\documentclass{amsbook} \newtheorem{theorem}{Theorem}[chapter] \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{section}{chapter} \numberwithin{equation}{chapter} \begin{document} \frontmatter \title{Minqi's Solutions to the Exercises of Walter Rudin (1976), Principles of Mathematical Analysis (3rd Edition)} \author{Minqi Pan} \date{\today} \maketitle \setcounter{page}{4} \tableofcontents \mainmatter \chapter{The Real and Complex Number Systems} \section{Exercise 1: Mixed Operations between Rational and Irrational Numbers Produce Irrational Numbers} If $r$ is rational ($r\ne 0$) and $x$ is irrational, prove that $r+x$ and $rx$ are irrational. \begin{proof} Denote $\alpha=r+x, \beta=rx$. Suppose $\alpha\in\mathbb{Q}$. Then $x=\alpha-r$ is also in $\mathbb{Q}$, because $\mathbb{Q}$ is a field that is closed with respect to additive inverse and addition. This contradicts with the presumption that $x\notin\mathbb{Q}$. Suppose $\beta\in\mathbb{Q}$. Then $x=\beta/r$ is also in $\mathbb{Q}$, because $r\ne 0$ and $\mathbb{Q}$ is a field that is closed with respect to multiplicative inverse and multiplication. This contradicts with the presumption that $x\notin\mathbb{Q}$. \end{proof} \section{Exercise 2: $\sqrt{12}\notin\mathbb{Q}$} Prove that there is no rational number whose square is $12$. \begin{lemma}\label{l1} $\forall n\in \mathbb{Z}, \forall p\in \{2, 3\}$, $n^2\equiv 0 \implies n\equiv 0 \pmod p.$ \end{lemma} \begin{proof} We check the squares of every element of $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$, the ring of integers modulo $2$. $0^2 = 0$ $1^2 = 1$ Thus $n^2=0$ implies $n=0$ in $\mathbb{Z}/2\mathbb{Z}$ by exhaustion. Similarly we check the squares of every element in $\mathbb{Z}/3\mathbb{Z}=\{0,1,2\}$, the ring of integers modulo $3$. $0^2 = 0$ $1^2 = 1$ $2^2 = 1$ Thus $n^2=0$ implies $n=0$ in $\mathbb{Z}/3\mathbb{Z}$ by exhaustion. \end{proof} \begin{lemma}\label{l2} There is no rational number whose square is $3$. \end{lemma} \begin{proof} We now show that the equation \begin{equation}\label{e1} p^2=3 \end{equation} is not satisfied by any rational $p$. If there was such a $p$, we could write $p=\frac{m}{n}$ where $m\equiv 0 \pmod{3}$ and $n\equiv 0 \pmod{3}$ are not both true. Let us assume this is done. Then \ref{e1} implies \begin{equation}\label{e2} m^2=3n^2, \end{equation} This shows that $m^2\equiv 0\pmod{3}$ and we have $m\equiv 0 \pmod{3}$ by Lemma~\ref{l1}. So $m^2\equiv 0\pmod{9}$. It follows that the right side of \ref{e2} is divisible by 9, so that $n^2\equiv 0\pmod{3}$, which implies that $n\equiv 0\pmod{3}$ again by Lemma~\ref{l1}. The assumption that \ref{e1} holds thus leads to the conclusion that both $m\equiv 0 \pmod{3}$ and $n\equiv 0 \pmod{3}$ are true, contrary to our choice of $m$ and $n$. Hence \ref{e1} is impossible for rational $p$. \end{proof} \begin{proof}[Main Proof] Suppose that there existed $p\in\mathbb{Q}$ such that $p^2=12$. Then there exists $m, n\in\mathbb{Z}$ such that $p=\frac{m}{n} \implies 12=\frac{m^2}{n^2} \implies m^2=2\times 6n^2 \implies m^2\equiv 0 \pmod{2}.$ So it follows from Lemma~\ref{l1} that $m\equiv 0 \pmod{2}.$ Therefore there exists $k\in\mathbb{Z}$ such that $m=2k$, and $4k^2=12n^2 \implies k^2=3n^2 \implies \left(\frac{k}{n}\right)^2=3.$ This contradicts with Lemma~\ref{l2}. \end{proof} \end{document}