Selected Solutions to Homework # 3
•
Appendix C # 6: Prove that 3

(4
n

1) for every
n
∈
Z
+
.
Proof: If
n
= 1, this is true. Assume that 3

(4
n

1) for some
n
≥
1.
Then, 4
n

1 = 3
m
for some
m
∈
Z
. Consider 4
n
+1

1:
4
n
+1

1 = 4
·
4
n

1 = 3
·
4
n
+ 4
n

1 = 3
·
4
n
+ 3
m
= 3(4
n
+
m
)
.
Therefore, 3

(4
n
+1

1). By the principle of mathematical induction,
3

(4
n

1) for every
n
≥
1. Q.E.D.
•
Appendix C # 17: We discussed this in class on 01/19/2011.
•
Section 1.2 #2: Prove that
b

a
if and only if (

b
)

a
.
Don’t forget to prove both implications!
Proof: If
b

a
, then
a
=
bc
for some
c
∈
Z
. So,
a
= (

b
)(

c
). Therefore,
(

b
)

a
.
Conversely, if (

b
)

a
, then
a
= (

b
)
c
for some
c
∈
Z
.
So,
a
=
b
(

c
). Therefore,
b

a
.
•
Section 1.2 #4:
(a) If
a

b
and
a

c
, then
a

(
b
+
c
).
(b) If
a

b
and
a

c
, then
a

(
br
+
ct
) for every
r, t
∈
Z
.
(This is an important exercise in lieu of one of the major conepts we
will study in chapter 6: see p. 135; this exercise proves that the set
S
=
{
n
∈
Z
:
a

n
}
is an
ideal
in the
ring
R
=
Z
.)
If you have questions about this problem, please stop by during office
hours or make an appointment to meet with me. If you did not get
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 Spring '10
 R.Bell
 Algebra, Greatest common divisor, Divisor, common divisor, important exercise

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