COT 3100
Spring’2001
Homework #2
Assigned: Jan. 24/25
Due: Feb. 07/08 in recitation
1. Let
A
be the set of prime numbers, let
B
be the set of positive even integers, and let
C
be the set of positive multiples of 3. The universal set is the set of positive integers.
Determine the following sets: list all of their elements if they have 10 or fewer elements;
list their 10 smallest elements if they have more than 10.
a)
A
∩
C
= {3}
b)
(
C
∪
B
) –
A
= {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, …}
c)
¬
(
B
∩
C
) = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11,
…}
d)
(
B
–
C
)
∪
(
C
–
B
) = {2, 3, 4, 8, 9, 10, 14, 15, 16, 20, …}
e)
A
∪
(
B
–
C
) = {2, 3, 4, 5, 7, 8, 10, 11, 13, 14, …}
2. (p. 138, # 12). For
U
=
Z
+
let
A
⊆
U
where
A
= {1, 2, 3, 4, 5, 7, 8, 10, 11, 14, 17, 18}.
a)
How many subsets of
A
contain six elements?
This is the number of combinations of size 6 from 12, C (12, 6)=12!/(6!
⋅
6!)=924.
b)
How many six-element subsets of
A
contain four even integers and two odd
integers?
We can choose four even integers from 6 available
{2, 4, 8, 10, 14, 18} in C (6, 4)=
6!/(4!
⋅
2!)=15 different ways. Two odd integers can be picked from six available in A
in C(6, 2)=15
different ways. By the product rule the number of six element subsets
with four even and two odd is 15
⋅
15=225.
c)
How many subsets of
A
contain only odd integers?
That is the number of subsets of a six-element set
2
6
=64.
3. Let
A
and
B
be arbitrary sets. Prove that
a)
(
A
∪
B
) = (
A
∩
B
)
if and only if
A
=
B
Proof
. We need to prove that two propositions are identical, i.e.
(
A
∪
B
) = (
A
∩
B
)
↔
A
=
B
. That is equivalent to prove two implications
(
A
∪
B
) = (
A
∩
B
)
→
A
=
B
and
A
=
B
→
(
A
∪
B
) = (
A
∩
B
).
Part 1.