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Appendix C Hints and Answers to Selected Even Exercises

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  <title>Hints and Answers to Selected Even Exercises</title>

</solutions>

I Basics
1 Preliminaries
1.4 Exercises

More Exercises

1.4.24.

1.4.24.a
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>

    <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.

  </p>

</hint>
\(f\) is one-to-one but not onto. \(f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.}\)
1.4.24.c
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>

    <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.

  </p>

</hint>
\(f\) is neither one-to-one nor onto. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)

1.4.26.

1.4.26.a
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>

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

  </p>

</hint>
\(f(n) = n + 1\text{.}\)

1.4.28.

1.4.28.a
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    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.
  </p>

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

1.4.30.

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

2 The Integers
2.4 Exercises

2.4.22. Fibonacci Numbers.

2.4.22.a

Hint.
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  <p>
    Use mathematical induction.
  </p>

</hint>
Use mathematical induction.

2.4.22.b

Hint.
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  <p>
    Use mathematical induction.
  </p>

</hint>
Use mathematical induction.

2.4.22.c

Hint.
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  <p>
    Show that <m>f_1 = 1</m>, <m>f_2 = 1</m>, and <m>f_{n + 2} = f_{n + 1} + f_n</m>.
  </p>

</hint>
Show that \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\text{.}\)

2.4.24.

Hint.
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  <p>
    Use the Fundamental Theorem of Arithmetic.
  </p>

</hint>
Use the Fundamental Theorem of Arithmetic.

2.4.28.

Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    Let <m>S = \{s \in {\mathbb N} : a \mid s</m>, <m>b \mid s \}</m>.
    Then <m>S \neq \emptyset</m>, since <m>|ab| \in S</m>.
    By the Principle of Well-Ordering, <m>S</m> contains a least element <m>m</m>.
    To show uniqueness, suppose that <m>a \mid n</m> and <m>b \mid n</m> for some <m>n \in {\mathbb N}</m>.
    By the division algorithm, there exist unique integers <m>q</m> and <m>r</m> such that <m>n = mq + r</m>, where <m>0 \leq r \lt m</m>.
    Since <m>a</m> and <m>b</m> divide both <m>m</m>, and <m>n</m>, it must be the case that  <m>a</m> and <m>b</m> both divide <m>r</m>.
    Thus, <m>r = 0</m> by the minimality of <m>m</m>.
    Therefore, <m>m \mid n</m>.
  </p>

</hint>
Let \(S = \{s \in {\mathbb N} : a \mid s\text{,}\) \(b \mid s \}\text{.}\) Then \(S \neq \emptyset\text{,}\) since \(|ab| \in S\text{.}\) By the Principle of Well-Ordering, \(S\) contains a least element \(m\text{.}\) To show uniqueness, suppose that \(a \mid n\) and \(b \mid n\) for some \(n \in {\mathbb N}\text{.}\) By the division algorithm, there exist unique integers \(q\) and \(r\) such that \(n = mq + r\text{,}\) where \(0 \leq r \lt m\text{.}\) Since \(a\) and \(b\) divide both \(m\text{,}\) and \(n\text{,}\) it must be the case that \(a\) and \(b\) both divide \(r\text{.}\) Thus, \(r = 0\) by the minimality of \(m\text{.}\) Therefore, \(m \mid n\text{.}\)

2.4.32.

Hint.
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  <p>
    Since <m>\gcd(a,b) = 1</m>, there exist integers <m>r</m> and <m>s</m> such that <m>ar + bs = 1</m>.
    Thus, <m>acr + bcs = c</m>.
    Since <m>a</m> divides both <m>bc</m> and itself, <m>a</m> must divide <m>c</m>.
  </p>

</hint>
Since \(\gcd(a,b) = 1\text{,}\) there exist integers \(r\) and \(s\) such that \(ar + bs = 1\text{.}\) Thus, \(acr + bcs = c\text{.}\) Since \(a\) divides both \(bc\) and itself, \(a\) must divide \(c\text{.}\)

2.4.34.

Hint.
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  <p>
    Every prime must be of the form 2, 3, <m>6n + 1</m>, or <m>6n + 5</m>.
    Suppose there are only finitely many primes of the form <m>6k + 5</m>.
  </p>

</hint>
Every prime must be of the form 2, 3, \(6n + 1\text{,}\) or \(6n + 5\text{.}\) Suppose there are only finitely many primes of the form \(6k + 5\text{.}\)

II Algebra (and Runestone)
1 Groups
1.5 Exercises

1.5.2.

1.5.2.a

Hint.
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  <p>
    Not a group.
  </p>

</hint>
Not a group.

1.5.2.c

Hint.
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  <p>
    A group.
  </p>

</hint>
A group.

1.5.6.

Hint.
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  <p>

    <md>
      \begin{array}{c|cccc} \cdot &amp; 1  &amp; 5  &amp; 7  &amp; 11 \\ \hline 1 &amp; 1 &amp; 5 &amp; 7 &amp; 11 \\ 5 &amp; 5 &amp; 1 &amp; 11 &amp; 7 \\ 7 &amp; 7 &amp; 11 &amp; 1 &amp; 5 \\ 11 &amp; 11 &amp; 7 &amp; 5 &amp; 1 \end{array}
    </md>

  </p>

</hint>
\begin{equation*} \begin{array}{c|cccc} \cdot & 1 & 5 & 7 & 11 \\ \hline 1 & 1 & 5 & 7 & 11 \\ 5 & 5 & 1 & 11 & 7 \\ 7 & 7 & 11 & 1 & 5 \\ 11 & 11 & 7 & 5 & 1 \end{array} \end{equation*}

1.5.8.

Hint.
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  <p>
    Pick two matrices.
    Almost any pair will work.
  </p>

</hint>
Pick two matrices. Almost any pair will work.

1.5.16.

Hint.
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  <p>
    Look at the symmetry group of an equilateral triangle or a square.
  </p>

</hint>
Look at the symmetry group of an equilateral triangle or a square.

1.5.18.

Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    Let
    <md>
      \sigma = \begin{pmatrix} 1 &amp; 2 &amp; \cdots &amp; n \\ a_1 &amp; a_2 &amp; \cdots &amp; a_n \end{pmatrix}
    </md>

    be in <m>S_n</m>.
    All of the <m>a_i</m>s must be distinct.
    There are <m>n</m> ways to choose <m>a_1</m>, <m>n-1</m> ways to choose <m>a_2</m>, <m>\ldots</m>, 2 ways to choose <m>a_{n - 1}</m>, and only one way to choose <m>a_n</m>.
    Therefore, we can form <m>\sigma</m> in <m>n(n - 1) \cdots 2 \cdot 1 = n!</m> ways.
  </p>

</hint>
Let
\begin{equation*} \sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & a_n \end{pmatrix} \end{equation*}
be in \(S_n\text{.}\) All of the \(a_i\)s must be distinct. There are \(n\) ways to choose \(a_1\text{,}\) \(n-1\) ways to choose \(a_2\text{,}\) \(\ldots\text{,}\) 2 ways to choose \(a_{n - 1}\text{,}\) and only one way to choose \(a_n\text{.}\) Therefore, we can form \(\sigma\) in \(n(n - 1) \cdots 2 \cdot 1 = n!\) ways.

1.5.46.

Hint.
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  <p>
    Look at <m>S_3</m>.
  </p>

</hint>
Look at \(S_3\text{.}\)

1.5.56.

Answer.
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  <p>

    <m>2</m>

  </p>

</answer>
\(2\)

1.5.58.

Answer.
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  <p>

    <m>n+1</m>

  </p>

</answer>
\(n+1\)

1.5.60.

1.5.60.a

Answer.
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  <p>

    <m>4</m>

  </p>

</answer>
\(4\)

1.5.60.b

1.5.60.b.i
Answer.
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  <p>

    <m>8</m>

  </p>

</answer>
\(8\)
1.5.60.b.ii
Answer.
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  <p>

    <m>12</m>

  </p>

</answer>
\(12\)

3 Runestone Testing
3.9 Multiple Choice Exercises

3.9.4. Multiple-Choice, Randomized, One Answer.

Hint 1.
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  <p>
    What did you see last time you went driving?
  </p>

</hint>
What did you see last time you went driving?
Hint 2.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    Maybe go out for a drive?
  </p>

</hint>
Maybe go out for a drive?

3.9.6. Multiple-Choice, Randomized, Multiple Answers.

Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    Do you know the acronym<ellipsis/><acro>ROY G BIV</acro> for the colors of a rainbow, and their order?
  </p>

</hint>
Do you know the acronym…ROY G BIV for the colors of a rainbow, and their order?

3.9.8. Multiple-Choice, Not Randomized, One Answer.

Hint 1.
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  <p>
    What did you see last time you went driving?
  </p>

</hint>
What did you see last time you went driving?
Hint 2.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>
    Maybe go out for a drive?
  </p>

</hint>
Maybe go out for a drive?

3.10 Parsons Exercises

3.19 Fill-In Exercises

3.19.2. Fill-In, New Markup Strings.

Hint.
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                        <hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

                          <p>
Do you really need a hint? Carefully reread the question.
                          </p>

                        </hint>
Do you really need a hint? Carefully reread the question.

3.21 Hodgepodge

3.21.2. With Tasks in an Exercises Division.

3.21.2.a True/False.

Hint.
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  <p>

    <m>P_n</m>, the vector space of polynomials with degree at most <m>n</m>, has dimension <m>n+1</m> by <xref ref="theorem-exponent-laws"/>.  [Cross-reference is just a demo, content is not relevant.]  What happens if we relax the defintion and remove the parameter <m>n</m>?

  </p>

</hint>
\(P_n\text{,}\) the vector space of polynomials with degree at most \(n\text{,}\) has dimension \(n+1\) by TheoremΒ 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)

3.22 Exercises that are Timed

Timed Exercises

3.22.2. Multiple-Choice, Not Randomized, One Answer.

Hint 1.
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                              <p>
What did you see last time you went driving?
                              </p>

                            </hint>
What did you see last time you went driving?
Hint 2.
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                            <hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

                              <p>
Maybe go out for a drive?
                              </p>

                            </hint>
Maybe go out for a drive?

3.30 Timed Chapter Exam

3.30.2. True/False.

Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">

  <p>

    <m>P_n</m>, the vector space of polynomials with degree at most <m>n</m>, has dimension <m>n+1</m> by <xref ref="theorem-exponent-laws"/>.  [Cross-reference is just a demo, content is not relevant.]  What happens if we relax the defintion and remove the parameter <m>n</m>?

  </p>

</hint>
\(P_n\text{,}\) the vector space of polynomials with degree at most \(n\text{,}\) has dimension \(n+1\) by TheoremΒ 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)