Is there a composite integer $n \geq 9$ such that $n \nmid (n-1)!$?  If we are not talking about composites then by Wilson's theorem we have $n \nmid (n-1)!$.
I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!  Here is what I am asked to prove:     If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.   Proof:  $n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0&lt;a,b&lt;n$.  Case 1: If $a=b$ then $n=a^{2}$. Now $n \mid (n-1)! \implies a \mid (n-1)!$, so $$\begin{aligned} (n-1)! &amp;\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &amp;\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &amp;\equiv 0 \pmod n \end{aligned}$$  Case 2: $0&lt;a&lt;b&lt;n$.  Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.  So this implies $(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$, Q.E.D.
Install on Second Hard Drive with startup boot option? In theory, what I want to do is very simple, but I've done some google research and read some posts on this site and I'm still confused.   I want to keep win7 on my primary C:\ drive, and install Ubuntu (12.10) on a secondary drive (that I have yet to buy or install. I just want to be certain that I'm successful when I do this). If at all possible I'd also like it if when I start the computer I have the option to boot either Win7 or Ubuntu, like an account log in type thing where I just pick which drive I'm running with that day. And just to reiterate, I do not want to partition my C:\ drive or alter, change, or affect my Win7 system in any way.  The reason I'm still fairly uncertain about all of this, is that a lot of the guides I've read have been for earlier versions of Ubuntu. Also the vast majority of guides are about partitioning one drive to fit both OS'. I want two separate OS running from two separate drives that I can pick at boot. The guides I did find mentioned that there is some issue with the grub being installed on C:\ but the rest of the OS being installed on the other drive. Is this still the case? How do I avoid that?  Also, if it helps, I've been running 12.04lts on my laptop for a few months now in and out of class and I love it.  Finally, if you could please point me to a guide or write your answer in the form of a guide for a two year old that'd be awesome. I'm still an Ubuntu novice and I don't want to mess up my win7. Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? How to control Windows 7 snap feature with two monitors? There is an  feature in Windows 7. However, it seems to not work when dragging a window to the edge that is adjoined to the second monitor. How to make it work properly? Maybe some regedit modification or something? How to Regex search/replace only first occurrence in a string in .NET? It seems the .NET Regex.Replace method automatically replaces all matching occurrences.  I could provide a MatchEvaluator delegate that returns the matched string after the first replacement, rendering no change, but that sounds very inefficient to me.  What is the most efficient way to stop after the first replacement? What is a reliable method to locate floating islands in Terraria? How to fix "Duplicate sources.list entry" warning? Every time I try to do some update (sudo apt-get update) it says:   Reading package lists ... ready W: Duplicate sources.list entry http://archive.canonical.com/ubuntu/ need /partner amd64 Packages (/var/lib/apt/lists/archive.canonical.com_ubuntu_dists_precise_partner_binary-amd64_Packages) W: Duplicate sources.list entry http://archive.canonical.com/ubuntu/ need /partner i386 Packages (/var/lib/apt/lists/archive.canonical.com_ubuntu_dists_precise_partner_binary-i386_Packages) W: You have to run apt-get update to correct these problems   sources.list:   ls /etc/apt/sources.list.d/ as requested:  milton@milton-HuronRiver-Platform:~$ ls /etc/apt/sources.list.d/ emesene-team-emesene-stable-precise.list precise-partner.list precise-partner.list.save upubuntu-com-chat-precise.list upubuntu-com-chat-precise.list.save Regex exec only returning first match When two accounts are merged, all the posts/votes are moved over to the "master" account, and the other account ("slave"?) is deleted. The merge is propagated network-wide, so a similar thing happens on each site.  But, this does not happen to the chat user. The "slave" chat user is kept, with a 404 parent profile link. I feel that the chat user should be merged as well, re attributing all the chat messages.  Thoughts? How can I get apt to use a mirror close to me, or choose a faster mirror? Downloading from the main Ubuntu archive is slow even when it's not release day, how can I get apt-get to automatically use a mirror that is close to me? Dirac delta function as a limit of sinc function I have a Dirichlet distribution from which I'm maximizing $\alpha$ by calculating the log likelihood with the following equation  $p\left(s|\alpha\right) = \sum\limits_{i=1}^O\log\left(\frac{n_i!}{\prod\limits_{j=1}^kn_{ij}!}\frac{\Gamma\left(\alpha_0\right)}{\Gamma\left(\sum\limits_{j=1}^k\left(n_{ij}+a_j\right)\right)} \prod\limits_{j=1}^k \frac{\Gamma\left(n_{ij}+\alpha_j\right)}{\Gamma\left(\alpha_j\right)}\right)$  I have a python script to calculate the log likelihood. It works for small values, but when $n_{ij}$ gets high $\Gamma\left(\sum\limits_{j=1}^k\left(n_{ij}+a_j\right)\right)$ and $\Gamma\left(n_{ij}+\alpha_j\right)$ python runs into overflow error (the max for math.gamma is between 171 and 172).   Is there a way to change the likelihood function so that $\Gamma\left(\sum\limits_{j=1}^k\left(n_{ij}+a_j\right)\right)$ and $\Gamma\left(n_{ij}+\alpha_j\right)$ don't run into overflow errors. Can I just divide the values in $\Gamma$ by for example 100, if I then compare the likelihood between $p(s|\alpha)$ with different $\alpha$ values do the best $\alpha$ values still have the highest likelihood? Windows update doesn't progress beyond 0 KB 0% I've just installed Ubuntu and I need either video, wireless, or some sort of other manufacturer (non-opensource) driver. How can I run the tool that will fetch all of these for me? I'd like graphical instructions.  Related: Why does instanceof return false for some literals? &quot;foo&quot; instanceof String //=&gt; false &quot;foo&quot; instanceof Object //=&gt; false  true instanceof Boolean //=&gt; false true instanceof Object //=&gt; false false instanceof Boolean //=&gt; false false instanceof Object //=&gt; false  12.21 instanceof Number //=&gt; false /foo/ instanceof RegExp //=&gt; true  // the tests against Object really don't make sense  Array literals and Object literals match... [0,1] instanceof Array //=&gt; true {0:1} instanceof Object //=&gt; true  Why don't all of them? Or, why don't they all not? And, what are they an instance of, then? It's the same in FF3, IE7, Opera, and Chrome. So, at least it's consistent. I am new to Stack  Exchange and I think it's very impressive.  One thing bothers me is that because it is so far down the line, I don't think beginners stand a chance of growing.  As time goes on, more and more questions are being answered and therefore if someone asks something with a really basic answer you can just comment. However, beginners can't comment until they have a reputation of 50. The only way they can build up is by answering questions. But if they answer a basic question long winded, they will be voted down because not so much effort was needed or because its already a duplicate and will lose the bit of reputation they have earned.  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