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PreTeXt Sample Book Abstract Algebra (SAMPLE ONLY)

Thomas W. Judson, Isaac Newton (Editor)

Appendix C Hints and Answers to Selected Even Exercises

View Source for solutions
<solutions divisional="hint answer solution" admit="even">
  <title>Hints and Answers to Selected Even Exercises</title>
</solutions>

I Basics
1 Preliminaries
1.4 Exercises

Warm-up

1.4.2.
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) \(A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\text{;}\) (d) \(A \times D = \emptyset\text{.}\)
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1.4.6.
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>
If \(x \in A \cup (B \cap C)\text{,}\) then either \(x \in A\) or \(x \in B \cap C\text{.}\) Thus, \(x \in A \cup B\) and \(A \cup C\text{.}\) Hence, \(x \in (A \cup B) \cap (A \cup C)\text{.}\) Therefore, \(A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)\text{.}\) Conversely, if \(x \in (A \cup B) \cap (A \cup C)\text{,}\) then \(x \in A \cup B\) and \(A \cup C\text{.}\) Thus, \(x \in A\) or \(x\) is in both \(B\) and \(C\text{.}\) So \(x \in A \cup (B \cap C)\) and therefore \((A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)\text{.}\) Hence, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
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1.4.10.
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>
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
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1.4.14.
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>
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
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More Exercises

1.4.18.
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. \(f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.}\) (c) \(f\) is neither one-to-one nor onto. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
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1.4.20.
Hint.
View Source for hint
<hint>
  <p>
    (a) <m>f(n) = n + 1</m>.
  </p>
</hint>
(a) \(f(n) = n + 1\text{.}\)
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1.4.22.
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 \(x, y \in A\text{.}\) 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. (b) 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.
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1.4.24.
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 \(y \in f(A_1 \cup A_2)\text{.}\) 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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
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1.4.28.
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 \(X = {\mathbb N} \cup \{ \sqrt{2}\, \}\) and define \(x \sim y\) if \(x + y \in {\mathbb N}\text{.}\)
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II Algebra (and Runestone)
1 Groups
1.5 Exercises

1.5.2.

Hint.
View Source for hint
<hint>
  <p>
    (a) Not a group; (c) a group.
  </p>
</hint>
(a) Not a group; (c) a group.
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1.5.6.

Hint.
View Source for hint
<hint>
  <p>
    <me>
      \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}
    </me>
  </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*}
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1.5.8.

Hint.
View Source for hint
<hint>
  <p>
    Pick two matrices.
    Almost any pair will work.
  </p>
</hint>
Pick two matrices. Almost any pair will work.
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1.5.16.

Hint.
View Source for hint
<hint>
  <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.
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1.5.18.

Hint.
View Source for hint
<hint>
  <p>
    Let
    <me>
      \sigma = \begin{pmatrix} 1 &amp; 2 &amp; \cdots &amp; n \\ a_1 &amp; a_2 &amp; \cdots &amp; a_n \end{pmatrix}
    </me>
    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.
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1.5.46.

Hint.
View Source for hint
<hint>
  <p>
    Look at <m>S_3</m>.
  </p>
</hint>
Look at \(S_3\text{.}\)
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1.5.56.

Answer.
View Source for answer
<answer>
  <p>
    <m>2</m>
  </p>
</answer>
\(2\)
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1.5.58.

Answer.
View Source for answer
<answer>
  <p>
    <m>n+1</m>
  </p>
</answer>
\(n+1\)
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1.5.60.

1.5.60.a
Answer.
View Source for answer
<answer>
  <p>
    <m>4</m>
  </p>
</answer>
\(4\)
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1.5.60.b
1.5.60.b.i
Answer.
View Source for answer
<answer>
  <p>
    <m>8</m>
  </p>
</answer>
\(8\)
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1.5.60.b.ii
Answer.
View Source for answer
<answer>
  <p>
    <m>12</m>
  </p>
</answer>
\(12\)
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3 Runestone Testing
3.9 Multiple Choice Exercises

3.9.4. Multiple-Choice, Randomized, One Answer.

Hint 1.
View Source for hint
<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>
What did you see last time you went driving?
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Hint 2.
View Source for hint
<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>
Maybe go out for a drive?
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3.9.6. Multiple-Choice, Randomized, Multiple Answers.

Hint.
View Source for hint
<hint>
  <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?
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3.9.8. Multiple-Choice, Not Randomized, One Answer.

Hint 1.
View Source for hint
<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>
What did you see last time you went driving?
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Hint 2.
View Source for hint
<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>
Maybe go out for a drive?
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3.10 Parsons Exercises

3.10.6. Parsons Problem, Mathematical Proof, Numbered Blocks.

Hint.
View Source for hint
<hint>
  <p>
    Dorothy will not be much help with this proof.
  </p>
</hint>
Dorothy will not be much help with this proof.
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3.19 Fill-In Exercises

3.19.2. Fill-In, New Markup Strings.

Hint.
View Source for hint
<hint>
  <p>
    Do you really need a hint?
    Carefully reread the question.
  </p>
</hint>
Do you really need a hint? Carefully reread the question.
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3.19.6. Fill-In, Dynamic Math with Simple Numerical Answer.

3.19.8. Fill-In, Dynamic Math with Interdependent Formula Checking.

3.20 Hodgepodge

3.20.2. With Tasks in an Exercises Division.

3.20.2.a True/False.
Hint.
View Source for hint
<hint>
  <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 Theorem 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)
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3.21 Exercises that are Timed

Timed Exercises

3.21.2. Multiple-Choice, Not Randomized, One Answer.
Hint 1.
View Source for hint
<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>
What did you see last time you went driving?
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Hint 2.
View Source for hint
<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>
Maybe go out for a drive?
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3.29 Timed Chapter Exam

3.29.2. True/False.

Hint.
View Source for hint
<hint>
  <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 Theorem 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)
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