[Tex/LaTex] Why are the inputs invalid in math mode? Why are the words running together

math-mode

Please help. I can't figure out why I keep getting a message saying "command \textgreater invalid in math mode on line.." I get the same message for textless. Also, does this have anything to do with why my words are running together? Please help.

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\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{hyperref}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

% set page and text layout
\linespread{1.8}
\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin =  0.1 in
\evensidemargin = 0.1 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in

% set theorem numbering
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}

% header information
\title{F18-311 Writing Project 1}
\author{kuyguk}
\date{\today}

\begin{document}

\maketitle

\section{The Division Algorithm}

% Don't worry that the numbers won't match up exactly like in the textbook.

\begin{theorem} (Division Algorithm) Let a and b be integers, with b \textgreater 0. Then there exist unique integers q and r such that \[a=bq+r\] where 0 $\leq r \leq b$.

\end{theorem}

\begin{proof}
Existence of q and r. Let 

\[S = a - bk : k \in \mathbb{Z} and a - bk \geq 0 .\] 
\newline $If 0 \in S$, then then b divides a, and we can let q=a/b and r=0. If $0 \notin S$, we can use the Well-Ordering Principle. We must first show that S is nonempty. If $a - b * 0 \in S.$ If a \textless 0, then a - b (2a) = a (1-2b) $\in$ S. Therefore, a = bq + r, r $\leq$ 0. Then \[ a - b (q + 1) = a - bq - b = r - b \textgreater 0. \]
\newline Since 0 $\notin$ S, r $\neq$ b and so r \textless b.

Uniqueness of q and r. Suppose there exist integers r, $r^\prime$, q, and $q^\prime$ such that \[a = bq + r, 0 \leq r \textless  b and a = bq^\prime + r^\prime , 0 \leq r^\prime \textless b.\] 

\end{proof}

Best Answer

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{hyperref}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

% set page and text layout
\linespread{1.8}
\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin =  0.1 in
\evensidemargin = 0.1 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in

% set theorem numbering
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}

% header information
\title{F18-311 Writing Project 1}
\author{kuyguk}
\date{\today}

\begin{document}

\maketitle

\section{The Division Algorithm}

% Don't worry that the numbers won't match up exactly like in the textbook.

\begin{theorem} (Division Algorithm) Let 
% not a and b
$a$ and $b$ be integers, with 
% not \textgreater
$b > 0$. Then there exist unique integers $q$ and $r$ such that \[a=bq+r\] where 
% the whole expression in math 0 $\leq r \leq b$.
$0 \leq r \leq b$.

\end{theorem}

\begin{proof}
Existence of $q$ and $r$. Let % never leave a blank line before display math
\[S = a - bk : k \in \mathbb{Z}
% and in text
\text{ and }a - bk \geq 0 
\text{.}\] 


% avoid forced line breaks\newline

$If 0 \in S$, then then $b$ divides $a$, and we can let 
% whole expression in math
$q=a/b$ and $r=0$. If $0 \notin S$, we can use the Well-Ordering Principle.
We must first show that $S$ is nonempty. If 
% . not in math
$a - b * 0 \in S$.
%whole expressions in math
 If $a < 0$, then $a - b (2a) = a (1-2b) \in S$.
Therefore, $a = bq + r$, $r \leq 0$. Then
\[ a - b (q + 1) = a - bq - b = r - b > 0. \]
Since $0 \notin S$, $r \neq b$ and so $r < b$.

Uniqueness of $q$ and $r$. Suppose there exist integers $r$, $r'$,
$q$, and $q'$ such that \[a = bq + r, 0 \leq r <  b \text{ and } a = bq' + r' , 0 \leq r' M b\text{}.\] 

\end{proof}

\end{document}
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