<solutions divisional="hint answer solution" admit="odd" xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<title>Hints and Answers to Selected Odd Exercises</title>
</solutions>
Appendix B Hints and Answers to Selected Odd Exercises
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I Basics
1 Preliminaries
1.4 Exercises
Warm-up
1.4.5.
1.4.5.a
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>
</p>
</hint>
\(A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\)
1.4.5.d
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>A \times D = \emptyset</m>
</p>
</hint>
\(A \times D = \emptyset\)
1.4.9.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<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, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
1.4.13.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<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>
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
1.4.17.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<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
1.4.37.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<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 \(x + y \in {\mathbb N}\text{.}\)
2 The Integers
2.4 Exercises
2.4.1.
Answer.
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<answer xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
The base case, <m>S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2</m> is true.
</p>
<p>
Assume that <m>S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6</m> is true.
Then
<md>
<mrow>1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2</mrow>
<mrow>& = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6,</mrow>
</md>
and so <m>S(k + 1)</m> is true.
Thus, <m>S(n)</m> is true for all positive integers <m>n</m>.
</p>
</answer>
The base case, \(S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2\) is true.
Assume that \(S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6\) is true. Then
\begin{align*}
1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2\\
& = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6,
\end{align*}
and so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.3.
Answer.
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<answer xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
The base case, <m>S(4): 4! = 24 \gt 16 =2^4</m> is true.
Assume <m>S(k): k! \gt 2^k</m> is true.
Then <m>(k + 1)! = k! (k + 1) \gt 2^k \cdot 2 = 2^{k + 1}</m>, so <m>S(k + 1)</m> is true.
Thus, <m>S(n)</m> is true for all positive integers <m>n</m>.
</p>
</answer>
The base case, \(S(4): 4! = 24 \gt 16 =2^4\) is true. Assume \(S(k): k! \gt 2^k\) is true. Then \((k + 1)! = k! (k + 1) \gt 2^k \cdot 2 = 2^{k + 1}\text{,}\) so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.9.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
Follow the proof in <xref ref="example-integers-binomial-theorem"/>.
</p>
</hint>
Follow the proof in ExampleΒ 2.1.4.
2.4.11.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
The base case, <m>S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x</m> is true.
Assume <m>S(k): (1 + x)^k -1 \geq kx</m> is true.
Then
<md>
<mrow>(1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1</mrow>
<mrow>& = (1 + x)^k + x(1 + x)^k - 1</mrow>
<mrow>& \geq kx + x(1 + x)^k</mrow>
<mrow>& \geq kx + x</mrow>
<mrow>& = (k + 1)x,</mrow>
</md>
so <m>S(k + 1)</m> is true.
Therefore, <m>S(n)</m> is true for all positive integers <m>n</m>.
</p>
</hint>
The base case, \(S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x\) is true. Assume \(S(k): (1 + x)^k -1 \geq kx\) is true. Then
\begin{align*}
(1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1\\
& = (1 + x)^k + x(1 + x)^k - 1\\
& \geq kx + x(1 + x)^k\\
& \geq kx + x\\
& = (k + 1)x,
\end{align*}
so \(S(k + 1)\) is true. Therefore, \(S(n)\) is true for all positive integers \(n\text{.}\)
II Algebra (and Runestone)
1 Groups
1.5 Exercises
1.5.1.
1.5.1.a
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}</m>
</p>
</hint>
\(3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\)
1.5.1.c
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>18 + 26 \mathbb Z</m>
</p>
</hint>
\(18 + 26 \mathbb Z\)
1.5.1.e
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>5 + 6 \mathbb Z</m>
</p>
</hint>
\(5 + 6 \mathbb Z\)
1.5.15.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
There is a nonabelian group containing six elements.
</p>
</hint>
There is a nonabelian group containing six elements.
1.5.17.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
The are five different groups of order 8.
</p>
</hint>
The are five different groups of order 8.
1.5.25.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<md>
<mrow>(aba^{-1})^n & = (aba^{-1})(aba^{-1}) \cdots (aba^{-1})</mrow>
<mrow>& = ab(aa^{-1})b(aa^{-1})b \cdots b(aa^{-1})ba^{-1}</mrow>
<mrow>& = ab^na^{-1}.</mrow>
</md>
</p>
</hint>
\begin{align*}
(aba^{-1})^n & = (aba^{-1})(aba^{-1}) \cdots (aba^{-1})\\
& = ab(aa^{-1})b(aa^{-1})b \cdots b(aa^{-1})ba^{-1}\\
& = ab^na^{-1}.
\end{align*}
1.5.31.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
Since <m>abab = (ab)^2 = e = a^2 b^2 = aabb</m>, we know that <m>ba = ab</m>.
</p>
</hint>
Since \(abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}\) we know that \(ba = ab\text{.}\)
1.5.35.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>H_1 = \{ id \}</m>, <m>H_2 = \{ id, \rho_1, \rho_2 \}</m>, <m>H_3 = \{ id, \mu_1 \}</m>, <m>H_4 = \{ id, \mu_2 \}</m>, <m>H_5 = \{ id, \mu_3 \}</m>, <m>S_3</m>.
</p>
</hint>
\(H_1 = \{ id \}\text{,}\) \(H_2 = \{ id, \rho_1, \rho_2 \}\text{,}\) \(H_3 = \{ id, \mu_1 \}\text{,}\) \(H_4 = \{ id, \mu_2 \}\text{,}\) \(H_5 = \{ id, \mu_3 \}\text{,}\) \(S_3\text{.}\)
1.5.41.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
The identity of <m>G</m> is <m>1 = 1 + 0 \sqrt{2}</m>.
Since <m>(a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}</m>, <m>G</m> is closed under multiplication.
Finally, <m>(a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)</m>.
</p>
</hint>
The identity of \(G\) is \(1 = 1 + 0 \sqrt{2}\text{.}\) Since \((a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}\text{,}\) \(G\) is closed under multiplication. Finally, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)
1.5.49.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
Since <m>a^4b = ba</m>, it must be the case that <m>b = a^6 b = a^2 b a</m>, and we can conclude that <m> ab = a^3 b a = ba</m>.
</p>
</hint>
Since \(a^4b = ba\text{,}\) it must be the case that \(b = a^6 b = a^2 b a\text{,}\) and we can conclude that \(ab = a^3 b a = ba\text{.}\)
1.5.59.
1.5.59.a
Answer.
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<answer xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>2</m>
</p>
</answer>
\(2\)
1.5.59.b
1.5.59.b.i
Answer.
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<answer xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>6</m>
</p>
</answer>
\(6\)
1.5.59.b.ii
Answer.
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<answer xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
<m>10</m>
</p>
</answer>
\(10\)
3 Runestone Testing
3.8 True/False Exercises
3.8.1. 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 \(n\text{?}\)
3.9 Multiple Choice Exercises
3.9.1. Multiple-Choice, Not Randomized, One Answer.
Hint 1.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<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?
3.9.3. Multiple-Choice, Not 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 acronymβ¦ROY G BIV for the colors of a rainbow, and their order?
3.9.5. Multiple-Choice, Randomized, One Answer but with Checkboxes.
Hint 1.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
What did you see last time you went driving?
</p>
</hint>
What did you see last time you went driving?
Hint 2.
View Source for hint
<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
Maybe go out for a drive?
</p>
</hint>
Maybe go out for a drive?
3.9.7. Mathematical Multiple-Choice, Not Randomized, Multiple Answers.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
You can take a derivative on any one of the choices to see if it is correct or not, rather than using techniques of integration to find <em>a single</em> correct answer.
</p>
</hint>
You can take a derivative on any one of the choices to see if it is correct or not, rather than using techniques of integration to find a single correct answer.
3.10 Parsons Exercises
3.10.1. Parsons Problem, Mathematical Proof.
Hint.
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<p>
Dorothy will not be much help with this proof.
</p>
</hint>
Dorothy will not be much help with this proof.
3.12 Matching Exercises
3.12.3. Cardsort Problem, Linear Algebra.
Hint.
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<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
For openers, a basis for a subspace must be a <em>subset</em> of the subspace.
</p>
</hint>
For openers, a basis for a subspace must be a subset of the subspace.
3.13 Clickable Area Exercises
3.13.3. Clickable Areas, Text in a Table.
Hint.
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<p>
Python boolean variables begin with capital latters.
</p>
</hint>
Python boolean variables begin with capital latters.
3.19 Fill-In Exercises
3.19.7. Fill-In, Dynamic Math with Formulas as Answers.
3.22 Exercises that are Timed
Timed Exercises
3.22.1. True/False.
Hint.
View Source for hint
<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 \(n\text{?}\)
3.28 Group Exercises
3.28.1. Multiple-Choice, Not Randomized, One Answer.
Hint 1.
View Source for hint
<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
What did you see last time you went driving?
</p>
</hint>
What did you see last time you went driving?
Hint 2.
View Source for hint
<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
3.30.1. Multiple-Choice, Not Randomized, One Answer.
Hint 1.
View Source for hint
<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
What did you see last time you went driving?
</p>
</hint>
What did you see last time you went driving?
Hint 2.
View Source for hint
<hint xmlns:pi="http://pretextbook.org/2020/pretext/internal">
<p>
Maybe go out for a drive?
</p>
</hint>
Maybe go out for a drive?

