Is it possible to define a group structure on arbitrary set? It is obvious for finite sets and also sets with cardinality |Q| and |R| and also we don't know that is there other cardinality betwen them or not... what can we say a bout sets with other cardinalities?
Is it (and if yes how?) possible to define an an binary operation $*$ for an arbitrary set $M$ such that $(M,*)$ is a group?  If $M$ is finite or countable infinite this is trivial, but is it also possible for any other sets? And if yes how can you construct $*$? If no can you provide a counterexample?
Clear up confusion about the meaning of entropy So I though, and was told, that entropy is the amount of disorder in a system. Specifically the example of heat flow and it flows to maximize entropy. To me this seemed odd. This seemed more ordered to me, from two things (hot and cold) to one thing (lukewarm).   So as I have read today on wikipedia        In recent years the long-standing use of term "disorder" to discuss   entropy has met with some criticism.[20][21][22]      When considered at a microscopic level, the term disorder may quite   correctly suggest an increased range of accessible possibilities; but   this may result in confusion because, at the macroscopic level of   everyday perception, a higher entropy state may appear more homogenous   (more even, or more smoothly mixed) apparently in diametric opposition   to its description as being "more disordered". Thus for example there   may be dissonance at equilibrium being equated with "perfect internal   disorder"; or the mixing of milk in coffee being described as a   transition from an ordered state to a disordered state.      It has to be stressed, therefore, that "disorder", as used in a   thermodynamic sense, relates to a full microscopic description of the   system, rather than its apparent macroscopic properties. Many popular   chemistry textbooks in recent editions increasingly have tended to   instead present entropy through the idea of energy dispersal, which is   a dominant contribution to entropy in most everyday situations.    So what this is saying basically many people mistake the macroscopic outcome of entropy in the same manner as I have above. Seemingly two dissimilar things combining and becoming one.   They say to understand it microscopically. But this still is odd to me, take for example the particles in a box, when they come to thermal equilibrium they are said to have the most entropy they can within their system (from wiki again)     Entropy – in thermodynamics, a parameter representing the state of   disorder of a system at the atomic, ionic, or molecular level; the   greater the disorder the higher the entropy.[7]   What is disorder in the molecular sense I guess is my question. When the particles all come to thermal equilibrium they all seem to be uniform, which I equate with ordering.  I feel disorder in the molecular sense would be fast moving particles, making a mind boggling amount of interactions with each other. Such as in a hot gas, hot gas seems super disordered, but put cold gas in thermal contact with it, let them come to thermal equilibrium, and now they are more ordered!? Why am I getting a NoClassDefFoundError in Java? I've been asked the following question:  What five odd integers from the set  $\{1, 3, 5, 7, 9, 11, 13, 15\}$ that when summed together equals to $30$? Note that any integer can be used more than once.   If my limited knowledge of maths is correct, there should be no answer, as no odd number of odd integers summed together can give an even number. Theorem  Let $\{a_n\}$ and $\{b_n\}$ be convergent real sequences. Assume that there exists a $N\in\mathbb{N}$ so  $a_n\le b_n$ (eq. 1)  for all $n\ge N$. Then  $\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$.  My attempt at proving  Let $a=\lim_{n\to\infty}a_n$ and $b=\lim_{n\to\infty}b_n$. $a$ and $b$ are real numbers because the limit of a sequence of real numbers is a real number (which is a corollary in my textbook). For proof by contradiction, assume  $b&lt;a$.  Let $\epsilon=\frac{a-b}{2}$. Then we can find $N_a,N_b\in\mathbb{N}$ so  $|a-a_n|\le\frac{\epsilon}{2}$ when $n\ge N_a$, (eq. 2)  $|b-b_n|\le\frac{\epsilon}{2}$ when $n\ge N_b$. (eq. 3)  In a textbook I found that I should do the following:  Choose a $n$ so $n\ge\max(N,N_a,N_b$), then eqs. (1-3) are satisfied. Then  $a-\epsilon\le a_n+\frac{\epsilon}{2}-\epsilon$  How do I get that last thing? Is it necessary to do a proof by contradiction or can one do a direct proof? Newsletter open and click tracking use http. Can i configure it that it uses https with our certificate? Can this cause block bounces? Evaluating the integral $\mathop{\lim}\limits_{n \to \infty} \int_{-1}^{1} f(t)\cos^{2}(nt) \ dt $ Given that $f\colon [-1,1] \to \mathbb{R}$ is a continuous function such that $ \int_{-1}^{1} f(t) \ dt =1$, how do I evaluate the limit of this integral: $$\lim_{n \to \infty} \int_{-1}^{1} f(t) \cos^{2}{nt} \,dt$$   What I did was to write $\cos^{2}{nt} = \frac{1+\cos{2nt}}{2}$ and substitute it in the integral so that I can make use of the given hypothesis of $\int_{-1}^{1} f(t) \ dt =1$. So the integral becomes,   \begin{align*} \int_{-1}^{1} f(t)\cos^{2}{nt} \ dt = \int_{-1}^{1} f(t) \biggl[\frac{1+\cos{2nt}}{2}\biggr] \ dt  &amp;  \\  \hspace{3cm} = \frac{1}{2}\int_{-1}^{1}f(t) \ dt + \int_{-1}^{1} \frac{f(t)\cos{2nt}}{2} \ dt \end{align*}  But I don't really know how I can evaluate the second integral and also I can't realize as to why that integral condition on $f$ has been assumed. Moreover without assuming that condition on $f$ is it possible to evaluate this integral? If yes, then what would the answer be? I have snippets of Html stored in a table. Not entire pages, no  tags or the like, just basic formatting.  I would like to be able to display that Html as text only, no formatting, on a given page (actually just the first 30 - 50 characters but that's the easy bit).  How do I place the "text" within that Html into a string as straight text?  So this piece of code.  &lt;b&gt;Hello World.&lt;/b&gt;&lt;br/&gt;&lt;p&gt;&lt;i&gt;Is there anyone out there?&lt;/i&gt;&lt;p&gt;   Becomes:  Hello World. Is there anyone out there? In Thor: The Dark World, Odin and Loki are discussing Loki's actions during The Avengers.  Loki argues that he would have ruled Earth as &quot;a benevolent god&quot;, and Odin replies:  Odin: We are not gods! We are born, we live, we die, just as humans do.  Now that's pretty definitive.  And Loki's snarky reply (&quot;Give or take five thousand years...&quot;) would seem to confirm that Asgardians have a limited lifespan, and will eventually die of old age.1  And we know that Asgardians can die in battle, because we've seen it happen over and over with supporting characters and Asgardian soldiers. Now, I've always been more of a fan of Thor stories (like Jason Aaron's Thor: God of Thunder arc) in which Thor and the Asgardians are gods.  Straight-up, genuine, higher-level-of-existence gods, not von-Däniken-esque wannabes playing on the tired old notion of magic just being super-advanced science.  BUT, despite my preferences, this exchange would seem to confirm once and for all that in the MCU, the Asgardians are not gods, but rather were mistaken for gods because of their ram-horns full of awesome-sauce and their magnificent bulging pecks. But wait!  Watch that movie for about 30 seconds longer and we come across this exchange:  Thor: She is ill. Odin: She is mortal. Illness is their defining trait.  Wait, what?  I thought Asgardians weren't immortal?  They can die in battle, and they naturally die of old age.  In my book, that makes them long-lived, but entirely &quot;mortal&quot;.  Surely Odin isn't splitting a hair so fine as to say that resistance to disease alone makes one &quot;immortal&quot;? So, if the Asgardians of the MCU aren't actually gods (grumblegrumblegrumble...), and they can't live forever (or even particularly long in a cosmic sense), and they can be killed in battle, in what sense are they &quot;immortal&quot; at all?  1 Fun fact: adjusted for a roughly 5,000 year lifespan, 1,048-year-old Loki is currently in the tail-end of his rebellious teenage phase. MT7610U - Unable to install wifi driver Loop-cut won't go all the way around How to fix Permission Denied (Public key) error? While trying to communicate between my desktop Ubuntu 12.10 and Fedora 19 desktop, I am facing the error of Permission Denied (Public key). I am able to login through ssh from fedora to Ubuntu, but not vice-versa. I even performed telnet on both machines via port 22 and both machines are listening, but not through ssh. What to do? Any idea?  My sshd_config file is as follows:  # Package generated configuration file # See the sshd_config(5) manpage for details  # What ports, IPs and protocols we listen for Port 22 # Use these options to restrict which interfaces/protocols sshd will bind to #ListenAddress :: #ListenAddress 0.0.0.0 Protocol 2 # HostKeys for protocol version 2 HostKey /etc/ssh/ssh_host_rsa_key HostKey /etc/ssh/ssh_host_dsa_key HostKey /etc/ssh/ssh_host_ecdsa_key #Privilege Separation is turned on for security UsePrivilegeSeparation yes  # Lifetime and size of ephemeral version 1 server key KeyRegenerationInterval 3600 ServerKeyBits 768  # Logging SyslogFacility AUTH LogLevel INFO  # Authentication: LoginGraceTime 120 PermitRootLogin yes StrictModes yes  RSAAuthentication yes PubkeyAuthentication yes #AuthorizedKeysFile    %h/.ssh/authorized_keys  # Don't read the user's ~/.rhosts and ~/.shosts files IgnoreRhosts yes # For this to work you will also need host keys in /etc/ssh_known_hosts RhostsRSAAuthentication no # similar for protocol version 2 HostbasedAuthentication no # Uncomment if you don't trust ~/.ssh/known_hosts for RhostsRSAAuthentication #IgnoreUserKnownHosts yes  # To enable empty passwords, change to yes (NOT RECOMMENDED) PermitEmptyPasswords no  # Change to yes to enable challenge-response passwords (beware issues with # some PAM modules and threads) ChallengeResponseAuthentication no  # Change to no to disable tunnelled clear text passwords PasswordAuthentication yes  # Kerberos options #KerberosAuthentication no #KerberosGetAFSToken no #KerberosOrLocalPasswd yes #KerberosTicketCleanup yes  # GSSAPI options GSSAPIAuthentication no GSSAPICleanupCredentials yes  X11Forwarding yes X11DisplayOffset 10 PrintMotd no PrintLastLog yes TCPKeepAlive yes #UseLogin no  #MaxStartups 10:30:60 #Banner /etc/issue.net  # Allow client to pass locale environment variables AcceptEnv LANG LC_*  Subsystem sftp /usr/lib/openssh/sftp-server  # Set this to 'yes' to enable PAM authentication, account processing, # and session processing. If this is enabled, PAM authentication will # be allowed through the ChallengeResponseAuthentication and # PasswordAuthentication.  Depending on your PAM configuration, # PAM authentication via ChallengeResponseAuthentication may bypass # the setting of "PermitRootLogin without-password". # If you just want the PAM account and session checks to run without # PAM authentication, then enable this but set PasswordAuthentication # and ChallengeResponseAuthentication to 'no'. UsePAM yes   This is sshd_config of ubuntu system. How do I batch convert many Word documents and have them saved as [originalfilename].pdf? How to design a product table for many kinds of product where each product has many parameters If one is asking for a person, what should s/he say&mdash;1 or 2?   Could I speak to Kristina? Could I speak with Kristina? Prove $$\det(e^A) = e^{\operatorname{tr}(A)}$$ for all matrices $A \in \mathbb{C}_{n×n}$. How to remove a watermark from a PDF file? I thought this would be a simple task, but it turned out the other way.  The watermark is the very same (overlapping, but transparent) image on every single page. I created the PDF file myself (so no copyright worries here) using PDFCreator 0.9.8.  I have already tried my friend's Adobe Acrobat Pro, but it didn't work. It tries to remove it, but it can't. I tried to remove header/footer, etc., but the watermark just won't disappear.    How can I remove the watermark? Fix terminal after displaying a binary file A finite summation of double binomial coefficients I find the following identity and have checked on Mathematica, while I have no idea how to prove it:  $$\sum_{j=0}^n(-1)^{n-j}\binom{p+j}{j}\binom{n+\beta}{n-j}=\binom{p-\beta}{n}, \quad \beta&gt;-1, \quad p&gt;\beta-1.$$  It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either. Is it possible to have f.lux (software that adjusts display's color temperature) on Ubuntu 14.04? f.lux is software which adjusts temperature of your display according to time of the day. Is it possible to make f.lux work in Ubuntu 14.04? If so, maybe you can share your knowledge or point me to a guide. Let $x*y=\gcd(x,y)$ on $D_n=\{x\in \mathbb N:x |n\}$. I have to prove that $(D_n,*)$ is a commutative monoid and I have a problem when finding the identity element.  $x*e=x$  So $gcd(x,e)=x \to e=xk,k\in \mathbb N^*$ and $e \leq n$  A good $e$ is $0$ but it's not in $D_n$