A monoid where square of all elements are 1 is abelian The following problem gives me a very hard time:     Let $M$ be a monoid with $a^2 = 1$ for $a \in M$. Show that $M$ is abelian.   It looks so simple as a monoid only needs to be associative and must have a neutral element (here $1$). So there are not much things to try. However, after some hours of trying I have to admit that I don't know what to try next.  Maybe somebody of you can give me a hint.  Kind regards!
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.
Why can't energy be created or destroyed? Undefined control sequence error on \fancy@reset, \f@ncyolh and \f@ncyorh when using newlfm class I updated to TeXLive 2017 and couldn't get newlfm.cls to work without errors.  Here is a minimal example:  \documentclass{newlfm} \begin{document} \begin{newlfm} \end{newlfm} \end{document}   LaTex is throwing undefined control sequence errors on \fancy@reset, \f@ncyolh, and \f@ncyorh.  What causes the errors?  What am I missing? Can I install any iOS version I wish by picking and choosing which firmware to restore? Why are so many forces explainable using inverse squares when space is three dimensional? Euclid Lemma proof reasoning This question is more concerned with understanding the reasoning behind mathematical proving rather than explaining this specific proof. I understand why this proof works.   This is the proof of Euclid Lemma from Wikipedia:      This   states that if $x$ and $y$ are relatively prime integers (i.e. they   share no common divisors other than 1) there exist integers $r$ and   $s$ such that      $rx+sy=1$. Let $a$ and $n$ be relatively prime, and assume that   $n|ab$. By Bézout's identity, there are $r$ and $s$ making      $rn+sa=1$. Multiply both sides by $b$:      $rnb+sab=b$. The first term on the left is divisible by $n$, and the   second term is divisible by $ab$, which by hypothesis is divisible by   $n$. Therefore their sum, $b$, is also divisible by $n$.   I am mostly concerned with this statement:      Multiply both sides by $b$:   My question is why this multiplication by $b$ happens. Is it because the person proving this observed that multiplying the equation by $b$ he would get a sum of two multiples of $n$? Or is there a mathematical rule by which we know that the next step is multiplying by $b$?  I am aware that this question might seem trivial to people here, but I just want to make sure. A word to describe "work that is ready to be done" Say I have the following CSS and HTML code:    #header {   height: 150px; } &lt;div id="header"&gt;   &lt;h1&gt;Header title&lt;/h1&gt;   Header content (one or multiple lines) &lt;/div&gt;     The header section is fixed height, but the header content may change.  I would like the content of the header to be vertically aligned to the bottom of the header section, so the last line of text "sticks" to the bottom of the header section.  So if there is only one line of text, it would be like:   ----------------------------- | Header title | | | | header content (resulting in one line) -----------------------------   And if there were three lines:   ----------------------------- | Header title | | header content (which is so | much stuff that it perfectly | spans over three lines) -----------------------------   How can this be done in CSS? Behaviour of return statement in catch and finally How to check whether a JavaScript variable defined in cross-browser way?  I ran into this problem when writing some JavaScript utilizing FireBug logging. I wrote some code like below:  function profileRun(f) {     // f: functions to be profiled     console.profile(f.constructor);     f();     console.profileEnd(f.constructor); }   It works fine in FireFox/FireBug, but it reports error in IE8 RC1. So, I'd like to do some checking whether console variable exists in the execution environment.  Below code works fine in FireFox, but not in IE8 RC1.   function profileRun(f) {     if (console != undefined) {         console.profile(f.constructor);     }      f();      if (console != undefined) {         console.profileEnd(f.constructor);     } }   However, if I do it this way. It works in IE8 RC1. Why?   function profileRun(f) {     if (window.console != undefined) {         console.profile(f.constructor);     }      f();      if (window.console != undefined) {         console.profileEnd(f.constructor);     } }   Is there any cross-browser way to check it? I have a Google Apps email (me@business.com) set to auto-forward all mail to my personal Gmail (me@gmail.com).  I have added the me@business.com address to my personal Gmail so I can send mail as me@business.com from within my personal email account.  It's a fairly common way to consolidate email accounts.  Here's my problem.  I want me@gmail.com to filter all emails being forwarded from me@business.com and apply the Business label.  So I tried it two different ways:  A) Use an Alias:   me@business.com forward to me+business@gmail.com Create filter "To:me+business@gmail.com" in me@gmail account Have filter apply Business label   B) Use Googlemail.com address:   me@business.com forward to me@googlemail.com Create filter "To:me@googlemail.com" in me@gmail account Have filter apply Business label   However, neither of these filters work.  I've tested it using a third email address, me@yahoo.com.  Whenever I send an email from me@yahoo.com to me@business.com, the email is properly forwarded to me@gmail.com, and I can reply to me@yahoo.com using the me@business.com address.  But Gmail does NOT apply the Business label.  This is case using either filtering method A or B.  But if I send an email from me@yahoo.com DIRECTLY to me+business@gmail.com or me@googlemail.com, the filter works properly and applies the label.  So there must be something going on with the forwarding of me@business.com to me@gmail.com.  Does anyone know how to get this working? Are PHP short tags acceptable to use? Here's the information :     There are four different pairs of   opening and closing tags which can be   used in PHP. Two of those, &lt;?php ?&gt;   and &lt;script language="php"&gt; &lt;/script&gt;,   are always available. The other two   are short tags and ASP style tags, and   can be turned on and off from the   php.ini configuration file. As such,   while some people find short tags and   ASP style tags convenient, they are   less portable, and generally not   recommended.   In my experience most servers do have short tags enabled. Typing  &lt;?=   is far more convenient than typing  &lt;?php echo    The programmers convenience is an important factor, so why are they not recommended? CodeIgniter: A Class/Library to help get meta tags from a web page? Removing all fraction symbols like “¼” and “½” from a string Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$  such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...   My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$  Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.   Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon &gt; 0 ,\exists \delta_1 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| &lt; \epsilon  $$  Since this statement must be true for all $\epsilon &gt; 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus  $$\forall \epsilon &gt; 0, \exists \delta_2 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_2 \implies |g(x)-g(x_0)| &lt; \epsilon $$  and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* &gt; 0, \forall \delta&gt;0: \exists x \in  \mathbb{R}, |x-x_0| &lt; \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$  My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon &gt; 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*&gt;0$, so statement 1 and 2 are still true for said $\epsilon$  I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.  Let $|x-x_0|&lt; \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| &lt; - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| &lt;0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation. Transit Visa in Hong Kong for 15 hour layover between India and USA? Find a recurrence relation for the number of $n$-digit binary sequences with no pair of consecutive $1$s Find a recurrence relation for the number of $n$-digit binary sequences with no pair of consecutive $1$s.  I know my base case:$$a_1 = 2$$ for either $1$ or $0$. Normally to construct the recurrent relation I would say fill in $n-1$ spaces then add a $1$ or $0$, which is simply:$$a_n = 2\cdot a_{n-1}$$However, I don't believe this is right, because this doesn't take into account that there can't be any consecutive $1$s.  Even when I use the base case of $2$ digits:$$a_2 = 3$$for either $01$, $10$, or $00$, I still get the recurrence relation of:$$a_n = 3\cdot a_{n-2}$$which also does not account for $01$ and $10$ occurring consecutively.   So how would I express this? How do I debug Node.js applications? Writing string at the same position using Console.Write in C# 2.0 $f(f(x)f(y))+f(x+y)=f(xy)$ Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$. Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)