MATH 135
Fall 2008
Assignment #4 Solutions
HandIn Problems
1. In each part, we must ensure that the remainder
r
satisfies 0
≤
r <

b

= 12.
(a) Since 261 = 21(12) + 9, then the quotient is 21 and the remainder is 9.
(b) Since

261 =

22(12) + 3, then the quotient is

22 and the remainder is 3.
(c) Since 261 = (

21)(

12) + 9, then the quotient is

21 and the remainder is 9.
(d) Since

261 = 22(

12) + 3, then the quotient is 22 and the remainder is 3.
2. We prove the result by contradiction.
Assume that
a, b, c
are consecutive terms in a geometric sequence.
Thus,
b
=
ar
and
c
=
ar
2
for some real number
r
.
(Since none of the three terms equals 0, then
a
= 0 and
r
= 0.)
So our quadratic equation becomes
ax
2
+
arx
+
ar
2
= 0.
Since
a
= 0, we can divide through by
a
to obtain
x
2
+
rx
+
r
2
= 0.
The discriminant is Δ =
r
2

4(1)(
r
2
) =

3
r
2
which is strictly less than 0 since
r
= 0.
Since the discriminant is negative, the quadratic equation cannot have real roots.
This is a contradiction, as we are told that the quadratic equation does have real roots.
Therefore,
a, b, c
cannot be consecutive terms in a geometric sequence.
3.
(a) Using the Euclidean Algorithm,
1989
=
7(251) + 232
251
=
1(232) + 19
232
=
12(19) + 4
19
=
4(4) + 3
4
=
1(3) + 1
3
=
3(1) + 0
so gcd(1989
,
251) = 1, the second last remainder.
(b) Reversing the Euclidean Algorithm above,
gcd(1989
,
251)
=
1
=
4

3
=
4

(19

4(4)) = 5(4)

1(19)
=
5(232

12(19))

1(19) = 5(232)

61(19)
=
5(232)

61(1(251)

1(232)) = 66(232)

61(251)
=
66(1989

7(251))

61(251) = 66(1989)

523(251)
so one integer solution to 1989
x
+ 251
y
= gcd(1989
,
251) is (
x, y
) = (66
,

523).
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4. We answer both parts together using the Extended Euclidean Algorithm.
1
0
686
0
1
511
1

1
175
1

2
3
161
2
3

4
14
1

35
47
7
11
73

98
0
2
Therefore, gcd(686
,
511) = 7.
Also,
one
solution
to
686
x
+
511
y
=
gcd(686
,
511)
is
(
x, y
) = (

35
,
47).
5. Since
ac

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

qac
= 0 or
c
(
b

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

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

b
.
6. We prove the contrapositive: “If
m

n
, then
m
2

n
2
.”
Since
m

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

n
2
by definition.
Since the contrapositive is true, then the original statement is true.
7. Let
d
= gcd(
a, b
).
Since
d

a
and
d

b
, then
nd

an
and
nd

nb
.
Since
d
= gcd(
a, b
), then there exist
x, y
∈
Z
such that
ax
+
by
=
d
.
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 Fall '08
 ANDREWCHILDS
 Algebra, Remainder, Natural number, Euclidean algorithm, Euclidean domain

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