relations.xml

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<!-- Logic and Proof for Teachers                                                                -->
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<chapter xml:id="relations" xmlns:xi="http://www.w3.org/2001/XInclude">

<title>Relations</title>

    <assemblage>
        <title>TEXAS STATE BOARD FOR EDUCATOR CERTIFICATION (SBEC): MATHEMATICS STANDARDS COVERED</title>
        <p><ul>

            <li>STANDARD II: PATTERNS AND ALGEBRA: The mathematics teacher understands and uses patterns, relations, functions, algebraic reasoning, analysis, and technology appropriate to teach the statewide curriculum (Texas Essential Knowledge and Skills [TEKS]) in order to prepare students to use mathematics.</li>

            <li>STANDARD V: MATHEMATICAL PROCESSES: The mathematics teacher understands and uses mathematical processes to reason mathematically, to solve mathematical problems, to make mathematical connections within and outside of mathematics, and to communicate mathematically.</li>

            <li>STANDARD VI: MATHEMATICAL PERSPECTIVES: The mathematics teacher understands the historical development of mathematical ideas, the interrelationship between society and mathematics, the structure of mathematics, and the evolving nature of mathematics and mathematical knowledge.</li>

            </ul></p>
    </assemblage>

    <introduction>
        <p>In our everyday lives we encounter many circumstances which necessitate relating objects, sets, or people. Hiring is determined by comparing abilities of applicants and purchases are made based on relative prices. Descriptions such as <q>is faster than,</q> <q>is the brother of,</q> and <q>is smaller than</q> are heard countless times each day.</p>

        <p>This concept of relating elements from different collections of objects is used extensively in mathematics. In previous courses, you have considered relations primarily from a very informal perspective; we will now approach the study of relations in a more formal way and give a formal definition of a relation below. A variety of special types of relations, including functions, will be considered in greater depth in this chapter.</p>
    </introduction>

    <section xml:id="relations-section-relations">
    	<title>Relations</title>

    	<p>For convenience, we restate the definition of the Cartesian product of sets with which you might already be familiar.</p>

    	<definition>
    		<p>The <em><term>Cartesian product</term></em> of two sets <m>A</m> and <m>B</m>, denoted <m>A \times B</m>, is the set consisting of all ordered pairs in which the first element comes from set <m>A</m> and the second element comes from set <m>B</m>; that is, <m>A \times B = \{ (a,b) \mid a \in A \ \text{and} \ b \in B\}</m>.</p>
    	</definition>
        <idx><h>Cartesian product</h></idx>

    	<definition>
    		<p>A <em><term>relation</term></em> <m>R</m> from a set <m>A</m> to a set <m>B</m> is a subset of <m>A \times B</m>. If <m>R</m> is a relation from <m>A</m> to <m>A</m>, we say <m>R</m> is a relation on <m>A</m>.</p>
    	</definition>
        <idx><h>Relation</h></idx>

    	<p>While a relation may be described in a variety of ways, it is really just a set of ordered pairs.</p>

    	<example xml:id="relations-example-relations">
    		<p>Let <m>A = \{a,b,c,d\}</m> and <m>B = \{1,2,3\}</m>. Consider the following sets.
    			<ol>

    				<li><m>R_1 =\{ (a,2), (b,3), (c,2), (d,3)\}</m></li>

    				<li><m>R_2 = \{ (a,1), (a, 2), (b,3) \}</m></li>

    				<li><m>R_3 = \{(a, 1), (b, 3), (c, 2) \}</m></li>

    				<li><m>R_4 = \{ (a,a), (b,b), (c,c), (d,d) \}</m></li>

    				<li><m>R_5 = \{ (d, a), (c, a), (d,d) \}</m></li>

    			</ol>
    		Each of the sets <m>R_1</m>, <m>R_2</m>, and <m>R_3</m> is a relation from <m>A</m> to <m>B</m> because the first elements in the ordered pairs are from <m>A</m> and the second elements are from <m>B</m>, while <m>R_4</m> and <m>R_5</m> are relations on <m>A</m> because both first and second elements are from <m>A</m>.</p>
    	</example>

    	<p>Although relations are actually sets of ordered pairs, such collections of ordered pairs may be indicated in many ways. Notice that a relation is essentially a pairing of elements according to some criteria. For example, we may <q>pair</q> a person's name with his or her social security number, age, height, or the names of cities lived in. Sometimes relations are specified simply by listing these pairs in set form as in the example above. When relations are described in this form, the rules or criteria for the pairing are often not known. For instance, in <m>R_1</m> of <xref ref="relations-example-relations" />, we know that <m>b</m> and <m>d</m> are both related to <m>3</m>, but we do not know why. Similarly, relations may be indicated in table or graphical form as in the below.</p>

    	<figure xml:id="figure-relations-picture">
            <caption>Pictorial representation <m>R</m> from <m>A</m> to <m>B</m></caption>
                		<image width="80%" source="images/relations-picture.jpg">
                <description>pictorial representation R from A to B</description>
            </image>
        </figure>

    	<example xml:id="relations-example-relation-picture">
    		<p><xref ref="figure-relations-picture" /> is a pictorial representation of of the relation <m>R = \{(1, 4), (6, 4), (3, 3), (10, 2)\}</m> from <m>A = \{1, 3, 6, 10\}</m> to <m>B = \{2, 3, 4\}</m>. The arrow diagram below is a different way of representing the same relation.</p>

	        <image width="40%" xml:id="relation-arrows">
                <description>arrow representation R from A to B</description>
                <latex-image>
                <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
                <![CDATA[
                \begin{tikzpicture}[scale=1]
                \draw[->] (0,0) node[left] {$10$} -- (3,2.5) node[right] {$2$};
                \draw[->] (0,1) node[left] {$6$} -- (3,0.45) node[right] {$4$}; 
                \draw[->] (0,2) node[left] {$3$} -- (3,1.5) node[right] {$3$};   
                \draw[->] (0,3) node[left] {$1$} -- (3,0.55);     
                \end{tikzpicture}]]>
                </latex-image>
            </image>

            <p>A third representation of <m>R</m> is given in table form below.</p>

            <table xml:id="table-relations-table">
                <title>A table representation <m>R</m> from <m>A</m> to <m>B</m></title>
                <tabular halign="center" top="medium">
                    <row bottom="medium">
                        <cell><m>a</m></cell><cell><m>b</m></cell>
                    </row>
                    <row>
                        <cell>1</cell><cell>4</cell>
                    </row>
                    <row>
                        <cell>3</cell><cell>3</cell>
                    </row>                <row>
                        <cell>6</cell><cell>4</cell>
                    </row>                <row bottom="medium">
                        <cell>10</cell><cell>2</cell>
                    </row>
               </tabular>
            </table>

    	</example>

    	<p>In many cases, listing all the ordered pairs in a relation is tedious or simply impossible. Under those circumstances the relation may be described in many different ways; associated ordered pairs are indicated by specifying a defining rule.</p>

    	<example xml:id="relations-example-relation-ordered-pairs">
    		<p>The equation <m>x + y = 4</m> describes a relation, <m>R</m>, consisting of an infinite set of ordered pairs whose components will satisfy the equation. Hence, the ordered pairs <m>(2, 2)</m>, <m>( 3/2, 5/2 )</m>, <m>(-7, 11)</m>, and <m>(0, 4)</m> are a few of the elements in the relation. Precisely stated,
    			<me>R(x,y) = \{ (x, y) \in \mathbb R \times \mathbb R \mid x + y = 4 \}</me>
    		That is, the relation described by the equation is actually the infinite set of ordered pairs of real numbers whose sum is equal to <m>4</m>. In the following figure, we have indicated all the ordered pairs which satisfy the linear equation; this is sometimes called the graphical representation of the equation.</p>

            <figure xml:id="figure-relations-graph">
                <caption>Graphical representation of <m>x + y = 4</m></caption>
                <image width="50%" source="images/relations-graph.png">
                    <description>pictorial representation R from A to B</description>
                </image>
            </figure>

    	</example>

        <example xml:id="relations-example-relation-set-builder">
            <p>Let <m>A = \{ 1, 2, 3, 4\}</m> and define a relation <m>R</m> on <m>A</m> by
                <me>R(x,y) = \{ (x, y) \in A \times A \mid x \leq y \}.</me>
            That is, <m>R</m> is the set of all ordered pairs whose components both come from the set <m>A</m> and in which the first component is less than or equal to the second. In this case we could enumerate <m>R</m> as follows:
                <me>R = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4) \}</me></p>
        </example>

        <p>Notice in <xref ref="relations-example-relation-set-builder" /> that it is merely inconvenient to have to list all the ordered pairs implied by the rule given.  However, if the set A were the set of all integers instead of the finite set <m>\{1, 2, 3, 4\}</m>, the same rule applied to A would yield an infinite set of ordered pairs that would be impossible to list. In such cases, it is convenient to use notation that refers to the relation by the criteria used for comparison. We say the relation <m>R</m> is <q>less than or equal to</q> or <q><m>\leq</m></q> rather than the actual set of ordered pairs, and instead of saying <m>(1, 3) \in R</m>, we say <m>1 \leq 3</m>.</p>

        <remark>
            <p>In general, we often substitute the notation <m>aRb</m> for <m>(a, b) \in R</m>.</p>
        </remark>

        <p>Then to determine whether two elements in the set are related, we simply substitute our familiar relation for <m>R</m> in the expression <m>aRb</m>. For example, instead of asking if the ordered pair <m>(3, 2)</m> is an element of the relation <q>is greater than,</q> we ask if <m>3</m> <q>is greater than</q> <m>2</m>; that is, we usually write <m>3 > 2</m> instead of writing <m>(3,2) \in</m> <q><m>></m></q>.</p>


        <p>In considering each of the examples given in this section, it is clear that a relation between sets <m>A</m> and <m>B</m> need not <q>use up</q> all of both sets. This idea leads us to consider those subsets of <m>A</m> and <m>B</m> whose elements are paired by the relation, namely the domain and range.</p> 

        <definition xml:id="relations-definition-domain-range">
            <p>If <m>R</m> is a relation from <m>A</m> to <m>B</m>, then the <term>domain</term> of <m>R</m>, <m>\domain(R)</m> is defined by
                <me>\domain(R) = \{a \in A \mid (a, b) \in R \text{ for some } b \in B\}.</me>
            The <term>range</term>, denoted by <m>\range(R)</m>, is defined by
                <me>\range(R) = \{b \in B \mid (a, b) \in R \text{ for some } a \in A\}.</me></p>
        </definition>
        <idx><h>Domain</h></idx>
        <idx><h>Range</h></idx>

        <p>By <xref ref="relations-definition-domain-range" />, it is clear that <m>\domain(R)</m> is a subset of <m>A</m> and <m>\range(R)</m> is a subset of <m>B</m>.</p>

        <p>Informally, we designate the domain of a relation <m>R</m> as the set of all elements of the set <m>A</m> that are actually paired with (or mapped to) at least one element in <m>B</m>. The range of a relation R may be described as the set of all elements in the set <m>B</m> to which at least one element of <m>A</m> is mapped. These concepts are defined formally in the definition below.</p>

        <example xml:id="relations-example-domain-range">
            <p>Consider the sets <m>R_1, R_2, \ldots, R_5</m> as defined in <xref ref="relations-example-relations" />.  Then
                <ol>

                    <li><m>\domain(R_1) = A</m> and <m>\range(R_1) = \{ 2, 3 \}</m>.</li>


                    <li><m>\domain(R_2) = \{ a, b\}</m> and <m>\range(R_2) = B</m>.</li>

                    <li><m>\domain(R_3) = \{a, b, c\}</m> and <m>\range(R_3) = B</m>.</li>

                    <li><m>\domain(R_4) = A</m> and <m>\range(R_4) = A</m>.</li>

                    <li><m>\domain(R_5) = \{c, d \}</m> and <m>\range(R_5) = \{ a, d \}</m>.</li>


                </ol></p>
        </example>

        <example xml:id="relations-example-unit-circle">
            <p>Consider the equation <m>x^2 + y^2 = 1</m> defined on <m>\mathbb R</m>.  Let
                <me>R = \{ (x, y) \in \mathbb R \times \mathbb R \} \mid x^2 + y^2 = 1 \}.</me>
            Then <m>\domain(R) = \{ x \mid -1 \leq x \leq 1 \}</m> and <m>\range(R) = \{ y \mid -1 \leq y \leq 1 \}</m> since substituting numbers  whose squares are greater than <m>1</m> for either variable yields an equation with only complex solutions.</p>
        </example>

        <p>We will investigate the concepts of domain and range in more detail in the study of functions. However, the reader may suspect that although these sets are sometimes easily found, often they will require some insight and work to determine!</p>

        <p>Earlier we considered the relation <q><m>\leq</m>.</q> If we were to consider the associated pairs in reverse order, we might describe this new list of ordered pairs by the relation <q><m>\geq</m>.</q> Thus we see that relations are in some sense reversible, and we formalize this concept in the definition that follows.</p>

        <definition xml:id="relations-definition-inverse-relation">
            <p>If <m>R</m> is a relation from <m>A</m> to <m>B</m>, then the <term>inverse relation</term> of <m>R</m>, denoted by <m>R^{-1}</m>, is defined by
                <me>R^{-1} = \{(b, a)  \mid (a, b) \in R \}.</me></p>
        </definition>
        <idx><h>Inverse relation</h></idx>

        <p>Thus, <m>R^{-1}</m> is a new relation from <m>B</m> to <m>A</m> and consists of ordered pairs from <m>R</m> with the components reversed.  Hence, the inverse relation of <m>R = \{ (2,3), (4, 7), (2, 9), (6,9) \}</m> is the relation <m>R^{-1} = \{ (3,2), (7,4), (9,2), (9, 6) \}</m>.</p>

        <p>Consider the relation in <xref ref="figure-relations-relation-example" />.  The inverse relation <m>R^{-1}</m> from <m>B</m> to <m>A</m> can be found by simply reversing the arrows (<xref ref="figure-relations-inverse-relation-example" />).  In this case, <m>\domain(R) = \range(R^{-1}</m> and <m>\range(R) = \domain(R^{-1})</m>.</p>

        <figure xml:id="figure-relations-relation-example">
            <caption>The relation <m>R</m> from <m>A</m> to <m>B</m></caption>
            <image width="80%" xml:id="relations-relation-example">
                <description>two overlapping circles with the common part shaded</description>
                <latex-image>
                <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
                <![CDATA[
                \begin{tikzpicture}[scale=1]
                \draw [very thick] (2,2) circle(1.5);  
                \draw [very thick] (6,2) circle(1.5); 
                \draw[->] (2,3) node [left] {$a$} -- (6,3) node [right] {$1$};
                \draw[->] (2,2.5) node [left] {$b$} -- (6,2.5) node [right] {$2$}; 
                \draw[->] (2,2.5) -- (6,2); 
                \draw[->] (2,1.5) node [left] {$d$} -- (6,1.5) node [right] {$4$};
                \draw[->] (2,1) node [left] {$e$} -- (6,2) node [right] {$3$};
                \node at (0,3) {$A$};
                \node [left] at (2,2) {$c$};
                \node at (8,3) {$B$};                                     
                \end{tikzpicture}]]>
                </latex-image>
            </image>
        </figure>



        <figure xml:id="figure-relations-inverse-relation-example">
            <caption>The relation <m>R^{-1}</m> from <m>B</m> to <m>A</m></caption>
            <image width="80%" xml:id="relations-inverse-relation-example">
                <description>two overlapping circles with the common part shaded</description>
                <latex-image>
                <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
               <![CDATA[
                \begin{tikzpicture}[scale=1]
                \draw [very thick] (2,2) circle(1.5);  
                \draw [very thick] (6,2) circle(1.5); 
                \draw[<-] (2,3) node [left] {$a$} -- (6,3) node [right] {$1$};
                \draw[<-] (2,2.5) node [left] {$b$} -- (6,2.5) node [right] {$2$}; 
                \draw[<-] (2,2.5) -- (6,2); 
                \draw[<-] (2,1.5) node [left] {$d$} -- (6,1.5) node [right] {$4$};
                \draw[<-] (2,1) node [left] {$e$} -- (6,2) node [right] {$3$};
                \node at (0,3) {$A$};
                \node [left] at (2,2) {$c$};
                \node at (8,3) {$B$};                                     
                \end{tikzpicture}]]>
                </latex-image>
            </image>
        </figure>


    </section>

     <section xml:id="relations-section-equiv-relations">
        <title>Equivalence Relations</title>

        <p>One particular kind of relation that plays a vital role in mathematics is an equivalence relation. Before defining an equivalence relation, we will consider definitions and examples of each of the properties involved.</p>

        <definition xml:id="relations-definition-reflexive">
            <p>A relation <m>R</m> is  <term>reflexive</term> if <m>(a, a) \in R</m> for every <m>a \in A</m>. That is for each <m>a \in A</m>, We have <m>aRa</m>.</p>
        </definition>
        <idx><h>Reflexive</h></idx>

        <p>It is important to realize that this definition involves a quantified expression, <q>for each <m>a \in A</m>.</q> Recall from logic that when such an expression is negated, the quantifier changes. Hence, it follows that a relation is not reflexive if there exists even one element in <m>A</m> that is not related to itself.</p>

        <example xml:id="relations-example-less-than-equal">
            <p>The relation <q><m>\leq</m></q> on <m>\mathbb Z</m> has the reflexive property since every integer is less than or equal to itself.</p>
        </example>

         <example xml:id="relations-example-equal">
            <p>The relation <q><m>=</m></q> on <m>\mathbb R</m> is reflexive since every real number is equal to itself.</p>
        </example>

         <example xml:id="relations-example-not-reflexive">
            <p>The relation <m>R = \{ (2, 3), (2,2), (3,2), (3,3) \}</m> is not a reflexive relation on the set <m>A = \{  1, 2, 3\}</m>  since <m>1 \in A</m>, but <m>(1,1) \notin R</m>. But <m>R</m> is reflexive when considered as a relation on the set <m>B = \{2, 3 \}</m>.</p>
        </example>

        <example xml:id="relations-example-power-set">
            <p>Let <m>S</m> be nonempty and <m>\mathcal P(S)</m> be the power set of <m>S</m>. Then the subset relation is reflexive on <m>\mathcal P(S)</m> since every set is a subset of itself.</p>
        </example>

        <definition xml:id="relations-definition-symmetric">
            <p>A relation <m>R</m> on <m>A</m> is  <term>symmetric</term> if whenever <m>(a, b) \in R</m>, then <m>(b,a) \in R</m>. Alternatively, if <m>aRb</m>, then <m>bRa</m>.</p>
        </definition>
        <idx><h>Symmetric</h></idx>

        <p>You should notice a major distinction in the nature of the definitions of these terms. The definition of reflexive is universal in that it must be true for all members of the set upon which it is defined. In contrast, the definition for the symmetric property is stated in the form of a conditional. (Remember from formal logic that a conditional, <m>p \rightarrow q</m>, is false only when <m>p</m> is true and <m>q</m> is false.) So to show a relation is not symmetric we must be able to find an ordered pair <m>(a,b)</m> in <m>R</m> such that <m>(b, a)</m> is not in <m>R</m>.</p>

        <example xml:id="relations-example-symmetric">
            <p>Let <m>A = \{ 1, 2, 3, 4 \}</m> and consider the given relations on <m>A</m>.  The relations <m>R = \{ (1, 2), (2, 1), (3, 4), (4,3)\}</m> and <m>S = \{ (1,1)\}</m> have the symmetric property. However, <m>T = \{ (3,3). (2,4), (4,2), (1, 2)  \}</m> is not symmetric since <m>(1, 2) \in T</m> but <m>(2, 1) \notin T</m>.</p>
        </example>

        <p>It is important to remember that to show a relation does <em>not</em> have a certain property, we need only provide a single counterexample.</p>

        <example xml:id="relations-example-not-symmetric">
            <p>The relation <q><m>\geq</m></q> is not symmetric on <m>\mathbb R</m> since <m>5 \geq 2</m> but <m>2 \not\geq 5</m>.</p>
        </example>

        <example xml:id="relations-example-rectangles">
            <p>Let <m>A</m> be the set of rectangles in the Cartesian plane, and let elements of <m>A</m> be related if they have the same area.  Then this relation is symmetric.</p>
        </example>

        <definition xml:id="relations-definition-transitive">
            <p>A relation <m>R</m> on <m>A</m> is <term>transitive</term> if whenever <m>(a, b) \in R</m> and <m>(b, c) \in R</m>, then <m>(a, c) \in R</m>.</p>
        </definition>
        <idx><h>Transitive</h></idx>

        <p>A relation <m>R</m> is not transitive if there exist <m>a</m>, <m>b</m>, and <m>c</m> in <m>A</m> such that <m>(a,b) \in R</m>  and <m>(b, c) \in R</m> but <m>(a, c) \notin R</m>.</p>

        <example xml:id="relations-example-less-than">
            <p>The relation <q><m>\lt</m></q> is transitive on <m>\mathbb R</m>.</p>
        </example>

        <example xml:id="relations-example-transitive">
            <p>Let <m>R = \{  (1, 1), (2, 2), (3,3) \}</m> on <m>\mathbb Z</m>. Then <m>R</m> is transitive since there do not exist integers <m>a</m>, <m>b</m>, and <m>c</m> such that <m>(a,b) \in R</m>  and <m>(b, c) \in R</m> but <m>(a, c) \notin R</m>.</p>
        </example>

        <example xml:id="relations-example-not-transitive">
            <p>Let <m>R = \{  (1, 2), (2, 3) \}</m> on <m>\mathbb Z</m>. Then <m>R</m> is not transitive since <m>(1, 2) \in R</m> and <m>(2, 3) \in R</m> but <m>(1,3) \notin R</m>.</p>
        </example>

        <definition xml:id="relations-definition-equivalence-relation">
            <p>A relation <m>R</m> on <m>A</m> which is reflexive, symmetric, and transitive is called an <term>equivalence relation</term> on <m>A</m>.  If two elements <m>a</m> and <m>b</m> are equivelent, we write <m>a \sim b</m>, unless of course there is already a notation in place such as <q><m>=</m>.</q></p>
        </definition>
        <idx><h>Equivalence relation</h></idx>

        <example xml:id="relations-example-not-equivalence-not-reflexive">
            <p>Let <m>A = \{ 1, 2, 3\}</m> and <m>R = \{  (1, 1), (2, 2) \}</m>. Then <m>R</m> is not an equivalence relation on <m>A</m>.  The relation is both transitive and symmetric but is not reflexive  since <m>(3, 3) \notin R</m>.</p>
        </example>


        <example xml:id="relations-example-not-equivalence-not-transitive">
            <p>Let <m>A = \{ 1, 2, 3\}</m> and
                <me>S = \{  (1, 1), (2, 2), (3,3), (1, 2), (2,1), (2, 3), (3,2) \}.</me>
            The relation <m>S</m> is both reflexive and symmetric but is not transitive, since <m>(1, 2) \in S</m> and <m>(2, 3) \in S</m> but <m>(1, 3) \notin S</m>.</p>
        </example>

        <example xml:id="relations-example-power-set-not-equivalence">
            <p>Let <m>A = \{a, b\}</m> and <m>\mathcal P(A)</m> be the power set of <m>A</m>. Then the subset relation is reflexive  and transitive  but is not symmetric since <m>\{a\} \subseteq \{a, b\}</m> but <m>\{a, b\} \not\subseteq \{a\}</m>.</p>
        </example>



        <example xml:id="relations-example-mod5">
            <p>Consider the set <m>\mathbb Z</m> and define <m>a \sim b</m> if <m>5</m> divides <m>a - b</m>, denoted by <m>5 \mid (a - b)</m>; i.e., there is no remainder when <m>a - b</m> is divided by <m>5</m>. Then <m>2 \sim 12</m>, since <m>5 \mid (2 - 12)</m> and <m>3 \not\sim 7</m>, since <m>5 \not\mid (3 - 7)</m>.  We have defined an eqiuivalence relation on <m>\mathbb Z</m>.  The proof is left as an exercise (<xref ref="relations-exercise-mod5" />).</p>
        </example>

        <example xml:id="relations-example-modn">
            <p>Consider the set <m>\mathbb Z</m>, and let <m>n</m> be a fixed integer that is not equal to zero.  define <m>a \sim b</m> if <m>n \mid (a - b)</m>.  The relation is a generalization of <xref ref="relations-example-modn" /> and is also an equivalence relation.</p>
        </example>

        <example>
            <p>Some other equivalence relations are <q>is the same age as</q> on the set of all people, <q>has the same area as</q> on the set of all rectangles in the Cartesian plane, and <q>lives on the same street as</q> on the set of all people living in a given city.</p>

        </example>

        <p>Consider the relation <q>is the same age as</q> on the set of all students in a given class. This relation groups the people in the class according to age. Each person in the class is in some group, even if it is a single member group. In addition, no person is in more than one group and all people in a specific group are the same age. These qualities are common to all equivalence relations, leading us to the following formal definition.</p>

        <definition xml:id="relations-definition-equivalence-class">
            <p>Let <m>\sim</m> be an equivalence relation on <m>A</m>. If <m>x \in A</m>, then the <term>equivalence class</term> of <m>x</m>, denoted by <m>\overline{x}</m>, is defined by <m>\overline{x} = \{ y \in A \mid x \sim y \}</m>.</p>
        </definition>
        <idx><h>Equivalence class</h></idx>

        <example xml:id="relations-example-equivalence-class">
            <p>Suppose <m>A = \{ 6, 7, 8, 9, 10\} </m> and let the relation <m>R</m> on <m>A</m> be defined by <m>aRb</m> if <m>a</m> and <m>b</m> have the same remainder when divided by <m>2</m>. This relation is an equivalence relation since it is reflexive, symmetric, and transitive. (Verification is left as an exercise.) Then by <xref ref="relations-definition-equivalence-class" />, 
                <me>\overline{6} = \overline{8} = \overline{10} = \{6, 8, 10\}</me>
            and 
                <me>\overline{7} = \overline{9} = \{7, 9\}.</me>
            Furthermore, <m>\overline{6}</m> and <m>\overline{7}</m> are disjoint sets whose union is <m>A</m>. So <m>A</m> has just partitioned into disjoint pieces where each piece can have different names.</p>
        </example>

        <p>The observations for the specific relation considered in <xref ref="relations-example-equivalence-class" /> lead us to generalize these concepts to any equivalence relation defined on an arbitrary set.</p>

        <theorem  xml:id="relations-theorem-equivalence-class">
            <statement>
                <p>Let <m>\sim</m> be an equivalence relation on <m>A</m>. Then:
                    <ol>

                        <li>Each equivalence class is non-empty.</li>

                        <li>Any two equivalence classes are either equal or disjoint</li>

                        <li>The set <m>A</m> is equal to the union of all the equivalence classes.</li>

                    </ol>
                </p>
            </statement>
        </theorem>

        <p>When given an equivalence relation on a set, we may find the classes generated by that relation by first choosing elements at random from the set. For example, suppose
            <me>A = \left\{ \frac{2}{3}, -\frac{1}{2}, -1, 0, \frac{4}{6}, \frac{0}{3}, - \frac{4}{8}, \frac{10}{15} \right\}.</me>
        and let the equivalence relation on be <q><m>=</m>.</q>  To find the equivalence classes determined by the relation, we may choose any element in <m>A</m>, say <m>4/6</m>.  Then <m>4/6 \in \overline{4/6}</m>, since <m>4/6 = 4/6</m>.  The other elements of <m>\overline{4/6}</m> are <m>2/3</m> and <m>10/15</m>, since they are the only elements of <m>A</m> equal to <m>4/6</m>.  Thus, <m>\overline{4/6} = \{ 4/6, 2/3, 10/15 \}</m>.  Part (2) of <xref ref="relations-theorem-equivalence-class" /> implies we might just as easily have chosen <m>2/3></m> or <m>10/15</m>, and we would have arrived at the same equivalence class.  That is, since <m>2/3 \in \overline{4/6}</m> and <m> 10/15 \in\overline{4/6}</m>, then <m>\overline{2/3} = \overline{4/6} = \overline{10/15}</m>. So we conclude that we may designate an equivalence class completely using any of its elements.</p> 

        <assemblage>
            <title>TAKS CONNECTION</title>
            <p>How might a student apply the concept of equivalence classes to answer the following question taken from the 2006 Texas Assessment of Knowledge and Skills (TAKS) Grade 5 Mathematics test?</p>

            <p>Stan was putting fruit into baskets. He wanted each basket to be more than  <m>7/10</m> full. Which fraction is more than <m>7/10</m>?</p>

            <p>Which one of the following numbers belongs in the region of the diagram marked by the question mark?
                <ul>

                    <li>A. <m>4/5</m></li>

                    <li>B. <m>1/2</m></li>

                    <li>C. <m>2/3</m></li>

                    <li>D. <m>3/5</m></li>

                </ul></p> 
        </assemblage>

        <example>
            <p>Let <m>A</m> be the set of all ordered pairs of real numbers, and define <m>(a, b) \sim (c, d)</m> if <m>a^2 + b^2 = c^2 + d^2)</m>. Before proceeding further, you should find some ordered pairs that are related and then verify that this indeed defines an equivalence relation. Then for any <m>(x, y) \in \mathbb R \times \mathbb R</m>, we have 
                <me>\overline{(x, y)} = \{ (s, t) \mid x^2 + y^2 = s^2 + t^2 \}.</me>
            These equivalence classes are represented geometrically by circles centered at the origin.</p>
        </example>

        <p>Before leaving this section, we introduce a visual method for representing finite relations. For rather small finite relations, these visual representations, called <term>digraphs</term>, can be very helpful in conceptualizing a relation and its properties.</p>
        <idx><h>Digraph</h></idx>

        <p>In general, consider a finite set <m>A</m> and a relation <m>R</m> defined on it. Digraphs are constructed by drawing a small dot representing each element of <m>A</m> and labeling that circle appropriately. These dots are called the <term>vertices</term> of the graph. At each dot, say <m>x</m>, draw a directed line segment from it to any other dot, labeled <m>y</m>, if and only if <m>xRy</m>. These directed line segments are called <term>edges</term>. Then notice that the dots represent <m>A</m> and the edges represent the ordered pairs in <m>R</m>.</p>

        <example xml:id="relations-example-digraph">
            <p>Consider the set <m>A = \{a, b, c, d \}</m>and the relation
                <me>R = \{ (a, a), (c, c), (a, b), (b, a), (b, d), (c , d), (b, c) \}.</me></p>

            <figure xml:id="figure-relations-digraph">
                <caption>The digraph for the relation <m>R</m> on <m>A</m></caption>
                <image width="50%" xml:id="relations-digraph">
                    <description>two overlapping circles with the common part shaded</description>
                    <latex-image>
                    <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
                   <![CDATA[
                    \begin{tikzpicture}[scale=1]
                    
                    \draw [fill=blue!20] (3,1) coordinate(A) circle [radius=0.1];
                    \draw [thick] (A) circle(0.1);  
                    \node [above right] at (A) {$a$};


                    \draw [fill=blue!20] (1,2) coordinate(B) circle [radius=0.1];
                    \draw [thick] (B) circle(0.1);  
                    \node [above right] at (B) {$b$};
     
                    \draw [fill=blue!20] (3,3) coordinate(C) circle [radius=0.1];
                    \draw [thick] (C) circle(0.1);  
                    \node [above right] at (C) {$c$}; 

                    \draw [fill=blue!20] (2,4) coordinate(D) circle [radius=0.1];
                    \draw [thick] (D) circle(0.1);  
                    \node [above right] at (D) {$d$};    

                    \draw[thick, ->] (3,0.9) to [out=270,in=270] (1,1.9);     
                    \draw[thick, ->] (1.1,2) to [out=0,in=90] (3,1.1);   
                    \draw[thick, ->] (1.1,2) to [out=0,in=270] (3,2.9);  
                    \draw[thick, ->] (3,3.1) to [out=90,in=0] (2.1,4);    
                    \draw[thick, ->] (1,2.1) to [out=90,in=195] (1.9,4);  
                    
                    \draw[thick, ->] (3.1,3) arc [radius=0.5, start angle=-180, end angle= 180];
                    \draw[thick, ->] (3.1,1) arc [radius=0.5, start angle=-180, end angle= 180];
                    \end{tikzpicture}]]>
                    </latex-image>
                </image>
            </figure>

            <p>Notice that there is an edge representing every ordered pair in <m>R</m> (<xref ref="figure-relations-digraph" />). For those elements in <m>A</m> that are related to themselves, there is an edge that looks like a loop. In this relation there are only two loops since only <m>a</m> and <m>c</m> are related to themselves. From this example it is reasonable to conclude that for an arbitrary finite relation <m>R</m> defined on <m>A</m>, the relation will have the reflexive property if and only if each vertex has a loop. Thus, we can quickly see from the digraph that <m>R</m> is not reflexive. We may also conclude from the digraph that <m>R</m> is not symmetric and not transitive. Why?</p>
        </example>


        <exercises>

            <exercise>
                <statement>
                    <p>Determine whether or not the following relations are reflexive, symmetric, or transitive. Which are equivalence relations on the given sets? Justify your thinking.
                        <ol>

                            <li><m>R = \{ (1,1), (3,3), (5,5) \}</m> on <m>A = \{ 1, 3, 5 \}</m></li>

                            <li><m>R = \{ (1,1), (2,2), (1,2) \}</m> on <m>A = \{ 1, 2 \}</m></li>

                            <li><m>R = \{ (1,1), (1,2), (2,1) \}</m> on <m>A = \{ 1, 2 \}</m></li>

                            <li><m>R = \{ (1,3), (2,3), (3,2), (3,1) \}</m> on <m>A = \{ 1, 2, 3 \}</m></li>

                            <li><m>R = \{ (1,1), (2,2), (3,3), (4,4), (1,3), (2,4) \}</m> on <m>A = \{ 1, 2, 3, 4 \}</m></li>

                            <li><m>R = \{ (3,4) \}</m> on <m>A = \{ 3, 4 \}</m></li>

                            <li><m>R = \{ (3,3) \}</m> on <m>A = \{ 3,4 \}</m></li>

                            <li><m>R = \{ (1,1), (2,2), (3,3) \}</m> on <m>A = \{ 1, 2, 3 \}</m></li>

                            <li><m>R = \{ (1,1), (2,2), (3,3)  \}</m> on <m>A = \{ 1, 2, 3, 4 \}</m></li>

                            <li><m>R = \{ (1,3), (3,1), (1,1), (3,3) \}</m> on <m>A = \{ 1, 3 \}</m></li>

                            <li><m>R = \{ (1,3), (3,1), (1, 1), (3,3) \}</m> on <m>A = \{ 1, 2, 3 \}</m></li>

                        </ol></p>
                </statement>
            </exercise>

            <exercise xml:id="relations-exercise-mod5">
                <statement>
                    <p>Prove the relation described in <xref ref="relations-example-mod5" /> is an equivalence relation.</p>
                </statement>
            </exercise>


            <exercise xml:id="relations-exercise-modn">
                <statement>
                    <p>Prove the relation described in <xref ref="relations-example-modn" /> is an equivalence relation.</p>
                </statement>
            </exercise>
<!--
            <exercise xml:id="exercise-relations-same-age">
                <statement>
                    <p>Prove the relation described in <xref ref="relations-example-equivalence-class" /> is an equivalence relation.</p>
                </statement>
            </exercise>
-->
            <exercise>
                <statement>
                    <p> Let <m>A = \{ 1, 2, 3, 4 \}</m>. For each of the parts below, find an example of a relation on the set that meets the conditions described.
                        <ol>

                            <li><m>R</m> is reflexive and symmetric but not transitive.</li>

                            <li><m>R</m> is reflexive and transitive but not symmetric.</li>

                            <li><m>R</m> is symmetric and transitive but not reflexive.</li>

                            <li><m>R</m> is reflexive but neither symmetric nor transitive.</li>
                        </ol></p>

                    </statement>
                </exercise>

            <exercise>
                <statement>
                    <p>Define <m>\sim</m> on <m>\mathbb R</m> by <m>A \sim b</m> if and only if <m>|a| = |b|</m>.  Prove <m>\sim</m> is an equivalence relation on <m>\mathbb R</m>.  For an arbitrary <m>t \in \mathbb R</m>, find <m>\overline{t}</m>.</p>
                </statement>
            </exercise>


        <exercisegroup xml:id="exercises-realtions-reflexive-symmetric-transitive">
            <title>Reflexive, Symmetric, and Transitive</title>
            <introduction><p>In <xref ref="exercises-realtions-reflexive-symmetric-transitive"/>, a relation <m>R</m> is defined on a given set.  In each case, prove or disprove: (a) <m>R</m> is reflexive. (b) <m>R</m> is symmetric, (c) <m>R</m> is transitive.  In each problem where <m>R</m> is an equivalence relation, find <m>\overline{a}</m>, where <m>a</m> is any element in the set on which the relation is defined.</p></introduction>
                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb N</m> by <m>aRb</m> if and only if <m>a = 10^kb</m> for some <m>k \in \mathbb Z</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb R</m> by <m>xRy</m> if and only if <m>x - y \in \mathbb Z</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb N</m> by <m>xRy</m> if and only if <m>2 \mid (x + y)</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                         <p>Define <m>R</m> on <m>\mathbb N</m> by <m>xRy</m> if and only if <m>3 \mid (x + y)</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb R \times \mathbb R</m> by <m>(a,b)R(c,d)</m> if and only if <m>a - c \in \mathbb Z</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb R \times \mathbb R</m> by <m>(a,b)R(c,d)</m> if and only if <m>a - c \in \mathbb Z</m> and <m>b - d \in \mathbb Z</m>.</p>
                    </statement>
                </exercise>

                <exercise>
                    <statement>
                        <p>Define <m>R</m> on <m>\mathbb Z \times \mathbb Z</m> by <m>R = \{ (a, b) \in \mathbb Z \times \mathbb Z \mid |a - b| \lt 5\}</m>.</p>
                    </statement>
                </exercise>
        </exercisegroup>

            <exercise>
            	<statement>
            		<p>Let <m>R_1</m> and <m>R_2</m> be relations on <m>A</m>.
            			<ol>

            				<li>If <m>R_1</m> and <m>R_2</m> are both reflexive, if <m>R_1 \cap R_2</m>reflexive?  What about <m>R_1 \cup R_2</m>?  Justify you answers.</li>


							<li>If <m>R_1</m> and <m>R_2</m> are both symmetric, if <m>R_1 \cap R_2</m>reflexive?  What about <m>R_1 \cup R_2</m>?  Justify you answers.</li>

							<li>If <m>R_1</m> and <m>R_2</m> are both transitive, if <m>R_1 \cap R_2</m>reflexive?  What about <m>R_1 \cup R_2</m>?  Justify you answers.</li>

						</ol></p>
				</statement>
			</exercise>

			<exercise>
				<statement>
					<p>Let <m>R</m> for a relation on a finite nonempty set <m>A</m>.  What can be said about the digraph of <m>R</m> if the relation is 
						<ol>

							<li>reflexive?</li>

							<li>not reflexive?</li>

							<li>symmetric?</li>

							<li>not symmetric?</li>

							<li>transitive?</li>

							<li>not transitive?</li>

						</ol>
					Jusitify your claim for each part.</p>
				</statement>
			</exercise>

			<exercise>
				<statement>
					<p>Let <m>A = \{ 1, 2, 3, 4, 5 \}</m> and <m>R = \{ (2, 3), (2, 4), (3, 5), (2, 5), (5, 5) \}</m>.
						<ol>

							<li>Draw a digraph of this relation.</li>

							<li>Determine the properties of <m>R</m>, explaining in each case how you can tell from your digraph.</li>

						</ol></p>
				</statement>
			</exercise>

			<exercise>
				<statement>
					<p>Draw a digraph of the relation <m>\lt</m> on the set <m>A = \{ 1, 2, 3, 4 \}</m></p>
				</statement>
			</exercise>


			<exercise>
				<statement>
					<p>Draw a digraph of the relation <m>\leq</m> on the set <m>A = \{ 1, 2, 3, 4 \}</m></p>
				</statement>
			</exercise>

			<exercise>
				<statement>
					<p>Draw a digraph of <m>\mathcal P(S)</m> under the relation <m>\subseteq</m>, where <m>S = \{ a,b \}</m></p>
				</statement>
			</exercise>


        </exercises>

    </section>


    <section xml:id="relations-section-functions">
        <title>Functions and Cardinality</title>

        <p>Another special type of relation is a function.</p>

         <definition xml:id="relations-definition-function">
            <p>A <term>function</term> from a set <m>A</m> to a set <m>B</m> is a relation from <m>A</m> to <m>B</m>, where each element of <m>A</m> is paired with exactly one element of <m>B</m>.  In other words, each input value results in exactly one output value.</p>
        </definition>
        <idx><h>Function</h></idx>

        <p>Most of the mathematics that you have learned or will teach is based around the idea of functions. When students are asked to find a rule that defines a pattern, students are being asked to define a function. When you examine a sequence of values, you are examining a function. When you go to the Coke machine to get a soda, you are employing a function. (The input is your money. The output is the soda.) The dosage of medicine given to you when you are sick is a result of a function. The decision making process is an illustration of a function. Your daily life is a function whose domain is time and whose range is the activity you are doing at that time. They are everywhere!</p>

        <p>You have studied functions in lots of places. Perhaps you looked at several definitions, graphical representations, the function notation, characteristics, etc. We will be looking specifically at two characteristics of functions and their applications.</p>

        <definition xml:id="relations-definition-one-to-one">
            <p>A function <m>f</m> from a set <m>A</m> to a set <m>B</m> is called <term>one-to-one</term> provided that each output results from exactly one input. That is, if <m>b \in \range(f)</m> with <m>f(a_1)= f(a_2) = b</m>, then <m>a_1 = a_2</m>.</p>

            <p>A function <m>f</m> from a set <m>A</m> to a set <m>B</m> is called <term>onto</term> provided that every element of <m>B</m> is an element of <m>\range(f)</m>. That is, for every <m>b \in B</m>, there exists <m>A \in A</m> such that <m>f(a) = b</m>.</p>

            <p>A function from a set <m>A</m> to a set <m>B</m> that is both one-to-one and onto is called a <term>one-to-one correspondence</term> or <term>bijection</term>.</p>
        </definition>
        <idx><h>One-to-one function</h></idx>
        <idx><h>Onto function</h></idx>
        <idx><h>One-to-one correspondence</h></idx>
        <idx><h>Bijection</h></idx>

        <p>You are probably wondering what you can possibly learn about functions that you have not already seen (maybe more than once!). We are going to use the definitions of one-to-one and onto to study sets. Primarily, we are going to study the cardinality of sets.</p>


        <definition xml:id="relations-definition-cardinality">
            <p>The <term>cardinality</term> of a set <m>A</m> is the number of elments in the set, denoted <m>|A|</m>.</p>
        </definition>
        <idx><h>Cardinality</h></idx>

        <p>At first this may not seem to difficult.  You simply need to count the elements in the set.  However, what if your sets are infinite?  Again you might be asking why this is a big deal. The answer of how many elements is in an infinite set is infinitely many, right? Would you believe that there are different sizes of infinity? This is the foundation of cardinality and the study of mathematician Georg Cantor (see <xref ref="sets-paragraphs-historical-note" />).</p>

        <definition xml:id="relations-definition-cardinality-types">
            <p>Two sets, <m>A</m> and <m>B</m> have the <term>same cardinality</term> if there is a one-to-one correspondence <m>f</m> from <m>A</m> to <m>B</m>.</p>

            <p>A set <m>A</m> is finite with cardinality <m>n</m> provided that there is a one-to-one correspondence <m>f</m> from <m>A</m> to the set <m>\{1, 2, 3, 4, ... , n\}</m>.</p>

            <p>The set of natural numbers is an infinite set with cardinality <m>\aleph_0</m> (aleph naught), the smallest of all infinities. We say that the set of natural numbers are <term>countably infinite</term>.</p>
        </definition>
        <idx><h>Countably infinite</h></idx>

        <p>We are not going to spend a great deal of time discussing the sizes of infinity by name, but we are going to discuss the most common number sets to see whether they have the same cardinality as the natural numbers. Let's start with the set of whole numbers.</p>

        <theorem>
        	<p>The set of whole numbers has the same cardinality as the natural numbers.</p>
        </theorem>

        <p>At first glance, you might be inclined to say that the set of whole numbers has one more element than the set of natural numbers and thus, it is impossible for them to be of the same <q>size.</q> Remember though that we are examining a different idea of <q>size</q>. To show that the set of whole numbers has the same cardinality as the natural numbers, we simply have to demonstrate that there is a one-to-one correspondence between the two sets.</p>

        <proof>
        	<p>Consider the function <m>f</m>, mapping the set of whole numbers to the set of natural numbers, defined by <m>f(w) = w + 1</m>. Notice that this function maps <m>0</m> to <m>1</m>, <m>1</m> to <m>2</m>, <m>2</m> to <m>3</m> and so on. Because the function is linear, it is obviously one-to-one. Moreover, it is onto because if <m>n</m> is a natural number, then
        		<me>f(n - 1) = (n - 1) + 1 = n.</me>
			(Note that since the natural numbers has a smallest element of <m>1</m>, the smallest value of<m>n - 1</m> is <m>0</m> which is the smallest whole number.)</p>

			<p>Therefore, since <m>f</m> is a one-to-one correspondence, the set of whole numbers has the same cardinality as the set of natural numbers. Thus the set of whole numbers is also countably infinite.</p>
		</proof>

		<theorem>
        	<p>The set of integers is countably infinite. That is, the set of integers has the same cardinality as the set of natural numbers.</p>
        </theorem>

        <p>You may have a bit more difficulty buying into this idea. After all, the set of natural numbers is a proper subset of the integers! How can this set possibly be the same <q>size</q> as the natural numbers? Remember, cardinality is a different way to determine <q>size.</q> The infinite sets that we are examining can not be counted in the ordinary way. Cardinality provides a tool that we can use to categorize the <q>size</q> of infinite sets.</p>

        <p>After you get past the initial thoughts of denial, we can begin to think about how to develop the ont-to-one correspondence necessary to show that our claim is true. We know that an argument similar to the one we created for the set of whole numbers will not work because we wwould not account for any of the negative numbers. So what if we did a back and forth trick between the positive and negative integers. We know that we have even and odd natural numbers. What if we used the even to cover the positive numbers and the odds to cover the negative numbers? For example, we can send <m>1</m> to <m>0</m>, <m>2</m> to <m>1</m>, <m>3</m> to <m>-1</m>, <m>4</m> to <m>2</m>, <m>5</m> to <m>-2</m>, and so on.</p>

        <proof>
        	<p>Consider the function <m>f</m>, mapping the natural numbers to the integers, defined by the following criteria.
        		<ul>

        			<li>If <m>n = 1</m>, then <m>f(n) = 0</m>.</li>

        			<li>If <m>n</m> is an even integer, then <m>f(n) = n/2</m>.</li>

        			<li>If n is an odd integer and <m>n \neq 1</m>, then <m>f(n) = -(n - 1)/2</m>.</li>

        		</ul>
        	Note that this function is one-to-one. We can examine the graph to verify, if necessary. Also the function is onto. To see this, let <m>x</m> be any integer. If <m>x</m> is <m>0</m>, then we know <m>f(1) = 0</m>. If <m>x</m> is positive, then <m>f(2x) = 2x/2 = x</m>. Notice that <m>2x</m> is an even natural number. If <m>x</m> is negative, then
        		<me>f(-2x+1)=-((-2x+1)-1)/2=-(-2x)/2=x.</me>
        	Notice <m>-2x+1</m> is an odd natural number. Thus in all three cases, <m>x</m> is mapped to by some natural number and <m>f</m> is onto. Therefore, <m>f</m> is a one-to-one correspondence and the set of integers have the same cardinality as the set of natural numbers, and is thus a countably infinite set.</p>
        </proof>

        <theorem>
        	<p>The set of rational numbers is countably infinite. That is, the set of rational numbers has the same cardinality as the set of natural numbers.</p>
        </theorem>

        <p>This set is a bit harder to work with than the previous ones because the process of listing the rational numbers is difficult. We will eliminate some of the difficulty by working only with positive rational numbers. You should be able to explain how to adapt the solution to the complete set once you see the pattern.</p>

        <p>Consider <xref ref="figure-relations-rationals-with-duplicates" />. Notice that eventually, if the process was allowed to continue indefinitely, all rational numbers would be listed. Some numbers however are represented more than once. In order to preserve the one-to-one requirement, we need to eliminate any numbers that are equivalent to a rational number previously listed. The image below has made that adjustment.</p> 

        <figure xml:id="figure-relations-rationals-with-duplicates">
            <caption>Listing the rational numbers</caption>
            <image width="50%" xml:id="relations-rationals-with-duplicates">
                <description>two overlapping circles with the common part shaded</description>
                <latex-image>
                <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
               <![CDATA[
                \begin{tikzpicture}[scale=1]
                
					\tikzstyle{keepstyle} =[]
					\node at (0,0) {$\vdots$};
					\node[keepstyle] (51) at (0,1) {$\frac{5}{1}$};
					\node[keepstyle] (41) at (0,2) {$\frac{4}{1}$};
					\node[keepstyle] (31) at (0,3) {$\frac{3}{1}$};
					\node[keepstyle] (21) at (0,4) {$\frac{2}{1}$};
					\node[keepstyle] (11) at (0,5) {$\frac{1}{1}$};
					\node at (1,0) {$\vdots$};
					\node[keepstyle] (52) at (1,1) {$\frac{5}{2}$};
					\node at (1,2) {$\frac{4}{2}$};
					\node[keepstyle] (32) at (1,3) {$\frac{3}{2}$};
					\node at (1,4) {$\frac{2}{2}$};
					\node[keepstyle] (12) at (1,5) {$\frac{1}{2}$};
					\node at (2,0) {$\vdots$};
					\node at (2,1) {$\frac{5}{3}$};
					\node[keepstyle] (43) at (2,2) {$\frac{4}{3}$};
					\node at (2,3) {$\frac{3}{3}$};
					\node[keepstyle] (23) at (2,4) {$\frac{2}{3}$};
					\node[keepstyle] (13) at (2,5) {$\frac{1}{3}$};
					\node at (3,0) {$\vdots$};
					\node at (3,1) {$\frac{5}{4}$};
					\node at (3,2) {$\frac{4}{4}$};
					\node[keepstyle] (34) at (3,3) {$\frac{3}{4}$};
					\node at (3,4) {$\frac{2}{4}$};
					\node[keepstyle] (14) at (3,5) {$\frac{1}{4}$};
					\node at (4,0) {$\vdots$};
					\node  at (4,1) {$\frac{5}{5}$};
					\node at (4,2) {$\frac{4}{5}$};
					\node at (4,3) {$\frac{3}{5}$};
					\node[keepstyle] (25) at (4,4) {$\frac{2}{5}$};
					\node[keepstyle] (15) at (4,5) {$\frac{1}{5}$};
					\node at (5,0) {$\vdots$};
					\node  at (5,1) {$\frac{5}{6}$};
					\node at (5,2) {$\frac{4}{6}$};
					\node at (5,3) {$\frac{3}{6}$};
					\node at (5,4) {$\frac{2}{6}$};
					\node[keepstyle] (16) at (5,5) {$\frac{1}{6}$};
					\node at (6,1) {$\cdots$};
					\node at (6,2) {$\cdots$};
					\node at (6,3) {$\cdots$};
					\node at (6,4) {$\cdots$};
					\node at (6,5) {$\cdots$};
                \end{tikzpicture}]]>
                </latex-image>
            </image>
        </figure>

        <p>Now we are going to develop our map. Unlike previous examples, we are not going to state our rule but we are going to let a picture do the talking. Remember that all we have to do is to demonstrate a one-to-one onto function.</p>

       <figure xml:id="figure-relations-rationals">
            <caption>Listing the rational numbers without repetition.</caption>
            <image width="50%" xml:id="relations-rationals">
                <description>two overlapping circles with the common part shaded</description>
                <latex-image>
                <!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
               <![CDATA[
                \begin{tikzpicture}[scale=1]
                
					\tikzstyle{keepstyle} =[rectangle, rounded corners, draw, fill=white]
					\node at (0,0) {$\vdots$};
					\node[keepstyle] (51) at (0,1) {$\frac{5}{1}$};
					\node[keepstyle] (41) at (0,2) {$\frac{4}{1}$};
					\node[keepstyle] (31) at (0,3) {$\frac{3}{1}$};
					\node[keepstyle] (21) at (0,4) {$\frac{2}{1}$};
					\node[keepstyle] (11) at (0,5) {$\frac{1}{1}$};
					\node at (1,0) {$\vdots$};
					\node[keepstyle] (52) at (1,1) {$\frac{5}{2}$};
					\node at (1,2) {$\frac{4}{2}$};
					\node[keepstyle] (32) at (1,3) {$\frac{3}{2}$};
					\node at (1,4) {$\frac{2}{2}$};
					\node[keepstyle] (12) at (1,5) {$\frac{1}{2}$};
					\node at (2,0) {$\vdots$};
					\node at (2,1) {$\frac{5}{3}$};
					\node[keepstyle] (43) at (2,2) {$\frac{4}{3}$};
					\node at (2,3) {$\frac{3}{3}$};
					\node[keepstyle] (23) at (2,4) {$\frac{2}{3}$};
					\node[keepstyle] (13) at (2,5) {$\frac{1}{3}$};
					\node at (3,0) {$\vdots$};
					\node at (3,1) {$\frac{5}{4}$};
					\node at (3,2) {$\frac{4}{4}$};
					\node[keepstyle] (34) at (3,3) {$\frac{3}{4}$};
					\node at (3,4) {$\frac{2}{4}$};
					\node[keepstyle] (14) at (3,5) {$\frac{1}{4}$};
					\node at (4,0) {$\vdots$};
					\node  at (4,1) {$\frac{5}{5}$};
					\node at (4,2) {$\frac{4}{5}$};
					\node at (4,3) {$\frac{3}{5}$};
					\node[keepstyle] (25) at (4,4) {$\frac{2}{5}$};
					\node[keepstyle] (15) at (4,5) {$\frac{1}{5}$};
					\node at (5,0) {$\vdots$};
					\node  at (5,1) {$\frac{5}{6}$};
					\node at (5,2) {$\frac{4}{6}$};
					\node at (5,3) {$\frac{3}{6}$};
					\node at (5,4) {$\frac{2}{6}$};
					\node[keepstyle] (16) at (5,5) {$\frac{1}{6}$};
					\node at (6,1) {$\cdots$};
					\node at (6,2) {$\cdots$};
					\node at (6,3) {$\cdots$};
					\node at (6,4) {$\cdots$};
					\node at (6,5) {$\cdots$};
					\draw [-latex,red, thick] (11) -- (12);
					\draw [-latex, red, thick] (12) -- (21);
					\draw [-latex, red, thick] (21) -- (31);
					\draw [-latex, red, thick] (31) -- (13);
					\draw [-latex, red, thick] (13) -- (14);
					\draw [-latex, red, thick] (14) -- (23);
					\draw [-latex, red, thick] (23) -- (32);
					\draw [-latex, red, thick] (32) -- (41);
					\draw [-latex, red, thick] (41) -- (51);
					\draw [-latex, red, thick] (51) -- (15);
					\draw [-latex, red, thick] (15) -- (16);
					\draw [-latex, red, thick] (16) -- (25);
					\draw [-latex, red, thick] (25) -- (34);
					\draw [-latex, red, thick] (34) -- (43);
					\draw [-latex, red, thick] (43) -- (52);
                \end{tikzpicture}]]>
                </latex-image>
            </image>
        </figure>

        <p>So <m>1</m> will map to <m>1</m>, <m>2</m> to <m>1/2</m>, <m>3</m> to <m>2/1</m>, <m>4</m> to <m>3/1</m>, <m>5</m> to <m>1/3</m>, etc. Notice that each rational number will eventually be the output for our function and will be the output for exactly one natural number. Thus, we have created a one-to-one correspondence between the natural numbers and the positive rational numbers. We can easily extend this idea by sending the even natural numbers to the positive rational numbers and the odd natural numbers to the negative rational numbers to show that there is a one-to-one correspondence from the natural numbers to the set of all rational numbers. Therefore, the set of rational numbers is countably infinite.</p>

        <theorem>
            <p>The set of real numbers is not countably infinite. That is, the set of real numbers does not have the same cardinality as the set of natural numbers.</p>
        </theorem>

        <p>The real numbers will provide an example of a set that is <q>larger</q> than the set of natural numbers, whole numbers, integers, and rational numbers. To demonstrate, we will actually show that even the interval (0, 1) is not countable infinite. In other words, the set of numbers greater than zero and less than one cannot be put into a one-to-one correspondence with the set of natural numbers. To show that this is true, we will use a proof by contradiction.</p>

        <p>We have to clear up a couple points before we begin. Frist, you whould realize that every number in the interval <m>(0, 1)</m> can be written as an infinite decimal. For example,
            <me>0.25 = 0.250000000000000000\ldots</me>
        Also recall from our discussions with geometric sequences and series, we discussed that <m>0.99999\ldots = 1</m>. You can prove this using the fact that <m>0.9999\ldots</m> is a geometric series with <m>r = 1/10</m> and <m>a = 9/10</m>. Since <m>r \lt 1</m>, the infinite sum of the series is <m>S=1/(1-r)</m>. With this in mind, anytime that we have this situation occur (a infinite number of repeating <m>9</m>'s), we are going to assume that we will equate this number with its equivalency in terms of repeated zeros. For example,
            <me>0.555999999999\ldots = 0.556000000000\ldots</me>. 
        This will ensure that our function is one-to-one.</p>

        <proof>
            <p>Consider the interval <m>(0,1)</m> and assume to the contrary that this set of numbers in countable infinite. Since the set is countable infinite, every number in the set can be listed in a one-to-one correspondence with the natural numbers. So we can say that there is a first element which maps to <m>1</m>, a second element which maps to <m>2</m>, so on and so forth. In order to list these values in an ordered way, let's call the first number <m>a_1</m> and denote it as
                <me>a_1 = 0.d_{11} d_{12} d_{13} d_{14} d_{15} d_{16} d_{17} d_{18} d_{19} \ldots,</me>
            where <m>d_{11}</m> is the digit in the first number in our list and in the first position beyond the decimal, <m>d_{12}</m> is the digit in the first number in the list and in the second position beyond the decimal, etc.</p>

            <p>Then the second number in the list would look like
                <me>a_2 = 0.d_{21} d_{22} d_{23} d_{24} d_{25} d_{26} d_{27} d_{28} d_{29} \ldots,</me>
            where <m>d_{21}</m> is the digit in the second number in our list and in the first position beyond the decimal, <m>d_{22}</m> is the digit in the second number in the list and in the second position beyond the decimal, etc.</p>

            <p>In general then, the ith number in the list would look like
                <me>a_i = 0.d_{i1} d_{i2} d_{i3} d_{i4} d_{i5} d_{i6} d_{i7} d_{i8} d_{i9} \ldots,</me>
            where <m>d_{i1}</m> is the digit in the <m>i</m>th number in our list and in the first position beyond the decimal, <m>d_{i2}</m> is the digit in the <m>i</m>th number in the list and in the second position beyond the decimal, etc.</p>

            <p>We now have the following list of all numbers that are in the interval <m>(0, 1)</m>,
                <md>
                    <mrow>a_1 &amp; = 0.d_{11} d_{12} d_{13} d_{14} d_{15} d_{16} d_{17} d_{18} d_{19} \ldots,</mrow>
                    <mrow>a_2 &amp; = 0.d_{21} d_{22} d_{23} d_{24} d_{25} d_{26} d_{27} d_{28} d_{29} \ldots,</mrow>
                    <mrow>a_3 &amp; = 0.d_{31} d_{32} d_{33} d_{34} d_{35} d_{36} d_{37} d_{38} d_{39} \ldots,</mrow>
                    <mrow>a_4 &amp; = 0.d_{41} d_{42} d_{43} d_{44} d_{45} d_{46} d_{47} d_{48} d_{49} \ldots,</mrow>
                    <mrow>a_5 &amp; = 0.d_{51} d_{52} d_{53} d_{54} d_{55} d_{56} d_{57} d_{58} d_{59} \ldots,</mrow>
                    <mrow>&amp; \cdots</mrow>
                    <mrow>a_i &amp; = 0.d_{i1} d_{i2} d_{i3} d_{i4} d_{i5} d_{i6} d_{i7} d_{i8} d_{i9} \ldots</mrow>
                </md></p>

            <p>The key here is the realization that based on our assumption, every single number in the interval <m>(0, 1)</m> is represented somewhere in this list. This is where we will obtain our contradiction.</p>

            <p>Consider the number
                <me>d = 0.d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} d_{8} d_{9} \ldots,</me>
            where <m>d_j</m> is the <m>j</m>th digit beyond the decimal and is determined based on the following criteria:
                <blockquote>If <m>d_{jj} \neq 5</m>, then <m>d_j = 5</m>; otherwise, <m>d_j = 2</m>.</blockquote>
            Notice that this means that by definition of the number <m>d</m>, <m>d_1 \neq d_{11}</m> and so the number <m>d</m> is not the number <m>a_1</m>. Similarly <m>d \neq a_2</m> because <m>d_2 \neq d_{22}</m>. Continuing this line of thought, <m>d</m> is not the same as any number in this list because <m>d_j \neq d_{jj}</m> for any natural number <m>j</m>. Thus <m></m> is a number between <m>0</m> and <m>1</m> that is not in our list. This contradicts our assumption that there exists a one-to-one correspondence between the set of numbers greater than <m>0</m> and less than <m>1</m> and the set of natural numbers.  Therefore the set of numbers in the interval <m>(0, 1)</m> is not countably infinite</p>
        </proof>

        <p>We say then that the set of numbers in the interval <m>(0, 1)</m> is <term>uncountable</term> and thus, so is the set of real numbers.</p>
        <idx><h>Uncountable</h></idx>

        
        <exercises>

        	<exercise>
        		<statement>
        			<p>Show that the set of even numbers is countably infinite.</p>
        		</statement>
        	</exercise>

        </exercises>



    </section>


</chapter>