<solutions divisional="hint answer solution">
<title>Hints and Answers to Selected Exercises</title>
</solutions>
Appendix D Hints and Answers to Selected Exercises
View Source for 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) \(A \times D = \emptyset\text{.}\)
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, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
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>
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
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>
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
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. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
1.4.20.
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.
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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
1.4.28.
2 The Integers
2.4 Exercises
2.4.1.
Answer.
View Source for answer
<answer>
<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\text{,}
\end{align*}
and so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.3.
Answer.
View Source for answer
<answer>
<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.
View Source for hint
<hint>
<p>
Follow the proof in <xref ref="example-integers-binomial-theorem"/>.
</p>
</hint>
Follow the proof in Example 2.1.4.
2.4.11.
Hint.
View Source for hint
<hint>
<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\text{,}
\end{align*}
so \(S(k + 1)\) is true. Therefore, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.17. Fibonacci Numbers.
Hint.
View Source for hint
<hint>
<p>
For <xref ref="fn_ineq"/> and <xref ref="fn_prod"/> use mathematical induction.
<xref ref="fn_formula"/> Show that <m>f_1 = 1</m>,
<m>f_2 = 1</m>, and <m>f_{n + 2} = f_{n + 1} + f_n</m>.
<xref ref="fn_ratio"/> Use part <xref ref="fn_formula"/>.
<xref ref="fn_coprime"/> Use part <xref ref="fn_prod"/> and <xref ref="exercise-integers-gcd-1"/>.
</p>
</hint>
For Item 2.4.17.a and Item 2.4.17.b use mathematical induction. Item 2.4.17.c Show that \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\text{.}\) Item 2.4.17.d Use part Item 2.4.17.c. Item 2.4.17.e Use part Item 2.4.17.b and Exercise 2.4.16.
2.4.19.
2.4.23.
Hint.
View Source for hint
<hint>
<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, \(m \mid n\text{.}\)
2.4.27.
Hint.
View Source for hint
<hint>
<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 \(c\text{.}\)
2.4.29.
Hint.
View Source for hint
<hint>
<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
1.5.1.
Hint.
1.5.2.
1.5.6.
Hint.
View Source for hint
<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*}
1.5.8.
1.5.15.
1.5.16.
1.5.17.
1.5.18.
Hint.
View Source for hint
<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.
1.5.25.
Hint.
View Source for hint
<hint>
<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}\text{.}
\end{align*}
1.5.31.
1.5.35.
Hint.
View Source for hint
<hint>
<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.
View Source for hint
<hint>
<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, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)
1.5.46.
1.5.49.
Hint.
View Source for hint
<hint>
<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 \(ab = a^3 b a = ba\text{.}\)
1.5.55.
1.5.56.
1.5.57.
1.5.58.
1.5.59.
1.5.59.a
1.5.59.b
1.5.59.b.i
1.5.59.b.ii
1.5.60.
1.5.60.a
1.5.60.b
1.5.60.b.i
1.5.60.b.ii
3 Runestone Testing
3.8 True/False Exercises
3.8.1. 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 \(n\text{?}\)
3.9 Multiple Choice Exercises
3.9.1. Multiple-Choice, Not Randomized, One Answer.
3.9.3. Multiple-Choice, Not Randomized, Multiple Answers.
3.9.4. Multiple-Choice, Randomized, One Answer.
3.9.5. Multiple-Choice, Randomized, One Answer but with Checkboxes.
3.9.6. Multiple-Choice, Randomized, Multiple Answers.
3.9.7. Mathematical Multiple-Choice, Not Randomized, Multiple Answers.
Hint.
View Source for hint
<hint>
<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.
3.9.8. Multiple-Choice, Not Randomized, One Answer.
3.10 Parsons Exercises
3.10.1. Parsons Problem, Mathematical Proof.
3.10.6. Parsons Problem, Mathematical Proof, Numbered Blocks.
3.12 Matching Exercises
3.12.3. Cardsort Problem, Linear Algebra.
3.13 Clickable Area Exercises
3.13.3. Clickable Areas, Text in a Table.
3.19 Fill-In Exercises
3.19.2. Fill-In, New Markup Strings.
3.19.6. Fill-In, Dynamic Math with Simple Numerical Answer.
3.19.7. Fill-In, Dynamic Math with Formulas as Answers.
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 \(n\text{?}\)
3.21 Exercises that are Timed
Timed Exercises
3.21.1. 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 \(n\text{?}\)
3.21.2. Multiple-Choice, Not Randomized, One Answer.
3.27 Group Exercises
3.27.1. Multiple-Choice, Not Randomized, One Answer.
3.29 Timed Chapter Exam
3.29.1. Multiple-Choice, Not Randomized, One Answer.
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 \(n\text{?}\)
