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\lecture{0}{September 19, 2012}{Aleksander M\k{a}dry}{Your name here}

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\section{A Section}
Blah Blah Blah

\subsection{A Subsection}

Blah Blah Blah

\begin{lemma}\label{lem:nonnegative}
For any $a$ and $b$,
\[
(a-b)^2\geq 0
\]
\end{lemma}

\begin{proof}
The lemma follows as a square of any (real) number is non-negative.
\end{proof}


\begin{theorem}\label{thm:earth}
For any $a$ and $b$, 
\[
a^2+b^2\geq 2ab.
\]
\end{theorem}

(Normally, theorems and lemmas are less trivial than the ones here.)

\begin{proof}
Here is the proof. 
Note that
\[
a^2+b^2-2ab=(a-b)^2\geq 0,
\]
where the last inequality follows from Lemma \ref{lem:nonnegative}.

So, after rearranging the terms we get the desired inequality. 
\end{proof}

By using theorem \ref{thm:earth} with $a=\sqrt{2}$ and $b=2$, we get 
 
\begin{corollary}
$2+4\geq 4\sqrt{2}$
\end{corollary}

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\subsection{Another Subsection}

Blah Blah Blah

\subsubsection{A Subsubsection!}

Blah Blah Blah\footnote{A footnote: Blah Blah Blah.}

\section*{Another Section (without number)}

Blah Blah Blah

\subsection{A Subsection of Another Section}

Blah Blah Blah


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