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)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 6pts, 5pts, 3pts, 10 G

Click to edit the document details