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Appendix C Hints and Answers to Selected Even Exercises

I Basics
1 Preliminaries
1.4 Exercises

More Exercises

1.4.24.

1.4.24.a
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.
\(f\) is neither one-to-one nor onto. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)

1.4.28.

1.4.28.a
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.
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
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

2.4.28.

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.
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.
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

1.5.6.

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*}

1.5.16.

Hint.
Look at the symmetry group of an equilateral triangle or a square.

1.5.18.

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 Runestone Testing
3.9 Multiple Choice Exercises

3.10 Parsons Exercises

3.19 Fill-In Exercises

3.21 Hodgepodge

3.22 Exercises that are Timed

Timed Exercises

3.30 Timed Chapter Exam

3.30.2. True/False.

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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