Suppose $x\in \mathbb{Q}$ such that $x^{2}=6$. Since  $x\in \mathbb{Q}$, there exists $m,n \in \mathbb{Z} $ where either $m$ or $n$ is odd such that $x=\frac{m}{n}$.  $\implies$ $x^2=(\frac{m}{n})^2=\frac{m^2}{n^2}=6$  $\implies$ $m^2=6n^2$, so $m^2$ is even. Hence, $m$ is even.  Since $m$ is even, $m=2k, k\in \mathbb{Z}$.  Then, $m^2=(2k)^2=4k^2=6n^2$  $\implies$ $n^2$ is even, so $n$ is even. But one of $m$ or $n$ must be odd, so $x\notin \mathbb{Q}$.  Therefore, $\sqrt{6}$ is irrational.  Does everything look alright here?
Suppose $\sqrt6 = \frac pq$ where $p$ and $q$ have no common factors.  $$6 = \frac {p^2}{q^2}$$  $$6p^2 = q^2$$  So $q^2$ and therefore $q$ is divisible by $6$.   $$p^2 = \frac {q^2}{6}$$  So $p^2$ and therefore $p$ is divisible by $6$.   S0, $p$ and $q$ have a common factor $6$. Contradiction.  Therefore $\sqrt6$ is irrational.  Does the proof work?
Probability of a painted cube being reassembled into itself Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where. Allow suppression of automatic linking of web addresses Strange behaviour of binomial coefficient's delimiters This question led to a new package:      While writing down my Combinatorics exercises I found that the binomial coefficient's delimiters weren't displayed as indicated inside the amsmath documentation.  By subsequent trials and errors I was able to reduce the volume of my document to this simpler and fully compilable MWE:  \documentclass[11pt]{article} % \usepackage{amsmath} % \makeatletter % \usepackage{lmodern} \usepackage{lipsum} % \makeatother % \begin{document} % \lipsum[1] \begin{equation} 2^k-\binom{k}{1}2^{k-1}+\binom{k}{2}2^{k-2} \end{equation} \lipsum[1] % \end{document}   Which gives (using TeXworks 0.44 on MikTeX 2.9.4521 updated 09/10/13) this un-expected result:        While the amsmath doc shows that with the loaded packages inside the MWE the result would actually be like this:        From here we can see that the delimiters on the first example are a little bit stretched vertically as they surpass the arguments inside the binom. coeff.  The more strange event is that when I comment the line \usepackage{lmodern} the result on delimiter's shape in my example looks exactly as the amsmath doc:  \documentclass[11pt]{article} % \usepackage{amsmath} % \makeatletter % %\usepackage{lmodern} \usepackage{lipsum} % \makeatother % \begin{document} % \lipsum[1] \begin{equation} 2^k-\binom{k}{1}2^{k-1}+\binom{k}{2}2^{k-2} \end{equation} \lipsum[1] % \end{document}   Output:        From here I deduce that the problem lies inside lmodern, but I'd like to keep using this package and fix this wierd "issue" with the math mode. Is there some clue that I've missed or this is in fact an update-process bug infecting amsmath's delimiter control? How can I get this symbol without using bbold? A solution I saw somewhere suggested to use   \usepackage{bbold} $\mathbb{1}$   which works well and I get what I want. However, now other symbols used to represent real numbers, natural numbers, etc look ugly.    I'd like them to look (without using bbold)    I'd like to only use that package to achieve that strange double one. Is there any way I can do that? For which $f,g \in k[t]$, $k[f,g]$ is integrally closed? Let $f,g \in k[t]$, $k$ is a field of characteristic zero, $\deg(f)=n \geq 2$, $\deg(g)=m \geq 2$.     Is it possible to characterize all such $f$ and $g$ for which $k[f,g]$ is integrally closed in its field of fractions $k(f,g)$?   I do not mind to further assume that $k(f,g)=k(t)$ (but I do not want to assume that $k[f,g]=k[t]$, which makes my question trivial).  Thank you very much!  Edit: (i)  is relevant.  (ii) If also $k(f,g)=k(t)$, then being integrally closed is equivalent to $k[f,g]=k[t]$. Indeed: (a) If $k[f,g]=k[t]$ then $k[f,g]$ is integrally closed (since $k[t]$ is). (b) If $k[f,g]$ is integrally closed then, in particular, since $t \in k(t)=k(f,g)$ (= the field of fractions of $k[f,g]$) and obviously $t$ is integral over $k[f,g]$ (for example, $t$ is a root of $f(T)-f \in k[f,g][T]$) hence $t \in k[f,g]$, so $k[f,g]=k[t]$.  For example: $f(t)=t^2+1$, $g(t)=t^3-4t$. We have, $t= \frac{t(t^2-4)}{t^2+1-5}=\frac{g}{f-5}$, so $k(f,g)=k(t)$. By Abhyankar-Moh-Suzuki theorem, $k[f,g] \neq k[t]$, so $k[f,g]$ is not integrally closed. Similarly, for every $f,g$ such that $k(f,g)=k(t)$ and one of the degrees does not divides the other, then $k[f,g]$ is not integrally closed.  (iii) If also $k(f,g)=k(t)$ and $k$ is algebraically closed, then being integrally closed is equivalent to $H(t)=(f(t),g(t))$ is injective and $H'(t)=(f'(t),g'(t)) \neq (0,0)$ for every $t \in k$. Indeed, $k[f,g]=k[t]$ is equivalent to the new conditions (it is important that $k$ is algebraically closed!!).  Edit: What if we replace $k$ by an integral domain $D$ (which is a $k$-algebra)? See also  question. My question isn't how they receive the energy to jump, but why.  When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals (else we would see a continuous spectrum).  Electrons can be released in the form of beta decay, thus proving that they are capable of traveling outside of orbitals contrary to the statement my teacher said that they stay within orbitals. Then, to add to the confusion, the older model of rings floating around a nucleus has, from what I can tell, been outdated, which would support this model. My teacher's explanation was that the electrons made a quantum jump of some kind. How do electrons move between orbitals or do we know how they jump, excluding the reason that energy causes them to jump, and why are positrons formed sometimes instead of electrons in Beta decay?  When I'm asking "how do electrons jump" I would like to know how an electron can jump between each orbital such as how it moves and how it knows where to jump since it appears to be a jump where the electron doesn't slow into a orbital position. Specifically how they jump what is this Atomic electron transition, I understand that they jump and that they do this through absorbing and releasing energy but what is this Atomic electron transition other than what is already on the wikipedia article . Is there any way I can 'undo' the factorial operation? JUst like you can do squares and square roots, can you do factorials and factorial roots (for lack of a better term)?   Here is an example: 5! = 120.  Is there a way I can work out the number that must be factorialed (??) to give the answer 120? I wrote a bash script as below:  #!/bin/bash echo "Enter a file name" read filename import -window root $HOME/Desktop/$filename.png   I expected this script to take a screen shot and save the image with the given filename on my desktop.  Problem:  My Ubuntu terminal is giving error:     import: Command Not Found xml serializing of saved game data I have some commands to in my rc.local. I know that they are failing. How can I get log file with messages produced by executing rc.local? Where is it located?  I have checked the /var/log/boot.log. I know my messages are not there because I know already what is the reason of failure. But I still want to make sure from log file.  Note, I don't want to run script again, I could but I don't want. I would rather analyse wht happened during startup.  Thanks for any help.  Ubuntu 12.04 Desktop (if it matters) List of chat rooms with host stackoverflow.com is not the same as the actual list of rooms on stackoverflow.com chat itself Generally chat.&lt;site&gt;.com redirects to chat.stackexchange.com/?tab=site&amp;host=&lt;site&gt;.com and go for same thing but here you can check your self:       Both are Different!  If one visits  and select site: StackOverflow, then it outputs  and they miss rooms that are only listed by .  So, why does this happen? Is this a bug?  Note: I found this while  for general chat-room at StackOverflow; see chat-discussion What is the principle of underdetermination? While studying I read about the principle of underdetermination of scientific theories. I made some researches online but I am more confused than before. I read about the Quine-Duhem holistic thesis, but I do not know if is the same as underdetermination. I need to exclude .git from my find search. To achive that, I am using the -path ./.git -prune switch:  $ find . -path ./.git -prune -o \( -type f -o -type l -o -type d \) | grep '.git' ./.git   However, even though this skips the contents of the .git directory, it lists the directory itself. It works when I add -path ./.git -prune -o -print -a  find . -path ./.git -prune -o -print -a \( -type f -o -type l -o -type d \) | grep '.git'   Why is this necessary. I thought that both commands should have same output. The second syntax is quite ugly. I am currently looking for a website or a tool that would allow me to compare the package state of a particular software in different Linux distributions.  For instance, which version of gimp is provided by Mint, Ubuntu, Debian Sid and Fedora 18?  An immediate interest would be to be able to avoid reinventing the wheel when packaging software (for instance re-use patches from other distros). For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , show $V= \ker T^n \oplus $ Im $T^n$ ZIP Code (US Postal Code) validation I thought people would be working on little code projects together, but I don't see them, so here's an easy one:  Code that validates a valid US Zip Code. I know there are ZIP code databases out there, but there are still uses, like web pages, quick validation, and also the fact that zip codes keep getting issued, so you might want to use weak validation.  I wrote a little bit about zip codes in a side project on my wiki/blog:    There is also a new, weird type of zip code.     I can do the javascript code, but it would be interesting to see how many languages we can get here. Why must the eigenvalue of the number operator be an integer? My apologies if this question has already been answered. Using the notation of Sakurai, we have an energy eigenket N and an eigenvalue n.  $$N | n \rangle = n | n \rangle$$  For the number operator $N = a^\dagger a $. My question is precisely why does n have to be an integer. I do understand that it must be positive definite due to the norm condition on $ a | n \rangle $:  $$\langle n | a^\dagger a | n \rangle \geq 0$$  I can see that we can show:  $$ a | n \rangle = \sqrt{n-1} | n-1 \rangle $$ and similarly for $a^\dagger | n \rangle$  Sakurai argues that the sequential operation of $a$ operators leads to the form which he does not write, but I will write here for clarity:  $$ a^k | n \rangle = \left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right) | n - k + 1 \rangle $$   Since  $$ a|0 \rangle = 0 $$  the sequence must terminate, but can only terminate if n is an integer. How can we make that jump? Is there a formal mathematical proof that the sequence $\left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right)$ will terminate if and only if n is an integer? Which LaTeX to install on Linux? Which LaTeX distribution would you recommend for using on Linux from command line? I am looking for one with most complete coverage of packages. Long boot delay on Ubuntu loading/splash screen following regular dist-upgrade on clean SSD install (18.04) Answer draft is lost if you click to sign up When writing an answer on a site where you don't have an account or are not logged in, below the answer entry textbox are options to create an account or continue as guest by providing a nickname and email. I had written an answer of several paragraphs on a network site, clicked &quot;Sign up with Google&quot; (resulting in association to my existing network-wide profile) but after doing so instead of showing my answer to be posted, only the first five or six words were retained.  I know that I had no account on this site previously because I just received the association bonus... I cannot tell whether the bug also affects totally new users or those with accounts but who are not logged in. This is a very unwelcoming user experience.  Please fix it. If the intended workflow is to click &quot;Post Answer&quot; first and associate the account afterwards, then the sign up options should not be shown until the answer is posted, and certainly not above the Post Answer button. This problem also occurs if you already have an account on the site, and click 'sign up with google' (which allows you to successfully log in).