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\handout{lec1-sup}{Supplement to Lecture 1}{September 15, 1998}

This handout contains two proofs that were presented in Lecture 1.
Each proof illustrates a common type of mathematical argument.  The
first shows that two sets are equal by proving that each contains the
other, and the second shows that $\sqrt{2}$ is irrational using a
proof by contradiction.

The proofs are presented in a stylized form which may make the
structure of the arguments more clear.  In this stylized form, an
assertion is made first, and then the justification follows in a
sequence of indented steps.

\section{DeMorgan's Law for Sets}

\begin{theorem}
If $A$ and $B$ are sets, then $\overline{A \cup B} = \overline{A} \cap
\overline{B}$.
\end{theorem}

This theorem is sometimes called DeMorgan's Law.  This name can be
confusing, however, since there is an analogous theorem involving
boolean variables instead of sets that is also called DeMorgan's Law.

The proof of the theorem uses the standard method for showing that two
sets are equal; namely, each set is shown to contain the other.

\bigskip

\begin{proof}
Goal: Show that $\overline{A \cup B} = \overline{A} \cap \overline{B}$.
\begin{enumerate}
\item $\overline{A \cup B} \subseteq \overline{A} \cap \overline{B}$
\begin{enumerate}
\item Let $x$ be an element of $\overline{A \cup B}$.
\item $x \not\in A \cup B$  by definition of complement.
\item $x \not\in A$ and $x \not\in B$ by definition of union.
\item $x \in \overline{A}$ and $x \in \overline{B}$ by definition of complement.
\item $x \in \overline{A} \cap \overline{B}$ by definition of intersection.
\item Q. E. D. by definition of set inclusion.
\end{enumerate}
\item $\overline{A} \cap \overline{B} \subseteq \overline{A \cup B}$
\begin{enumerate}
\item Let $x$ be an element of $\overline{A} \cap \overline{B}$.
\item $x \in \overline{A}$ and $x \in \overline{B}$ by definition of intersection.
\item $x \not\in A$ and $x \not\in B$ by definition of complement.
\item $x \not\in A \cup B$ by definition of union.
\item $x \in \overline{A \cup B}$ by definition of complement.
\item Q. E. D. by definition of set inclusion.
\end{enumerate}
\item Q. E. D. since two-way set inclusion implies equality.
\end{enumerate}
\end{proof}

\section{The Square Root of 2 is Irrational}

\begin{theorem}
$\sqrt{2}$ is irrational.
\end{theorem}

The following argument is a ``proof by contradiction''.  The idea is
to assume that the theorem is false (i.e., $\sqrt{2}$ is rational) and
follow the consequences of this assumption until a contradiction is
reached.  This contradiction implies that the theorem must actually be
true (i.e., $\sqrt{2}$ is irrational).

\bigskip

\begin{proof}
Goal:  Show that $\sqrt{2}$ being rational implies a false conclusion.
\begin{enumerate}
\item Assume that $\sqrt{2}$ is rational.
\item False conclusion.
\begin{enumerate}
\item There exist $a$ and $b$ such that $\sqrt{2} = a / b$ where 
\begin{itemize}
\item $a$ is an integer,
\item $b$ is a positive integer, and
\item $a$ and $b$ share no common factor
\end{itemize}
by definition of a rational number and by the reducibility of
fractions to lowest terms.
\item $2b^2 = a^2$ by algebra.
\item $a$ is even.  (See note below.)
\item There exists an integer $c$ such that $a = 2c$ by definition of
even.
\item $2b^2 = 4c^2$ by substitution.
\item $b^2 = 2c^2$ by algebra.
\item $b$ is even.  (See same note below.)
\item $a$ and $b$ share 2 as a common factor.
\item Q. E. D since (a) contradicts (h).
\end{enumerate}
\item Q. E. D. by contradiction.
\end{enumerate}
\end{proof}

Note: In two steps indicated above, there are logical jumps that might
require some consideration.  One could expand the proof to justify
these jumps more minutely.  Ultimately, however, there is an upper
bound on the level of rigor attainable in this course, since we have not
defined any axioms.  Therefore, our goal is only to write proofs where
the reasoning proceeds in small, plausible steps.
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