\documentclass[11pt]{amsart} \usepackage{geometry} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \geometry{a4paper} \newtheorem*{theorem*}{Theorem} \newtheorem*{lemma*}{Lemma} \newtheorem*{proposition*}{Proposition} \title{$$\sqrt{2}$$ is not a rational number} \author{Minqi Pan} \date{\today} \begin{document} \maketitle \begin{lemma*} $\forall n\in \mathbb{N}$, $n^2$ is even implies that $n$ is even. \end{lemma*} \begin{proof} Suppose that $n$ is odd, then $n=2k+1$ for some $k\in \mathbb{N}$. Thus, $n^2=4k^2+4k+1=2(2k^2+2k)+1$ which means $n^2$ is odd. The lemma thus holds by contraposition. \end{proof} \begin{proposition*} $$\sqrt{2}$$ is not a rational number. \end{proposition*} \begin{proof} Suppose that $$\sqrt{2}$$ is a rational number, then $\sqrt{2}=\frac{m}{n}$ for some $m\in \mathbb{N}$, $n\in \mathbb{N}$ and $(m, n)$ are co-prime integers. Also, $2=\frac{m^2}{n^2}$ which means that $$\label{1} m^2=2n^2$$ thus $m^2$ is even. And by Lemma, $m$ is even. Thus there exists $k\in \mathbb{N}$ such that $$\label{2} m=2k$$ Now substitute equation~\ref{2} into equation~\ref{1}, $4k^2=2n^2$ That is, $2k^2=n^2$ Thus $n^2$ is even. By Lemma, $n$ is even. Thus $n$ and $m$ are both even, contradicting the proposition that $(m, n)$ are co-prime integers. \end{proof} \end{document}