This lesson is about writing correctly.

Editing poor work of someone else

In the “Schrijfgids”, you find how to write mathematics (and LaTeX) in a proper fashion. In the following exercise, you have to correct a fellow student.

Exercise 1

  1. Open ‘TeXworks’ on your computer.
  2. Paste the following code in an empty document and save it as ‘poorwork1.tex’.

    
    \documentclass{scrartcl}
    \usepackage[dutch]{babel}
    \usepackage{amsmath, amssymb, amsthm}
    % Environment voor stellingen.
    \theoremstyle{plain}                            
    \newtheorem{stelling}{Stelling}                          
    % Eigen comamndo's
    %\DeclareMathOperator{\ggd}{ggd}
    
    \begin{document}
    
    \begin{stelling}
    Er zijn oneindig veel priemgetallen.
    \end{stelling}
    
    \begin{proof}
    ${P}$ = de verzameling van alle priemgetallen.
    Stel nu: ${P}$ is eindig, zeg ${P}= \{ p_{1},\dots ,p_{r}\}$.
    Neem het getal $N = p_{1}p_{2}\cdots p_{r}+1$. $N\notin{P} \implies N$ is geen priemgetal. Laat $p > 1$ de kleinste priemdeler van $N$ zijn. 
    $p<N \implies p\in{P}$. $p$ deelt zowel $p_{1}p_{2}\cdots p_{r}$ als $N$ en daarom ook het verschil $N~-~p_{1}p_{2}\cdots p_{r}=1$.
    Dit kan alleen als $p=1$. $\bot$~{Tegenspraak}. 
    
    Conclusie: ${P}$ is niet eindig. 
    \end{proof}
    
    \end{document}
  3. Typeset the document. Adapt the code such that the exercise is written according to the rules in the “Schrijfgids”.

Exercise 2

  1. Open ‘TeXworks’ on your computer.
  2. Paste the following code in an empty document and save it as ‘poorwork2.tex’.

    \documentclass{scrartcl}
    
    \usepackage[dutch]{babel}
    \usepackage{amsmath, amssymb, amsthm}
    % Environment voor stellingen.
    \theoremstyle{plain}                            
    \newtheorem{stelling}{Stelling}                          
    % Eigen comamndo's
    %\DeclareMathOperator{\ggd}{ggd}
    
    \begin{document}
    	
    \begin{stelling}
    Het getal $\sqrt2$ is niet rationaal.
    \end{stelling}
    
    \bigskip
    
    \begin{proof}
    $\sqrt2$ rationaal. Dan geldt $\sqrt2=\frac{m}{n}$ met $m,n \in \mathbb{Z}$ en $n\not=0$ en $ggd(m,n)=1$.
    
    \bigskip
    
    $\sqrt2 = \frac{m}{n}$
    
    \bigskip
    
    $2 = (\frac{m}{n})^2 = \frac{m^2}{n^2}$
    
    \bigskip
    
    $2n^{2} = m^2$
    
    \bigskip
    $m^2$ is even, dus $m$ is even, oftewel: $m=2k, k \in \mathbb{Z}$. 
    
    \bigskip
    Substitutie van $m=2k$ geeft:
    
    \bigskip
    $2n^2 =(2k)^2 = 4k^2$
    
    \bigskip
    $n^2=2k^2$
    
    \bigskip
    $n^2$ is even, dus $n$ is even.
    
    \bigskip
    Dus: $m$ is even en $n$ is even: $ggd(m,n)\geq2$. Tegenspraak!
    
    \bigskip
    Dus $\sqrt2$ is niet rationaal
    
    \bigskip
    \end{proof}
    
    \end{document}
  3. Typeset the document. Adapt the code such that the exercise is written according to the rules in the “Schrijfgids”.