PiAndPi2AreIrrational 2004-10-13 03:20:29 2008-01-10 13:12:04 Theorem pi % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm}{Theorem} \begin{thm} $\pi$ and $\pi^2$ are irrational. \end{thm} \begin{proof} For any strictly positive integer $n$ ,$x\in (0,1)$ we define: $$f=f(x)=\frac{x^n(1-x)^n}{n!}=\frac{1}{n!}\sum_{m=n}^{2n}c_mx^m$$ where $c_m$ are integers. For $0<x<1$ we have \begin{equation} \label{firseq} 0<f(x)<\frac{1}{n!} \end{equation} For a contradiction, suppose $\pi^2$ is rational, so that $\pi^2=\frac{a}{b}$, where $a,b$ are positive integers. For $x\in (0,1)$ let us define $$G(x)=b^n[\pi^{2n}f(x)-\pi^{2n-2}f''(x)+\pi^{2n-4}f^{(4)}(x)-...+(-1)^nf^{(2n)}(x)].$$ We have that $f(0)=0$ and $f^{(m)}(0)=0$ if $m<n$ or $m>2n$. But, if $n \leq m \leq 2n$, then $$f^{(m)}(0)=\frac{m!}{n!}c_m,$$ an integer. Hence $f(x)$ and all its derivates take integral values at $x=0$.Since $f(1-x)=f(x)$, the same is true at $x=1$ so that $G(0)$ and $G(1)$ are integers. We have \begin{eqnarray*} \frac{d}{dx}[G'(x)\sin{\pi x}-\pi G(x)\cos{\pi x}] &=& [G''(x)+\pi^2G(x)]\sin{\pi x} \\ &=& b^n\pi^{2n+2}f(x)\sin{\pi x} \\ &=& \pi^2a^n \sin{\pi x}f(x). \end{eqnarray*} Hence $$\pi\int_0^1a^n \sin{\pi x}f(x)dx=[\frac{G'(x)\sin{\pi x}}{\pi}-G(x)\cos{\pi x}]_0^1$$ $$=G(0)+G(1),$$ witch is an integer. But by equation \ref{firseq}, $$0<\pi\int_0^1a^n \sin{\pi x}f(x)dx<\frac{\pi a^n}{n!}<1.$$ For a large enough $n$, we obtain a contradiction. For any integer $n$, if $a^n$ is irrational then a is irrational \PMlinkexternal{(proof)}{http://planetmath.org/?op=getobj&from=objects&id=5779}, and since $\pi^2$ is irrational $\sqrt{\pi^2}=\pi$ is also irrational. \end{proof} The irrationality of $\pi$ was Proved by Lambert in 1761. The above proof is not the original proof due to Lambert. \begin{thebibliography}{99} \bibitem a G.H.Hardy and E.M.Wright \emph{An Introduction to the Theory of Numbers}, Oxford University Press, 1959 \end{thebibliography} \subsection*{See also} \begin{itemize} \item The MacTutor History of Mathematics Archive, \PMlinkexternal{A history of Pi}{http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Pi_through_the_ages.html} \item The MacTutor History of Mathematics Archive, \PMlinkexternal{Johann Heinrich Lambert}{http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Lambert.html} \item \PMlinkexternal{Irrationality proofs}{http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.html} \end{itemize}