\include{preamble}
\setcounter{try}{16}



\begin{document}
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\centerline{\bf Math 365 -- Wednesday 2/13/19 -- Section 5.1: Proof by Induction}

\begin{try} For each of the following, outline a proof by induction. Namely, 
\begin{enumerate}[\qquad $\bullet$]
\item Define $P(n)$. 
\item Prove the base case.
\item To prepare for the inductive step, write down $P(n+1)$. 
\item Make your inductive hypothesis (fix $n \in \NN$ and assume $P(n)$), and prove $P(n+1)$. Be sure to mark where you use the IH (stands for ``Induction Hypothesis''). 
\item Make a conclusion.
\end{enumerate}
Where indicated, also write a ``final draft'' version of your proof.
\smallskip
\begin{enumerate}[(a)]
\item Show for $n \in \NN$ and $r \neq 1$
$$\sum_{i=0}^n r^i = \frac{r^{n+1} -1}{r-1}.$$
\textbf{Also give a final draft proof.}

\item For $n = 1,2,3$, calculate $\sum_{i = 1}^n 2i-1$ (the sum of the first $n$ odd numbers). Notice that in each case, $\displaystyle \sum_{i = 1}^n 2i-1 = n^2$. Show that this is true in general. 

\item Show $n^3 + 2n$ is a multiple of 3 for all $n \in \NN$. 
\textbf{Also give a final draft proof.}


\item Show $n! < n^n$ for $n>1$. 


\item Suppose $A_1, A_2, \dots A_N$ and $B_1, B_2, \dots, B_N$ are sets such that 
$A_i \subseteq B_i$ for all  $1 \leq i \leq N$.
Then
$$\bigcup_{i = 1}^N A_i \subseteq \bigcup_{i = 1}^N B_i.$$
\textbf{Also give a final draft proof.}

\item Suppose $A_1, A_2, \dots A_N$ and $B$ are sets. Then 
\begin{align*}
(A_1 - B) \cap (A_2 -B) &\cap \cdots \cap (A_N - B)
	= (A_1 \cap A_2\cap \cdots \cap A_N )- B.
\end{align*}

\end{enumerate}

\end{try}

\begin{try}~
\begin{enumerate}[(a)]
\item Suppose that you know that a golfer plays the first hole of a golf course with an infinite number of holes and that if this golfer plays one hole, then the golfer goes on to play the next hole. Use ``proof by induction'' reasoning to explain why this golfer must play every hole on the course.
\item Find the error in the following faulty proof that all horses are the same color.

\begin{proof}
Let $P(n)$ be the proposition that all the horses in a set of $n$ horses are the same color. First,  $P(1)$ is true since in any group of 1 horse, all the horses must be the same color. Now, fix $n \ge 1$ and assume that $P(n)$ is true, so that all the horses in any set of $n$ horses are the same color. Consider any $n+1$ horses; number these as horses $1, 2, 3, \dots, n, n + 1$. Now the first $n$ of these horses all must have the same color, and the last $n$ of these must also have the same color. Because the set of the first $n$ horses and the set of the last $n$ horses overlap, all $n + 1$ must be the same color. Thus $P (n)$ is true  for all $n \ge 1$. 
\end{proof}

 \item The following is an example of why checking the base case is important.

\smallskip

\noindent Let $P(n)$ be the statement $\displaystyle \sum_{i=1}^n i = \frac{\left(n^2 + \frac{1}{2}\right)^2}{2}$.


\begin{enumerate}[(a)]
\item Show that $P(n)$ implies $P(n+1)$. Namely, assume $\sum_{i=1}^n = \frac{\left(n^2 + \frac{1}{2}\right)^2}{2}$ for some $n \ge 1$, and \underline{use that to show} that $\sum_{i=1}^{n+1} = \frac{\left((n+1)^2 + \frac{1}{2}\right)^2}{2}$. 

%\begin{proof}
%\end{proof}

\item Check the base case, $P(1)$. 


%\begin{ans}
%\end{ans}


\item Prove that $P(n)$ is actually \underline{false for all $n \in \ZZ_{\ge 1}$}. 

%\begin{proof}
%\end{proof}

\end{enumerate}
\end{enumerate}
\end{try}



\pagebreak
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\noindent Fill in the proof by induction that 
$$ \sum_{i=1}^n i = \frac{n(n+1)}{2}.$$

\begin{proof}
\begin{enumerate}[]
\item \emph{Define $P(n)$}:  $P(n)$ is\dots\smallskip

\fbox{\begin{minipage}{6in}
\vspace{.6in}~
\end{minipage}}\smallskip

\item \emph{Basis case}: $P(1)$ is\dots\smallskip

\fbox{\begin{minipage}{6in}
\vspace{1in}~
\end{minipage}}\smallskip

\item \emph{Goal}: Fix $n \geq 1$, assume $P(n)$, and use it to prove $P(n+1)$, which is\dots \smallskip

\fbox{\begin{minipage}{6in}
\vspace{.3in}
$P(n+1)$:
\vspace{.3in}
\end{minipage}}\smallskip

\item \emph{Inductive step}:  Fix $n \geq 1$, assume 
$$ \sum_{i=1}^n i = \frac{n(n+1)}{2}.$$
Then\dots
\smallskip

\fbox{\begin{minipage}{6in}
\vspace{2in}~
\end{minipage}}\smallskip

\item \emph{Conclusion}:
\smallskip

\fbox{\begin{minipage}{6in}
\vspace{1in}~
\end{minipage}}\smallskip


\end{enumerate}
\end{proof}

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