deletions | additions       diff --git a/begin_itemize_item_textbf_sqrt__.tex b/begin_itemize_item_textbf_sqrt__.tex  index 969bab4..d26fcf7 100644  --- a/begin_itemize_item_textbf_sqrt__.tex  +++ b/begin_itemize_item_textbf_sqrt__.tex 
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\[(\frac{p}{q})^2 = 2\]  \[\frac{p^2}{q^2} = 2 \]  \[ p^2 = 2q^2\] where p^2, 2, q^2 are integers  \\We have  $2|(2q^2)$ so $2|p^2$ will equal $2|p * p$ Since 2 is prime $2|p$.  \\Thus, $p = 2p_1$, for some integer p_1.  \[ (2p_1)^2 = 2q^2\] 
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\[(\frac{p}{q})^2 = 5\]  \[\frac{p^2}{q^2} = 5 \]  \[ p^2 = 5q^2\] where p^2, 5, q^2 are integers  \\We have  $5|(5q^2)$ so $5|p^2$ will equal $5|p * p$ Since 5 is prime $5|p$.  \\Thus, $p = 5p_1$, for some integer p_1.  \[ (5p_1)^2 = 5q^2\] 
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\[(\frac{p}{q})^2 = k\]  \[\frac{p^2}{q^2} = k \]  \[ p^2 = kq^2\] where p^2, k, q^2 are integers  \\We have  $k|(kq^2)$ so $k|p^2$ will equal $k|p * p$ k has to be prime so $k|p$.  \\Thus, $p = kp_1$, for some integer p_1.  \[ (kp_1)^2 = kq^2\]