SB Exercises

Exercises 1.4 Exercises

View Source for exercises

  <exercises xml:id="exercises-sets" xml:base="exercises/sets.xml">
    <title>Exercises</title>

    <subexercises>
        <title>Warm-up</title>
        <introduction>
            <p>
                This is a meaningless subdivision of the exercises for the sake of testing output.
            </p>
        </introduction>

        <exercise number="1">
            <statement>
                <p>Suppose that<md>
                    <mrow>A &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},</mrow>
                    <mrow>B &amp; = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},</mrow>
                    <mrow>C &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of 5}\}.</mrow>
                </md></p>
                <p>Describe each of the following sets.</p>
                <ol cols="2">
                    <li><p><m>A \cap B</m></p></li>
                    <li><p><m>B \cap C</m></p></li>
                    <li><p><m>A \cup B</m></p></li>
                    <li><p><m>A \cap (B \cup C)</m></p></li>
                </ol>
            </statement>
            <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
            <!-- <hint>
                <p>(a) <m>A \cap B = \{ 2 \}</m>; (b) <m>B \cap C = \{ 5 \}</m>.</p>
            </hint> -->
        </exercise>

        <exercise number="2">
            <statement>
                <p>If <m>A = \{ a, b, c \}</m>, <m>B = \{ 1, 2, 3 \}</m>, <m>C = \{ x \}</m>, and <m>D = \emptyset</m>, list all of the elements in each of the following sets.</p>
                <ol cols="2">
                    <li><p><m>A \times B</m></p></li>
                    <li><p><m>B \times A</m></p></li>
                    <li><p><m>A \times B \times C</m></p></li>
                    <li><p><m>A \times D</m></p></li>
                </ol>
            </statement>
            <hint>
                <p>(a) <m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>; (d) <m>A \times D = \emptyset</m>.</p>
            </hint>
        </exercise>

        <exercise number="3">
            <statement>
                <p>Find an example of two nonempty sets <m>A</m> and <m>B</m> for which <m>A \times B = B \times A</m> is true.</p>
            </statement>
        </exercise>

        <exercise number="4">
            <statement>
                <p>Prove  <m>A \cup \emptyset = A</m> and <m>A \cap \emptyset = \emptyset</m>.</p>
            </statement>
        </exercise>

        <exercise number="5">
            <statement>
                <p>Prove <m>A \cup B = B \cup A</m> and <m>A \cap B = B \cap A</m>.</p>
            </statement>
        </exercise>

        <exercise number="6">
            <statement>
                <p>Prove <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
            </statement>
            <hint>
                <p>If <m>x \in A \cup (B \cap C)</m>, then either <m>x \in A</m> or <m>x \in B \cap C</m>.  Thus,   <m> x \in A \cup B</m> and <m>A \cup C</m>.  Hence, <m> x \in (A \cup B) \cap (A \cup C)</m>. Therefore, <m>  A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)</m>.  Conversely, if <m>x \in (A \cup B) \cap (A \cup C)</m>, then <m>x \in A \cup B</m> and <m>A \cup C</m>. Thus, <m>x \in A</m> or <m>x</m> is in both <m>B</m> and <m>C</m>. So <m>x \in A \cup (B \cap C)</m> and therefore <m>(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)</m>. Hence, <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
            </hint>
        </exercise>

        <exercise number="7">
            <statement>
                <p>Prove <m>A \cap (B \cup C) = (A \cap B) \cup (A \cap C)</m>.</p>
            </statement>
        </exercise>

        <exercise number="8">
            <statement>
                <p>Prove  <m>A \subset B</m> if and only if <m>A \cap B = A</m>.</p>
            </statement>
        </exercise>

        <exercise number="9">
            <statement>
                <p>Prove <m>(A \cap B)' = A' \cup B'</m>.</p>
            </statement>
        </exercise>

        <exercise number="10">
            <statement>
                <p>Prove  <m>A \cup B = (A \cap B) \cup (A \setminus B) \cup (B \setminus A)</m>.</p>
            </statement>
            <hint>
                <p><m>(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B</m>.</p>
            </hint>
        </exercise>

        <exercise number="11">
            <statement>
                <p>Prove  <m>(A \cup B) \times C = (A \times C ) \cup (B \times C)</m>.</p>
            </statement>
        </exercise>

        <exercise number="12">
            <statement>
                <p>Prove  <m>(A \cap B) \setminus B = \emptyset</m>.</p>
            </statement>
        </exercise>

        <exercise number="13">
            <statement>
                <p>Prove  <m>(A \cup B) \setminus B = A \setminus B</m>.</p>
            </statement>
        </exercise>

        <exercise number="14">
            <statement>
                <p>Prove  <m>A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)</m>.</p>
            </statement>
            <hint>
                <p><m>A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)</m>.</p>
            </hint>
        </exercise>
    </subexercises>

    <subexercises>
        <title>More Exercises</title>
        <introduction>
            <p>
                This is a meaningless subdivision of the exercises for the sake of testing output.
            </p>
        </introduction>

        <exercise number="15">
            <statement>
                <p>Prove  <m>A \cap (B \setminus C) = (A \cap B) \setminus (A \cap C)</m>.</p>
            </statement>
        </exercise>

        <exercise number="16">
            <statement>
                <p>Prove  <m>(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)</m>.</p>
            </statement>
        </exercise>

        <exercise number="17">
            <statement>
                <p>Which of the following relations <m>f: {\mathbb Q} \rightarrow {\mathbb Q}</m> define a mapping? In each case, supply a reason why <m>f</m> is or is not a mapping.</p>
                <ol cols="2">
                    <li><p><m>\displaystyle f(p/q) = \frac{p+ 1}{p - 2}</m></p></li>
                    <li><p><m>\displaystyle f(p/q) = \frac{3p}{3q}</m></p></li>
                    <li><p><m>\displaystyle f(p/q) = \frac{p+q}{q^2}</m></p></li>
                    <li><p><m>\displaystyle f(p/q) = \frac{3 p^2}{7 q^2} - \frac{p}{q}</m></p></li>
                </ol>
            </statement>
            <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
            <!-- <hint>
                <p>(a) Not a map since <m>f(2/3)</m> is undefined; (b) this is a map; (c) not a map, since <m>f(1/2) = 3/4</m> but <m>f(2/4)=3/8</m>; (d) this is a map.</p>
            </hint> -->
        </exercise>

        <exercise number="18">
            <statement>
                <p>Determine which of the following functions are one-to-one and which are onto.  If the function is not onto, determine its range.</p>
                <ol>
                    <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = e^x</m></p></li>
                    <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(n) = n^2 + 3</m></p></li>
                    <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = \sin x</m></p></li>
                    <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(x) = x^2</m></p></li>
                </ol>
            </statement>
            <hint>
                <p>(a) <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.  (c) <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.</p>
            </hint>
        </exercise>

        <exercise number="19">
            <statement>
                <p>Let <m>f :A \rightarrow B</m> and <m>g : B \rightarrow C</m> be invertible mappings; that is, mappings such that <m>f^{-1}</m> and <m>g^{-1}</m> exist. Show that <m>(g \circ f)^{-1} =f^{-1} \circ g^{-1}</m>.</p>
            </statement>
        </exercise>

        <exercise number="20">
            <statement>
                <ol>
                    <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is one-to-one but not onto.</p></li>
                     <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is onto but not one-to-one.</p></li>
                </ol>
            </statement>
            <hint>
                <p>(a) <m>f(n) = n + 1</m>.</p>
            </hint>
        </exercise>

        <exercise number="21">
            <statement>
                <p>Prove the relation defined on <m>{\mathbb R}^2</m> by <m>(x_1, y_1 ) \sim (x_2, y_2)</m> if <m>x_1^2 + y_1^2 = x_2^2 + y_2^2</m> is  an equivalence relation.</p>
            </statement>
        </exercise>

        <exercise number="22">
            <statement>
                <p>Let <m>f : A \rightarrow B</m> and <m>g : B \rightarrow C</m> be maps.</p>
                <ol>
                    <li><p>If <m>f</m> and <m>g</m> are both one-to-one functions, show that <m>g \circ f</m> is one-to-one.</p></li>
                    <li><p>If <m>g \circ f</m> is onto, show that <m>g</m> is onto.</p></li>
                    <li><p>If <m>g \circ f</m> is one-to-one, show that <m>f</m> is one-to-one.</p></li>
                    <li><p>If <m>g \circ f</m> is one-to-one and <m>f</m> is onto, show that <m>g</m> is one-to-one.</p></li>
                    <li><p>If <m>g \circ f</m> is onto and <m>g</m> is one-to-one, show that <m>f</m> is onto.</p></li>
                </ol>
            </statement>
            <hint>
                <p>(a) Let <m>x, y \in A</m>. Then <m>g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))</m>.  Thus, <m>f(x) = f(y)</m> and <m>x = y</m>,  so <m>g \circ f</m> is one-to-one. (b) Let <m>c \in C</m>, then <m>c = (g \circ f)(x) = g(f(x))</m> for some <m>x \in A</m>. Since <m>f(x) \in B</m>, <m>g</m> is onto.</p>
            </hint>
        </exercise>

        <exercise number="23">
            <statement>
                <p>Define a function on the real numbers by
                    <me>f(x) = \frac{x + 1}{x - 1}.</me>
                What are the domain and range of <m>f</m>? What is the inverse of <m>f</m>?  Compute <m>f \circ f^{-1}</m> and <m>f^{-1} \circ f</m>.</p>
            </statement>
            <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
            <!-- <hint>
                <p><m>f^{-1}(x) = (x+1)/(x-1)</m>.</p>
            </hint> -->
        </exercise>

        <exercise number="24">
            <statement>
                <p>Let <m>f: X \rightarrow Y</m> be a map with <m>A_1, A_2 \subset X</m> and <m>B_1, B_2 \subset Y</m>.</p>
                <ol>
                    <li><p>Prove <m>f( A_1 \cup A_2 ) = f( A_1) \cup f( A_2 )</m>.</p></li>
                    <li><p>Prove <m>f( A_1 \cap A_2 ) \subset f( A_1) \cap f( A_2 )</m>.  Give an example in which equality fails.</p></li>
                    <li><p>Prove <m>f^{-1}( B_1 \cup B_2 ) = f^{-1}( B_1) \cup f^{-1}(B_2 )</m>, where <me>f^{-1}(B) = \{ x \in X : f(x) \in B \}.</me></p></li>
                    <li><p>Prove <m>f^{-1}( B_1 \cap B_2 ) = f^{-1}( B_1) \cap f^{-1}( B_2 )</m>.</p></li>
                    <li><p>Prove <m>f^{-1}( Y \setminus B_1 ) = X \setminus f^{-1}( B_1)</m>.</p></li>
                </ol>
            </statement>
            <hint>
                <p>(a) Let <m>y \in f(A_1 \cup A_2)</m>.  Then there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Hence, <m> y \in f(A_1)</m> or <m>f(A_2) </m>.  Therefore, <m> y \in f(A_1) \cup f(A_2)</m>. Consequently, <m> f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)</m>. Conversely, if <m>y \in f(A_1) \cup f(A_2)</m>, then <m> y \in f(A_1)</m> or <m>f(A_2)</m>.  Hence, there exists an <m>x \in A_1</m> or there exists an <m>x \in A_2</m> such that <m>f(x) = y</m>.  Thus, there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Therefore, <m> f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)</m>, and <m>f(A_1 \cup A_2) = f(A_1) \cup f(A_2)</m>.</p>
            </hint>
        </exercise>

        <exercise number="25">
            <statement>
                <p>Determine whether or not the following relations are equivalence relations on the given set.  If the relation is an equivalence relation, describe the partition given by it.  If the relation is not an equivalence relation, state why it fails to be one.</p>
                <ol cols="2">
                    <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>x \geq y</m></p></li>
                    <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>mn &gt; 0</m></p></li>
                    <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>|x - y| \leq 4</m></p></li>
                    <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>m \equiv n \pmod{6}</m></p></li>
                </ol>
            </statement>
            <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
            <!-- <hint>
                <p>(a) NThe relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.</p>
            </hint> -->
        </exercise>

        <exercise number="26">
            <statement>
                <p>Define a relation <m>\sim</m> on <m>{\mathbb R}^2</m> by stating that <m>(a, b) \sim (c, d)</m> if and only if <m>a^2 + b^2 \leq c^2 + d^2</m>. Show that <m>\sim</m> is reflexive and transitive but not symmetric.</p>
            </statement>
        </exercise>

        <exercise number="27">
            <statement>
                <p>Show that an <m>m \times n</m> matrix gives rise to a well-defined  map from <m>{\mathbb R}^n</m> to <m>{\mathbb R}^m</m>.</p>
            </statement>
        </exercise>

        <exercise number="28">
            <statement>
                <p>Find the error in the following argument by providing a counterexample. <q>The reflexive property is redundant in the axioms for an equivalence relation.  If <m>x \sim y</m>, then <m>y \sim x</m> by the symmetric property.  Using the transitive property, we can deduce that <m>x \sim x</m>.</q> </p>
            </statement>
            <hint>
                <p>Let <m>X = {\mathbb N} \cup \{ \sqrt{2}\, \}</m> and define <m>x \sim y</m> if <m>x + y \in {\mathbb N}</m>.</p>
            </hint>
        </exercise>

        <exercise number="29">
            <title>Projective Real Line</title>
            <statement>
                <p>Define a relation on <m>{\mathbb R}^2 \setminus  \{ (0,0) \}</m> by letting <m>(x_1, y_1) \sim (x_2, y_2)</m> if there exists a nonzero real number <m>\lambda</m> such that <m>(x_1, y_1)  = ( \lambda x_2, \lambda y_2)</m>.  Prove that <m>\sim</m> defines an equivalence relation on <m>{\mathbb R}^2 \setminus (0,0)</m>.  What are the corresponding  equivalence classes?  This equivalence relation defines the projective line, denoted by  <m>{\mathbb P}({\mathbb R}) </m>, which is very important in geometry.</p>
            </statement>
        </exercise>
    </subexercises>

</exercises>

Warm-up

View Source for subexercises

<subexercises>
    <title>Warm-up</title>
    <introduction>
        <p>
            This is a meaningless subdivision of the exercises for the sake of testing output.
        </p>
    </introduction>

    <exercise number="1">
        <statement>
            <p>Suppose that<md>
                <mrow>A &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},</mrow>
                <mrow>B &amp; = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},</mrow>
                <mrow>C &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of 5}\}.</mrow>
            </md></p>
            <p>Describe each of the following sets.</p>
            <ol cols="2">
                <li><p><m>A \cap B</m></p></li>
                <li><p><m>B \cap C</m></p></li>
                <li><p><m>A \cup B</m></p></li>
                <li><p><m>A \cap (B \cup C)</m></p></li>
            </ol>
        </statement>
        <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
        <!-- <hint>
            <p>(a) <m>A \cap B = \{ 2 \}</m>; (b) <m>B \cap C = \{ 5 \}</m>.</p>
        </hint> -->
    </exercise>

    <exercise number="2">
        <statement>
            <p>If <m>A = \{ a, b, c \}</m>, <m>B = \{ 1, 2, 3 \}</m>, <m>C = \{ x \}</m>, and <m>D = \emptyset</m>, list all of the elements in each of the following sets.</p>
            <ol cols="2">
                <li><p><m>A \times B</m></p></li>
                <li><p><m>B \times A</m></p></li>
                <li><p><m>A \times B \times C</m></p></li>
                <li><p><m>A \times D</m></p></li>
            </ol>
        </statement>
        <hint>
            <p>(a) <m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>; (d) <m>A \times D = \emptyset</m>.</p>
        </hint>
    </exercise>

    <exercise number="3">
        <statement>
            <p>Find an example of two nonempty sets <m>A</m> and <m>B</m> for which <m>A \times B = B \times A</m> is true.</p>
        </statement>
    </exercise>

    <exercise number="4">
        <statement>
            <p>Prove  <m>A \cup \emptyset = A</m> and <m>A \cap \emptyset = \emptyset</m>.</p>
        </statement>
    </exercise>

    <exercise number="5">
        <statement>
            <p>Prove <m>A \cup B = B \cup A</m> and <m>A \cap B = B \cap A</m>.</p>
        </statement>
    </exercise>

    <exercise number="6">
        <statement>
            <p>Prove <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
        </statement>
        <hint>
            <p>If <m>x \in A \cup (B \cap C)</m>, then either <m>x \in A</m> or <m>x \in B \cap C</m>.  Thus,   <m> x \in A \cup B</m> and <m>A \cup C</m>.  Hence, <m> x \in (A \cup B) \cap (A \cup C)</m>. Therefore, <m>  A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)</m>.  Conversely, if <m>x \in (A \cup B) \cap (A \cup C)</m>, then <m>x \in A \cup B</m> and <m>A \cup C</m>. Thus, <m>x \in A</m> or <m>x</m> is in both <m>B</m> and <m>C</m>. So <m>x \in A \cup (B \cap C)</m> and therefore <m>(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)</m>. Hence, <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
        </hint>
    </exercise>

    <exercise number="7">
        <statement>
            <p>Prove <m>A \cap (B \cup C) = (A \cap B) \cup (A \cap C)</m>.</p>
        </statement>
    </exercise>

    <exercise number="8">
        <statement>
            <p>Prove  <m>A \subset B</m> if and only if <m>A \cap B = A</m>.</p>
        </statement>
    </exercise>

    <exercise number="9">
        <statement>
            <p>Prove <m>(A \cap B)' = A' \cup B'</m>.</p>
        </statement>
    </exercise>

    <exercise number="10">
        <statement>
            <p>Prove  <m>A \cup B = (A \cap B) \cup (A \setminus B) \cup (B \setminus A)</m>.</p>
        </statement>
        <hint>
            <p><m>(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B</m>.</p>
        </hint>
    </exercise>

    <exercise number="11">
        <statement>
            <p>Prove  <m>(A \cup B) \times C = (A \times C ) \cup (B \times C)</m>.</p>
        </statement>
    </exercise>

    <exercise number="12">
        <statement>
            <p>Prove  <m>(A \cap B) \setminus B = \emptyset</m>.</p>
        </statement>
    </exercise>

    <exercise number="13">
        <statement>
            <p>Prove  <m>(A \cup B) \setminus B = A \setminus B</m>.</p>
        </statement>
    </exercise>

    <exercise number="14">
        <statement>
            <p>Prove  <m>A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)</m>.</p>
        </statement>
        <hint>
            <p><m>A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)</m>.</p>
        </hint>
    </exercise>
</subexercises>

This is a meaningless subdivision of the exercises for the sake of testing output.

🔗

1.

View Source for exercise

<exercise number="1">
    <statement>
        <p>Suppose that<md>
            <mrow>A &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},</mrow>
            <mrow>B &amp; = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},</mrow>
            <mrow>C &amp; = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of 5}\}.</mrow>
        </md></p>
        <p>Describe each of the following sets.</p>
        <ol cols="2">
            <li><p><m>A \cap B</m></p></li>
            <li><p><m>B \cap C</m></p></li>
            <li><p><m>A \cup B</m></p></li>
            <li><p><m>A \cap (B \cup C)</m></p></li>
        </ol>
    </statement>
    <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
    <!-- <hint>
        <p>(a) <m>A \cap B = \{ 2 \}</m>; (b) <m>B \cap C = \{ 5 \}</m>.</p>
    </hint> -->
</exercise>

Suppose that

\begin{aligned} A & = {x : x \in N and x is even}, \\ B & = {x : x \in N and x is prime}, \\ C & = {x : x \in N and x is a multiple of 5} . \end{aligned}

🔗

Describe each of the following sets.

🔗

$A \cap B$
🔗

🔗
$B \cap C$
🔗

🔗
$A \cup B$
🔗

🔗
$A \cap (B \cup C)$
🔗

🔗

🔗

2.

View Source for exercise

<exercise number="2">
    <statement>
        <p>If <m>A = \{ a, b, c \}</m>, <m>B = \{ 1, 2, 3 \}</m>, <m>C = \{ x \}</m>, and <m>D = \emptyset</m>, list all of the elements in each of the following sets.</p>
        <ol cols="2">
            <li><p><m>A \times B</m></p></li>
            <li><p><m>B \times A</m></p></li>
            <li><p><m>A \times B \times C</m></p></li>
            <li><p><m>A \times D</m></p></li>
        </ol>
    </statement>
    <hint>
        <p>(a) <m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>; (d) <m>A \times D = \emptyset</m>.</p>
    </hint>
</exercise>

A = {a, b, c},

B = {1, 2, 3},

C = {x},

and

D = \emptyset,

list all of the elements in each of the following sets.

🔗

$A \times B$
🔗

🔗
$B \times A$
🔗

🔗
$A \times B \times C$
🔗

🔗
$A \times D$
🔗

🔗

Hint.

View Source for hint

<hint>
    <p>(a) <m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>; (d) <m>A \times D = \emptyset</m>.</p>
</hint>

(a)

A \times B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)};

(d)

A \times D = \emptyset .

🔗

3.

View Source for exercise

<exercise number="3">
    <statement>
        <p>Find an example of two nonempty sets <m>A</m> and <m>B</m> for which <m>A \times B = B \times A</m> is true.</p>
    </statement>
</exercise>

Find an example of two nonempty sets

A

and

B

for which

A \times B = B \times A

is true.

🔗

4.

View Source for exercise

<exercise number="4">
    <statement>
        <p>Prove  <m>A \cup \emptyset = A</m> and <m>A \cap \emptyset = \emptyset</m>.</p>
    </statement>
</exercise>

Prove

A \cup \emptyset = A

and

A \cap \emptyset = \emptyset .

🔗

5.

View Source for exercise

<exercise number="5">
    <statement>
        <p>Prove <m>A \cup B = B \cup A</m> and <m>A \cap B = B \cap A</m>.</p>
    </statement>
</exercise>

Prove

A \cup B = B \cup A

and

A \cap B = B \cap A .

🔗

6.

View Source for exercise

<exercise number="6">
    <statement>
        <p>Prove <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
    </statement>
    <hint>
        <p>If <m>x \in A \cup (B \cap C)</m>, then either <m>x \in A</m> or <m>x \in B \cap C</m>.  Thus,   <m> x \in A \cup B</m> and <m>A \cup C</m>.  Hence, <m> x \in (A \cup B) \cap (A \cup C)</m>. Therefore, <m>  A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)</m>.  Conversely, if <m>x \in (A \cup B) \cap (A \cup C)</m>, then <m>x \in A \cup B</m> and <m>A \cup C</m>. Thus, <m>x \in A</m> or <m>x</m> is in both <m>B</m> and <m>C</m>. So <m>x \in A \cup (B \cap C)</m> and therefore <m>(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)</m>. Hence, <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
    </hint>
</exercise>

Prove

A \cup (B \cap C) = (A \cup B) \cap (A \cup C) .

🔗

Hint.

View Source for hint

<hint>
    <p>If <m>x \in A \cup (B \cap C)</m>, then either <m>x \in A</m> or <m>x \in B \cap C</m>.  Thus,   <m> x \in A \cup B</m> and <m>A \cup C</m>.  Hence, <m> x \in (A \cup B) \cap (A \cup C)</m>. Therefore, <m>  A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)</m>.  Conversely, if <m>x \in (A \cup B) \cap (A \cup C)</m>, then <m>x \in A \cup B</m> and <m>A \cup C</m>. Thus, <m>x \in A</m> or <m>x</m> is in both <m>B</m> and <m>C</m>. So <m>x \in A \cup (B \cap C)</m> and therefore <m>(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)</m>. Hence, <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
</hint>

x \in A \cup (B \cap C),

then either

x \in A

x \in B \cap C .

Thus,

x \in A \cup B

and

A \cup C .

Hence,

x \in (A \cup B) \cap (A \cup C) .

Therefore,

A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C) .

Conversely, if

x \in (A \cup B) \cap (A \cup C),

then

x \in A \cup B

and

A \cup C .

Thus,

x \in A

x

is in both

B

and

C .

x \in A \cup (B \cap C)

and therefore

(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C) .

Hence,

A \cup (B \cap C) = (A \cup B) \cap (A \cup C) .

🔗

7.

View Source for exercise

<exercise number="7">
    <statement>
        <p>Prove <m>A \cap (B \cup C) = (A \cap B) \cup (A \cap C)</m>.</p>
    </statement>
</exercise>

Prove

A \cap (B \cup C) = (A \cap B) \cup (A \cap C) .

🔗

8.

View Source for exercise

<exercise number="8">
    <statement>
        <p>Prove  <m>A \subset B</m> if and only if <m>A \cap B = A</m>.</p>
    </statement>
</exercise>

Prove

A \subset B

if and only if

A \cap B = A .

🔗

9.

View Source for exercise

<exercise number="9">
    <statement>
        <p>Prove <m>(A \cap B)' = A' \cup B'</m>.</p>
    </statement>
</exercise>

Prove

(A \cap B)^{'} = A^{'} \cup B^{'} .

🔗

10.

View Source for exercise

<exercise number="10">
    <statement>
        <p>Prove  <m>A \cup B = (A \cap B) \cup (A \setminus B) \cup (B \setminus A)</m>.</p>
    </statement>
    <hint>
        <p><m>(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B</m>.</p>
    </hint>
</exercise>

Prove

A \cup B = (A \cap B) \cup (A ∖ B) \cup (B ∖ A) .

🔗

Hint.

View Source for hint

<hint>
    <p><m>(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B</m>.</p>
</hint>

(A \cap B) \cup (A ∖ B) \cup (B ∖ A) = (A \cap B) \cup (A \cap B^{'}) \cup (B \cap A^{'}) = [A \cap (B \cup B^{'})] \cup (B \cap A^{'}) = A \cup (B \cap A^{'}) = (A \cup B) \cap (A \cup A^{'}) = A \cup B .

🔗

11.

View Source for exercise

<exercise number="11">
    <statement>
        <p>Prove  <m>(A \cup B) \times C = (A \times C ) \cup (B \times C)</m>.</p>
    </statement>
</exercise>

Prove

(A \cup B) \times C = (A \times C) \cup (B \times C) .

🔗

12.

View Source for exercise

<exercise number="12">
    <statement>
        <p>Prove  <m>(A \cap B) \setminus B = \emptyset</m>.</p>
    </statement>
</exercise>

Prove

(A \cap B) ∖ B = \emptyset .

🔗

13.

View Source for exercise

<exercise number="13">
    <statement>
        <p>Prove  <m>(A \cup B) \setminus B = A \setminus B</m>.</p>
    </statement>
</exercise>

Prove

(A \cup B) ∖ B = A ∖ B .

🔗

14.

View Source for exercise

<exercise number="14">
    <statement>
        <p>Prove  <m>A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)</m>.</p>
    </statement>
    <hint>
        <p><m>A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)</m>.</p>
    </hint>
</exercise>

Prove

A ∖ (B \cup C) = (A ∖ B) \cap (A ∖ C) .

🔗

Hint.

View Source for hint

<hint>
    <p><m>A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)</m>.</p>
</hint>

A ∖ (B \cup C) = A \cap (B \cup C)^{'} = (A \cap A) \cap (B^{'} \cap C^{'}) = (A \cap B^{'}) \cap (A \cap C^{'}) = (A ∖ B) \cap (A ∖ C) .

🔗

More Exercises

View Source for subexercises

<subexercises>
    <title>More Exercises</title>
    <introduction>
        <p>
            This is a meaningless subdivision of the exercises for the sake of testing output.
        </p>
    </introduction>

    <exercise number="15">
        <statement>
            <p>Prove  <m>A \cap (B \setminus C) = (A \cap B) \setminus (A \cap C)</m>.</p>
        </statement>
    </exercise>

    <exercise number="16">
        <statement>
            <p>Prove  <m>(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)</m>.</p>
        </statement>
    </exercise>

    <exercise number="17">
        <statement>
            <p>Which of the following relations <m>f: {\mathbb Q} \rightarrow {\mathbb Q}</m> define a mapping? In each case, supply a reason why <m>f</m> is or is not a mapping.</p>
            <ol cols="2">
                <li><p><m>\displaystyle f(p/q) = \frac{p+ 1}{p - 2}</m></p></li>
                <li><p><m>\displaystyle f(p/q) = \frac{3p}{3q}</m></p></li>
                <li><p><m>\displaystyle f(p/q) = \frac{p+q}{q^2}</m></p></li>
                <li><p><m>\displaystyle f(p/q) = \frac{3 p^2}{7 q^2} - \frac{p}{q}</m></p></li>
            </ol>
        </statement>
        <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
        <!-- <hint>
            <p>(a) Not a map since <m>f(2/3)</m> is undefined; (b) this is a map; (c) not a map, since <m>f(1/2) = 3/4</m> but <m>f(2/4)=3/8</m>; (d) this is a map.</p>
        </hint> -->
    </exercise>

    <exercise number="18">
        <statement>
            <p>Determine which of the following functions are one-to-one and which are onto.  If the function is not onto, determine its range.</p>
            <ol>
                <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = e^x</m></p></li>
                <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(n) = n^2 + 3</m></p></li>
                <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = \sin x</m></p></li>
                <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(x) = x^2</m></p></li>
            </ol>
        </statement>
        <hint>
            <p>(a) <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.  (c) <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.</p>
        </hint>
    </exercise>

    <exercise number="19">
        <statement>
            <p>Let <m>f :A \rightarrow B</m> and <m>g : B \rightarrow C</m> be invertible mappings; that is, mappings such that <m>f^{-1}</m> and <m>g^{-1}</m> exist. Show that <m>(g \circ f)^{-1} =f^{-1} \circ g^{-1}</m>.</p>
        </statement>
    </exercise>

    <exercise number="20">
        <statement>
            <ol>
                <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is one-to-one but not onto.</p></li>
                 <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is onto but not one-to-one.</p></li>
            </ol>
        </statement>
        <hint>
            <p>(a) <m>f(n) = n + 1</m>.</p>
        </hint>
    </exercise>

    <exercise number="21">
        <statement>
            <p>Prove the relation defined on <m>{\mathbb R}^2</m> by <m>(x_1, y_1 ) \sim (x_2, y_2)</m> if <m>x_1^2 + y_1^2 = x_2^2 + y_2^2</m> is  an equivalence relation.</p>
        </statement>
    </exercise>

    <exercise number="22">
        <statement>
            <p>Let <m>f : A \rightarrow B</m> and <m>g : B \rightarrow C</m> be maps.</p>
            <ol>
                <li><p>If <m>f</m> and <m>g</m> are both one-to-one functions, show that <m>g \circ f</m> is one-to-one.</p></li>
                <li><p>If <m>g \circ f</m> is onto, show that <m>g</m> is onto.</p></li>
                <li><p>If <m>g \circ f</m> is one-to-one, show that <m>f</m> is one-to-one.</p></li>
                <li><p>If <m>g \circ f</m> is one-to-one and <m>f</m> is onto, show that <m>g</m> is one-to-one.</p></li>
                <li><p>If <m>g \circ f</m> is onto and <m>g</m> is one-to-one, show that <m>f</m> is onto.</p></li>
            </ol>
        </statement>
        <hint>
            <p>(a) Let <m>x, y \in A</m>. Then <m>g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))</m>.  Thus, <m>f(x) = f(y)</m> and <m>x = y</m>,  so <m>g \circ f</m> is one-to-one. (b) Let <m>c \in C</m>, then <m>c = (g \circ f)(x) = g(f(x))</m> for some <m>x \in A</m>. Since <m>f(x) \in B</m>, <m>g</m> is onto.</p>
        </hint>
    </exercise>

    <exercise number="23">
        <statement>
            <p>Define a function on the real numbers by
                <me>f(x) = \frac{x + 1}{x - 1}.</me>
            What are the domain and range of <m>f</m>? What is the inverse of <m>f</m>?  Compute <m>f \circ f^{-1}</m> and <m>f^{-1} \circ f</m>.</p>
        </statement>
        <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
        <!-- <hint>
            <p><m>f^{-1}(x) = (x+1)/(x-1)</m>.</p>
        </hint> -->
    </exercise>

    <exercise number="24">
        <statement>
            <p>Let <m>f: X \rightarrow Y</m> be a map with <m>A_1, A_2 \subset X</m> and <m>B_1, B_2 \subset Y</m>.</p>
            <ol>
                <li><p>Prove <m>f( A_1 \cup A_2 ) = f( A_1) \cup f( A_2 )</m>.</p></li>
                <li><p>Prove <m>f( A_1 \cap A_2 ) \subset f( A_1) \cap f( A_2 )</m>.  Give an example in which equality fails.</p></li>
                <li><p>Prove <m>f^{-1}( B_1 \cup B_2 ) = f^{-1}( B_1) \cup f^{-1}(B_2 )</m>, where <me>f^{-1}(B) = \{ x \in X : f(x) \in B \}.</me></p></li>
                <li><p>Prove <m>f^{-1}( B_1 \cap B_2 ) = f^{-1}( B_1) \cap f^{-1}( B_2 )</m>.</p></li>
                <li><p>Prove <m>f^{-1}( Y \setminus B_1 ) = X \setminus f^{-1}( B_1)</m>.</p></li>
            </ol>
        </statement>
        <hint>
            <p>(a) Let <m>y \in f(A_1 \cup A_2)</m>.  Then there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Hence, <m> y \in f(A_1)</m> or <m>f(A_2) </m>.  Therefore, <m> y \in f(A_1) \cup f(A_2)</m>. Consequently, <m> f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)</m>. Conversely, if <m>y \in f(A_1) \cup f(A_2)</m>, then <m> y \in f(A_1)</m> or <m>f(A_2)</m>.  Hence, there exists an <m>x \in A_1</m> or there exists an <m>x \in A_2</m> such that <m>f(x) = y</m>.  Thus, there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Therefore, <m> f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)</m>, and <m>f(A_1 \cup A_2) = f(A_1) \cup f(A_2)</m>.</p>
        </hint>
    </exercise>

    <exercise number="25">
        <statement>
            <p>Determine whether or not the following relations are equivalence relations on the given set.  If the relation is an equivalence relation, describe the partition given by it.  If the relation is not an equivalence relation, state why it fails to be one.</p>
            <ol cols="2">
                <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>x \geq y</m></p></li>
                <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>mn &gt; 0</m></p></li>
                <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>|x - y| \leq 4</m></p></li>
                <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>m \equiv n \pmod{6}</m></p></li>
            </ol>
        </statement>
        <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
        <!-- <hint>
            <p>(a) NThe relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.</p>
        </hint> -->
    </exercise>

    <exercise number="26">
        <statement>
            <p>Define a relation <m>\sim</m> on <m>{\mathbb R}^2</m> by stating that <m>(a, b) \sim (c, d)</m> if and only if <m>a^2 + b^2 \leq c^2 + d^2</m>. Show that <m>\sim</m> is reflexive and transitive but not symmetric.</p>
        </statement>
    </exercise>

    <exercise number="27">
        <statement>
            <p>Show that an <m>m \times n</m> matrix gives rise to a well-defined  map from <m>{\mathbb R}^n</m> to <m>{\mathbb R}^m</m>.</p>
        </statement>
    </exercise>

    <exercise number="28">
        <statement>
            <p>Find the error in the following argument by providing a counterexample. <q>The reflexive property is redundant in the axioms for an equivalence relation.  If <m>x \sim y</m>, then <m>y \sim x</m> by the symmetric property.  Using the transitive property, we can deduce that <m>x \sim x</m>.</q> </p>
        </statement>
        <hint>
            <p>Let <m>X = {\mathbb N} \cup \{ \sqrt{2}\, \}</m> and define <m>x \sim y</m> if <m>x + y \in {\mathbb N}</m>.</p>
        </hint>
    </exercise>

    <exercise number="29">
        <title>Projective Real Line</title>
        <statement>
            <p>Define a relation on <m>{\mathbb R}^2 \setminus  \{ (0,0) \}</m> by letting <m>(x_1, y_1) \sim (x_2, y_2)</m> if there exists a nonzero real number <m>\lambda</m> such that <m>(x_1, y_1)  = ( \lambda x_2, \lambda y_2)</m>.  Prove that <m>\sim</m> defines an equivalence relation on <m>{\mathbb R}^2 \setminus (0,0)</m>.  What are the corresponding  equivalence classes?  This equivalence relation defines the projective line, denoted by  <m>{\mathbb P}({\mathbb R}) </m>, which is very important in geometry.</p>
        </statement>
    </exercise>
</subexercises>

This is a meaningless subdivision of the exercises for the sake of testing output.

🔗

15.

View Source for exercise

<exercise number="15">
    <statement>
        <p>Prove  <m>A \cap (B \setminus C) = (A \cap B) \setminus (A \cap C)</m>.</p>
    </statement>
</exercise>

Prove

A \cap (B ∖ C) = (A \cap B) ∖ (A \cap C) .

🔗

16.

View Source for exercise

<exercise number="16">
    <statement>
        <p>Prove  <m>(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)</m>.</p>
    </statement>
</exercise>

Prove

(A ∖ B) \cup (B ∖ A) = (A \cup B) ∖ (A \cap B) .

🔗

17.

View Source for exercise

<exercise number="17">
    <statement>
        <p>Which of the following relations <m>f: {\mathbb Q} \rightarrow {\mathbb Q}</m> define a mapping? In each case, supply a reason why <m>f</m> is or is not a mapping.</p>
        <ol cols="2">
            <li><p><m>\displaystyle f(p/q) = \frac{p+ 1}{p - 2}</m></p></li>
            <li><p><m>\displaystyle f(p/q) = \frac{3p}{3q}</m></p></li>
            <li><p><m>\displaystyle f(p/q) = \frac{p+q}{q^2}</m></p></li>
            <li><p><m>\displaystyle f(p/q) = \frac{3 p^2}{7 q^2} - \frac{p}{q}</m></p></li>
        </ol>
    </statement>
    <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
    <!-- <hint>
        <p>(a) Not a map since <m>f(2/3)</m> is undefined; (b) this is a map; (c) not a map, since <m>f(1/2) = 3/4</m> but <m>f(2/4)=3/8</m>; (d) this is a map.</p>
    </hint> -->
</exercise>

Which of the following relations

f : Q \to Q

define a mapping? In each case, supply a reason why

f

is or is not a mapping.

🔗

$f (p / q) = \frac{p + 1}{p - 2}$
🔗

🔗
$f (p / q) = \frac{3 p}{3 q}$
🔗

🔗
$f (p / q) = \frac{p + q}{q^{2}}$
🔗

🔗
$f (p / q) = \frac{3 p^{2}}{7 q^{2}} - \frac{p}{q}$
🔗

🔗

🔗

18.

View Source for exercise

<exercise number="18">
    <statement>
        <p>Determine which of the following functions are one-to-one and which are onto.  If the function is not onto, determine its range.</p>
        <ol>
            <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = e^x</m></p></li>
            <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(n) = n^2 + 3</m></p></li>
            <li><p><m>f: {\mathbb R} \rightarrow {\mathbb R}</m> defined by <m>f(x) = \sin x</m></p></li>
            <li><p><m>f: {\mathbb Z} \rightarrow {\mathbb Z}</m> defined by <m>f(x) = x^2</m></p></li>
        </ol>
    </statement>
    <hint>
        <p>(a) <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.  (c) <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.</p>
    </hint>
</exercise>

Determine which of the following functions are one-to-one and which are onto. If the function is not onto, determine its range.

🔗

$f : R \to R$ defined by $f (x) = e^{x}$
🔗

🔗
$f : Z \to Z$ defined by $f (n) = n^{2} + 3$
🔗

🔗
$f : R \to R$ defined by $f (x) = \sin x$
🔗

🔗
$f : Z \to Z$ defined by $f (x) = x^{2}$
🔗

🔗

Hint.

View Source for hint

<hint>
    <p>(a) <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.  (c) <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.</p>
</hint>

(a)

f

is one-to-one but not onto.

f (R) = {x \in R : x > 0} .

(c)

f

is neither one-to-one nor onto.

f (R) = {x : - 1 \leq x \leq 1} .

🔗

19.

View Source for exercise

<exercise number="19">
    <statement>
        <p>Let <m>f :A \rightarrow B</m> and <m>g : B \rightarrow C</m> be invertible mappings; that is, mappings such that <m>f^{-1}</m> and <m>g^{-1}</m> exist. Show that <m>(g \circ f)^{-1} =f^{-1} \circ g^{-1}</m>.</p>
    </statement>
</exercise>

Let

f : A \to B

and

g : B \to C

be invertible mappings; that is, mappings such that

f^{- 1}

and

g^{- 1}

exist. Show that

(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1} .

🔗

20.

View Source for exercise

<exercise number="20">
    <statement>
        <ol>
            <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is one-to-one but not onto.</p></li>
             <li><p>Define a function <m>f: {\mathbb N} \rightarrow {\mathbb N}</m> that is onto but not one-to-one.</p></li>
        </ol>
    </statement>
    <hint>
        <p>(a) <m>f(n) = n + 1</m>.</p>
    </hint>
</exercise>

Define a function $f : N \to N$ that is one-to-one but not onto.
🔗

🔗
Define a function $f : N \to N$ that is onto but not one-to-one.
🔗

🔗

Hint.

View Source for hint

<hint>
    <p>(a) <m>f(n) = n + 1</m>.</p>
</hint>

(a)

f (n) = n + 1 .

🔗

21.

View Source for exercise

<exercise number="21">
    <statement>
        <p>Prove the relation defined on <m>{\mathbb R}^2</m> by <m>(x_1, y_1 ) \sim (x_2, y_2)</m> if <m>x_1^2 + y_1^2 = x_2^2 + y_2^2</m> is  an equivalence relation.</p>
    </statement>
</exercise>

Prove the relation defined on

R^{2}

(x_{1}, y_{1}) \sim (x_{2}, y_{2})

x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}

is an equivalence relation.

🔗

22.

View Source for exercise

<exercise number="22">
    <statement>
        <p>Let <m>f : A \rightarrow B</m> and <m>g : B \rightarrow C</m> be maps.</p>
        <ol>
            <li><p>If <m>f</m> and <m>g</m> are both one-to-one functions, show that <m>g \circ f</m> is one-to-one.</p></li>
            <li><p>If <m>g \circ f</m> is onto, show that <m>g</m> is onto.</p></li>
            <li><p>If <m>g \circ f</m> is one-to-one, show that <m>f</m> is one-to-one.</p></li>
            <li><p>If <m>g \circ f</m> is one-to-one and <m>f</m> is onto, show that <m>g</m> is one-to-one.</p></li>
            <li><p>If <m>g \circ f</m> is onto and <m>g</m> is one-to-one, show that <m>f</m> is onto.</p></li>
        </ol>
    </statement>
    <hint>
        <p>(a) Let <m>x, y \in A</m>. Then <m>g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))</m>.  Thus, <m>f(x) = f(y)</m> and <m>x = y</m>,  so <m>g \circ f</m> is one-to-one. (b) Let <m>c \in C</m>, then <m>c = (g \circ f)(x) = g(f(x))</m> for some <m>x \in A</m>. Since <m>f(x) \in B</m>, <m>g</m> is onto.</p>
    </hint>
</exercise>

Let

f : A \to B

and

g : B \to C

be maps.

🔗

If $f$ and $g$ are both one-to-one functions, show that $g \circ f$ is one-to-one.
🔗

🔗
If $g \circ f$ is onto, show that $g$ is onto.
🔗

🔗
If $g \circ f$ is one-to-one, show that $f$ is one-to-one.
🔗

🔗
If $g \circ f$ is one-to-one and $f$ is onto, show that $g$ is one-to-one.
🔗

🔗
If $g \circ f$ is onto and $g$ is one-to-one, show that $f$ is onto.
🔗

🔗

Hint.

View Source for hint

<hint>
    <p>(a) Let <m>x, y \in A</m>. Then <m>g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))</m>.  Thus, <m>f(x) = f(y)</m> and <m>x = y</m>,  so <m>g \circ f</m> is one-to-one. (b) Let <m>c \in C</m>, then <m>c = (g \circ f)(x) = g(f(x))</m> for some <m>x \in A</m>. Since <m>f(x) \in B</m>, <m>g</m> is onto.</p>
</hint>

(a) Let

x, y \in A .

Then

g (f (x)) = (g \circ f) (x) = (g \circ f) (y) = g (f (y)) .

Thus,

f (x) = f (y)

and

x = y,

g \circ f

is one-to-one. (b) Let

c \in C,

then

c = (g \circ f) (x) = g (f (x))

for some

x \in A .

Since

f (x) \in B,

g

is onto.

🔗

23.

View Source for exercise

<exercise number="23">
    <statement>
        <p>Define a function on the real numbers by
            <me>f(x) = \frac{x + 1}{x - 1}.</me>
        What are the domain and range of <m>f</m>? What is the inverse of <m>f</m>?  Compute <m>f \circ f^{-1}</m> and <m>f^{-1} \circ f</m>.</p>
    </statement>
    <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
    <!-- <hint>
        <p><m>f^{-1}(x) = (x+1)/(x-1)</m>.</p>
    </hint> -->
</exercise>

Define a function on the real numbers by

f (x) = \frac{x + 1}{x - 1} .

What are the domain and range of

f ?

What is the inverse of

f ?

Compute

f \circ f^{- 1}

and

f^{- 1} \circ f .

🔗

24.

View Source for exercise

<exercise number="24">
    <statement>
        <p>Let <m>f: X \rightarrow Y</m> be a map with <m>A_1, A_2 \subset X</m> and <m>B_1, B_2 \subset Y</m>.</p>
        <ol>
            <li><p>Prove <m>f( A_1 \cup A_2 ) = f( A_1) \cup f( A_2 )</m>.</p></li>
            <li><p>Prove <m>f( A_1 \cap A_2 ) \subset f( A_1) \cap f( A_2 )</m>.  Give an example in which equality fails.</p></li>
            <li><p>Prove <m>f^{-1}( B_1 \cup B_2 ) = f^{-1}( B_1) \cup f^{-1}(B_2 )</m>, where <me>f^{-1}(B) = \{ x \in X : f(x) \in B \}.</me></p></li>
            <li><p>Prove <m>f^{-1}( B_1 \cap B_2 ) = f^{-1}( B_1) \cap f^{-1}( B_2 )</m>.</p></li>
            <li><p>Prove <m>f^{-1}( Y \setminus B_1 ) = X \setminus f^{-1}( B_1)</m>.</p></li>
        </ol>
    </statement>
    <hint>
        <p>(a) Let <m>y \in f(A_1 \cup A_2)</m>.  Then there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Hence, <m> y \in f(A_1)</m> or <m>f(A_2) </m>.  Therefore, <m> y \in f(A_1) \cup f(A_2)</m>. Consequently, <m> f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)</m>. Conversely, if <m>y \in f(A_1) \cup f(A_2)</m>, then <m> y \in f(A_1)</m> or <m>f(A_2)</m>.  Hence, there exists an <m>x \in A_1</m> or there exists an <m>x \in A_2</m> such that <m>f(x) = y</m>.  Thus, there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Therefore, <m> f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)</m>, and <m>f(A_1 \cup A_2) = f(A_1) \cup f(A_2)</m>.</p>
    </hint>
</exercise>

Let

f : X \to Y

be a map with

A_{1}, A_{2} \subset X

and

B_{1}, B_{2} \subset Y .

🔗

Prove $f (A_{1} \cup A_{2}) = f (A_{1}) \cup f (A_{2}) .$
🔗

🔗
Prove $f (A_{1} \cap A_{2}) \subset f (A_{1}) \cap f (A_{2}) .$ Give an example in which equality fails.
🔗

🔗
Prove $f^{- 1} (B_{1} \cup B_{2}) = f^{- 1} (B_{1}) \cup f^{- 1} (B_{2}),$ where

$f^{- 1} (B) = {x \in X : f (x) \in B} .$

🔗

🔗
Prove $f^{- 1} (B_{1} \cap B_{2}) = f^{- 1} (B_{1}) \cap f^{- 1} (B_{2}) .$
🔗

🔗
Prove $f^{- 1} (Y ∖ B_{1}) = X ∖ f^{- 1} (B_{1}) .$
🔗

🔗

Hint.

View Source for hint

<hint>
    <p>(a) Let <m>y \in f(A_1 \cup A_2)</m>.  Then there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Hence, <m> y \in f(A_1)</m> or <m>f(A_2) </m>.  Therefore, <m> y \in f(A_1) \cup f(A_2)</m>. Consequently, <m> f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)</m>. Conversely, if <m>y \in f(A_1) \cup f(A_2)</m>, then <m> y \in f(A_1)</m> or <m>f(A_2)</m>.  Hence, there exists an <m>x \in A_1</m> or there exists an <m>x \in A_2</m> such that <m>f(x) = y</m>.  Thus, there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Therefore, <m> f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)</m>, and <m>f(A_1 \cup A_2) = f(A_1) \cup f(A_2)</m>.</p>
</hint>

(a) Let

y \in f (A_{1} \cup A_{2}) .

Then there exists an

x \in A_{1} \cup A_{2}

such that

f (x) = y .

Hence,

y \in f (A_{1})

f (A_{2}) .

Therefore,

y \in f (A_{1}) \cup f (A_{2}) .

Consequently,

f (A_{1} \cup A_{2}) \subset f (A_{1}) \cup f (A_{2}) .

Conversely, if

y \in f (A_{1}) \cup f (A_{2}),

then

y \in f (A_{1})

f (A_{2}) .

Hence, there exists an

x \in A_{1}

or there exists an

x \in A_{2}

such that

f (x) = y .

Thus, there exists an

x \in A_{1} \cup A_{2}

such that

f (x) = y .

Therefore,

f (A_{1}) \cup f (A_{2}) \subset f (A_{1} \cup A_{2}),

and

f (A_{1} \cup A_{2}) = f (A_{1}) \cup f (A_{2}) .

🔗

25.

View Source for exercise

<exercise number="25">
    <statement>
        <p>Determine whether or not the following relations are equivalence relations on the given set.  If the relation is an equivalence relation, describe the partition given by it.  If the relation is not an equivalence relation, state why it fails to be one.</p>
        <ol cols="2">
            <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>x \geq y</m></p></li>
            <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>mn &gt; 0</m></p></li>
            <li><p><m>x \sim y</m> in <m>{\mathbb R}</m> if <m>|x - y| \leq 4</m></p></li>
            <li><p><m>m \sim n</m> in <m>{\mathbb Z}</m> if <m>m \equiv n \pmod{6}</m></p></li>
        </ol>
    </statement>
    <!-- commented out for testing, so that only even hints, answers, solutions may appear in the "solutions" -->
    <!-- <hint>
        <p>(a) NThe relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.</p>
    </hint> -->
</exercise>

Determine whether or not the following relations are equivalence relations on the given set. If the relation is an equivalence relation, describe the partition given by it. If the relation is not an equivalence relation, state why it fails to be one.

🔗

$x \sim y$ in $R$ if $x \geq y$
🔗

🔗
$m \sim n$ in $Z$ if $m n > 0$
🔗

🔗
$x \sim y$ in $R$ if $| x - y | \leq 4$
🔗

🔗
$m \sim n$ in $Z$ if $m \equiv n (\mod 6)$
🔗

🔗

🔗

26.

View Source for exercise

<exercise number="26">
    <statement>
        <p>Define a relation <m>\sim</m> on <m>{\mathbb R}^2</m> by stating that <m>(a, b) \sim (c, d)</m> if and only if <m>a^2 + b^2 \leq c^2 + d^2</m>. Show that <m>\sim</m> is reflexive and transitive but not symmetric.</p>
    </statement>
</exercise>

Define a relation

\sim

R^{2}

by stating that

(a, b) \sim (c, d)

if and only if

a^{2} + b^{2} \leq c^{2} + d^{2} .

Show that

\sim

is reflexive and transitive but not symmetric.

🔗

27.

View Source for exercise

<exercise number="27">
    <statement>
        <p>Show that an <m>m \times n</m> matrix gives rise to a well-defined  map from <m>{\mathbb R}^n</m> to <m>{\mathbb R}^m</m>.</p>
    </statement>
</exercise>

Show that an

m \times n

matrix gives rise to a well-defined map from

R^{n}

R^{m} .

🔗

28.

View Source for exercise

<exercise number="28">
    <statement>
        <p>Find the error in the following argument by providing a counterexample. <q>The reflexive property is redundant in the axioms for an equivalence relation.  If <m>x \sim y</m>, then <m>y \sim x</m> by the symmetric property.  Using the transitive property, we can deduce that <m>x \sim x</m>.</q> </p>
    </statement>
    <hint>
        <p>Let <m>X = {\mathbb N} \cup \{ \sqrt{2}\, \}</m> and define <m>x \sim y</m> if <m>x + y \in {\mathbb N}</m>.</p>
    </hint>
</exercise>

Find the error in the following argument by providing a counterexample. “The reflexive property is redundant in the axioms for an equivalence relation. If

x \sim y,

then

y \sim x

by the symmetric property. Using the transitive property, we can deduce that

x \sim x .

”

🔗

Hint.

View Source for hint

<hint>
    <p>Let <m>X = {\mathbb N} \cup \{ \sqrt{2}\, \}</m> and define <m>x \sim y</m> if <m>x + y \in {\mathbb N}</m>.</p>
</hint>

Let

X = N \cup {\sqrt{2}}

and define

x \sim y

x + y \in N .

🔗

29. Projective Real Line.

View Source for exercise

<exercise number="29">
    <title>Projective Real Line</title>
    <statement>
        <p>Define a relation on <m>{\mathbb R}^2 \setminus  \{ (0,0) \}</m> by letting <m>(x_1, y_1) \sim (x_2, y_2)</m> if there exists a nonzero real number <m>\lambda</m> such that <m>(x_1, y_1)  = ( \lambda x_2, \lambda y_2)</m>.  Prove that <m>\sim</m> defines an equivalence relation on <m>{\mathbb R}^2 \setminus (0,0)</m>.  What are the corresponding  equivalence classes?  This equivalence relation defines the projective line, denoted by  <m>{\mathbb P}({\mathbb R}) </m>, which is very important in geometry.</p>
    </statement>
</exercise>

Define a relation on

R^{2} ∖ {(0, 0)}

by letting

(x_{1}, y_{1}) \sim (x_{2}, y_{2})

if there exists a nonzero real number

λ

such that

(x_{1}, y_{1}) = (λ x_{2}, λ y_{2}) .

Prove that

\sim

defines an equivalence relation on

R^{2} ∖ (0, 0) .

What are the corresponding equivalence classes? This equivalence relation defines the projective line, denoted by

P (R),

which is very important in geometry.

🔗

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