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Evaluation of Proof Exercise}}\vspace{0.3\baselineskip}
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Following  the instructions for 
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{\SMALL (linked)}
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\sq
\EA 
\vsp{8}
\framebox{\parbox[c][\totalheight][c]{490pt}{%
\begin{center} 
{\large \textbf{The Importance of the Base Step}}
\end{center}
{\normalsize
Most of the work done in an induction proof  
 is usually in proving the inductive step. This was certainly
the case in Proposition 4.2 (pg.~175). However, the basis step is an essential part of the
proof.  As this Exercise illustrates,  an induction proof is incomplete without the Base Step.}
}}   % ends \parbox and \framebox
\vsp{15}\noindent 
\sqmb
Let $P\left( n\right)$ be  \cstd{the open sentence in the variable $n\in\mathbb{N}$}
\begin{equation} 
\label{eq:Pn} 
\tag{$P\left( n\right)$}
\sum_{j=1}^n j = \frac{n^2 + n+ 1}{2} .
\end{equation}
\qp{1} 
Below the dotted line, 
complete the proof that  $P\left(  n\right)$ implies $P\left( n+1\right)$ 
for each $n\in\mathbb{N}$. 
Just delete the ``for you $\ldots$''  parts and replacing them with 
the needed algebra (add/delete  rows as needed). 
\qp{2}
Is $P\left( 1\right)$ true?  Is $P\left( 2\right)$ true? 
Using Progress Check 4.3 (see pages 177 and 511), for what $n\in\mathbb{N}$ 
is $P\left( n\right)$ true?
Explain how
this shows that the basis step is an essential part of a proof by induction.
\StartYourSolutionBelowHere
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\qp{1} 
\textbf{Solution for Part 1}.
\begin{proof} 
Let $P\left( n\right)$ be the open sentence in the variable $n\in\mathbb{N}$
\begin{equation*} 
\sum_{j=1}^n j = \frac{n^2 + n+ 1}{2}  .
\end{equation*}
Fix $n\in\mathbb{N}$. Let  $P\left( n\right)$ be true 
\cstd{think of as the inductive hypothesis}. 
Thus 
\begin{equation}
\tag{IH}
\label{eq:IH} 
\sum_{j=1}^n j = \frac{n^2 + n+ 1}{2} .
\end{equation} 
We shall show that $P\left( n+1 \right)$ is true 
\cstd{think of as the inductive conclusion}, i.e., 
we shall show that 
\begin{equation}
\tag{IC}
\label{eq:IC} 
\sum_{j=1}^{n+1} j = \frac{\left( n+1\right)^2 + \left( n+1\right) + 1}{2} .
\end{equation}

Using \eqref{eq:IH} and algebra we get 
\begin{align*}
\label{eq:TheCalulation}
\sum_{j=1}^{n+1}  j ~
&= 
\left[ \sum_{j=1}^n j \right] + \left( n+1\right) 
\intertext{and by  \eqref{eq:IH} we get} 
&= 
\left[ \frac{n^2+n+1}{2} \right] + \left( n+1\right) 
\intertext{and now by algebra \cstd{remember, look at where you are going for guidance on what algebra to do next}}
&= 
\textnormal{for you ... getting a common demoninator is a good start}
\\
&= 
\textnormal{for you  ... now keep doing simple algebra ...} 
\\
&= 
\textnormal{for you ... use as many lines as you need ... } 
\\
&= 
\textnormal{for you ... until you get the Right Hand Side of \eqref{eq:IC} ... } 
\\
&= 
\frac{\left( n+1\right)^2 + \left( n+1\right) + 1}{2} .
\end{align*}
We have just show that \eqref{eq:IC} holds.  

Thus, for each $n\in\mathbb{N}$, 
if $P\left( n\right)$ is true then $P\left( n+1\right)$ is true. 
\end{proof} 



\qp{2}
\textbf{Solution for Part 2}. 


DELETE  this whole line and then put your answer to part 2  here.

 




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