deletions | additions       diff --git a/begin_itemize_item_textbf_sqrt__.tex b/begin_itemize_item_textbf_sqrt__.tex  index 04b1fac..eb8fe6a 100644  --- a/begin_itemize_item_textbf_sqrt__.tex  +++ b/begin_itemize_item_textbf_sqrt__.tex 
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Since $2p_1 = p$  \[2|p^2 = 2|q^2\]  \[2|p = 2|q\]  $\therefore $\therefore$  By 2|p$ $2|p$  and $2|q$ $gcd(p,q) \ne 1$ Proof by contradiction.