MATH 135
Winter 2008
Midterm Solutions
1. (a) Write down the converse of “If
x
is a perfect square, then
x
≥
0”.
(b) Write down the contrapositive of “If
b >
8, then
f
(
b
)
<
0 or
f
(
b
) = 99”.
(c) Determine (with brief justiﬁcation) the
second smallest
positive integer
s
such that
70
x
+ 42
y
=
s
has a solution
x, y
∈
Z
.
(d) Could mathematical induction be used to prove that

cos(
x
)
 ≤
1 for all
x
∈
R
?
Brieﬂy explain why or why not.
(e) With the universe of discourse as the rationals, state (but do not prove) the following
using quantiﬁers.
“There is no smallest positive rational number”.
Solution:
(a) The converse is “If
x
≥
0, then
x
is a perfect square”.
(b) The contrapositive is “If
f
(
b
)
≥
0 and
f
(
b
)
6
= 99, then
b
≤
8”.
(c) By inspection, gcd(70
,
42) = 14.
(Alternately, the gcd could be determined using the Extended Euclidean Algorithm).
By the Linear Diophantine Equation Theorem 2.31 Part 1, 70
x
+ 42
y
=
s
has a solution
if and only if gcd(70
,
42)

s
. The smallest positive integer
s
for which 14

s
is 14 and the
second smallest positive integer
s
is 14
·
2 = 28.
(d) Mathematical induction cannot be used to prove that

cos(
x
)
 ≤
1 for all
x
∈
R
.
It can be used to prove that a statement
P
(
x
) is true for all
x
∈
P
, but not for all
x
∈
R
.
(e) Original Solution:
∀
x
∃
y,
0
< y < x
.
A student rightly pointed out that this implies that implies that all rational numbers
x
are
positive, and the universe of discourse was the rational numbers, not the positive rationals.
Corrected Solution:
∀
x, x >
0
→ ∃
y,
0
< y < x
.
2. Prove that if 8
ab

3
a
+
ab
3
is even, then
a
is even or
b
is odd.
Solution:
Suppose that 8
ab

3
a
+
ab
3
is even and
a
is odd. We must show that
b
is odd.
8
ab
is even, and since 3 and
a
are both odd, 3
a
is odd. Hence 8
ab

3
a
is odd.
Since 8
ab

3
a
+
ab
3
is even and 8
ab

3
a
is odd, then
ab
3
must be odd and
b
3
must be odd.
Therefore
b
is odd.
3. A sequence
{
y
n
}
is given by
•
y
1
= 4,
•
y
2
= 16,
•
For each
n
∈
P
, n
≥
3
, y
n
= 2
y
n

1
+ 3
y
n

2
.