\documentclass[letterpaper,11pt]{report}

\input{../share/100B.tex}

\begin{document}

\claim{$p^2 = 2$ has no solutions for $p\in\Q$}

\proof{Suppose $\exists p\in \Q \st p^2 = 2$ Then $p=\frac{a}{b}, a,b
  \in \N$ Furthermore, assume $a$ and $b$ have no common factors.
  \begin{eqnarray*}
        \left(\frac{a}{b}\right)^2 = 2 \\
        a^2 = 2b^2 \\
        2|a \Rightarrow a=2k,k\in\N \\
        (2k)^2 = 2b^2 \Rightarrow 2k^2=b^2 \\
        2|b
  \end{eqnarray*}
  So $2|b$ and $2|a$, violating the assumption they share no common
  factors}

\end{document}
