I know Euclid's clever way of proving that there are infinitely many primes but while I was studying, I also saw a sentence on the book: "Euler-phi function can also be used to show that there are infinitely many primes." but could not figure out how. Could you please help me to understand it?  Regards
I need something explained or corrected:  In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:  Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 &lt; A \le M$:  Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$  So the $gcd(A,M) &gt; 1.$ Thus $\phi(M) = 1$ ...?   Which is not even and contradicts the theorem that $\phi(N)$ is even for $N&gt;2.$ Therefore there exists an infinite number of primes.  I get confused by the statement "Thus $\phi(M) = 1$"....  Did i possibly copy this proof down wrong? or am I missing something? Thank you in advance.   Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear.   Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it.
Converting FLV to MP4 Using FFMPEG and Preserving the Quality Im trying to use ffmpeg to convert my flv files to mp4 to play them on iOS devices but the converted video has a much worse quality than the original one.  Here is the command i use:  ffmpeg -i input.flv -ar 22050 output.mp4   I would really appreciate if someone could provide me with the best settings for flv to mp4 conversion. Argument list too long error for rm, cp, mv commands Does the War Caster feat grant ranged spell attacks in melee range without disadvantage? An opportunity attack is described this way (PHB, p. 195):     You can make an opportunity attack when a hostile creature that you can see moves out of your reach. To make the opportunity attack, you use your reaction to make one melee attack against the provoking creature.   The last benefit of the War Caster feat says (PHB, p. 170):        When a hostile creature’s movement provokes an opportunity attack from you, you can use your reaction to cast a spell at the creature, rather than making an opportunity attack. The spell must have a casting time of 1 action and must target only that creature.      Without the Crossbow Expert feat, all ranged attacks (including ranged spell attacks) made when an enemy is adjacent suffer this penalty (PHB, p. 195):     You have disadvantage on a ranged attack roll if you are within 5 feet of a hostile creature that can see you and that isn’t incapacitated.   As an opportunity attack normally grants a melee attack, does it seem reasonable to assume that the target remains at melee range for the spell attack granted by War Caster? If so, does this require ranged spell attack rolls to be made with disadvantage?  The trigger for an OA is a creature moving "out of your reach". This suggests to me that the creature is out of the 5' disadvantage zone, but it seems like that would preclude making a melee spell attack.  Do characters with the War Caster feat get the best of both worlds: being  allowed to make either a melee spell attack or a non-disadvantaged ranged spell attack? How to edit command line in full screen editor in ZSH? How to put small numbers outside the pie slices without making a mess I'm new to LaTeX and need some help with my figures. I would like to to put small numbers outside the pie slices without making a mess. Having found this thread , I tried it out, but didn't get all the way through with it. Right now, my pie looks like this:    I used the following code:  \usepackage{tikz,stackengine} \makeatletter  \tikzstyle{chart}= [legend label/.style={font={\scriptsize},anchor=west,align=left}, legend box/.style={rectangle, draw, minimum size=5pt}, axis/.style={black,thin,-&gt;}, axis label/.style={anchor=east,font={\tiny}}]  \tikzstyle{bar chart}=[ chart, bar width/.code={     \pgfmathparse{##1/2}     \global\let\bar@w\pgfmathresult}, bar/.style={thin, draw=black}, bar label/.style={font={\bf\small},anchor=north}, bar value/.style={font={\footnotesize}}, bar width=.75,]  \tikzstyle{pie chart}= [chart, slice/.style={line cap=round, line join=round, thin, draw=black}, pie title/.style={font={}}, slice type/.style 2 args={     ##1/.style={fill=##2},     values of ##1/.style={}}]  \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground}  \newcommand{\pye}[3][]{     \begin{scope}[#1]         \pgfmathsetmacro{\curA}{90}         \pgfmathsetmacro{\r}{1}         \def\c{(0,0)}         \node[pie title] at (270:1.3) {#2};         \foreach \v/\s in{#3}{             \pgfmathsetmacro{\deltaA}{\v/100*360}             \pgfmathsetmacro{\nextA}{\curA + \deltaA}             \pgfmathsetmacro{\midA}{(\curA+\nextA)/2}             \path[slice,\s] \c             -- +(\curA:\r)             arc (\curA:\nextA:\r)             -- cycle;             \pgfmathsetmacro{\d}{max((\deltaA * -(.5/50) + 1) , .5)} %%%              %\pgfmathprintnumberto{#3}{\fnum}  %\ifthenelse{\lengthtest{#3 pt &lt; 10pt}} %{\path (O) -- ++ (\midangle:\radius) %node[inner sep=0,align=left,anchor=west,rotate=\midangle](nums#3)    {~~\fnum};} %{\path (O) -- ++(\midangle:\innerpos)  %node {\scalefont{#3}\beforenumber\fnum\afternumber};}           %%%                         \begin{pgfonlayer}{foreground}                 \path \c -- node[pos=\d,pie values,values of \s]{$\v\%$} +    (\midA:\r);             \end{pgfonlayer}             \global\let\curA\nextA}     \end{scope}}  \newcommand{\legend}[2][]{     \begin{scope}[#1]         \path         \foreach \n/\s in {#2}         {++(0,-5pt) node[\s,legend box] {} +(2pt,0) node[legend label]     {\n}};     \end{scope}}      \begin{document}         \begin{center}      \begin{figure}[H]     \centering \begin{tikzpicture} [pie chart, slice type={A}{black!0}, slice type={B}{black!10}, slice type={C}{black!20}, slice type={D}{black!30}, slice type={E}{black!40}, slice type={F}{black!50}, slice type={G}{black!60}, slice type={H}{black!70}, pie values/.style={font={\small}}, scale=2]  \pye{1994}{6.1/A, 2.9/B, 0.2/C, 54.3/D, 25.2/E, 1.4/F, 6.7/G, 3.2/H} \pye[xshift=2.2cm,values of coltello/.style={pos=1.1}]% {2005}{6.8/A, 3.0/B, 0.5/C, 52.3/D, 25.6/E, 1.8/F, 7.1/G, 2.8/H} \pye[xshift=4.4cm,values of caffe/.style={pos=1.1}]% {2015}{7.2/A, 3.1/B, 0.6/C, 50.6/D, 26.5/E, 2.3/F, 7.5/G, 2.2/H}  \legend[shift={(0cm,-1.5cm)}]{{A}/A, {B}/B}  \legend[shift={(1.4cm,-1.5cm)}]{{C}/C, {D}/D}  \legend[shift={(2.8cm,-1.5cm)}]{{E}/E, {F}/F} \legend[shift={(4.2cm,-1.5cm)}]{{G}/G, {H}/H} \end{tikzpicture} \end{figure}  \end{document}   This looks quite the way I want it to, only the small numbers should be placed outside the pies.  Thanks in advance :) How do you disable anonymous login? When would you use the Builder Pattern? What are some common, real world examples of using the Builder Pattern?  What does it buy you?  Why not just use a Factory Pattern? A Riemann integrable function must have infinitely many points of continuity Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ Merge two rows in SQL Separate long math text under sum symbol into different lines? I just saw a user named: .yahoo.co.jpaqwsykcj3aulh3h1k0cy6nzs3isj  Could new users be named according to their user-number, or something other than "Unknown"? Perhaps User1234567 instead? It would be less awkward to address them in answers and comments.   Additionally, I often times have to check profiles to make sure users aren't abusing the system, since I find many of the same questions from "unknown (google)" from time to time. That, or I'll mistakenly think I've helped the user on a similar issue in the past, and assume the user has knowledge that he or she does not have. The  contains a link to the new code of conduct,  . As of 2018-08-07T05:27:38+00, . Also, I cannot find any other link to the conduct yet, neither in the help center, nor in the site navigation, which is a bummer, given that we're invited to read the new CoC:     Really, please, take a moment to read the new Code of Conduct now. It’s not long, it’s not wordy-and-legalese-y and in our long held tradition, we feel that it does a good job of assuming that the vast majority of folks that use our sites always have the best intentions at heart.    Is the code of conduct not yet published?  Edit: As mentioned in the comments,  announces the conduct and links to , so maybe it's not public yet? Side effects of changing blackboard bold font with \setmathfont Let $x = (a, b), y = (c, d) \in \mathbb{Z}^2$. What is the condition on $a, b, c, d$ so that ${x, y}$ is a basis?   My answer: $ad\neq bc$ and $gcd(a, c) = gcd(b, d) = 1$.  The first condition ensures that they aren't the same vector; the second ensures that we can actually "get" all of the integer values/lattice points.  Is this correct?  Thanks. If $F: \mathbb R^2 \to \mathbb R$ and $F_x$ (partial derivative of $F$ wrt $x$) and $F_y$ exist at $(x_0,y_0)$ then the function is continuous at that point. Is this true? If not what could be a counter-example? Efficient evaluation of functions defined by NIntegrate Why do some Unicode text messages crash the iPhone OS? Recently a new "glitch" has surfaced over the Internet over the last few weeks where if an iPhone user is sent:    effective.        Power            لُلُصّبُلُلصّبُررً ॣ ॣh ॣ ॣ  冗   via iMessage it crashes the phone OS temporarily and force restarts the receiving phone.  Why does this work? $f(x)=x/1$ is not injective? In chapter 3 of Atiyah and Macdonalds Intro to Comm Algebra there is the statement  "We also have a ring homomorphism $f:A\rightarrow S^{-1}A$ defined by $f(x)=x/1$. This is not in general injective." (Note: $A$ denotes an integral domain and $S$ a multiplicatively closed subset of $A$.)  How could this possibly not be injective? If $f(x)=f(y)$ then $x/1=y/1$ so $1x=1y$ hence $x=y$. What am I missing here? Living potato clock?