Description
Abstract Algebra An Introduction 3rd Edition Solution Manual
Chapter
1
Arithmetic
in
Revisited
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
b y
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
b y
Theorem
1.1
this
representation
is unique,
it must
be
that
q
is the
quotient
and
rc
the
remainder
on
dividing
ac
b y
bc
.
6.
When
q
is divided
b y
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.
8.
9.
1
0.
≤
‘
‘
‘
‘
‘
‘
‘)
‘
≥
≤
‘
≤
‘
‘
‘
‘
‘
Z
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1.2
Divisibility
1.
(a)
8.
(b)
6.
(c)
1.
(d)
11.
(e)
9.
(f)
17.
(g)
592.
(h)
6.
2.
3.
4.
5.
Since
a

b
, w e
have
b
=
ak
for
some
integer
k
, and
a
6
=
0.
Since
b

a
, w e
have
a
=
bl
for
some
integer
l
, and
b
6
= 0.
Thus
a
=
bl
= (
ak
)
l
=
a
(
kl
).
Since
a
6
= 0,
divide
through
b y
a
to
get
1 =
kl
.
But
this
means
that
k
=
±
1 and
l
=
±
1, so
that
a
=
±
b
.
6.
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.
9.
No,
ab
need
not
divide
c
. F o r
one
example,
note
that
4

12
and
6

12,
but
4
·
6 =
24
does
not
divide
12.
10.
11.
12.
(a)
False.
(
ab,a
) is always
at
least
a
since
a

ab
and
a

a
.
(b)
False.
F o r
example,
(2
,
3)
= 1 and
(2
,
9)
= 1,
but
(3
,
9)
= 3.
(c)
False.
F o r
example,
let
a
= 2,
b
= 3, and
c
= 9.
Then
(2
,
3)
= 1 = (2
,
9),
but
(2
·
3
,
9)
= 3.
11.
≠
≤
a
Arithmetic
in
Z
Revisited
2
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.