deletions | additions       diff --git a/begin_itemize_item_textbf_sqrt__.tex b/begin_itemize_item_textbf_sqrt__.tex  index f95939b..32bb6fb 100644  --- a/begin_itemize_item_textbf_sqrt__.tex  +++ b/begin_itemize_item_textbf_sqrt__.tex 
...
\[k|p = k|q\]  $\therefore$ By $k|p$ and $k|q$ $gcd(p,q) \ne 1$  Proof by contradiction.   \item \textbdf{Why does the proof break down for perfect squares?}  This proof breaks down for perfect squares because the square root of a perfect square lets say $\sqrt{n^2}$ will always be an integer n.  \item \textbf{Show that $\log _2(3)$ is not a rational number.}  \\