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\begin{document} \pagestyle{empty} \begin{flushleft}
\scs{Name(s):}
\scs{Math 301 \hfill Induction, Round Two \hfill March 27, 2015}
\prob{Find the flaw(s) in the following induction proof.
\begin{proposition}
All dogs are the same color.
\end{proposition}
\begin{proof}[Proof by induction]
The proposition may be interpreted as saying that every set of $n$ dogs contains dogs of only one color.
We procede by induction on the number $n$.
\vsp
Base case: $n=1$.
This means that we have a set of just one dog.
Obviously, if there's just one dog, then all dogs in the set are the same color.
\vsp
Iductive step: let $k \in \NN$ and suppose that any set of $k$ dogs consists of dogs of just one color.
Let $A$ be a set of $k+1$ dogs.
Let $a$ be one of the dogs in $A$.
Then the set $A-\{a\}$ is a set of $k$ dogs and hence they are all the same color (by the inductive hypothesis).
Let $b$ be another dog in $A$.
Similarly, $A-\{b\}$ is a set of $k$ dogs and thus all of these dogs are the same color.
Because $a \in A-\{b\}$ and $b \in A - \{a\}$ it follows that dog $a$ and dog $b$ are the same color as all the other dogs in $A$.
Hence $A$ consists of dogs of only one color.
\vsp
Therefore, by induction, all dogs are the same color.
\end{proof}
}
\vfill
\prob{
Use induction to prove that for any non-negative integer $n$,
any set with $n$ elements has $2^n$ subsets.
\vspace{2in}
\vfill
}
\newpage
\prob{
Prove that for any $n \in \NN$, $\displaystyle
1(2) + 2(3) + 3(4) + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}. $
}
\vfill
\prob{
Prove that for any $n \in \NN$, $\displaystyle
\frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \dots + \frac{1}{n^2} \leq 2-\frac{1}{n}. $
}
\vfill
\end{flushleft}
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