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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}
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  • also you math variables should be in math so ... Let $a$ and $b$ be integers... and $r^\prime$ is better to be input as $r'$ Commented Nov 19, 2018 at 8:07
  • for the text in math: Do something like \[a = bq + r, 0 \leq r < b\ \text{and}\ a = bq^\prime + r^\prime , 0 \leq r^\prime < b.\] or \text{ and } or use the command \wedge. In any case, you might want to define a custom command so that you may switch later on. Commented Nov 19, 2018 at 8:09
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    as the command name says, those commands are meant for usage within text. Just do > and < in your formulae Commented Nov 19, 2018 at 8:14

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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 % 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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  • If $a < 0$$, then a - b (2a) = a (1-2b) \in S$. Commented Nov 19, 2018 at 8:18
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You should read some newer documentation, because there are ways to set up the margins not like that, for example, with package geometry or with KOMA-Script packages like typearea.

Also, if you want to put things in different paragraphts, just leave a blank line in between.

In any case, when something is math, it should be math. And math is written between $..$ in text and \[ .. \] in displaystyle.

\section{The Division Algorithm} \begin{theorem}[Division Algorithm] Let $a$ and $b$ be integers, with $b > 0$. 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 \[ S = a - bk : k \in \mathbb{Z} \text{ and } a - bk \geq 0. % instead of `\text{ and }` you can use `\ \text{and} \ ` or `\quad \text{and} \quad` \] 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 $a < 0$, 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 \[ a = bq + r, \ 0 \leq r < b \quad\text{and}\quad a = bq' + r' , \ 0 \leq r^' < b. \qedhere \] \end{proof}

Note that I just changed the things related to math mode. If a variable is math mode, you should put it in math mode, even if it's just a letter. And if an expression is math you have to put the whole expression in math so that the program takes care of the typesetting.

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