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\centerline{\bf Homework 8, MORALLY Due April 1} 
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\begin{enumerate}


\item
(30 points) 
Recall what induction is:

{\bf From }
\begin{itemize}
\item
$P(0)$
\item
$(\forall n\ge 0)[P(n)\goes P(n+1)]$
\end{itemize}

{\bf You have}

$(\forall n)[P(n)]$. 

\bigskip

Induction is great for proving theorems of the form $(\forall n\in\N)[P(n)]$. 


We want to prove theorems of the form 

$$(\forall z\in \Z)[P(z)].$$

\bigskip

Give a scheme similar to the one above that will have the conclusion $(\forall z\in\Z)[P(z)]$. 


\newpage



\item 
(35 points) 
Consider the recurrence

$T(1)=10$

$T(2)=30$.

$$
(\forall n\ge 3)\biggl [T(n) = 
\biggl (\floor {\frac{n}{a}}\biggr ) + T\biggl(\floor{\frac{n}{b}}\biggr ) + cn\biggr ].$$

Find an infinite number of triples $(a,b,c)$ of POSITIVE rationals such that

$$(\forall n\ge 1)[T(n)\le 100n].$$


\newpage

\item
(35 point) 
Consider the following recurrence:

$T(0)=1$

$T(1)=6$

$T(2)=21$

$(\forall n\ge 3)[T(n) = 2T(n-1) + 4T(\floor{\sqrt{n}}) + 5n.$

Show that, for all $n\ge 1$, $T(n)\equiv 1\pmod 5$. 

{\it Hint:} Use Strong Induction.


\end{enumerate}

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