   proof:  given $a_n=a^{\frac{1}{n}}$ for $a\in\mathbb{R},a&gt;1$.  for $n\in\mathbb{N}^*,n+1&gt;n\Rightarrow \frac{1}{n+1}&lt;\frac{1}{n}$ and because $a&gt;1$ we gets $a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}$, since $\frac{1}{n+1}&gt;0$, then $1=a^{0}&lt;a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}\Rightarrow 1&lt;a_{n+1}&lt;a_{n}$.  then we proof that $\lim\limits_{n\to+\infty}a_n=1$ we need to show that $\forall\epsilon&gt;0,\exists N,\forall n&gt;N,|a_n-1|&lt;\epsilon$  then for $\epsilon&gt;0$, choose $N=\frac{1}{\log_a(\epsilon+1)}$, since $a&gt;1$ we gets that $$\forall n&gt;N,1&lt;a^{\frac{1}{n}}&lt;a^\frac{1}{N}\Rightarrow0&lt;a^{\frac{1}{n}}-1&lt;a^{\frac{1}{N}}-1\Rightarrow |a^{\frac{1}{n}}-1|&lt;|a^{\frac{1}{N}}-1|=|a^{\log_a(\epsilon+1)}-1|=|\epsilon+1-1|=\epsilon\Rightarrow |a^{\frac{1}{n}}-1|&lt;\epsilon$$  wich implies that $\lim\limits_{n\to+\infty}a_n=1\square$  my proof to the limit is correct? We are supposed to prove that the zero vector is unique with the additive identity $x+O=x$ and then by creating another additive identity with $O'$, then $x+O'=x$.  We set them equal $x+O=x+O'$ and use the vector cancelling theorem to get $O=O'$, but how does this prove that the zero vector is unique? Find Java classes implementing an interface Some time ago, I came across a piece of code, that used some piece of standard Java functionality to locate the classes that implemented a given interface. I know the functions were hidden in some non-logical place, but they could be used for other classes as the package name implied. Back then I did not need it, so I forgot about it, but now I do, and I can't seem to find the functions again. Where can these functions be found?  Edit: I'm not looking for any IDE functions or anything, but rather something that can be executed within the Java application. What is the problem with nested environments using \BODY (environ package)? This question led to a new package:      I want to define some environments, which place their content into multiple different other environments. To access the contents I used the environ package, which introduces the \BODY command. However at some point pdflatex seemed to be stuck in an infinite loop.  I reduced it to a minimal example:  \documentclass{article} \usepackage{environ}  \NewEnviron{assertion}{Assertion: \BODY} \NewEnviron{outerassertion}{\begin{assertion}\BODY\end{assertion}}  \begin{document}  \begin{assertion} test \end{assertion}  \begin{outerassertion} test2 \end{outerassertion} \end{document}   Any ideas how to work around this problem? I guess, the problem is with the \BODY command since everything works fine as long as I use it in only one of the new environments.