Abstract Algebra: An
Interactive Approach,
2nd Edition
by
William Paulsen
** Immediate Download
** All Chapters included
K25510 SM_Cover.indd 1 11/12/15 11:35 am
,Answers to Even-Numbered
Problems
Section 0.1
2) q = 15, r = 12
4) q = −21, r = 17
6) q = 87, r = 67
8) q = −1, r = 215
10) 1 + n < 1 + (n − 1)2 = n2 + 2(1 − n) < n2
12) If (n − 1)2 + 3(n − 1) + 4 = 2k, then n2 + 3n + 4 = 2(k + n + 1).
14) If 4n−1 − 1 = 3k, then 4n − 1 = 3(4k + 1).
16) (1 + x)n = (1 + x)(1 + x)n−1 ≥ (1 + x)(1 +(n − 1)x) = 1 + nx + x2(n − 1) ≥
1 + nx
18) (n − 1)2 + (2n − 1) = n2.
20) (n − 1)2((n − 1) + 1)2/4 + n3 = n2(n + 1)2/4.
22) (n − 1)/((n − 1) + 1) + 1/(n(n + 1)) = n/(n + 1).
24) 4 · 100 + (−11) · 36 = 4.
26) (−6) · 464 + 5 · 560 = 16.
28) (−2) · 465 + 9 · 105 = 15.
30) (−54) · (487) + (−221) · (−119) = 1.
32) Let c = gcd(a, b). Then c is the smallest positive element of the set A =
all integers of the form au + bv. If we multiply all element of A by d, we get
the set of all integers of the form dau + dbv, and the smallest positive element
of this set would be dc. Thus, gcd(da, db) = dc.
34) Since both x/gcd(x, y) and y/gcd(x, y) are both integers, we see that
(x ·y)/gcd(x, y) is a multiple of both x and y. If lcm(x, y) = ax = by is
smaller then (x · y)/gcd(x, y), then (x ·y)/lcm(x, y) would be greater than
gcd(x, y). Yet (x ·y)/lcm(x, y) = y/a = x/b would be a divisor of both x and
y.
36) 2 · 3 · 23 · 29.
38) 7 · 29 · 31.
40) 3 · 132 · 101.
42) u = −222222223, v = 1777777788.
44) 34 · 372 · 3336672.
Section 0.2
2) If a/b = c/d, so that ad = bc, then ab(c2 +d2) = abc2 +abd2 = a2cd+b2cd =
cd(a2 + b2). Thus, ab/(a2 + b2) = cd/(c2 + d2).
4) a) One-to-one, 3x + 5 = 3y + 5 ⇒ x = y. b) Onto, f ((y − 5)/3) = y.
6) a) One-to-one, x/3 − 2/5 = y/3 − 2/5 ⇒ x = y. b) Onto, f (3y + 6/5) = y.
1
,2 Answers to Even-Numbered Problems
8) a) One-to-one, if x > 0, y < 0 then y = 3x > 0. b) Onto, if y ≥ 0,
f (y/3) = y. If y < 0, f (y) = y.
10) a) Not one-to-one f (1) = f (2) = 1. b) Onto, f (2y − 1) = y.
12) a) One-to-one, if x even, y odd, then y = 2x + 2 is even. b) Not onto,
f (x) ̸= 3.
14) a) Not one-to-one f (5) = f (8) = 24. b) Not onto, f (x) ̸= 1.
16) Suppose f were one-to-one, and let B̃ = f (A), so that f˜ : A → B̃ would
be a bijection. By lemma 0.5, |A| = | B̃ |, but | B˜ | ≤ |B| < |A|.
18) Suppose f were not one-to-one. Then there is a case where f (a1) = f (a2),
and we can consider the set A˜ = A — {a1 }, and the function f˜ : A→
˜ B would
still be onto. But A | ˜| < B
| |so by Problem 17 f˜ cannot be onto. Hence, f is
one-to-one.
20) x4 + 2x2.
22) x3 − 3x{+ 2.
3x + 14 if x is even,
24) f (x) =
6x + 2 if x is odd.
26) If f (g(x)) = f (g(y)), then since f is one-to-one, g(x) = g(y). Since g is
onto, x = y.
28) There is some c ∈ C such that f (y) ̸= c for all y ∈ B. Then f (g(x)) ̸= c
since g(x) ∈ B.
30) If x even and y odd, f (x) = f (y) means y = x + 8 is even. Onto is proven
{
x + 3 if x is even,
by finding the inverse: f −1(x) =
x∗−z)5 = ifx +
x is
y+ odd.
z−
32) Associative, (x ∗ y) ∗ z = x ∗ (y 2.
34) Not associative, (x ∗ y) ∗ z = x − y − z, x ∗ (y ∗ z) = x − y + z.
36) Yes.
38) Yes.
40) Yes.
42) f (x) is both one-to-one and onto.
Section 0.3
2) 55
4) 25
6) 36
8) 7
10) 10
12) 91
14) 43
16) 223
18) 73
20) 1498
22) 3617
24) 3875
26) First find 0 ≤ q ≤ u · v such that q ≡ x(mod u) and q ≡ y(mod v). Then
find k so that k ≡ q(mod u · v) and k ≡ z(mod w).
28) 12
, Answers to Even-Numbered Problems 3
30) 4
32) 35
34) 17
36) 30
38) 51
40) 3684623194282304903214
42) 21827156424272739145155343596495185185220332
44) 1334817563332517248
Section 0.4
2) Since 1 +2⌊an⌋ is an integer, 1 +2⌊an⌋−an will have the same denominator
as an. Thus, the numerator an+1 is the denominator of an. Note that the
fractions will already be in lowest terms.
4) Since the sequence begins b0 = 0, b1 = 1, b2 = 1, b3 = 2,. . . we see
that the equations are true for n = 1. Assume both equations are true for
the previous n, that is, b2n−2 = bn−1 and b2n−1 = bn−1 + bn. Then by
the recursion formula, b2n = bn−1 + (bn−1 + bn— ) 2(bn−1 mod (bn−1 + bn)).
But (bn−1 mod (bn−1 + bn)) = bn−1, since bn−1 + bn > bn−1. So b2n = bn.
Then we can compute b2n+1 = bn−1 + bn + bn − 2(bn−1 + bn mod bn). But
(bn−1 + bn mod bn) = (bn−1 mod bn), and bn+1 = bn−1 + bn − 2(bn−1 mod bn).
Thus, b2n+1 = bn + bn+1.
6) a2n+1 = b2n+1/b2n+2 = (bn + bn+1)/bn+1 = (bn/bn+1) + 1 = an + 1.
8) If ai = aj for i > j, then because an+1 is determined solely on an, a2i−j =
ai. In fact, the sequence will repeat forever, so there would be only a finite of
rational numbers in the sequence. But this contradicts that every rational is
in the sequence, which is an infinite set.
10) In computing the long division of p/q, the remainders at each stage is
given by the sequence in Problem 9. Since this sequence eventually repeats,
the digits produced by the long division algorithm will eventually repeat.
12) If p3/q3 = 2 with p and q coprime, then 2|p, but replacing p = 2r shows
2|q too.
14) If p2/q2 = 5 with p and q coprime, then 5|p, but replacing p = 5r shows
5|q too.
16) If p3/q3 = 3 with p and q coprime, then 3|p, but replacing p = 3r shows
3|q too.
18) If 1/a were rational, then a = 1/a−1 would be rational.
20) Given x and y, choose any irrational z, and find a rational q between x —z
and y — z. Then q + z is irrational by Problem 19.
√ √
22) x2 = 5 + 2 6, and 6 is irrational, so x2 is too. If x were rational, then
x2 would be rational.
√ √
24) 2 — 2 and 2 are both irrational, but the sum is 2.
√ √
26) a6 = 2 + 4, a102 = 2 + 8.
,4 Answers to Even-Numbered Problems
Section 1.1
2) 12 steps.
4) y = y · e = y · (x · y′) = (y · x) · y′ = e · y′ = y′, so y = y′.
6) x = a−1 · b.
8) After 2 flips, Terry will be facing towards the audience again, so it would
be a rotation.
10) FlipRt·Spin = Spin·FlipLft. Other answers are possible.
12) (FlipRt·Spin)2 ̸= Stay·Stay. Other answers are possible.
14) FlipRt, FlipRt, and Spin cannot be expressed in terms of RotLft or RotRt.
Section 1.2
2)
0 1 2 3 4 5
0 0 1 2 3 4 5 4)
1 1 2 3 4 5 0 0 2 4 6
2 2 3 4 5 0 1 0 0 2 4 6
3 3 4 5 0 1 2 2 2 4 6 0
4 4 5 0 1 2 3 4 4 6 0 2
5 5 0 1 2 3 4 6 6 0 2 4
6)
0 3 6 9 12 15 18 21
0 0 3 6 9 12 15 18 21 8)
3 3 6 9 12 15 18 21 0 1 2 4 5 7 8
6 6 9 12 15 18 21 0 3 1 1 2 4 5 7 8
9 9 12 15 18 21 0 3 6 2 2 4 8 1 5 7
12 12 15 18 21 0 3 6 9 4 4 8 7 2 1 5
15 15 18 21 0 3 6 9 12 5 5 1 2 7 8 4
18 18 21 0 3 6 9 12 15 7 7 5 1 8 4 2
21 21 0 3 6 9 12 15 18 8 8 7 5 4 2 1
12)
10) 1 5 7 11 13 17 19 23
1 3 5 9 11 13 1 1 5 7 11 13 17 19 23
1 1 3 5 9 11 13 5 5 1 11 7 17 13 23 19
3 3 9 1 13 5 11 7 7 11 1 5 19 23 13 17
5 5 1 11 3 13 9 11 11 7 5 1 23 19 17 13
9 9 13 3 11 1 5 13 13 17 19 23 1 5 7 11
11 11 5 13 1 9 3 17 17 13 23 19 5 1 11 7
13 13 11 9 5 3 1 19 19 23 13 17 7 11 1 5
23 23 19 17 13 11 7 5 1
14) Since f (x) = f (x), x ∼ x. If f (x) = f (y), then f (y) = f (x), so y ∼ x.
Finally, if f (x) = f (y) and f (y) = f (z), then f (x) = f (z), so x ∼ z.
16) 15.
18) 11.
20) 5.
22) 97.
, Answers to Even-Numbered Problems 5
24) 491.
26) 353.
Section 1.3
2) Yes, this is a group.
4) Not closed, no identity, hence no inverses.
6) Yes, this is a group.
8) No additive inverses.
10) Not closed, no identity, hence no inverses.
12) 2 has no inverse.
14) Yes, this is a group, e = −3.
16) If e1 and e2 are two identity elements, then e1 = e1 · e2 = e2.
18) Note that (x · y) · (y−1 · x−1) = x · x−1 = e, so y−1 · x−1 is the inverse of
x·y
20) Consider the set{ a· x | x ∈S }which is a subset of S, yet must contain
the same number of elements.
22) To show x· y = y· x, start with (x· y)2 = x · y · x · y = e.
24) If a3 = e then (a−1)3 = e. Furthermore, if a ̸ = e, then a−1 ̸= a. So the
non-identity solutions pair off, and with the identity we have an odd number
of solutions.
26)
· a b c d
a b a d c
b a b c d
c d c a b
d c d b a
28) 9 → 6, 27 → 18, 81 → 54, 243 → 162, 5 → 4, 25 → 20, 125 → 100.
Conjecture (p − 1)n/p.
∗ | = |Z ∗ | · |Z ∗|.
30) If m and n are coprime, |Zmn m n
Section 2.1
2) 1, 3, 5, 9, 11, and 13.
4) 1, 5, 7, 11, 13, 17, 19, and 23.
6) 2, 6, 7, and 8.
8) 3 and 5.
10) No generators
12) No generators
14) 96.
16) 320.
18) 1210.
20) 1680.
Interactive Approach,
2nd Edition
by
William Paulsen
** Immediate Download
** All Chapters included
K25510 SM_Cover.indd 1 11/12/15 11:35 am
,Answers to Even-Numbered
Problems
Section 0.1
2) q = 15, r = 12
4) q = −21, r = 17
6) q = 87, r = 67
8) q = −1, r = 215
10) 1 + n < 1 + (n − 1)2 = n2 + 2(1 − n) < n2
12) If (n − 1)2 + 3(n − 1) + 4 = 2k, then n2 + 3n + 4 = 2(k + n + 1).
14) If 4n−1 − 1 = 3k, then 4n − 1 = 3(4k + 1).
16) (1 + x)n = (1 + x)(1 + x)n−1 ≥ (1 + x)(1 +(n − 1)x) = 1 + nx + x2(n − 1) ≥
1 + nx
18) (n − 1)2 + (2n − 1) = n2.
20) (n − 1)2((n − 1) + 1)2/4 + n3 = n2(n + 1)2/4.
22) (n − 1)/((n − 1) + 1) + 1/(n(n + 1)) = n/(n + 1).
24) 4 · 100 + (−11) · 36 = 4.
26) (−6) · 464 + 5 · 560 = 16.
28) (−2) · 465 + 9 · 105 = 15.
30) (−54) · (487) + (−221) · (−119) = 1.
32) Let c = gcd(a, b). Then c is the smallest positive element of the set A =
all integers of the form au + bv. If we multiply all element of A by d, we get
the set of all integers of the form dau + dbv, and the smallest positive element
of this set would be dc. Thus, gcd(da, db) = dc.
34) Since both x/gcd(x, y) and y/gcd(x, y) are both integers, we see that
(x ·y)/gcd(x, y) is a multiple of both x and y. If lcm(x, y) = ax = by is
smaller then (x · y)/gcd(x, y), then (x ·y)/lcm(x, y) would be greater than
gcd(x, y). Yet (x ·y)/lcm(x, y) = y/a = x/b would be a divisor of both x and
y.
36) 2 · 3 · 23 · 29.
38) 7 · 29 · 31.
40) 3 · 132 · 101.
42) u = −222222223, v = 1777777788.
44) 34 · 372 · 3336672.
Section 0.2
2) If a/b = c/d, so that ad = bc, then ab(c2 +d2) = abc2 +abd2 = a2cd+b2cd =
cd(a2 + b2). Thus, ab/(a2 + b2) = cd/(c2 + d2).
4) a) One-to-one, 3x + 5 = 3y + 5 ⇒ x = y. b) Onto, f ((y − 5)/3) = y.
6) a) One-to-one, x/3 − 2/5 = y/3 − 2/5 ⇒ x = y. b) Onto, f (3y + 6/5) = y.
1
,2 Answers to Even-Numbered Problems
8) a) One-to-one, if x > 0, y < 0 then y = 3x > 0. b) Onto, if y ≥ 0,
f (y/3) = y. If y < 0, f (y) = y.
10) a) Not one-to-one f (1) = f (2) = 1. b) Onto, f (2y − 1) = y.
12) a) One-to-one, if x even, y odd, then y = 2x + 2 is even. b) Not onto,
f (x) ̸= 3.
14) a) Not one-to-one f (5) = f (8) = 24. b) Not onto, f (x) ̸= 1.
16) Suppose f were one-to-one, and let B̃ = f (A), so that f˜ : A → B̃ would
be a bijection. By lemma 0.5, |A| = | B̃ |, but | B˜ | ≤ |B| < |A|.
18) Suppose f were not one-to-one. Then there is a case where f (a1) = f (a2),
and we can consider the set A˜ = A — {a1 }, and the function f˜ : A→
˜ B would
still be onto. But A | ˜| < B
| |so by Problem 17 f˜ cannot be onto. Hence, f is
one-to-one.
20) x4 + 2x2.
22) x3 − 3x{+ 2.
3x + 14 if x is even,
24) f (x) =
6x + 2 if x is odd.
26) If f (g(x)) = f (g(y)), then since f is one-to-one, g(x) = g(y). Since g is
onto, x = y.
28) There is some c ∈ C such that f (y) ̸= c for all y ∈ B. Then f (g(x)) ̸= c
since g(x) ∈ B.
30) If x even and y odd, f (x) = f (y) means y = x + 8 is even. Onto is proven
{
x + 3 if x is even,
by finding the inverse: f −1(x) =
x∗−z)5 = ifx +
x is
y+ odd.
z−
32) Associative, (x ∗ y) ∗ z = x ∗ (y 2.
34) Not associative, (x ∗ y) ∗ z = x − y − z, x ∗ (y ∗ z) = x − y + z.
36) Yes.
38) Yes.
40) Yes.
42) f (x) is both one-to-one and onto.
Section 0.3
2) 55
4) 25
6) 36
8) 7
10) 10
12) 91
14) 43
16) 223
18) 73
20) 1498
22) 3617
24) 3875
26) First find 0 ≤ q ≤ u · v such that q ≡ x(mod u) and q ≡ y(mod v). Then
find k so that k ≡ q(mod u · v) and k ≡ z(mod w).
28) 12
, Answers to Even-Numbered Problems 3
30) 4
32) 35
34) 17
36) 30
38) 51
40) 3684623194282304903214
42) 21827156424272739145155343596495185185220332
44) 1334817563332517248
Section 0.4
2) Since 1 +2⌊an⌋ is an integer, 1 +2⌊an⌋−an will have the same denominator
as an. Thus, the numerator an+1 is the denominator of an. Note that the
fractions will already be in lowest terms.
4) Since the sequence begins b0 = 0, b1 = 1, b2 = 1, b3 = 2,. . . we see
that the equations are true for n = 1. Assume both equations are true for
the previous n, that is, b2n−2 = bn−1 and b2n−1 = bn−1 + bn. Then by
the recursion formula, b2n = bn−1 + (bn−1 + bn— ) 2(bn−1 mod (bn−1 + bn)).
But (bn−1 mod (bn−1 + bn)) = bn−1, since bn−1 + bn > bn−1. So b2n = bn.
Then we can compute b2n+1 = bn−1 + bn + bn − 2(bn−1 + bn mod bn). But
(bn−1 + bn mod bn) = (bn−1 mod bn), and bn+1 = bn−1 + bn − 2(bn−1 mod bn).
Thus, b2n+1 = bn + bn+1.
6) a2n+1 = b2n+1/b2n+2 = (bn + bn+1)/bn+1 = (bn/bn+1) + 1 = an + 1.
8) If ai = aj for i > j, then because an+1 is determined solely on an, a2i−j =
ai. In fact, the sequence will repeat forever, so there would be only a finite of
rational numbers in the sequence. But this contradicts that every rational is
in the sequence, which is an infinite set.
10) In computing the long division of p/q, the remainders at each stage is
given by the sequence in Problem 9. Since this sequence eventually repeats,
the digits produced by the long division algorithm will eventually repeat.
12) If p3/q3 = 2 with p and q coprime, then 2|p, but replacing p = 2r shows
2|q too.
14) If p2/q2 = 5 with p and q coprime, then 5|p, but replacing p = 5r shows
5|q too.
16) If p3/q3 = 3 with p and q coprime, then 3|p, but replacing p = 3r shows
3|q too.
18) If 1/a were rational, then a = 1/a−1 would be rational.
20) Given x and y, choose any irrational z, and find a rational q between x —z
and y — z. Then q + z is irrational by Problem 19.
√ √
22) x2 = 5 + 2 6, and 6 is irrational, so x2 is too. If x were rational, then
x2 would be rational.
√ √
24) 2 — 2 and 2 are both irrational, but the sum is 2.
√ √
26) a6 = 2 + 4, a102 = 2 + 8.
,4 Answers to Even-Numbered Problems
Section 1.1
2) 12 steps.
4) y = y · e = y · (x · y′) = (y · x) · y′ = e · y′ = y′, so y = y′.
6) x = a−1 · b.
8) After 2 flips, Terry will be facing towards the audience again, so it would
be a rotation.
10) FlipRt·Spin = Spin·FlipLft. Other answers are possible.
12) (FlipRt·Spin)2 ̸= Stay·Stay. Other answers are possible.
14) FlipRt, FlipRt, and Spin cannot be expressed in terms of RotLft or RotRt.
Section 1.2
2)
0 1 2 3 4 5
0 0 1 2 3 4 5 4)
1 1 2 3 4 5 0 0 2 4 6
2 2 3 4 5 0 1 0 0 2 4 6
3 3 4 5 0 1 2 2 2 4 6 0
4 4 5 0 1 2 3 4 4 6 0 2
5 5 0 1 2 3 4 6 6 0 2 4
6)
0 3 6 9 12 15 18 21
0 0 3 6 9 12 15 18 21 8)
3 3 6 9 12 15 18 21 0 1 2 4 5 7 8
6 6 9 12 15 18 21 0 3 1 1 2 4 5 7 8
9 9 12 15 18 21 0 3 6 2 2 4 8 1 5 7
12 12 15 18 21 0 3 6 9 4 4 8 7 2 1 5
15 15 18 21 0 3 6 9 12 5 5 1 2 7 8 4
18 18 21 0 3 6 9 12 15 7 7 5 1 8 4 2
21 21 0 3 6 9 12 15 18 8 8 7 5 4 2 1
12)
10) 1 5 7 11 13 17 19 23
1 3 5 9 11 13 1 1 5 7 11 13 17 19 23
1 1 3 5 9 11 13 5 5 1 11 7 17 13 23 19
3 3 9 1 13 5 11 7 7 11 1 5 19 23 13 17
5 5 1 11 3 13 9 11 11 7 5 1 23 19 17 13
9 9 13 3 11 1 5 13 13 17 19 23 1 5 7 11
11 11 5 13 1 9 3 17 17 13 23 19 5 1 11 7
13 13 11 9 5 3 1 19 19 23 13 17 7 11 1 5
23 23 19 17 13 11 7 5 1
14) Since f (x) = f (x), x ∼ x. If f (x) = f (y), then f (y) = f (x), so y ∼ x.
Finally, if f (x) = f (y) and f (y) = f (z), then f (x) = f (z), so x ∼ z.
16) 15.
18) 11.
20) 5.
22) 97.
, Answers to Even-Numbered Problems 5
24) 491.
26) 353.
Section 1.3
2) Yes, this is a group.
4) Not closed, no identity, hence no inverses.
6) Yes, this is a group.
8) No additive inverses.
10) Not closed, no identity, hence no inverses.
12) 2 has no inverse.
14) Yes, this is a group, e = −3.
16) If e1 and e2 are two identity elements, then e1 = e1 · e2 = e2.
18) Note that (x · y) · (y−1 · x−1) = x · x−1 = e, so y−1 · x−1 is the inverse of
x·y
20) Consider the set{ a· x | x ∈S }which is a subset of S, yet must contain
the same number of elements.
22) To show x· y = y· x, start with (x· y)2 = x · y · x · y = e.
24) If a3 = e then (a−1)3 = e. Furthermore, if a ̸ = e, then a−1 ̸= a. So the
non-identity solutions pair off, and with the identity we have an odd number
of solutions.
26)
· a b c d
a b a d c
b a b c d
c d c a b
d c d b a
28) 9 → 6, 27 → 18, 81 → 54, 243 → 162, 5 → 4, 25 → 20, 125 → 100.
Conjecture (p − 1)n/p.
∗ | = |Z ∗ | · |Z ∗|.
30) If m and n are coprime, |Zmn m n
Section 2.1
2) 1, 3, 5, 9, 11, and 13.
4) 1, 5, 7, 11, 13, 17, 19, and 23.
6) 2, 6, 7, and 8.
8) 3 and 5.
10) No generators
12) No generators
14) 96.
16) 320.
18) 1210.
20) 1680.
