Solution Manual for Abstract Algebra: An Interactive Approach 2nd Edition | William Paulsen | Verified Solutions | Accurate Answers | A+ Grades

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.