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\theoremstyle{definition}
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\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}
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\author{Logan Axon}
\title{Problem 2 of the Worksheet from October 17}
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\begin{definition}
An integer $a$ is {\em even} if there is an integer $b$ such that $a=2b$.
The integer $a$ is {\em odd} if there is an integer $b$ such that $a = 2b+1$.
\end{definition}
\begin{proposition}
Let $a \in \ZZ$.
Then $a$ is odd if and only if $a^3$ is odd.
\end{proposition}
As with all proofs of ``if and only if'' statements we actually need to prove two things:
\begin{enumerate}
\item ($\Rightarrow$) If $a$ is odd, then $a^3$ is odd;
\item ($\Leftarrow$) If $a^3$ is odd, then $a$ is odd.
\end{enumerate}
This means coming up with two proofs, each of which can be a direct proof, a contrapositive proof, or a proof by contradiction.
In this case we prove first implication directly and the second by proving the contrapositive.
\begin{proof}
First we show that if $a$ is odd, then $a^3$ is odd.
Let $a$ be an odd integer.
By definition there is an integer $b$ such that $a = 2b+1$.
Hence $a^3 = (2b+1)^3 = 8b^3 + 12b^2 + 6b + 1 = 2(4b^3 + 6B^2 + 3b) + 1$.
The number $4b^3 + 6B^2 + 3b$ is an integer and thus, by definition, $a^3$ is odd.
Now we show that if $a^3$ is odd, then $a$ is odd.
We prove the contrapositive: if $a$ is even, then $a^3$ is even.
Let $a$ be an even integer.
By definition there is an integer $b$ such that $a = 2b$.
Hence $a^3 = (2b)^3 = 8b^3 = 2(4b^3)$.
The number $4b^3$ is an integer and thus, by definition, $a^3$ is even.
\end{proof}
In my original proof I did not use the contrapositive for the second half of the proof.
I assumed that $a^3$ was odd and tried to use that to show that $a$ must also be odd.
I applied the definition of odd and found that $a = \sqrt[3]{a^3} = \sqrt[3]{2b+1}$ for some integer $b$.
I then tried to find an expression for $\sqrt[3]{2b+1}$ in terms of $b$, but made mistakes.
My original proof is attached.
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