<solutions divisional="hint answer solution" admit="even">
<title>Hints and Answers to Selected Even Exercises</title>
</solutions>
Appendix C Hints and Answers to Selected Even Exercises
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1 Preliminaries
1.4 Exercises
Warm-up
1.4.2.
Hint.
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<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) \(A \times D = \emptyset\text{.}\)
1.4.6.
Hint.
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<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, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
1.4.10.
Hint.
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<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>
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
1.4.14.
Hint.
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<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>
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
More Exercises
1.4.18.
Hint.
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<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. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
1.4.20.
1.4.22.
Hint.
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<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.
1.4.24.
Hint.
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<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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
1.4.28.
3 Groups
3.5 Exercises
3.5.2.
3.5.6.
Hint.
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<hint>
<p><me>\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}</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*}
3.5.8.
3.5.16.
3.5.18.
Hint.
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<hint>
<p>Let
<me>\sigma =
\begin{pmatrix}
1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & 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.
3.5.46.
3.5.60.
3.5.60.a
3.5.60.b
3.5.60.b.i
3.5.60.b.ii
5 Runestone Testing
5.9 Multiple Choice Exercises
5.9.4. Multiple-Choice, Randomized, One Answer.
5.9.6. Multiple-Choice, Randomized, Multiple Answers.
5.9.8. Multiple-Choice, Not Randomized, One Answer.
5.10 Parsons Exercises
5.10.6. Parsons Problem, Mathematical Proof, Numbered Blocks.
5.19 Fill-In Exercises
5.19.2. Fill-In, New Markup Strings.
5.19.6. Fill-In, Dynamic Math with Simple Numerical Answer.
5.19.8. Fill-In, Dynamic Math with Interdependent Formula Checking.
5.20 Hodgepodge
5.20.2. With Tasks in an Exercises Division.
5.20.2.a True/False.
Hint.
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<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 3.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)
