COT3100
Summer’2001
Assignment #2 (Solution key).
Assigned: 05/22,
due: 05/31 in recitation.
Total 100 pts.
1.
Let
P
be the set of prime numbers (integers greater then 1 divisible only by 1 and
itself),
A
be the set of even integers and let
B
be set of positive multiples of 5. The
universal is the set of positive integers. Determine the following sets (list all
elements if there are 10 or fewer elements; list their 10 smallest elements if they
have more than 10)
In accordance with what we are given we have
P
= {2, 3, 5, 7, 11, 13, 17, 19, 23, }
A
={2, 4, 6, 8, 10, …}
B
= {5, 10, 15, 20, 25, 30, 35, …}
U
= {1, 2, 3, …}
a)
(2pts)
P
∩
A
P
∩
A
= {2}
b)
(2pts) (
A
∪
B
)

P
(
A
∪
B
)

P
= {4, 6, 8, 10, 12, 14, 15, 16, 18, 20, …}
c)
(2pts)
¬
(
A
∩
B
)
¬
(
A
∩
B
) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, …}
d)
(2pts) (
A

B
)
∪
(
B

A
)
{2, 4, 5, 6, 8, 12, 14, 15, 16, 18, …}
e)
(2pts) (
B

A
)
∪
P
{2, 3, 5, 7, 11, 13, 15, 17, 19, 23, …}
2. Let
A
,
B
and
C
be any three sets. Prove or disprove the following propositions without
using Venn diagrams or membership tables:
a)
(10pts)(
A
∩
B
)
⊆
C
→
(
A
⊆
C
∧
B
⊆
C
)
The proposition is false. It can be disproved by the following example:
A
= {1, 2},
B
={2, 3},
C
= {2, 4}.
For these sets we have the condition (
A
∩
B
)
⊆
C
satisfied, since
A
∩
B
={2}
⊆{2, 4}.
But never the less neither
A
= {1, 2} nor
B
={2, 3}is a subset of
C
={2, 4}.
b)