Cfrgtkky

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.

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}

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 quadtext{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.

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 breaksnewline



$If 0 in S$, 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 btext{}.] 



end{proof}



end{document}