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\title{Math 335 Portfolio}
\author{Jean Marie Linhart}
\date{January 2019}
\begin{document}
\maketitle
\section{Induction Proofs}
\subsection{Ordinary Induction}
\begin{exercise} Prove, for all natural numbers $n$, that
\begin{equation} \sum_{k=0}^n k = 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}
\label{eq:Pn}\end{equation}
\end{exercise}
\begin{proof}
We prove this by induction on $n\in\N$. In the base case, $n=0$, and \eqref{eq:Pn} becomes
$$\sum_{k=0}^n k = \sum_{k=0}^0 k = 0 = \dfrac{0(1)}{2} = \dfrac{n(n+1)}{2}$$
Now, let $n>0$ be arbitrary, and assume \eqref{eq:Pn}.
We show $\displaystyle \sum_{k=0}^{n+1} k = \dfrac{(n+1)(n+2)}{2}$.
To that end note
\begin{align*}
\sum_{k=0}^{n+1} k &= \left(\sum_{k=0}^{n} k\right) + (n+1) &\mbox{(sum definition)}\\
&= \frac{n(n+1)}{2} + (n+1) &\mbox{(induction hypothesis)}
\\
&= \frac{n(n+1)}{2} + \frac{2n+2}{2} &\mbox{(common denominator)}
\\
&= \frac{n^2 +n}{2} + \frac{2n+2}{2} &\mbox{(distribute)}
\\
&= \frac{n^2 +3n + 2}{2} &\mbox{(combine like terms)}
\\
&= \frac{(n+1)(n+2)}{2} & \mbox{(factor the numerator)}\\
\end{align*}
In all cases, \eqref{eq:Pn} is true, so $\forall n\in \N$,
$\Disp \sum_{k=0}^n k = \dfrac{n(n+1)}{2}$
\end{proof}
\end{document}