2
MATH 55 FINAL SOLUTIONS (yellow)
Do not write your answers on this sheet.
Instead please write your name, your
student ID, your TA’s name, your section time, “yellow,” and all your answers in
your blue books.
IMPORTANT: Write your answers to problems 1–12 on the first
page or two of your blue book, without the calculations you did to get the answers.
To guarantee full credit and to qualify for partial credit, you should show your work on
later
pages of the blue book.
Total: 200 pts., 170 minutes.
(1)
(5 pts. each) For each of (a)(e) below: If the statement is true (always), write
TRUE. Otherwise write FALSE. (Please do not use the abbreviations T and F, since
in handwriting they are sometimes indistiguishable.) No explanations are required
in this problem.
(a) Exactly half of the 3element subsets of
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
contain the number
5
.
FALSE. There are
(
7
3
)
= 35 threeelement subsets of
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
. Since 35
is odd, it cannot be that exactly 35
/
2 of them contain 5. In fact, the number of
threeelement subsets of
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
equals the number of twoelement subsets
of
{
1
,
2
,
3
,
4
,
6
,
7
}
, which is
(
6
2
)
= 15.
(b) If
A
,
B
, and
C
are pairwise disjoint sets (possibly infinite), and

A
∪
B

=

A
∪
C

, then

B

=

C

.
FALSE. Let
A
=
{
2
,
3
,
4
, . . .
}
,
B
=
{
0
,
1
}
, and
C
=
{
1
}
. These are pairwise
disjoint. The function
f
(
n
) =

1
,
if
n
= 0
n
+ 1
,
if
n >
0
defines a bijection between
A
∪
B
and
A
∪
C
, so

A
∪
B

=

A
∪
C

. But

B

= 2 =
1 =

C

. Thus we have a counterexample.
(c) If a function
f
(
n
)
is
O
(2
n
)
as
n
→ ∞
, then
f
(
n
)
is also
o
(
n
!)
as
n
→ ∞
.
TRUE. First, lim
n
→∞
2
n
/n
! = 0, by the Ratio Test. Since
f
(
n
) is
O
(2
n
), there
exists
c >
0 such that

f
(
n
)
 ≤
c
2
n
for sufficiently large
n
. Thus, for sufficiently large
n
, the ratio
f
(
n
)
/n
! is sandwiched between

c
2
n
/n
! and
c
2
n
/n
!. Both expressions
on the right tend to 0 as
n
→ ∞
, so lim
n
→∞
f
(
n
)
/n
! = 0.
This means that
f
(
n
) =
o
(
n
!), by definition.
(d) For any predicate
P
(
x
)
, the propositions
¬∃
xP
(
x
)
and
∀
xP
(
x
)
are logically
equivalent.
FALSE. Here is a counterexample. Suppose that
P
(
x
) is true for all
x
. Then
¬∃
xP
(
x
) is false (provided that the universe of discourse is nonempty), but
∀
xP
(
x
)
is true. So the propositions are not logically equivalent in this case.