MATH 135, Fall 2010
Solution of Assignment #4
Problem 1
.
Let
a,b,c
∈
Z
. Prove that if
ac

bc
and
c
6
= 0, then
a

b
.
Solution.
Since
ac

bc
, then
bc
=
qac
for some integer
q
.
Regrouping,
bc

qac
= 0 or
c
(
b

qa
) = 0.
Since
c
6
= 0, then
b

qa
= 0 or
b
=
qa
.
Therefore by deﬁnition,
a

b
.
Problem 2
.
(a) Let
m,n
∈
P
. Prove that if
m
2
6 
n
3
, then
m
6 
n
. (Hint: Use the contrapositive law.)
(b) Show that the converse of the above statement is false; i.e., the statement that if
m
6 
n
,
then
m
2
6 
n
3
, is false. (Hint: Use a counterexample.)
Solution.
(a) We prove the contrapositive: “If
m

n
, then
m
2

n
3
.”
Since
m

n
, then
n
=
qm
for some
q
∈
Z
.
Thus,
n
3
= (
qm
)
3
=
q
3
m
3
.
Since
q
3
m
∈
Z
, then
m
2

n
3
by deﬁnition.
Since the contrapositive is true, then the original statement is true.
(b) Let
m
= 8 and
n
= 4. It is easy to see that
m
6 
n
. However,
m
2
= 64 = (4)
3
=
n
3
. Thus,
m
2
±
±
n
2
.
Problem 3
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 Spring '10
 Mesta
 Greatest common divisor, Euclidean algorithm, gcd, nd, common divisor, Euclidean domain

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