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September 5, 2002} \hfill }
\begin{document}
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%THE FOLLOWING MACROS WERE USED FOR THE 1987 CDC PAPER
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\def\rarr{\rightarrow}
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\def\ckad{$C^{k}$-admissible}
\def\mysys{\Sigma=(M,U,f)}
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\def\vkk{(V,\kappa,K)}
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%\def\kk{\kappa(K)}
\def\kv{\kappa(V)}
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\def\skve{\sigma_{{\cal V},\eta}^{k}}
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\def\rtildeoveoet{\tilde \rho_{{\cal V},\eta_{1},\eta_{2}}^{1}}
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\def\delkvj{\delta_{{\cal V}_{j}}^{k}}
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\begin{center}
{\Large{\bf EXAMPLES OF EQUIVALENCE
RELATIONS}}
\end{center}




$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$
\begin{minipage}{8cm}
{\small{\it 
\baselineskip=9pt
``There's glory for you!''

``I don't know what you mean by `glory,'$\,$'' Alice said. 

Humpty Dumpty smiled contemptuously. ``Of course you don't--till I tell
you. I meant `there's a nice knock-down argument for you'!$\,$''

``But `glory' doesn't mean `a nice knock-down argument,'$\,$'' Alice objected. 

``When I use a word,'' Humpty Dumpty said in a rather a scornful tone,
``it means just what I choose it to mean--neither more nor less.''

``The question is,'' said Alice, ``whether you can make words mean
different things.''

``The question is,'' said Humpty Dumpty, ``which is to be master--that's
all.''

Alice was too much puzzled to say anything, so after a minute Humpty
Dumpty began again. ``They've a temper, some of them---particularly
verbs, they're the proudest---adjectives you can do anything with, but
not verbs--however, I can manage the whole lot! Impenetrability!
That's what I say!''

``Would you tell me, please,'' said Alice, ``what that \mbox{means}?'' 

``Now you talk like a reasonable child,'' said Humpty Dumpty, looking
very much pleased. ``I meant by `impenetrability' that we've had enough
of that subject, and it would be just as well if you'd mention what
you meant to do next, as I suppose you don't intend to stop here all
the rest of your life.''

``That's a great deal to make one word mean,'' Alice said in a
thoughtful tone.

``When I make a word do a lot of work like that,'' said Humpty Dumpty,
``I always pay it extra.''

}}
\end{minipage} 



%\bigbreak\noindent In this note I will discuss the notion of
%{\em equivalence relation}, which is one of the most important basic 
%concepts of mathematics. The discussion will be divided into three parts:
%
%\begin{itemize}
%\item[1.] why equivalence relations are important,
%\item[2.] the precise mathematical definition of ``equivalence relation''
%and related concepts,
%\item[3.] lots of examples. 
%\end{itemize}
%
%\bigbreak\noindent{\bf 3.\ Examples of equivalence relations.} 

\bigbreak\noindent{\bf EXAMPLE 1.} In how many ways can we express
the number $13$ as the sum of two natural numbers?  

The answer probably seems obvious to you. Here are all the ways to do
it:
$$
13=0+13\,,\quad 13=1+12\,,\quad 13=2+11\,,\quad13=3+10\,,
$$
$$
\quad 13=4+9\,,\quad 13=5+8\,,\quad 13=6+7\,.\quad
$$
So there are seven ways, and that's it. Right? (This seems to be what 
most students believed when this problem was first discussed in class.)

Now, suppose I were to tell you that there are other solutions, such as, for 
example $13=9+4$.  You would probably argue that is ``the same'' 
as $13=4+9$. But then I would ask  you {\bf exactly what do you mean
by ``the same''?} 

The answer is that 

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{11cm} {\Large{\bf The words ``the same''
mean whatever we make them mean (within reason), and we can choose to
make them mean different things depending on what is most convenient
to us.}}
\end{minipage}}}\end{center}

In the specific case of our problem of counting the solutions
of $13=x+y$, $x\in\nn$, $y\in\nn$, we have the following two options:

\medbreak\noindent{OPTION 1.} We define a {\em solution} of
$13=x+y$, $x\in\nn$, $y\in\nn$, to be an ordered pair 
$(x,y)\in\nn\times\nn$ such that $x+y=13$. 

\medbreak\noindent{OPTION 2.} We let $A$ be the set of all ordered pairs 
$(x,y)\in\nn\times\nn$ such that $x+y=13$, and define a {\em solution} of
$13=x+y$, $x\in\nn$, $y\in\nn$, to be an \mbox{$R$-equivalence} class, 
relative to the equivalence relation $R$ on $A$ such that, if 
$(x,y)\in A\times A$ and $(x',y')\in A\times A$, then 
$(x,y)R(x',y')$ if and only if 
$(x=x'\wedge y=y')\vee (x=y'\wedge y=x')$. (In other words, a
``solution'' in this sense is the same as a ``solution'' in the sense
of Option 1, except that now we are treating a pair $(x',y')$ obtained
from a pair $(x,y)$ by switching the order of the components as being
``the same solution'' as $(x,y)$. For example, the pair $(9,4)$ is ``the same
solution'' as the pair $(4,9)$ in the sense of Option 2, but not in the sense
of Option 1.

\medbreak\noindent{WHAT IS THE ``RIGHT'' ANSWER, THEN?} The answer is:
\begin{itemize}
\item[1.] If you define ``solution'' as in Option 1, then 
the problem of expressing $13$ as the sum of two natural numbers
has $14$ solutions. 
\item[2.] If you define ``solution'' as in Option 2, then 
the problem of expressing $13$ as the sum of two natural numbers
has $7$ solutions. 
\end{itemize}

\medbreak\noindent Which of the two is ``right''? I do not understand the
question. Most students seemed to think that the answer was 
``obviously'' Option 2. If you believe that, then PLEASE THINK IT OVER 
AGAIN.  And let me give you one example showing that Option 1 may be 
reasonable in some situations. Suppose you and a friend have lots
of one-dollar bills and no coins, and you have to pay a \$13 check.
In how many ways can you divide up the \$13 between the two of you?
Obviously, the solution where you pay \$4 and your friend pays 
\$9 is {\bf not} ``the same'' as the solution where you pay 
\$9 and your friend pays \$4$\,$!


\bigbreak\noindent{\bf EXAMPLE 2.} Suppose we want to deal with integers,
but when we consider an integer $n$ we are only interested in the remainder
of dividing $n$ by $7$. That is, as far as we are concerned, the integers
$4$, $11$, $18$, $-3$, $-10$ are all ``the same,'' in the sense 
that when we divide them by $7$ we get the same remainder.  

We cannot really 
write ``$4=11$,'' because $4$ and $11$ are not at all the same number. 
The precise mathematical way to do what we want is to introduce 
an equivalence relation, called ``congruence modulo $7$.'' 

\medbreak\noindent{DEFINITION.} Let $a$, $b$ be integers. We say that 
$a$ is {\em congruent to $b$ modulo $7$}, and write 
``$a\equiv_7 b$'', or ``$a=b\,\,{\rm mod}\,7$'', if 
$a-b$ is divisible by $7$. \proofend

\medbreak\noindent{FORMAL VERSION OF THE DEFINITION:} 

$\forall a\in\zz,\forall b\in\zz,(a\equiv_7 b\bimply 
7|a-b)$ 

\noindent or, if you prefer

$\forall a\in\zz,\forall b\in\zz,(a\equiv_7 b\bimply 
\exists k\in\zz,a-b=7.k)\,\,\,.$ 

\medbreak\noindent{PROBLEM.} Prove that $\equiv_7$ is an equivalence relation.

\medbreak\noindent{PROBLEM.} Determine the equivalence classes for 
$\equiv_7$.

\medbreak\noindent{SOLUTION.} By the Division theorem (book, page 255,
Theorem 31.1)), given any integer $a$ there exist $q,r\in\zz$ such
that $a=7q+r$ and $0\le r<$.  This means that $a\equiv_7 r$. So by
Proposition 12.10 (page 83), $[a]_{\equiv_7}=[r]_{\equiv_7}$. It
follows that if $a$ is an arbitrary integer then the equivalence class
$[a]_{\equiv_7}=[r]_{\equiv_7}$, where $r$ is one of the integers
$0,1,2,3,4,5,6$, So every equivalence class is one of the following
seven classes: $[0]_{\equiv_7}$, $[1]_{\equiv_7}$, $[2]_{\equiv_7}$,
$[3]_{\equiv_7}$, $[4]_{\equiv_7}$, $[5]_{\equiv_7}$,
$[6]_{\equiv_7}$.  Furthermore, these seven classes are all different.
(Proof: suppose $r\in\zz$, $s\in\zz$, $0\le r<7$ $0\le s<7$, and
$r\neq s$. Suppose $[r]_{\equiv_7}=[s]_{\equiv_7}$.  Proposition 12.9
implies that $r\in[r]_{\equiv_7}$and $s\in[s]_{\equiv_7}$. Since
$[r]_{\equiv_7}=[s]_{\equiv_7}$, we have $r\in[r]_{\equiv_7}$ and
$s\in[r]_{\equiv_7}$.  Then Proposition 12.11 tells us that
$r\equiv_7 s$, so $7|r-s|$. But this is impossible, because
$0<|r-s|<7$.) Hence {\bf there are seven equivalence classes,
namely,  $[0]_{\equiv_7}$, $[1]_{\equiv_7}$, $[2]_{\equiv_7}$,
$[3]_{\equiv_7}$, $[4]_{\equiv_7}$, $[5]_{\equiv_7}$, and
$[6]_{\equiv_7}$.}

\bigbreak\noindent{\bf EXAMPLE 3.} {\em (A generalization of Example 2.)}

\medbreak\noindent{DEFINITION.} Fix a
positive integer $n$.  Let $a$, $b$ be integers. We say that 
$a$ is {\em congruent to $b$ modulo $n$}, and write 
``$a\equiv_n b$'', or ``$a=b\,\,{\rm mod}\,n$'', if 
$a-b$ is divisible by $n$. \proofend

\medbreak\noindent{FORMAL VERSION OF THE DEFINITION:} 

$\forall n\in\zz,(n>0\imply(\forall a\in\zz,\forall b\in\zz,
(a\equiv_n b\bimply n|a-b)))$ 

\noindent or, if you prefer

$\forall n\in\zz,(n>0\imply(\forall a\in\zz,\forall b\in\zz,
(a\equiv_n b\bimply 
\exists k\in\zz,a-b=n.k)))\,\,\,.$ 

\medbreak\noindent{PROBLEM.} Prove that for every positive integer
$n$  $\equiv_n$ is an equivalence relation.

\medbreak\noindent{PROBLEM.} Determine the equivalence classes for 
$\equiv_n$.

\bigbreak\noindent{\bf EXAMPLE 4.} {\em Dominoes} are rectangles 
divided into two squares, on each one of which is written an
integer between $0$ and $6$. So, mathematically, we can think of
a ``domino'' as an ordered pair $(a,b)$ such that 
$a\in\{0,1,2,3,4,5,6\}$ and $b\in\{0,1,2,3,4,5,6\}$. However, we do 
not care which of the two numbers is first, so we want to treat
the pairs $(3,5)$ and $(5,3)$, for example, as being ``the same.'' 

We can do all this in a precise way by first defining a {\em pre-domino}
to be an ordered pair $(a,b)$ such that 
$a\in\{0,1,2,3,4,5,6\}$ and $b\in\{0,1,2,3,4,5,6\}$. (Therefore,
if we use $PD$ to denote the set of all pre-dominoes, then
$$
PD=\{0,1,2,3,4,5,6\}\times\{0,1,2,3,4,5,6\}\,.
$$
In particular, $|PD|=49$. That is, there are $49$ pre-dominoes.) 

We then define an equivalence relation $R_{dom}$ on the set PD of pre-dominoes
by declaring that two pre-dominoes $(a,b)$, $(c,d)$ satisfy 
$(a,b)R_{dom}(c,d)$ if either ($a=b$ and $c=d$) or ($a=d$ and $b=c$). 

The equivalence classes determined by this relation are the {\em dominoes}.

\medbreak\noindent{PROBLEM.} Prove that $R_{dom}$ 
is an equivalence relation.

\medbreak\noindent{PROBLEM.} How many dominoes are there?

\bigbreak\noindent{\bf EXAMPLE 5.} {\em (A generalization of Example 4.)}  
We would like to do for a general positive integer 
$n$ the same thing that we did in Example 4 for $n=6$.

\medbreak\noindent{DEFINITION.}  Fix a positive integer $n$.  A {\em
$n$-pre-domino} is an ordered pair $(a,b)\in\zz\times\zz$
such that  $0\le a\le n$ and $0\le b\le n$. 

We then define  an equivalence relation---called  $R_{dom,n}$---on 
the set $PD_n$ of $n$-pre-dominoes
by stipulating that two $n$-pre-dominoes $(a,b)$, $(c,d)$ satisfy 
\mbox{$(a,b)R_{dom,n}(c,d)$} if either ($a=b$ and $c=d$) or ($a=d$ and $b=c$). 

The equivalence classes determined by this relation are the {\em $n$-dominoes}.

\medbreak\noindent{PROBLEM.} Prove that $R_{dom,n}$ 
is an equivalence relation.

\medbreak\noindent{PROBLEM.} How many $n$-dominoes are there?

\bigbreak\noindent{\bf EXAMPLE 6.} Consider an ordinary chessboard.
The chessboard has $64$ squares, forming an $8\times 8$ square array.
Let $Sq$ be the set of squares, so $|Sq|=64$. Mathematically, we
can think of each square as an ordered pair of integers
between $1$ and $8$, so the set $Sq$ of squares is the 
Cartesian \mbox{product} $\{1,2,3,4,5,6,7,8\}\times\{1,2,3,4,5,6,7,8\}$. 
(The $(i,j)$ pair corresponds to the square in the $i$-row,
from left to right, and the $j$-th column, from bottom to top.)

Two squares $\alpha$, $\beta$ are {\em bishop-equivalent} if 
one can go from $\alpha$ to $\beta$ by one or several 
steps in each of which one moves diagonally. 

\medbreak\noindent{PROBLEM.} Prove that bishop-equivalence is
an equivalence relation.

\medbreak\noindent{PROBLEM.} Determine the equivalence classes.
How many equivalence classes are there?

\bigbreak\noindent{\bf EXAMPLE 7.} Consider an ordinary chessboard,
as in Example 6. A {\em knight move} consists
of moving from a square up or down two squares and then
left or right one square, or 
moving left or right two squares and then
up or down one square. 

Two squares $\alpha$, $\beta$ are {\em knight-equivalent} if 
one can go from $\alpha$ to $\beta$ by one or several 
steps in each of which one executes a knight move.

\medbreak\noindent{PROBLEM.} Prove that knight-equivalence is
an equivalence relation.

\medbreak\noindent{PROBLEM.} Determine the equivalence classes.
How many equivalence classes are there?

\bigbreak\noindent{\bf EXAMPLE 8.} (A generalization of Example 7.)  
Fix three positive integers $n$, $k$, $\ell$. 
Consider a chessboard, as in Examples 6, 7,  except
that now the board is $n\times n$.  A {\em $k$-$\ell$-knight move} consists
of moving from a square up or down $k$ squares and then
left or right $\ell$ squares, or 
moving left or right $k$ squares and then
up or down $\ell$ squares. 

Two squares $\alpha$, $\beta$ are {\em $k$-$\ell$-knight-equivalent} if
one can go from $\alpha$ to $\beta$ by one or several 
steps in each of which one executes a $k$-$\ell$-knight move.   

\medbreak\noindent{PROBLEM.} Prove that $k$-$\ell$-knight-equivalence is
an equivalence relation.

\medbreak\noindent{PROBLEM.} Determine the equivalence classes.
How many equivalence classes are there? (If you find it too difficult \
to do this for completely general $n$, $k$, $\ell$, try at least
a few examples, showing in particular that the answer isn't always 
as simple as that of Example 6.)

\bigbreak\noindent{\bf EXAMPLE 9.} {\em Constructing (or inventing, if you
prefer) the rational numbers.}  Suppose we know how to talk about
sets, and about integers, but we have not yet discovered (or ``invented,''
or ``created'') the rational numbers. How could we do it?

We could try to define a ``rational number'' to be an ordered pair
$(m,n)$ of integers such that $n\neq 0$. Then we could say that 
we will use ``$\,\,\frac{m}{n}\,\,$'' as an alternative notation for
``$(m,n)$'', and we we would have a set of objects that are of the 
desired form.  The trouble with this definition is that it would
imply that, for example, $\frac{4}{3}$ and $\frac{8}{6}$ are different 
objects, whereas we want to be  able to write
$$
\frac{4}{3}=\frac{8}{6}\,,
$$ 
i.e., to treat $\frac{4}{3}$ and $\frac{8}{6}$ as being ``the same
number.''

The correct way to do this is to introduce an equivalence relation.
To begin with, we define a {\em pre-rational number} to be an ordered
pair $(m,n)$ of integers such that $n\neq 0$. We use $Pre\qq$ to
denote the set of all pre-rational numbers. Then 
$$
Pre\qq=\{(m,n)\in\zz\times\zz:n\neq 0\}\,.
$$
We define a relation $R_{\qq}$ on $Pre\qq$ by letting
$$
(m,n)R_\qq(m',n')\bimply mn'=m'n\,,\qquad{\rm if}\quad
(m,n),(m',n')\in Pre_\qq\,.
$$

\medbreak\noindent THEOREM. $R_\qq$ is an equivalence relation.

\medbreak\noindent PROOF. We prove that $R_\qq$ is transitive.
We leave the proofs of the other two properties (reflexitivity and 
transitivity) to  you. 

\noindent We want to prove that 
$$
\forall x,\forall y,\forall z,((x\in Pre\qq
\wedge y\in Pre\qq\wedge z\in Pre\qq\wedge xR_\qq y\wedge yR_\qq z)
\imply xR_\qq z)\,.
$$

Let $x,y,z$ be arbitrary.

$\qquad$ Assume that $x\in Pre\qq$, $y\in Pre\qq$, $z\in Pre_\qq$,
$xR_\qq y$ and $yR_\qq z$. 

$\qquad$ Then we can pick $m,n,m',n',m'',n''\in\zz$ such that
$x=(m,n)$,

$\qquad$ $\qquad$ $y=(m',n')$, $z=(m'',n'')$, $n\neq 0$, $n'\neq 0$,
$n''\neq 0$.

$\qquad$ Then $mn'=m'n$ and $m'n''=m''n'$.

$\qquad$ Since $mn'=m'n$, we have $mn'n''=m'nn''$. So
$mn'n''=m'n''n$. Since 

$\qquad$ $m'n''=m''n'$, we can conclude that 
$mn'n''=m''n'n$. Since $n'\neq 0$, we 

$\qquad$ can cancel $n'$ in the previous
equation, and get $mn''=m''n$. But this 

$\qquad$ says that $xR_\qq z$. 

Therefore $(x\in Pre\qq\wedge y\in Pre\qq\wedge z\in Pre_\qq\wedge
xR_\qq y\wedge yR_\qq z)\imply xR_\qq z$. 

\noindent So $\forall x,\forall y,\forall z,((x\in Pre\qq
\wedge y\in Pre\qq\wedge z\in Pre\qq\wedge xR_\qq y\wedge yR_\qq z)
\imply xR_\qq z)$.

\noindent Therefore $R_\qq$ is transitive. \proofend

\medbreak\noindent DEFINITION. A {\em rational number} is an equivalence
class of the relation $R_\qq$ on $Pre\qq$. \proofend

\medbreak\noindent NOTATION. If $m,n\in\zz$ and $n\neq 0$, we use
$\frac{m}{n}$ to denote the $R_\qq$-equivalence class of the
pair $(m,n)$. That is,
$$
\frac{m}{n}=[(m,n)]_{R_\qq}\,.
$$

\medbreak\noindent NOTATION. We use $\qq$ to denote the set of all rational
numbers. 

\bigbreak\noindent With these notations we can write, for example, 
$$
\frac{4}{3}=\frac{8}{6}\,,
$$
because $(4,3)R_\qq(8,6)$ (since $4\times 6=3\times 8$), 
so $[(4,3)]_{R_\qq}=[(8,6)]_{R_\qq}$, i.e., $\frac{4}{3}=\frac{8}{6}$.

\medbreak\noindent PROBLEM. Complete the proof that $R_\qq$ is an equivalence
relation, by proving that $R_\qq$ is reflexive and symmetric. 

\medbreak\noindent THE ARITHMETICAL OPERATIONS FOR RATIONAL NUMBERS.
We would like to define what is meant by the sum and the product of two
rational numbers. We will do it for the sum, and leave it up to you
to define the product and to justify the definition.

To define addition of rational numbers, we let $r$, $s$ be two 
fixed but arbitrary rational numbers and seek to define 
$r+s$. Since $r$ and $s$ are rational numbers, each of of them
is an $R_\qq$-equivalence class. So we can pick integers 
$m,n,m',n'$ such that $n\neq 0\neq n'$,  $r=[(m,n)]_{R_qq}$ and 
$s=[(m',n')]_{R_qq}$ (that is, equivaently, $r=\frac{m}{n}$ and 
$s=\frac{m'}{n'}$).  We would like to define the 
sum $r+s$ to be given by the expression that we are all
used to from high-school arithmetic, where we learned
that 
$$
\frac{m}{n}+\frac{m'}{n'}=\frac{mn'+m'n}{nn'}\,.
$$
The trouble with this is the following: $r$ is the equivalence
class of lots of different pairs $(m,n)$, and 
$s$ is the equivalence class of lots of different pairs $(m',n')$.
Suppose that we choose two different pairs 
$(m_1,n_1)$, $(m_2,n_2)$ such that 
$$
r=[(m_1,n_1)]_{R_\qq}=[(m_2,n_2)]_{R_\qq}\,,
$$
and two different pairs $(m_1',n_1')$, $(m_2',n_2')$ such that 
$$
s=[(m_1',n_1')]_{R_\qq}=[(m_2',n_2')]_{R_\qq}\,.
$$
Then, using $(m_1,n_1)$ and $(m_1',n_1')$, we would conclude that 
$r+s$ is the equivalence class of the pair $(m_1n_1'+m_1'n_1,n_1n_1')$.
On the other hand,  using 
$(m_2,n_2)$ and $(m_2',n_2')$, we would conclude that 
$r+s$ is the equivalence class of the pair $(m_2n_2'+m_2'n_2,n_2n_2')$.
Suppose it turned out that these two equivalence classes
are not the same, i.e., that 
$$
\neg\,\Big((m_1n_1'+m_1'n_1,n_1n_1')R_\qq(m_2n_2'+m_2'n_2,n_2n_2')\Big)\,,
$$
or, if you prefer, that 
$$ 
\Big[(m_1n_1'+m_1'n_1,n_1n_1')\Big]_{R_\qq}\neq
\Big[(m_2n_2'+m_2'n_2,n_2n_2')\Big]_{R_\qq}\,,
$$
or 
$$
\frac{m_1n_1'+m_1'n_1}{n_1n_1'}\neq \frac{m_2n_2'+m_2'n_2}{n_2n_2'}\,.
$$
This would mean that ``the rational number $r+s$'' 
defined using $(m_1,n_1)$  and  $(m_1',n_1')$ is not the same 
as ``the rational number $r+s$'' 
defined using $(m_2,n_2)$  and  $(m_2',n_2')$.  In that case, 
we would not have a unique, well defined sum $r+s$.

Fortunately, this does not happen, because the following theorem
is true:

\medbreak\noindent THEOREM. Let $r,s$ be rational numbers.
Let the pairs $(m_1,n_1)\in Pre\qq$, \ $(m_2,n_2)\in Pre\qq$ \ be 
\ such that 
\ $r=[(m_1,n_1)]_{R_\qq}=[(m_2,n_2)]_{R_\qq}$, \  and \ let 
\mbox{$(m_1',n_1')\in Pre\qq$}, $(m_2',n_2')\in Pre\qq$ be such that 
$s=[(m_1',n_1')]_{R_\qq}\!=\![(m_2',n_2')]_{R_\qq}$. Then
$$
(m_1n_1'+m_1'n_1,n_1n_1')R_\qq(m_2n_2'+m_2'n_2,n_2n_2')\,.
$$

\medbreak\noindent PROOF. The fact that 
$[m_1,n_1]_{R_\qq}=[m_2,n_2]_{R_\qq}$ implies that 
$$
m_1n_2=m_2n_1\,.
$$
The fact that 
$[m_1',n_1']_{R_\qq}=[m_2',n_2']_{R_\qq}$ implies that 
$$
m_1'n_2'=m_2'n_1'\,.
$$
We want to prove that 
$(m_1n_1'+m_1'n_1,n_1n_1')R_\qq(m_2n_2'+m_2'n_2,n_2n_2')$, i.e., that 
\begin{equation}\label{biginte}
(m_1n_1'+m_1'n_1)n_2n_2'=(m_2n_2'+m_2'n_2)n_1n_1'\,.
\end{equation}
But 
\begin{eqnarray*}
(m_1n_1'+m_1'n_1)n_2n_2'&=&m_1n_1'n_2n_2'+m_1'n_1n_2n_2'\\
&=&(m_1n_2)n_1'n_2'+(m_1'n_2')n_1n_2\\
&=&(m_2n_1)n_1'n_2'+(m_2'n_1')n_1n_2\\
&=&m_2n_2'n_1n_1'+m_2'n_2n_1n_1'\\
&=&(m_2n_2'+m_2'n_2)n_1n_1'\,,
\end{eqnarray*}
so (\ref{biginte}) holds, and our proof is complete. \proofend

\medbreak\noindent DEFINITION. Let $r,s$ be rational numbers.  Then
the {\em sum} of $r$ and $s$ is the rational number $r+s$
given by $r+s=\frac{mn'+m'n}{nn'}$,
if $(m,n)\in Pre\qq$, $(m',n')\in Pre\qq$ are such that 
$r=\frac{m}{n}$ and $s=\frac{m}{n}$.\proofend

\medbreak\noindent The definition is justified because 
the previous theorem tells us that, no matter how we choose
the members $(m,n)$, $(m'n')$ of the equivalence classes $r$, $s$,
the resulting rational number $\frac{mn'+m'n}{nn'}$ will be the same.

\medbreak\noindent PROBLEM. Use ${\bf 0}$ to denote the rational 
number $\frac{0}{1}$, i.e., the equivalence class 
$[(0,1)]_{R_\qq}$.  Prove that addition of rational numbers,
as defined above, has the following properties:

$$
\forall r\in\qq,\forall s\in\qq,r+s=s+r
$$
(commutative law of addition),
$$
\forall r\in\qq,\forall s\in\qq,\forall t\in\qq,
(r+s)+t=r+(s+t)\,,
$$
(associative law of addition),
$$
\forall r\in\qq,r+{\bf 0}=r\,,
$$
and
$$
\forall r\in\qq,\exists s\in\qq,r+s={\bf 0} 
$$
(existence of additive inverses).


\medbreak\noindent PROBLEM. Define {\em multiplication of rational numb}ers.
That is, follow the model of the definition of addition given earlier, 
to define the operation of multiplication. This will require that
you state and prove a theorem, similar to the one we stated 
and proved before we defined the sum of two rational numbers,
needed to justify your definition by showing that the rational
number $r.s$ that you defined does not depend on how you choose
members $(m,n)$, $(m',n')$ of the equivalence classes $r$, $s$. 

\medbreak\noindent PROBLEM. Use ${\bf 1}$ to denote the rational 
number $\frac{1}{1}$, i.e., the equivalence class 
$[(1,1)]_{R_\qq}$.  Using the definition of multiplication of 
rational numbers given in the previous problem, prove that 
multiplication has the following properties:
$$
\forall r\in\qq,\forall s\in\qq,r.s=s.r
$$
(commutative law of multiplication),
$$
\forall r\in\qq,\forall s\in\qq,\forall t\in\qq,
(r.s).t=r.(s.t)\,,
$$
(associative law of multiplication),
$$
\forall r\in\qq,r.{\bf 1}=r\,,
$$
$$
\forall r\in\qq,(x\neq{\bf 0}\imply\exists s\in\qq,r+s={\bf 0})
$$
(existence of multiplicative inverses),
and  
$$
\forall r\in\qq,\forall s\in\qq,\forall t\in\qq,
(r+s).t=r.t+s.t\,,
$$
(distributive law of multiplication with respect to addition).

\baselineskip=11pt

\bigbreak\noindent{\bf EXAMPLE 10}. {\bf What is a ``curve''?}   Let us talk
about curves in the plane $\rr^2$, or in three-dimensional space $\rr^3$. 

In your Calculus courses, you have been exposed to ``curves'' 
in many ways. Sometimes a ``curve'' is given by an equation;
for example, the equation $x^2+y^2=R^2$ describes a circle
$C_R$ of radius $R$ and center at $(0,0)$. Sometimes a ``curve'' is given 
by ``parametric equations''; for example, the equations 
$x=R\cos\,t$, $y=R\sin\,t$, $0\le t\le 2\pi$, are parametric
equations of $C_R$. Sometimes a ``curve'' is ``oriented,''
whatever that means; for example, the circle $C_R$ oriented counterclockwise 
and the circle $C_R$ oriented clockwise are not ``the same''
as oriented curves. The distinction between the two ``oriented
versions'' of $C_R$ matters. For example, if ${\bf F}$ is the vector 
field $-y\,{\bf i}+x\,{\bf j}$, then the integral
$\int_{C_R}{\bf F}\cdot d{\bf r}$ is equal to $2\pi R^2$ if 
$C_R$ is oriented counterclockwise, and to $-2\pi R^2$ if 
$C_R$ is oriented clockwise. So one should really be talking about
{\em integrals along oriented curves}, not just along curves. 

To make all this mathematically precise, we look first at the set
$PC(\rr^2)$ of {\em parametrized curves in $\rr^2$}. Various
mathematically interesting concepts of ``curve in $\rr^2$'' can be
obtained by considering different equivalence relations on
$PC(\rr^2)$.  (A similar thing can be done for curves in $\rr^3$.)
Here, we will just consider two such concepts, giving rise to the
notions of {\em oriented curve} and {\em unoriented curve}.

\medbreak\noindent DEFINITION.  A {\em parametrized curve}
in $\rr^2$ is a continuous $\rr^2$-valued function ${\bf r}$
defined on some closed bounded interval $[a,b]$. \proofend

\medbreak\noindent In other words, a parametrized curve is exactly a 
pair of parametric equations $x=f(t)$, $y=g(t)$, where
$f$ and $g$ are continuous real functions on an interval 
$[a,b]$. In vector form, we can write these parametric equations
as 
$$
{\bf r}(t)=f(t)\,{\bf i}+g(t)\,{\bf j}\,,\,\,\,\,a\le t\le b\,.
$$

We use $PC(\rr^2)$ to denote the set of all parametrized
curves in $\rr^2$.

Now suppose that ${\bf r}_1:[a_1,b_1]\mapsto\rr^2$ and
${\bf r}_2:[a_2,b_2]\mapsto\rr^2$ are two parametrized curves.
We will want to say that ${\bf r}_1$ and ${\bf r}_2$ are ``equivalent
up to an orientation-preserving reparametrization''
if we can go from ${\bf r}_1$ to ${\bf r}_2$, and 
viceversa, by just ``changing time without changing
the direction of time.'' Precisely,

\medbreak\noindent DEFINITION.  Let ${\bf r}_1:[a_1,b_1]\mapsto\rr^2$ and
${\bf r}_2:[a_2,b_2]\mapsto\rr^2$ be two parametrized curves
in $\rr^2$. 
We say that ${\bf r}_1$ is {\em equivalent to ${\bf r}_2$ 
up to an orientation-preserving reparametrization}
if there exist continuous functions 
$\varphi:[a_1,b_1]\mapsto[a_2,b_2]$ and $\psi:[a_2,b_2]\mapsto[a_1,b_1]$, 
such that $\varphi(\psi(t))=t$ whenever $t\in[a_2,b_2]$, 
$\psi(\varphi(t))=t$ whenever $t\in[a_1,b_1]$, 
${\bf r}_1(\psi(t))={\bf r}_2(t)$ for all $t\in[a_2,b_2]$,
$\varphi(a_1)=a_2$,  and $\varphi(b_1)=b_2$.\proofend

\medbreak\noindent THEOREM.  Equivalence up to 
an orientation-preserving reparametrization is an equivalence
relation on $PC(\rr^2)$. 

\medbreak\noindent PROOF.  I will prove that the relation is symmetric,
and leave it up to you to do the proofs of reflexitvity
and transitivity. 

Suppose that ${\bf r}_1:[a_1,b_1]\mapsto\rr^2$ is equivalent to ${\bf
r}_2:[a_2,b_2]\mapsto\rr^2$ up to an orientation-preserving
reparametrization.  We want to prove that ${\bf r}_2$ is equivalent to
${\bf r}_1$ up to an orientation-preserving reparametrization.

To prove this, what we need is to construct a pair of continuous functions 
$\Phi:[a_2,b_2]\mapsto[a_1,b_1]$, $\Psi:[a_1,b_1]\mapsto[a_2,b_2]$, 
such that $\Phi(\Psi(t))=t$ whenever $t\in[a_1,b_1]$, 
$\Psi(\Phi(t))=t$ whenever $t\in[a_2,b_2]$, 
${\bf r}_2(\Psi(t))={\bf r}_1(t)$ for all $t\in[a_1,b_1]$,
$\Phi(a_2)=a_1$,  and $\Phi(b_2)=b_1$.

We know that there exist continuous functions 
$\varphi:[a_1,b_1]\mapsto[a_2,b_2]$ and $\psi:[a_2,b_2]\mapsto[a_1,b_1]$, 
such that $\varphi(\psi(t))=t$ whenever $t\in[a_2,b_2]$, 
$\psi(\varphi(t))=t$ whenever $t\in[a_1,b_1]$, 
${\bf r}_1(\psi(t))={\bf r}_2(t)$ for all $t\in[a_2,b_2]$,
$\varphi(a_1)=a_2$, and $\varphi(b_1)=b_2$.

Take $\Psi=\varphi$ and $\Phi=\psi$. Then 
$\Psi(\Phi(t))=t$ whenever $t\in[a_2,b_2]$ (because
$\varphi(\psi(t))=t$ whenever $t\in[a_2,b_2]$), and
$\Phi(\Psi(t))=t$ whenever $t\in[a_1,b_1]$ (because
$\psi(\varphi(t))=t$ whenever $t\in[a_1,b_1]$).
Let us prove that ${\bf r}_2(\Psi(t))={\bf r}_1(t)$ for all 
$t\in[a_1,b_1]$. If $t\in[a_1,b_1]$, then $t=\psi(\varphi(t)$,
and then it follows that ${\bf r}_1(t)={\bf r}_1(\psi(\varphi(t)))
={\bf r}_2(\varphi(t))$, using the fact that 
${\bf r}_1(\psi(s))={\bf r}_2(s)$ for all $s\in[a_2,b_2]$,
with $s=\varphi(t)$.   But then 
${\bf r}_1(t)={\bf r}_2(\Psi(t))$, which is exactly what 
we wanted to prove. Finally, let us prove that 
$\Phi(a_2)=a_1$ and $\Phi(b_2)=b_1$. We have 
$\Phi(a_2)=\psi(a_2)=\psi(\varphi(a_1))=a_1$, so 
$\Phi(a_2)=a_1$. Similarly, $\Phi(b_2)=b_1$. So $\Phi$ and 
$\Psi$ have all the required properties, and then
${\bf r}_2$ is equivalent to
${\bf r}_1$ up to an orientation-preserving reparametrization.
This concludes the proof that our relation is symmetric.\proofend

\medbreak\noindent DEFINITION.  An {\em oriented curve in $\rr^2$} is
an equivalence class of the equivalence relation ``equivalence up to
an orientation-preserving reparametrization.'' \proofend


\medbreak\noindent DEFINITION.  Let ${\bf r}_1:[a_1,b_1]\mapsto\rr^2$ and
${\bf r}_2:[a_2,b_2]\mapsto\rr^2$ be two parametrized curves
in $\rr^2$. 
We say that ${\bf r}_1$ is {\em equivalent to ${\bf r}_2$ 
up to a reparametrization} if there exist continuous functions 
$\varphi:[a_1,b_1]\mapsto[a_2,b_2]$, $\psi:[a_2,b_2]\mapsto[a_1,b_1]$, 
such that $\varphi(\psi(t))=t$ whenever $t\in[a_2,b_2]$, 
$\psi(\varphi(t))=t$ whenever $t\in[a_1,b_1]$, 
${\bf r}_1(\psi(t))={\bf r}_2(t)$ for all $t\in[a_2,b_2]$.
and either $\varphi(a_1)=a_2$ and  $\varphi(b_1)=b_2$  
or $\varphi(a_1)=b_2$ and  $\varphi(b_1)=a_2$.\proofend

\medbreak\noindent THEOREM.  Equivalence up to 
reparametrization is an equivalence
relation on $PC(\rr^2)$. 

\medbreak\noindent PROOF.  Exercise for the reader. 

\medbreak\noindent DEFINITION.  An {\em unoriented curve in $\rr^2$} is
an equivalence class of the equivalence relation ``equivalence up to
reparametrization.'' \proofend

\medbreak\noindent If this sounds too abstract to you, here is an example.
Consider the circle $C_R$ with equation $x^2+y^2=R^2$. 
Here are two parametrizations of $C_R$:
\begin{itemize}
\item[1.] ${\bf r}_1$ is the parametrization given by ${\bf
r}_1(t)=R\cos\,t\,{\bf i}+R\sin\,t\,{\bf j}$, $0\le t\le 2\pi$.
\item[2.] ${\bf r}_2$ is the parametrization given 
by ${\bf r}_2(t)=R\cos\,t\,{\bf i}-R\sin\,t\,{\bf j}$, $0\le t\le 2\pi$. 
\end{itemize}
Let us show that ${\bf r}_1$ and ${\bf r}_2$ are equivalent up to
reparametrization.  For this purpose, we must construct a
pair of continuous
functions $\varphi:[0,2\pi]\mapsto[0,2\pi]$,
$\psi:[0,2\pi]\mapsto[0,2\pi]$, 
such that $\varphi(\psi(t))=t$ whenever $t\in[0,2\pi]$, 
$\psi(\varphi(t))=t$ whenever $t\in[0,2\pi]$, 
${\bf r}_1(\psi(t))={\bf r}_2(t)$ for all $t\in[0,2\pi]$,
and either $\varphi(0)=0$ and  $\varphi(2\pi)=2\pi$  
or $\varphi(0)=2\pi$ and  $\varphi(2\pi)=0$.

Let us define $\varphi(t)=2\pi-t$, $\psi(t)=2\pi-t$, for
$0\le t\le 2\pi$. Then 
$$
\varphi(\psi(t))=2\pi-\psi(t)=2\pi-(2\pi-t)=t
$$
whenever $t\in[0,2\pi]$, and 
$$
\psi(\varphi(t))=2\pi-\varphi(t)=2\pi-(2\pi-t)=t
$$ 
whenever $t\in[0,2\pi]$. Furthermore, if $0\le t\le 2\pi$, then
\begin{eqnarray*}
{\bf r}_1(\psi(t))&=&R\cos\,\psi(t)\,{\bf i}+R\sin\,\psi(t)\,{\bf j}\\
&=&R\cos(2\pi-t)\,{\bf i}+R\sin(2\pi-t)\,{\bf j}\\
&=&R\cos(-t)\,{\bf i}+R\sin(-t)\,{\bf j}\\
&=&R\cos\,t\,{\bf i}-R\sin\,t\,{\bf j}\\
&=&{\bf r}_2(t)\,. 
\end{eqnarray*}
Finally, the condition that ``either $\varphi(0)=0$ and
$\varphi(2\pi)=2\pi$ or $\varphi(0)=2\pi$ and $\varphi(2\pi)=0$'' is
clearly true, because $\varphi(0)=2\pi$ and $\varphi(2\pi)=0$.

So ${\bf r}_1$ and ${\bf r}_2$ are equivalent under reparametrization.
In other words, {\bf ${\bf r}_1$ and ${\bf r}_2$ define the same unoriented
curve.} 

Are ${\bf r}_1$ and ${\bf r}_2$ equivalent under
orientation-preserving reparametrization? In other words, do {\bf
${\bf r}_1$ and ${\bf r}_2$ define the same oriented curve?}
The answer is ``no,'' but this is not completely trivial. 

First of all, notice that the $\varphi$ and $\psi$ that we have
defined above do not count for showing that ${\bf r}_1$ and 
${\bf r}_2$ are equivalent under orientation-preserving reparametrization,
because in the definition of ``equivalence  under orientation-preserving 
reparametrization'' the function $\varphi$ has to satisfy 
$\varphi(0)=0$,  and $\varphi(2\pi)=2\pi$, whereas the 
$\varphi$ that we have constructed satisfies
$\varphi(0)=2\pi$,  and $\varphi(2\pi)=0$. (That is, this 
$\varphi$ ``changes the direction of time from forward to backward.'') 

This observation, however, does not suffice to prove that 
the parametrized curves ${\bf r}_1$ and ${\bf r}_2$ are not equivalent under
orientation-preserving reparametrization. Just because one particular
choice of $\varphi$ and $\psi$ does not work, it does not mean that
there is not other choice that works. To prove that 
${\bf r}_1$ and ${\bf r}_2$ are not equivalent under
orientation-preserving reparametrization we have to show that
there do not exist $\varphi$, $\psi$ that satisfy the conditions
of the definition of ``\mbox{equivalence}  under
orientation-preserving reparametrization.'' That is, we have to show that
\begin{itemize}
\item[(*)] {\em there does  not exist a pair of  continuous functions
$\varphi:[0,2\pi]\mapsto[0,2\pi]$, $\psi:[0,2\pi]\mapsto[0,2\pi]$, 
such that 
$$
\varphi(\psi(t))=t\quad{\rm whenever}\quad t\in[0,2\pi]\,,
$$
$$
\psi(\varphi(t))=t\quad{\rm whenever}\quad t\in[0,2\pi]\,, 
$$
$$
{\bf r}_1(\psi(t))={\bf r}_2(t)\quad{\rm for\,\,\,all}\quad t\in[0,2\pi]\,,
$$
$$
\varphi(0)=0\,,
$$
and  
$$
\varphi(2\pi)=2\pi\,.
$$}
\end{itemize}
If we want to prove (*), we find ourselves in the familiar ``problematic''
situation that we have encountered earlier many times. {\bf Proving that
an object of a certain kind exists is easy, because all we have to 
do is exhibit one such object. But proving that
an object of a certain kind does not exist is hard, because 
if we try to find one and fail, that is  not enough to conclude
that the object does not exist. The reason for our failure
to find it might just have been that we weren't smart enough to look
for it in the right place.}

So {\large{\bf we have to think!!!}}  Let us think, then. We want to prove
that a certain thing (in our case, a certain pair of functions)
does not exist. The most obvious way to get started is to imagine 
that the thing does exist, and explore that possibility, hoping 
that we will get to something impossible. In other words, 
we should try to do a proof by contradiction. 

So let us suppose that there exists a pair $(\varphi,\psi)$ satisfying
all the conditions of (*). Pick one such pair $(\varphi,\psi)$. 

Then the condition that $\varphi(0)=0$, together with the fact that
$\psi(\varphi(t))=t$, implies $\psi(0)=\psi(\varphi(0))=0$.
So $\psi(0)=0$. Then $\psi(t)$ is close to $0$ if
$t$ is close to $0$. (This can be made precise using the 
``$\varepsilon$-$\delta$ definition of limit'' and the definition
of continuous function.) But, if 
$s$ is close to $0$ and $s\in[0,2\pi]$, then 
${\bf r}_1(s)=R\cos\,s\,{\bf i}+R\sin\,s\,{\bf j}$, so the 
${\bf j}$-component of ${\bf r}_1(s)$ is $R\sin\,s$, which is 
positive if $s$ is positive and close to $0$. In particular,
since $\psi(t)$ is close to $0$ if $t$ is close to $0$, and
$\psi(t)$ is  positive, we can conclude that {\em the
${\bf j}$-component of ${\bf r}_1(\psi(t))$ is
positive if $t$ is positive and close to $0$}.  On the other hand,
${\bf r}_1(\psi(t))={\bf r}_2(t)=R\cos\,t\,{\bf i}-R\sin\,t\,{\bf j}$.
So {\em the ${\bf j}$-component of ${\bf r}_1(\psi(t))$ is
negative if $t$ is positive and close to $0$}.  We have thus reached
a contradiction, and this shows that the
pair $(\varphi,\psi)$ does not exist. 

Summarizing:  
\begin{itemize}
\item[1.]  {\bf ${\bf r}_1$ and ${\bf r}_2$ define the same unoriented
curve.} 
\item[2.]  {\bf ${\bf r}_1$ and ${\bf r}_2$ do not define the same oriented
curve.} 
\end{itemize}

\bigbreak\noindent{\bf  EXAMPLE 11.} A curve in the plane 
divides the plane into several pieces. Consider, for example, the
following three curves $C_1$, $C_2$, $C_3$:

\vskip 3cm



Clearly, $C_1$ divides the plane into two pieces, $C_2$ divides it
into three pieces, and $C_3$ divides it into five pieces.  (In case
you want names and equations, $C_1$ is a {\em circle}, given by
$x^2+y^2=1$, $C_2$ is a {\em lemniscate}, given by $x^4-x^2+y^2=0$,
and $C_3$ is a {\em four-petal rose}, given in polar coordinates by
$r=\cos\,2\varphi$ and in Cartesian coordinates by
$(x^2+y^2)^3=(x^2-y^2)^2$.)

{\bf When we say that ``a curve divides the plane into a certain number of 
pieces,'' what does this mean, precisely?}  What is a ``piece.''
The answer, as you have probably guessed by now, is that a 
``piece'' is an equivalence class of some equivalence relation.
Here are the precise definitions:

\medbreak\noindent DEFINITION.  Let $A$ be a subset of the plane
$\rr^2$.  (We could equally well consider sets of points of $\rr^3$,
or of $\rr^4$, or of $\rr^n$ for any $n$, but here we will just stick to
the case $n=2$.) We say that a point $P\in A$ is {\em connected to a
point $Q\in A$ by a path in $A$} if there exists a continuous curve
$\xi:[0,1]\mapsto\rr^2$ such that $\xi(0)=P$, $\xi(1)=Q$, and
$\xi(t)\in A$ for every $t\in[0,1]$. \proofend

\medbreak\noindent THEOREM.  Let $A$ be a subset of the plane $\rr^2$.
Let $R$ be the relation on $A$ given by
$$
R=\{(P,Q)\in A\times A: P\,\,{\rm is\,\,connected\,\,to}\,\,
Q\,\,{\rm by\,\,a\,\,path\,\,in}\,\,A\}\,.
$$
Then $R$ is an equivalence relation. 

\medbreak\noindent PROOF. I will prove that $R$ is symmetric,
and leave it up to you to prove that $R$ is reflexive and transitive.

Suppose $P\in A$, $Q\in A$, and $P$ is connected to $Q$ by a path in $A$.
We want to prove that $Q$ is connected to $P$ by a path in $A$.
The idea is to do this by picking a path in $A$ that goes from 
$P$ to $Q$ and following it backwards, thereby obtaining a
path in $A$ that goes from $Q$ to $P$. 

To do this precisely, pick a continuous curve
$\xi:[0,1]\mapsto\rr^2$ such that $\xi(0)=P$, $\xi(1)=Q$, and
$\xi(t)\in A$ for every $t\in[0,1]$. Define another curve
$\eta:[0,1]\mapsto\rr^2$ by letting $\eta(t)=\xi(1-t)$ for 
$0\le t\le 1$. Then $\eta$ is continuous, $\eta(0)=\xi(1)=Q$,
$\eta(1)=\xi(0)=P$, and $\eta(t)\in A$ for all $t\in[0,1]$, because
if $t\in[0,1]$ then $\eta(t)=\xi(1-t)$, and $\xi(1-t)\in A$. 

So $Q$ is connected to $P$ by a path in $A$, and our proof that
$R$ is symmetric is complete.\proofend

\medbreak\noindent DEFINITION.  Let $A$ be a subset of the plane
$\rr^2$. Let $R$ be the equivalence relation on $A$ of the previous
theorem. Then the equivalence classes of $R$ are the 
{\em connected components} of $A$.\proofend

\medbreak\noindent COMMENT.  The notion of ``connected component''
is the precise mathematical translation of ``piece'' that we were looking for. 
To apply it to the three curves considered earlier, we look in each case
at the plane $\rr^2$ minus the curve. So we let 
$A_1=\rr^2-C_1$, $A_2=\rr^2-C_2$, $A_3=\rr^2-C_3$. Now that 
we have a precise definition of ``piece,'' we can prove rigorously
that $A_1$ has two ``pieces,'' $A_2$ has three, and $A_3$ has five.

\medbreak\noindent PROBLEM.  Prove that $A_1$ has exactly two 
connected components.

\medbreak\noindent PROBLEM.  Prove that $A_2$ has exactly three 
connected components.

\medbreak\noindent PROBLEM.  Prove that $A_3$ has exactly five
connected components.

\bigbreak\noindent{\bf EXAMPLE 12.} Some surfaces in three-dimensional
space have two ``sides.''  Other surfaces have only one ``side.''  An
example of a surface with only one side is the {\bf M\"obius
strip}. You can build a M\"obius strip as follows: cut a rectangular
band of paper with two very long sides and two very short sides;
label the vertices $a$, $b$, $c$, $d$ as shown in the picture;
then glue the $ab$ side to the $cd$ side after twisting
the band, in such a way that $a$ is glued to $d$ and $b$ to $c$.


\begin{minipage}{12cm}
$\qquad$
\vskip 2cm
$\qquad$
\end{minipage}

{\bf When we say that ``a surface has a certain number of 
sides,'' what does this mean, precisely?}  What is a ``side''
of a surface?

If you have guessed that a ``side'' of a surface is an equivalence
class of some equivalence relation, then you have guessed right!  Here
are the precise definitions:

\medbreak\noindent DEFINITION.  Let $S$ be a surface in $\rr^3$. 
A {\em unit normal} to $S$ is an ordered pair $(P,{\bf v})$ consisting
of (a) a point $P$ of $S$, and (b) a vector ${\bf v}$ such that
$|{\bf v}|=1$ and ${\bf v}$ is perpendicular to the surface at 
$P$. \proofend

\medbreak\noindent NOTATION.  Let $S$ be a surface in $\rr^3$. 
We use $N(S)$ to denote the set of all unit normals to $S$. 
(In mathematical jargon, $N(S)$ is called the 
``unit normal bundle'' of $S$.) \proofend


\medbreak\noindent DEFINITION.  Let $S$ be a surface in $\rr^3$. 
A unit normal $(P,{\bf v})\in N(S)$ is said to be {\em connected 
in $N(S)$ to a unit normal $(Q,{\bf w})\in N(S)$ 
by a path in $N(S)$} if there exists a continuous curve
$\xi:[0,1]\mapsto\rr^6$ such that $\xi(0)=(P,{\bf v})$, 
$\xi(1)=(Q,{\bf w})$, and
$\xi(t)\in N(S)$ for every $t\in[0,1]$. \proofend

\medbreak\noindent COMMENT. Why $\rr^6$? That's because the unit
normals to $S$ are pairs $(P,{\bf v})$, and $P$ and ${\bf v}$ have
three components each, so the pair $(P,{\bf v})$ has six components.

\medbreak\noindent THEOREM. Let $S$ be a surface in $\rr^3$. 
Let $R$ be the relation on $N(S)$ given by
$$
R=\Big\{\,\Big((P,{\bf v}),(Q,{\bf w})\Big):
(P,{\bf v})\,\,{\rm is\,\,connected\,\,
to\,\,}(Q,{\bf w})\,\,{\rm by\,\,a\,\,path\,\,in}\,\,N(S)\,\Big\}\,.
$$
Then $R$ is an equivalence relation on $S$.

\medbreak\noindent DOES THIS THEOREM NEED A PROOF? Not really, because
this is just a special case of the theorem of Example 11.  (In Example
11 we only talked about subsets of $\rr^2$, but we could equally well
have worked with subsets of $\rr^3$ or $\rr^4$ or even $\rr^6$.)

\medbreak\noindent DEFINITION.  Let $S$ be a surface in $\rr^3$. 
A {\em side} of $S$ is an equivalence class of the relation
$R$ of the previous theorem.\proofend

\medbreak\noindent COMMENT. Using the terminology of Example 11,
{\bf the sides of $S$ are the connected components of $N(S)$}. 


\medbreak\noindent ANOTHER COMMENT. How should we think of unit
normals and the relation defined above? You should visualize this as
follows: a unit normal is just a little person (of height equal to
$1$) standing on the surface, perpendicularly to it. A path in the set
of unit normals is what you get when our little person walks on the
surface. Two unit normals are connected by a path if our little person
can walk from one to the other.  A surface such as a M\"obius strip
has only one side because our little person can start walking at some
point on the surface and after a long walk she can come back to
the point where she started, but ``upside down.'' A spherical
surface has two sides because if our little person starts walking
at some point on the surface and is on the ``outside'' of the surface,
then no matter how she walks she can never return to the same
position but finding herself now on the ``inside.'' 

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