Proof: If no prime less than or equal to $\sqrt{n}$ divides $n$, then $n$ is a prime
why only square root approach to check number is prime
I am interested in theoretical and practical considerations. Chatroom owners can star their own messages I was able to star my own message by pinning it and unpinning it.  Pinning a message adds a star, even to your own messages, but when the pin is removed, the star remains. Let $(\Omega, \mathcal F,P)$ be a probability spaces and $H$ be a Hilbert space. By a random operator $A$ from $H$ to $H$ we mean a linear continuous mapping from $H$ into the Frechet space $L_0^H (\Omega, \mathcal F,P)$ of all $H$-valued random variables. My question is: How Can we define $|A|$ and the adjoint $A^*$? How do I delete a board in Trello? Why can't you use the keyword 'this' in a static method in .Net? I'm trying to use the this keyword in a static method, but the compiler won't allow me to use it.  Why not? Minimize the sum of distances between two point and a circle Cardinality of the set of bijective functions on $\mathbb{N}$? Setting "checked" for a checkbox with jQuery In what situations is octal base used? As an Indian citizen, can I apply and have both a B1 and F1 visa? Why does Type I error always occur in a NHST? Or, is NHST too weak to tell the truth? Recently I have noticed that Type I error of a test is too much persistent. Even if all the values are almost equal (variance is extremely low) and tested to that value (true mean), still there is nearly 5% Type I error of the test. I am frightened and curious to rethink, are the people who have banned (or skeptic about) NHST (null hypothesis significance testing) truly correct? Let's have a look at the following r code:       n=10000 t1err=0 set.seed(1235) for (i in 1:n){    x=rnorm(100, 10, 20)    if (((t.test(x, mu=10))$p.value)&lt;=0.05) (t1err=t1err+1)  } cat("Type I error rate in percentage is", (t1err/n)*100,"%")   It will produce roughly 5% as Type I error. Seems ok. But let's see this:      n=10000   t1err=0   set.seed(1235)   for (i in 1:n){      x=rnorm(100, 10, 0.000001)      if (((t.test(x, mu=10))$p.value)&lt;=0.05) (t1err=t1err+1)    }   cat("Type I error rate in percentage is", (t1err/n)*100,"%")   Each values are extremely close to 10. Still it produces about 5% Type I error. What conclusion we may draw from it? Or, am I missing something? MS Access to Oracle easy Conversion / Migration I want to utilize beamer's \uncover command within ChemFig's reaction scheme environment, i.e.:  \schemestart     \chemname[-1ex]{\chemfig[][scale=1]{*5(-=(-=[:-60]O)-O-=)}}{Furfural}     \uncover&lt;2-&gt;{         \arrow(furfural--mf){-&gt;[*{0}HDO][]}         \chemname[-1ex]{\chemfig{*5(-=(-)-O-=)}}{2-Methylfuran}     } \schemestop   However, this results in numerous "undefined control sequence" errors.  Is there a way to use beamer's \uncover command within ChemFig?  Here is a full MWE:  \documentclass[10pt]{beamer} \usepackage{chemfig}  \begin{document}  \begin{frame} \frametitle{Stuff}  \schemestart     \chemname[0ex]{\chemfig[][scale=1]{*5(-=(-=[:-60]O)-O-=)}}{Furfural}     \uncover&lt;2-&gt;{         \arrow(furfural--mf){-&gt;[*{0}\textbf{HDO}][]}         \chemname[0ex]{\chemfig{*5(-=(-)-O-=)}}{2-Methylfuran}     } \schemestop  \end{frame}  \end{document} Cannot inherit from multiple classes defining __slots__? I finally got fed up when wanting to read about bash's read and it's -s option with man bash. I found the right spot eventually (around line 4500), but it was a frustrating as usual, since both /read and even /\s-s\s searches have way too many matches.  So, the question is: How can I read long man pages efficiently, or get same information in other ways, locally? As a specific example, how to reach the relevant documentation after seeing read -s pwd in a shell script? A good answer could be a shell script snippet, or hint about some tool and how it is used, or something else entirely, as long as it helps in finding the right spot to read.  Note: I'm not tagging with  because I want the question to be about man page reading in general, even though that quite possibly is the most commonly encountered humongous man page. When did 5 foot squares become standard in D&D? single word that means which can be constructed at home Why is Linux reporting "free" memory strangely? This is a  about how Unix operating systems report memory usage.   Similar Questions:                    I have production server that is running Debian 6.0.6 Squeeze  #uname -a  Linux debsrv 2.6.32-5-xen-amd64 #1 SMP Sun Sep 23 13:49:30 UTC 2012 x86_64 GNU/Linux   Every day cron executes backup script as root:  #crontab -e  0 5 * * * /root/sites_backup.sh &gt; /dev/null 2&gt;&amp;1   #nano /root/sites_backup.sh  #!/bin/bash str=`date +%Y-%m-%d-%H-%M-%S` tar pzcf /home/backups/sites/mysite-$str.tar.gz /var/sites/mysite/public_html/www mysqldump -u mysite -pmypass mysite | gzip -9 &gt; /home/backups/sites/mysite-$str.sql.gz cd /home/backups/sites/ sha512sum mysite-$str* &gt; /home/backups/sites/mysite-$str.tar.gz.DIGESTS cd ~   Everything works perfectly, but I notice that Munin's memory graph shows increase of cache and buffers after backup.  Then I just download backup files and delete them. After deletion Munin's memory graph returns cache and buffers to the state that was before backup.  Here's Munin graph:  Externally hosted image was a dead link. Which fruits and vegetables should be kept in a fridge, and which outside? First, my apologies if this has already been asked/answered.  I wasn't able to find this question via search.  My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2.  In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof.  As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.  If we write equations (3) as $q = p - (p^2 - 2)x$, we can write (4) as   $$ q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. $$  Here, we need a rational $x &gt; 0$, chosen such that the expression in $[...]$ is positive.  Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need   $$ x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, $$  or, for $p \in B$, $x &lt; 1/\left(p + \sqrt{2}\right)$ or $x &gt; 1/\left(p - \sqrt{2}\right)$.  Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$.  Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.  The Question:  Why does Rudin choose $x = 1/(p + 2)$ specifically?  Is it just to make the expressions work out clearly algebraically?  Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof?  Is there a simpler derivation for the number $q$ that I am missing?