% Induction.tex
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% Subject:
%   Introduction to Induction.
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% Bruce M. Bolden
% September 17, 1998
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% Updated:  
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\begin{document}

\large

\section{Mathematical Induction}

What is {\em mathematical induction}?



\lecpb

\subsection{Terminology}

\noindent {\bf Proposition:}  A {\em proposition} is a statement
which, as given, is either true or false.

\bigskip

\noindent {\bf Theorem:}  A {\em theorem} is a proposition
which has been proved to be true about a mathematical system.

\bigskip

\noindent {\bf Proof:}  A {\em proof} is a sequence of
statements each of which is:
\begin{enumerate}
\item  true by assumption or
\item  a previously proven theorem or
\item  a previously agre-upon definition or axiom or
\item  a statement which is implied by the statements which precede it.
\end{enumerate}

\bigskip

\noindent {\bf Predicate:}  A statement which would be a
proposition except for the fact that it includes variables
whose values are not specified is called a {\em predicate}.


\lecpb

\subsection{Mathematical Symbols and Notation}

\begin{tabular}{ll}
LHS           &  Left hand side   \\
RHS           &  Right hand side  \\
$\forall$     &  for all          \\
$\rightarrow$ &  implies          \\
$\vdash$      &  proves           \\

\end{tabular}

\medskip

\noindent  Summation notation:

\[  \sum_{i=1}^{n} i = 1 + 2 + \cdots + n \]


\lecpb

\subsection{Mathematical Induction Defined}

\noindent {\bf Axiom of Mathematical Induction:}

\[  P(1), \forall k \left[ P(k) \rightarrow P(k+1) \right]
     \vdash \forall n P(n) \]

We can prove $\forall n P(n)$ provided that we can
prove two things:
\begin{enumerate}
\item  $P(1)$ is true.
\item Whenever $P(k)$ is true, $P(k+1)$ is also true.
\end{enumerate}

Part 1 of a proof by mathematical induction is usually called
the {\em basis step}, or the basis for induction, and part 2
is called the {\em induction step}.

\lecpb

The induction step is the part of the process which usually
creates the most difficulty.  In part 2 we have to prove
$ \forall k [ P(k) \rightarrow P(k+1)] $, which means that
we assume $P(k)$ to be true and then conclude that $P(k+1)$
is also true.  The assumption that $P(k)$ is true is called
the {\em induction hypothesis}.  This provokes the following
question---Aren't we assuming that the statement we are
trying to prove is true?  No!  We have only assumed that
$k$ was a number such that $P(k)$ is true, and we use that
to try to show that $P(k+1)$ must also be true.


\lecpb

\noindent {\bf Ladder Climbing Analogy}

It has been said that showing that $P(n)$ is true for a
particular $n$ is like climbing a ladder, once the two parts
of an indcution proof have been completed.  Part 1 tells us
we can get on the ladder at rung 1.  Part 2 tells us that if
we find ourselves on any rung of the ladder, we can climb up
to the next one.  So, to prove that we can reach rung 4, we 
note that according to part 1 we can get on at rung 1, and part
2 tells us that from rung 1 we can step up to rung 2 and then
to 3 and 4.


\lecpb

\subsection{Example {\theexampleNumber}}
\addtocounter{exampleNumber}{1}

Prove that the sum of the first $n$ integers is $n(n+1)/2$
for all $n$.  Symbolically this can be expressed as

\[  \sum_{i=1}^{n} i = \frac{n(n+1)}{2}\]

\noindent {\large \bf Solution}

\noindent {\bf Basis Step:}  Let $n = 1$.

\begin{eqnarray*}
  \sum_{i=1}^{1} i & = & \frac{1(1+1)}{2}  \\
%
  1 & = & \frac{1(2)}{2}  \\
%
  1 & = & 1
\end{eqnarray*}


\noindent {\bf Induction Step:}  Prove $\forall k \left[ P(k)
\rightarrow P(k+1) \right]$, by assuming

\[ P(k): 1 + 2 + \cdots + k = \frac{k(k+1)}{2}  \]

\noindent  We must get to

\[ P(k+1): 1 + 2 + \cdots + k + 1 = \frac{(k+1)(k+2)}{2}  \]

\noindent  By the induction hypothesis,

\[ 1 + 2 + \cdots + k + 1 = \frac{(k)(k+1)}{2} + k + 1  \]

\noindent  Factoring out a common factor of $k + 1$, we get

\[ 1 + 2 + \cdots + k + 1 = (k + 1)\left(\frac{k}{2} + 1
\right)  \]

\noindent  Elementary algebra gives us

\[  \frac{k}{2} + 1 = \frac{(k +2)}{2} \]

\noindent  Substituting into the previous expression yields,

\[ 1 + 2 + \cdots + k + 1 = \frac{(k+1)(k+2)}{2}  \]

\noindent as required.


\lecpb

\subsection{Example {\theexampleNumber}}
\addtocounter{exampleNumber}{1}

\[  \sum_{i=1}^{n} i = \frac{n(n+1)}{2}\]

\noindent {\bf Basis Step:}  Let $n = 1$.  Same as previous
example.

\noindent {\bf Induction Step:}  Substitute $n+1$ for $n$
into original expression.

\[  \sum_{i=1}^{n+1} i = \frac{(n+1)((n+1)+1)}{2}\]

\[  \sum_{i=1}^{n} i  + (n+1) = \frac{(n+1)(n+2)}{2}\]

\noindent  Substitute original induction hypothesis into LHS

\[  \frac{n(n+1)}{2}  + (n+1) = \frac{(n+1)(n+2)}{2}\]

\noindent  Factor out $(n + 1)$ on LHS

\[  (n+1) (1 + \frac{n}{2})   = \frac{(n+1)(n+2)}{2}\]

\noindent  Dividing both sides by $(n + 1)$,

\[  (1 + \frac{n}{2})     = \frac{(n+2)}{2}\]

\noindent  Elementary algebra $1 = \frac{2}{2}$

\[  (\frac{2}{2} + \frac{n}{2})   = \frac{(n+2)}{2}\]

\noindent  A little more elementary algebra yields,

\[  \frac{(n+2)}{2}  = \frac{(n+2)}{2}\]

\noindent as required.


\lecpb

\subsection{Example {\theexampleNumber}}
\addtocounter{exampleNumber}{1}

Prove that the sum of the first $n$ integers multiplied by 2 is
$n(n+1)$ for all $n$.  Symbolically this can be expressed as

\[  \sum_{i=1}^{n} 2i = {n(n+1)}\]

\noindent {\large \bf Solution}

\noindent {\bf Basis Step:}  Let $n = 1$.

\begin{eqnarray*}
  \sum_{i=1}^{1} 2i & = & {1(1+1)}  \\
%
  2 & = & {1(2)}  \\
%
  2 & = & 2
\end{eqnarray*}

\noindent {\bf Induction Step:}  Substitute $n+1$ for $n$
into original expression.

\[  \sum_{i=1}^{n+1} 2i = (n+1)((n+1)+1)  \]

\noindent  Use the following fact about summation notation to
simplify the LHS:

\[  \sum_{i=1}^{n+1} f(i) = \sum_{i=1}^{n} f(i) + f(n+1)  \]

\[  \sum_{i=1}^{n} 2i + 2(n+1) = (n+1)(n+2)  \]

\noindent  Substitute original induction hypothesis into LHS

\[  {n(n+1)} + 2(n+1) = (n+1)(n+2)  \]

\noindent  Factor out $(n + 1)$ on LHS,
 
\[  (n+2)(n+1) = (n+1)(n+2)  \]

\noindent as required.


\lecpb

\subsection{Example {\theexampleNumber}}
\addtocounter{exampleNumber}{1}

Prove that the sum of the first $n$ {\em odd} integers is
$n^2$ for all $n$.

\noindent {\large \bf Solution}

\noindent {\bf Basis Step:}  Let $n = 1$.

\[  \sum_{i=1}^{1} i^2 = (1)^2  \]

\noindent {\bf Induction Step:}  Substitute $n+1$ for $n$
into original expression.

Assume the sum of the first $k$ odd integers is $k^2$, and
attempt to prove that the sum of the first $k+1$ odd integers
is $(k+1)^2$.  Using summation notation:

\[  \sum_{i=1}^{k} (2i-1) = k^2  \]

\noindent  Recall

\[  \sum_{i=1}^{n+1} f(i) = \sum_{i=1}^{n} + f(n+1)  \]

So, we need to show that

\[  \forall k \left(
       \left[ \sum_{i=1}^{k} (2i-1) = k^2 \right]
           \rightarrow 
       \left[ \sum_{i=1}^{k+1} (2i-1) = (k+1)^2 \right]
       \right)  
\]

\noindent  Assume that the LHS is true for a fixed value
of $k$---this is our induction hypothesis.

\noindent  Need to show that

\begin{eqnarray*}
\sum_{i=1}^{k+1} (2i-1) & = & (k+1)^2 \\
%
\sum_{i=1}^{k+1} (2i-1) & = & \sum_{i=1}^{k} (2i-1) + 2(k+1) - 1 \\
%
\sum_{i=1}^{k+1} (2i-1) & = & k^2 + 2k + 1 \\
%
\sum_{i=1}^{k+1} (2i-1) & = & (k + 1)^{2} 
\end{eqnarray*}



\end{document}


\lecpb

Useful rules:

\[  \sum_{i=1}^{n+1} f(i) = \sum_{i=1}^{n} f(i) + f(n+1)  \]
