\newlec

\hdc

Rewrite the following using as few English words as possible.

\begin{enumerate}
\item No multiple of 15 plus any multiple of 6 equals 100.
\vskip1em
\item Whenever three divides both the sum and difference of two integers, it also divides each of these integers.
\end{enumerate}

\vskip3em
\soln
\begin{enumerate}
\item $\forall m,n \in \mathbb{Z}, (15m + 6n \neq 100)$
\item $\forall m,n \in \mathbb{Z}, ((3 \mid (m+n) \wedge 3\mid(m-n))\Rightarrow 3 \mid m \wedge 3 \mid n)$
\end{enumerate}

\ic{This is the 10 minute mark}

\newpage

\hdc

Write the following statements in (mostly) plain English.
\begin{enumerate}
\item \small{$\forall m \in \mathbb{Z},((\exists k \in \mathbb{Z}, m = 2k) \Rightarrow (\exists \ell \in \mathbb{Z}, 7m^{2} + 4 = 2\ell))$ }
\vskip1em
\item $n \in \mathbb{Z} \Rightarrow (\exists m \in \mathbb{Z}, m > n)$
\end{enumerate}

\vskip3em
\soln
\begin{enumerate}
\item If $m$ is an even integer, then $7m^{2} + 4$ is even. 
\item There is no greatest integer. (Alternatively, for every integer, there exists a greater integer).
\end{enumerate}

\ic{This is the 20 minute mark}

\newpage

\textbf{Contrapositive}
\vskip1em
\note Proofs are not always easy to discover. Sometimes you can convert a given problem to an easier equivalent problem.
\vskip1em
\ex $7 \nmid n \Rightarrow 14 \nmid n \equiv 14 \mid n \Rightarrow 7 \mid n$
\vskip1em
\defn The \textit{contrapositive} of $H \Rightarrow C$ is $\neg C \Rightarrow \neg H$.
\vskip1em
\note $H \Rightarrow C \equiv \neg C \Rightarrow \neg H$. This follows since

\begin{align*}
H \Rightarrow C &\equiv \neg H \vee C\\
&\equiv C \vee \neg H\\
&\equiv \neg(\neg C) \vee \neg H\\
&\equiv \neg C \Rightarrow \neg H
\end{align*}
or by using a Truth table

\begin{table}[ht!]
%\caption{default}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$H$ & $C$ & $H \Rightarrow C$ & $\neg C$ & $\neg H$ & $\neg C \Rightarrow \neg H$\\
\hline
T & T & T & F & F & T\\
T & F & F & T & F & F\\
F & T & T & F & T & T\\
F & F & T & T & T & T\\
\hline
\end{tabular}
\end{center}
%\label{default}
\end{table}%
Since the third and sixth columns are equal, their headings are logically equivalent.

\ic{This is the 32-37 minute mark}

\ex Let $x \in \mathbb{R}$. Prove $x^{3} - 5x^{2} + 3x \neq 15 \Rightarrow x \neq 5$.
\vskip1em
\pf
We prove the contrapositive. Let $x=5$. Then 
\begin{align*}
x^{3} - 5x^{2} + 3x &= (5)^{3} - 5(5)^{2} + 3(5)\\
&= 5^{3} - 5^{3} + 15\\
&= 15.
\end{align*}
\qed
\vskip1em
\ex Suppose $a,b \in \mathbb{R}$ and $ab \in \mathbb{R} - \mathbb{Q}$ (the set of irrational numbers). Show either $a \in \mathbb{R} - \mathbb{Q}$ or $b \in \mathbb{R} - \mathbb{Q}$.

\vskip1em
\pf Proceed by the contrapositive. Suppose that $a$ is rational and $b$ is rational. Then $\exists k,\ell,m,n \in \mathbb{Z}$ such that $a = \tfrac{k}{\ell}$ and $b = \tfrac{m}{n}$ with $\ell,n\neq 0$. Then
\[
ab = \tfrac{km}{\ell n} \in \mathbb{Q}
\]
as required. \qed

\ic{This is the 50 minute mark.}

