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PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

Thomas W. Judson, Isaac Newton (Editor)

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Appendix B Hints and Answers to Selected Odd Exercises

2 The Integers
2.4 Exercises

2.4.1.

Answer.
The base case, S(1):[1(1+1)(2(1)+1)]/6=1=12 is true.
Assume that S(k):12+22+⋯+k2=[k(k+1)(2k+1)]/6 is true. Then
12+22+⋯+k2+(k+1)2=[k(k+1)(2k+1)]/6+(k+1)2=[(k+1)((k+1)+1)(2(k+1)+1)]/6,
and so S(k+1) is true. Thus, S(n) is true for all positive integers n.

2.4.3.

Answer.
The base case, S(4):4!=24>16=24 is true. Assume S(k):k!>2k is true. Then (k+1)!=k!(k+1)>2k⋅2=2k+1, so S(k+1) is true. Thus, S(n) is true for all positive integers n.

2.4.11.

Hint.
The base case, S(0):(1+x)0−1=0≥0=0⋅x is true. Assume S(k):(1+x)k−1≥kx is true. Then
(1+x)k+1−1=(1+x)(1+x)k−1=(1+x)k+x(1+x)k−1≥kx+x(1+x)k≥kx+x=(k+1)x,
so S(k+1) is true. Therefore, S(n) is true for all positive integers n.

2.4.19.

Hint.
Use the Fundamental Theorem of Arithmetic.

2.4.23.

Hint.
Let S={s∈N:a∣s, b∣s}. Then S≠∅, since |ab|∈S. By the Principle of Well-Ordering, S contains a least element m. To show uniqueness, suppose that a∣n and b∣n for some n∈N. By the division algorithm, there exist unique integers q and r such that n=mq+r, where 0≤r<m. Since a and b divide both m, and n, it must be the case that a and b both divide r. Thus, r=0 by the minimality of m. Therefore, m∣n.

2.4.27.

Hint.
Since gcd(a,b)=1, there exist integers r and s such that ar+bs=1. Thus, acr+bcs=c. Since a divides both bc and itself, a must divide c.

2.4.29.

Hint.
Every prime must be of the form 2, 3, 6n+1, or 6n+5. Suppose there are only finitely many primes of the form 6k+5.

3 Groups
3.5 Exercises

3.5.1.

Hint.
(a) 3+7Z={…,−4,3,10,…}; (c) 18+26Z; (e) 5+6Z.

3.5.15.

Hint.
There is a nonabelian group containing six elements.

3.5.17.

Hint.
The are five different groups of order 8.

3.5.25.

Hint.
(aba−1)n=(aba−1)(aba−1)⋯(aba−1)=ab(aa−1)b(aa−1)b⋯b(aa−1)ba−1=abna−1.

3.5.31.

Hint.
Since abab=(ab)2=e=a2b2=aabb, we know that ba=ab.

3.5.35.

Hint.
H1={id}, H2={id,ρ1,ρ2}, H3={id,μ1}, H4={id,μ2}, H5={id,μ3}, S3.

3.5.41.

Hint.
The identity of G is 1=1+02. Since (a+b2)(c+d2)=(ac+2bd)+(ad+bc)2, G is closed under multiplication. Finally, (a+b2)−1=a/(a2−2b2)−b2/(a2−2b2).

3.5.49.

Hint.
Since a4b=ba, it must be the case that b=a6b=a2ba, and we can conclude that ab=a3ba=ba.

3.5.55.

Answer.
1

3.5.57.

Answer.
n

3.5.59.

3.5.59.a
Answer.
2
3.5.59.b
3.5.59.b.i
Answer.
6
3.5.59.b.ii
Answer.
10

5 Runestone Testing
5.7 True/False Exercises

5.7.1. True/False.

Hint.
Pn, the vector space of polynomials with degree at most n, 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?

5.8 Multiple Choice Exercises

5.8.1. Multiple-Choice, Not Randomized, One Answer.

Hint 1.
What did you see last time you went driving?
Hint 2.
Maybe go out for a drive?

5.8.3. Multiple-Choice, Randomized, One Answer.

Hint 1.
What did you see last time you went driving?
Hint 2.
Maybe go out for a drive?

5.8.5. Mathematical Multiple-Choice, Not Randomized, Multiple Answers.

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.

5.9 Parsons Exercises

5.9.1. Parsons Problem, Mathematical Proof.

Hint.
Dorothy will not be much help with this proof.

5.11 Matching Exercises

5.11.3. Matching Problem, Linear Algebra.

Hint.
For openers, a basis for a subspace must be a subset of the subspace.

5.12 Clickable Area Exercises

5.12.3. Clickable Areas, Text in a Table.

Hint.
Python boolean variables begin with capital latters.

5.16 Hodgepodge>

5.16.1. With Tasks in an Exercises Division.

5.16.1.a True/False.
Hint.
Pn, the vector space of polynomials with degree at most n, 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?

5.17 Exercises that are Timed

Timed Exercises

5.17.1. True/False.
Hint.
Pn, the vector space of polynomials with degree at most n, 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?
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