All known proofs that the circle cannot be squared are based on Lindemann's theorem that $\pi$ is not analgebraic number.  But this seems to be a case of using an atomic bomb to kill a fly.    What really has to be proved is that $\pi$ is not the root of an algebraic equation with rational coefficients whose degree is a power of $2$.  Lindemann proves it for all degrees, which is much much more than what is needed for the classical construction problem.    It is not too hard to prove that $\pi$ is not the root of a quadratic equation, and perhaps some clever induction argument could carry the day.  The point is that a direct proof for degree $2^n$ has not only never been published, it does not even seem to have been noticed that this much less powerful result is all that is needed.
Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass?  Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients?  The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof?  I just want a direct proof that $\pi$ is not constructible without appealing to transcendence.
Is there an idiom available, that is exactly opposite to "Cake walk" or "Child's play"? If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. A scene I made is not rendering, at all. This grey screen stays the same and nothing changes.  Here are my settings:  [ Do we need inertial frames in Lagrangian mechanics? How to have a one-to-many relationship with a privileged child? "Most importantly" or "more importantly"? Which of the following two sentences is more correct?   "A picture says a thousand words, more importantly in a fraction of a second"    OR   "A picture says a thousand words, most importantly in a fraction of a second"   Nothing important, just my own quotes! Do also suggest, if I can put it in a better way to express what I want to. Missing SOAP action header Im trying to get the session ID of another user in the same org by sending the following request by using the data from my enterprise wsdl and its an developer edition.  &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:urn="urn:enterprise.soap.sforce.com"&gt;    &lt;soapenv:Header&gt;       &lt;urn:organizationId&gt;xxxxxxx&lt;/urn:organizationId&gt;    &lt;/soapenv:Header&gt;    &lt;soapenv:Body&gt;       &lt;urn:login&gt;          &lt;urn:username&gt;xxxxxxxxx&lt;/urn:username&gt;          &lt;urn:password&gt;xxxxxxxxx&lt;/urn:password&gt;       &lt;/urn:login&gt;    &lt;/soapenv:Body&gt; &lt;/soapenv:Envelope&gt;   But when i send this request i get      Missing SOAP ACTION header  but in my wsdl i dont have the action ,its blank    how to overcome this problem ?I got totaly struck over here since there is no action though its asking for action header.Following method handles the http request.   public string   processrrequest(string ep,string message,string method) {      HttpRequest req = new HttpRequest();     req.setEndpoint(ep);     req.setMethod(method);     req.setHeader('Content-Type', 'text/xml');     req.setHeader('SOAPAction','');     req.setBody(message);     system.debug('Request::'+req.getbody());     Http http = new Http();     HttpResponse res =http.send(req);     //Dom.Document doc  = res.getbodyDocument();     system.debug('Response::'+res.getbody());     system.debug('Response2::'+res.getStatus());     Dom.Document doc= new Dom.Document();     doc=res.getBodyDocument();     system.debug('doc::'+doc);     return 'hai'; } Can I see why a package is installed? I had internet connection, but when I upgraded my VGA and restarted my PC then everything for eth0 has gone.     When I used ifdown:     ifdown: interface eth0 not configured   Is there something I could do?    EDIT: Content of /etc/network/interfaces    auto lo iface lo inet loopback   When I ifconfig I don't have these lines:    inet addr:192.168.1.5  Bcast:192.168.1.255  Mask:255.255.255.0 inet6 addr: fe80::219:5bff:fe5e:a5e/64 Scope:Link   In the Terminal typed:  sudo gedit /etc/network/interfaces   and then added these lines in the bottom:  auto eth0 iface eth0 inet dhcp   then I typed:    sudo ifdown eth0 sudo ifup eth0   when I send the second command I received the below message:    No DHCPOFFERS received No working leases in persistent database - sleeping   Any help? Prove $x^3 + 3 =4y(y+1)$ has no integer solutions Help Understanding How Long It Will Take To Heat My Wire Can I combine extension tubes and macro filters at once to give better macros? Determining the characteristic of a field If $\mathbb{F}=\{ 0,1,x,y\}$ is a field with four elements, why is the characteristic 2?   In a field with four elements, the characteristic is 2 or 3(since it has to be a prime). Is there any way to determine precisely the characteristic of this field, and in general any finite field? Vertical spaces between lines in cases environment \documentclass{article} \usepackage{amsmath} \begin{document} \begin{equation*}   X_i =    \begin{cases}     \left[\begin{smallmatrix}       0 \\ 1 \\ 2     \end{smallmatrix}\right]     &amp;     \text{if $i$ is even;}     \\     \left[\begin{smallmatrix}       2 \\ 1 \\ 0     \end{smallmatrix}\right]     &amp;     \text{otherwise.}   \end{cases} \end{equation*} \end{document}   This is the LaTeX code that I wrote based on a cases environment. There's no space between the matrices on the right. What's the best way to add some vertical spaces to separate them?    p.s. This is a simplified form of my original writing, so I cannot just accept suggestions like 'use a matrix transpose to transform the column vector into a row vector.' Why are there no byte or short literals in Java? I can create a literal long by appending an L to the value; why can't I create a literal short or byte in some similar way? Why do I need to use an int literal with a cast?  And if the answer is "Because there was no short literal in C", then why are there no short literals in C?  This doesn't actually affect my life in any meaningful way; it's easy enough to write (short) 0 instead of 0S or something. But the inconsistency makes me curious; it's one of those things that bother you when you're up late at night. Someone at some point made a design decision to make it possible to enter literals for some of the primitive types, but not for all of them. Why? Consider the following figure from Faraway's Linear Models with R (2005, p. 59).    The first plot seems to indicate that the residuals and the fitted values are uncorrelated, as they should be in a homoscedastic linear model with normally distributed errors. Therefore, the second and third plots, which seem to indicate dependency between the residuals and the fitted values, suggest a different model.  But why does the second plot suggest, as Faraway notes, a heteroscedastic linear model, while the third plot suggest a non-linear model?  The second plot seems to indicate that the absolute value of the residuals is strongly positively correlated with the fitted values, whereas no such trend is evident in the third plot. So if it were the case that, theoretically speaking, in a heteroscedastic linear model with normally distributed errors  $$ \mbox{Cor}\left(\mathbf{e},\hat{\mathbf{y}}\right) = \left[\begin{array}{ccc}1 &amp; \cdots &amp; 1 \\ \vdots &amp; \ddots &amp; \vdots \\ 1 &amp; \cdots &amp; 1\end{array}\right] $$  (where the expression on the left is the variance-covariance matrix between the residuals and the fitted values) this would explain why the second and third plots agree with Faraway's interpretations.  But is this the case? If not, how else can Faraway's interpretations of the second and third plots be justified? Also, why does the third plot necessarily indicate non-linearity? Isn't it possible that it is linear, but that the errors are either not normally distributed, or else that they are normally distributed, but do not center around zero? Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$ Question: If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that: $$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$  No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help! What does $\bigcup_{i=0}^{\infty}[i, i+2]$ mean? I think I get it but I wanted to be sure before attempting the question:  Question  Describe the following set, and prove your answers correct. (Here brackets denote intervals on $\mathbb{R}$.)  $\bigcup_{i=0}^{\infty}[i, i+2]$  Does it mean: {$0, 1, 2, 3, 4, 5...,\infty$} $\cup$ {$2, 3, 4, 5, 6, 7,...,\infty$}?  If so, what is the difference between:  $\bigcup_{i=0}^{\infty}[i, i+2]$ and $\bigcup_{i=0}^{\infty}(i, i+2)$ Why doesn't logging out of Stack Overflow log me out of Gmail? Why does logging in to Stack Overflow keep me signed in to my Gmail email after logging out and regardless of the Stay signed in checkbox?  Steps to reproduce:   Browse to  and confirm you are logged out Browse to  and confirm you are logged out Click login Click Google Enter Email/Password and confirm Stay signed in is unchecked Click Sign in Click logout (up top) Click Logout (down below) Browse to  and confirm you are still logged in Click Sign out &lt;- Shouldn't clicking Logout on Stack Overflow be doing the same thing as this? Browse to  and confirm you are logged out     Answers are making sense but my frustration is that clicking Logout (twice) on StackOverflow essentially does nothing. If I proceed to click login and Gmail I am immediately authenticated because I still have an active Gmail cookie. In order to effectively logout I need to click Logout twice on Stack Overflow and then manually browse to  and Sign out. Shape of the universe?