当我完成定理部分时，下一行没有正确排列。
这体现在：
\newtheorem*{them}{\textcolor{OliveGreen}{Theorem}}
\begin{them}
\textcolor{OliveGreen}{Pythagoras}\\
There no rational number whose square is 2.
\begin{pf}
Any rational number $r=\frac{p}{q}$ where $p.q\in\mathbb{Z}, q\neq 0$. in particular one can assume that p.q have no common factors.\\
(by contradiction) supposed $\exist r$ such that $r^2=l$\\
$\frac{p^2}{q^2}=2$, then\\
\begin{equation*}
\begin{split}
p^2 =2q^2 &\Longrightarrow \text{p is even} \\
&\Longrightarrow p=2k \qquad \text{for some } k\in\mathbb{Z}\\
&\Longrightarrow p^2=(2k)^2=4k^2\\
&\Longrightarrow 2q^2=4k^2\\
&\Longrightarrow q^2 = 2k^2\\
&\Longrightarrow \text{q is even}\\
&\Longrightarrow \text{both p.q are even}\\
&\Longrightarrow \text{p and q have a common factor, namely 2. which is a contradiction}.\\
\end{split}
\end{equation*}
\Longrightarrow \text{conclude that no rational number squared equals to 2}
\end{pf}
\end{them}
$\mathbb{N} =\{1,2,3,...\}$\\
$\mathbb{Z} =\{...-2,-1,0, 1,2...\}$\\