Solutions Manual
Abstract Algebra An Introduction 3rd Edition
By
homas Hungerford,
David Leep
( All Chapters Included - 100% Verified Solutions )
1
, Chapter 1
Arithmetic in Z Revisited
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
1.1 The Division Algorithm
1. (a) q = 4, r = 1. (b) q = 0, r = 0. (c) q = −5, r = 3.
2. (a) q = −9, r = 3. (b) q = 15, r = 17. (c) q = 117, r = 11.
3. (a) q = 6, r = 19. (b) q = −9, r = 54. (c) q = 62720, r = 92.
4. (a) q = 15021, r = 132. (b) q = −14940, r = 335. (c) q = 39763, r = 3997.
5. Suppose a = bq + r, with 0 ≤ r < b. Multiplying this equation through by c gives ac = (bc)q + rc.
Further, since 0 ≤ r < b, it follows that 0 ≤ rc < bc. Thus this equation expresses ac as a multiple
of bc plus a remainder between 0 and bc − 1. Since by Theorem 1.1 this representation is unique,
it must be that q is the quotient and rc the remainder on dividing ac by bc.
6. When q is divided by c, the quotient is k, so that q = ck. Thus a = bq + r = b(ck) + r = (bc)k + r.
Further, since 0 ≤ r < b, it follows (since c ≥ 1) than 0 ≤ r < bc. Thus a = (bc)k + r is the unique
representation with 0 ≤ r < bc, so that the quotient is indeed k.
7. Answered in the text.
8. Any integer n can be divided by 4 with remainder r equal to 0, 1, 2 or 3. Then either n = 4k,
4k + 1, 4k + 2 or 4k + 3, where k is the quotient. If n = 4k or 4k + 2 then n is even. Therefore if
n is odd then n = 4k + 1 or 4k + 3.
9. We know that every integer a is of the form 3q, 3q + 1 or 3q + 2 for some q. In the last case
a3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8 = 9k + 8 where k = 3q3 + 6q2 + 4q. Other cases are similar.
10. Suppose a = nq + r where 0 ≤ r < n and c = nq' + r' where 0 < r' < n. If r = r' then a – c =
n(q – q') and k = q – q' is an integer. Conversely, given a – c = nk we can substitute to find:
(r – r') = n(k – q + q'). Suppose r ≥ r' (the other case is similar). The given inequalities imply
that 0 ≤ (r – r') < n and it follows that 0 ≤ (k – q + q') < 101 and we conclude that k – q + q' = 0.
Therefore r – r' = 0, so that r = r' as claimed.
Not For Sale
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
2
,2
Not For Sale
11. Given integers a and c with c ≠ 0. Apply Theorem 1.1 with b = |c| to get a = |c| . q + r where 0
0
Arithmetic in Z Revisited
≤ r < |c|. Let q = q0 if c > 0 and q = –q0 if c < 0. Then a = cq + r as claimed. The uniqueness is
proved as in Theorem 1.1.
1.2 Divisibility
1. (a) 8. (d) 11. (g) 592.
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
(b) 6. (e) 9. (h) 6.
(c) 1. (f) 17.
2. If b | a then a = bx for some integer x. Then a = (–b)(–x) so that (–b) | a. The converse follows
similarly.
3. Answered in the text.
4. (a) Given b = ax and c = ay for some integers x, y, we find b + c = ax + ay = a(x + y).
Since x + y is an integer, conclude that a | (b + c).
(b) Given x and y as above we find br + ct = (ax)r + (ay)t = a(xr + yt) using the associative
and distributive laws. Since xr + yt is an integer we conclude that a | (br + ct).
5. Since a | b, we have b = ak for some integer k, and a 6= 0. Since b | a, we have a = bl for some
integer l, and b 6= 0. Thus a = bl = (ak)l = a(kl). Since a 6= 0, divide through by a to get 1 = kl.
But this means that k = ±1 and l = ±1, so that a = ± b.
6. Given b = ax and d = cy for some integers x, y, we have bd = (ax)(cy) = (ac)(xy). Then ac | bd
because xy is an integer.
7. Clearly (a, 0) is at most |a| since no integer larger than |a| divides a. But also |a| | a, and |a| | 0
since any nonzero integer divides 0. Hence |a| is the gcd of a and 0.
8. If d = (n, n + 1) then d | n and d | (n + 1). Since (n + 1) – n = 1 we conclude that d | 1. (Apply
Exercise 4(b).) This implies d = 1, since d > 0.
9. No, ab need not divide c. For one example, note that 4 | 12 and 6 | 12, but 4 · 6 = 24 does not
divide 12.
10. Since a | a and a | 0 we have a | (a, 0). If (a, 0) = 1 then a | 1 forcing a = ±1.
11. (a) 1 or 2 (b) 1, 2, 3 or 6. Generally if d = (n, n + c) then d | n and d | (n + c).
Since c is a linear combination of n and n+c, conclude that d | c.
12. (a) False. (ab, a) is always at least a since a | ab and a | a.
(b) False. For example, (2, 3) = 1 and (2, 9) = 1, but (3, 9) = 3.
(c) False. For example, let a = 2, b = 3, and c = 9. Then (2, 3) = 1 = (2, 9), but (2 · 3, 9) = 3.
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
3
, 1.2 Divisibility 3
13. (a) Suppose c | a and c | b. Write a = ck and b = cl. Then a = bq + r can be rewritten
ck = (cl)q + r, so that r = ck − clq = c(k − lq). Thus c | r as well, so that c is a common
divisor of b and r.
(b) Suppose c | b and c | r. Write b = ck and r = cl, and substitute into a = bq + r to get
a = ckq + cl = c(kq + l). Thus c | a, so that c is a common divisor of a and b.
(c) Since (a, b) is a common divisor of a and b, it is also a common divisor of b and r, by part (a).
If (a, b) is not the greatest common divisor (b, r) of b and r, then (a, b) > (b, r). Now, consider
(b, r). By part (b), this is also a common divisor of (a, b), but it is less than (a, b). This is a
contradiction. Thus (a, b) = (b, r).
14. By Theorem 1.3, the smallest positive integer in the set S of all linear combinations of a and b is
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
exactly (a, b).
(a) (6, 15) = 3 (b) (12, 17)=1.
15. (a) This is a calculation.
(b) At the first step, for example, by Exercise 13 we have (a, b) = (524, 148) = (148, 80) = (b, r).
The same applies at each of the remaining steps. So at the final step, we have (8, 4) = (4, 0);
putting this string of equalities together gives
(524, 148) = (148, 80) = (80, 68) = (68, 12) = (12, 8) = (8, 4) = (4, 0).
But by Example 4, (4, 0) = 4, so that (524, 148) = 4.
(c) 1003 = 56 · 17 + 51, 56 = 51 · 1 + 5, 51 = 5 · 10 + 1, 5 = 1 · 5 + 0. Thus (1003, 56) = (1, 0) = 1.
(d) 322 = 148 · 2 + 26, 148 = 26 · 5 + 18, 26 = 18 · 1 + 8, 18 = 8 · 2 + 2, 8 = 2 · 4 + 0, so that
(322, 148) = (2, 0) = 2.
(e) 5858 = 1436 · 4 + 114, 1436 = 114 · 12 + 68, 114 = 68 · 1 + 46, 68 = 46 · 1 + 22, 46 = 22 · 2 + 2,
22 = 2 · 11 + 0, so that (5858, 1436) = (2, 0) = 2.
(f) 68 = 148 − (524 − 148 · 3) = −524 + 148 · 4.
(g) 12 = 80 − 68 · 1 = (524 − 148 · 3) − (−524 + 148 · 4) · 1 = 524 · 2 − 148 · 7.
(h) 8 = 68 − 12 · 5 = (−524 + 148 · 4) − (524 · 2 − 148 · 7) · 5 = −524 · 11 + 148 · 39.
(i) 4 = 12 − 8 = (524 · 2 − 148 · 7) − (−524 · 11 + 148 · 39) = 524 · 13 − 148 · 46.
(j) Working the computation backwards gives 1 = 1003 · 11 − 56 · 197.
16. Let a = da1 and b = db1. Then a1 and b1 are integers and we are to prove: (a1, b1) = l. By
Theorem 1.3 there exist integers u, v such that au + bv = d. Substituting and cancelling we find
that a1u + b1v = l. Therefore any common divisor of a1 and b1 must also divide this linear
combination, so it divides 1. Hence (a1, b1) = 1.
17. Since b | c, we know that c = bt for some integer t. Thus a | c means that a | bt. But then Theorem
1.4 tells us, since (a, b) = 1, that a | t. Multiplying both sides by b gives ab | bt = c.
18. Let d = (a, b) so there exist integers x, y with ax + by = d. Note that cd | (ca, cb) since cd
divides ca and cb. Also cd = cax + cby so that (ca, cb) | cd. Since these quantities are positive we
get cd = (ca, cd).
19. Let d = (a, b). Since b + c = aw for some integer w, we know c is a linear combination of a and b
so that d | c. But then d | (b, c ) = 1 forcing d = 1. Similarly (a, c ) = 1.
Not For Sale
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
4
Abstract Algebra An Introduction 3rd Edition
By
homas Hungerford,
David Leep
( All Chapters Included - 100% Verified Solutions )
1
, Chapter 1
Arithmetic in Z Revisited
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
1.1 The Division Algorithm
1. (a) q = 4, r = 1. (b) q = 0, r = 0. (c) q = −5, r = 3.
2. (a) q = −9, r = 3. (b) q = 15, r = 17. (c) q = 117, r = 11.
3. (a) q = 6, r = 19. (b) q = −9, r = 54. (c) q = 62720, r = 92.
4. (a) q = 15021, r = 132. (b) q = −14940, r = 335. (c) q = 39763, r = 3997.
5. Suppose a = bq + r, with 0 ≤ r < b. Multiplying this equation through by c gives ac = (bc)q + rc.
Further, since 0 ≤ r < b, it follows that 0 ≤ rc < bc. Thus this equation expresses ac as a multiple
of bc plus a remainder between 0 and bc − 1. Since by Theorem 1.1 this representation is unique,
it must be that q is the quotient and rc the remainder on dividing ac by bc.
6. When q is divided by c, the quotient is k, so that q = ck. Thus a = bq + r = b(ck) + r = (bc)k + r.
Further, since 0 ≤ r < b, it follows (since c ≥ 1) than 0 ≤ r < bc. Thus a = (bc)k + r is the unique
representation with 0 ≤ r < bc, so that the quotient is indeed k.
7. Answered in the text.
8. Any integer n can be divided by 4 with remainder r equal to 0, 1, 2 or 3. Then either n = 4k,
4k + 1, 4k + 2 or 4k + 3, where k is the quotient. If n = 4k or 4k + 2 then n is even. Therefore if
n is odd then n = 4k + 1 or 4k + 3.
9. We know that every integer a is of the form 3q, 3q + 1 or 3q + 2 for some q. In the last case
a3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8 = 9k + 8 where k = 3q3 + 6q2 + 4q. Other cases are similar.
10. Suppose a = nq + r where 0 ≤ r < n and c = nq' + r' where 0 < r' < n. If r = r' then a – c =
n(q – q') and k = q – q' is an integer. Conversely, given a – c = nk we can substitute to find:
(r – r') = n(k – q + q'). Suppose r ≥ r' (the other case is similar). The given inequalities imply
that 0 ≤ (r – r') < n and it follows that 0 ≤ (k – q + q') < 101 and we conclude that k – q + q' = 0.
Therefore r – r' = 0, so that r = r' as claimed.
Not For Sale
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
2
,2
Not For Sale
11. Given integers a and c with c ≠ 0. Apply Theorem 1.1 with b = |c| to get a = |c| . q + r where 0
0
Arithmetic in Z Revisited
≤ r < |c|. Let q = q0 if c > 0 and q = –q0 if c < 0. Then a = cq + r as claimed. The uniqueness is
proved as in Theorem 1.1.
1.2 Divisibility
1. (a) 8. (d) 11. (g) 592.
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
(b) 6. (e) 9. (h) 6.
(c) 1. (f) 17.
2. If b | a then a = bx for some integer x. Then a = (–b)(–x) so that (–b) | a. The converse follows
similarly.
3. Answered in the text.
4. (a) Given b = ax and c = ay for some integers x, y, we find b + c = ax + ay = a(x + y).
Since x + y is an integer, conclude that a | (b + c).
(b) Given x and y as above we find br + ct = (ax)r + (ay)t = a(xr + yt) using the associative
and distributive laws. Since xr + yt is an integer we conclude that a | (br + ct).
5. Since a | b, we have b = ak for some integer k, and a 6= 0. Since b | a, we have a = bl for some
integer l, and b 6= 0. Thus a = bl = (ak)l = a(kl). Since a 6= 0, divide through by a to get 1 = kl.
But this means that k = ±1 and l = ±1, so that a = ± b.
6. Given b = ax and d = cy for some integers x, y, we have bd = (ax)(cy) = (ac)(xy). Then ac | bd
because xy is an integer.
7. Clearly (a, 0) is at most |a| since no integer larger than |a| divides a. But also |a| | a, and |a| | 0
since any nonzero integer divides 0. Hence |a| is the gcd of a and 0.
8. If d = (n, n + 1) then d | n and d | (n + 1). Since (n + 1) – n = 1 we conclude that d | 1. (Apply
Exercise 4(b).) This implies d = 1, since d > 0.
9. No, ab need not divide c. For one example, note that 4 | 12 and 6 | 12, but 4 · 6 = 24 does not
divide 12.
10. Since a | a and a | 0 we have a | (a, 0). If (a, 0) = 1 then a | 1 forcing a = ±1.
11. (a) 1 or 2 (b) 1, 2, 3 or 6. Generally if d = (n, n + c) then d | n and d | (n + c).
Since c is a linear combination of n and n+c, conclude that d | c.
12. (a) False. (ab, a) is always at least a since a | ab and a | a.
(b) False. For example, (2, 3) = 1 and (2, 9) = 1, but (3, 9) = 3.
(c) False. For example, let a = 2, b = 3, and c = 9. Then (2, 3) = 1 = (2, 9), but (2 · 3, 9) = 3.
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
3
, 1.2 Divisibility 3
13. (a) Suppose c | a and c | b. Write a = ck and b = cl. Then a = bq + r can be rewritten
ck = (cl)q + r, so that r = ck − clq = c(k − lq). Thus c | r as well, so that c is a common
divisor of b and r.
(b) Suppose c | b and c | r. Write b = ck and r = cl, and substitute into a = bq + r to get
a = ckq + cl = c(kq + l). Thus c | a, so that c is a common divisor of a and b.
(c) Since (a, b) is a common divisor of a and b, it is also a common divisor of b and r, by part (a).
If (a, b) is not the greatest common divisor (b, r) of b and r, then (a, b) > (b, r). Now, consider
(b, r). By part (b), this is also a common divisor of (a, b), but it is less than (a, b). This is a
contradiction. Thus (a, b) = (b, r).
14. By Theorem 1.3, the smallest positive integer in the set S of all linear combinations of a and b is
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
exactly (a, b).
(a) (6, 15) = 3 (b) (12, 17)=1.
15. (a) This is a calculation.
(b) At the first step, for example, by Exercise 13 we have (a, b) = (524, 148) = (148, 80) = (b, r).
The same applies at each of the remaining steps. So at the final step, we have (8, 4) = (4, 0);
putting this string of equalities together gives
(524, 148) = (148, 80) = (80, 68) = (68, 12) = (12, 8) = (8, 4) = (4, 0).
But by Example 4, (4, 0) = 4, so that (524, 148) = 4.
(c) 1003 = 56 · 17 + 51, 56 = 51 · 1 + 5, 51 = 5 · 10 + 1, 5 = 1 · 5 + 0. Thus (1003, 56) = (1, 0) = 1.
(d) 322 = 148 · 2 + 26, 148 = 26 · 5 + 18, 26 = 18 · 1 + 8, 18 = 8 · 2 + 2, 8 = 2 · 4 + 0, so that
(322, 148) = (2, 0) = 2.
(e) 5858 = 1436 · 4 + 114, 1436 = 114 · 12 + 68, 114 = 68 · 1 + 46, 68 = 46 · 1 + 22, 46 = 22 · 2 + 2,
22 = 2 · 11 + 0, so that (5858, 1436) = (2, 0) = 2.
(f) 68 = 148 − (524 − 148 · 3) = −524 + 148 · 4.
(g) 12 = 80 − 68 · 1 = (524 − 148 · 3) − (−524 + 148 · 4) · 1 = 524 · 2 − 148 · 7.
(h) 8 = 68 − 12 · 5 = (−524 + 148 · 4) − (524 · 2 − 148 · 7) · 5 = −524 · 11 + 148 · 39.
(i) 4 = 12 − 8 = (524 · 2 − 148 · 7) − (−524 · 11 + 148 · 39) = 524 · 13 − 148 · 46.
(j) Working the computation backwards gives 1 = 1003 · 11 − 56 · 197.
16. Let a = da1 and b = db1. Then a1 and b1 are integers and we are to prove: (a1, b1) = l. By
Theorem 1.3 there exist integers u, v such that au + bv = d. Substituting and cancelling we find
that a1u + b1v = l. Therefore any common divisor of a1 and b1 must also divide this linear
combination, so it divides 1. Hence (a1, b1) = 1.
17. Since b | c, we know that c = bt for some integer t. Thus a | c means that a | bt. But then Theorem
1.4 tells us, since (a, b) = 1, that a | t. Multiplying both sides by b gives ab | bt = c.
18. Let d = (a, b) so there exist integers x, y with ax + by = d. Note that cd | (ca, cb) since cd
divides ca and cb. Also cd = cax + cby so that (ca, cb) | cd. Since these quantities are positive we
get cd = (ca, cd).
19. Let d = (a, b). Since b + c = aw for some integer w, we know c is a linear combination of a and b
so that d | c. But then d | (b, c ) = 1 forcing d = 1. Similarly (a, c ) = 1.
Not For Sale
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
4
