%EXERCISE THREE
          
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\begin{center}
	\textbf{Exercise Three: Proclamations and Proofs}
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\begin{thm}
The product of any two even integers is an even integer.
\end{thm}

\begin{proof} 
Suppose $m$ and $n$ are even integers.

By the definition of even, $m=2j$ and $n=2k$ where $j$ and $k$ are integers.

Therefore, $mn=(2j)(2k)=2(2jk)$ by basic properties of multiplication. 

Since the integers are closed under multiplication, $jk$ is also an integer.

And this same property of closure also tells us that $2jk$ is an integer. 

So, $mn$ is twice another integer ($2jk$), meaning $mn$ is even, as desired.

\end{proof}

\begin{dfn}
An integer $k$ is \emph{odd} if $k=2j+1$ where $j$ is an integer.\\
\end{dfn}

\begin{thm}
The sum of any two odd integers is an even integer.
\end{thm}

\begin{proof}
Suppose $m$ and $n$ are odd integers.

By the definition of odd, $m=2j+1$ and $n=2k+1$ where $j$ and $k$ are integers.

Therefore, $m+n=(2j+1)+(2k+1)=2j+2k+2=2[(j+k)+1])$ by basic properties of 
addition and multiplication. 

Since the integers are closed under addition, $j+k$ is also an integer. 

And this same property of closure also tells us that $(j+k)+1$ is an integer. 

So, $m+n$ is twice another integer [$(j+k)+1$], meaning $m+n$ is even, as desired.
\end{proof}

\begin{thm}
Every even integer greater than two is the sum of two primes.
\end{thm}

This is the famous Goldbach Conjecture. As of this writing, it is unproven.
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