EXAM 1 SOLUTIONS
Problem 1.
If

A

= 4
and

B

= 5
, then

A
×
B

= 9
.
Proof.
This is false. The product has size 4
·
5 = 20. It is straightforward to construct a counterexample
if you want.
±
Problem 2.
Let
x
∈
R
. The sentence “If
x >
3
, then
x
2
<
2
.” is a statement.
Proof.
This is false. The sentence is an open sentence, but is not a statement. It would become a
statement if it was quantiﬁed like “
∀
x
∈
R
, if
x >
3 then
x
2
<
2.”
±
Problem 3.
The converse of the statement “If
x
∈
S
then
x
∈
T
” is the statement “If
x /
∈
T
then
x /
∈
S
.”
Proof.
This is false. That is the contrapositive. The converse would be “If
x
∈
T
then
x
∈
S
.”
±
Problem 4.
The negation of the statement “
∀
x
∈
R
, x
2
≥
1
and
x
3
<
2
” is the statement
∃
x
∈
R
such that
x
2
<
1
and
x
3
≥
2
.
Proof.
This is false. We need to change the word “and” to the word “or” using DeMorgan’s law.
±
Problem 5.
If
A
is a set and
P
(
A
)
is the power set of
A
, then
A
⊆ P
(
A
)
.
Proof.
This is false in general. The set
A
=
{
1
}
is a counterexample.
±
Problem 6.
For
i
∈
Z
, deﬁne
A
i
=
{
i

2
,i

1
,i
+ 2
,i
+ 3
}
. Then the set
{
A
3
,A
5
,A
11
,A
13
}
is a
partition of
{
n
∈
Z
: 1
≤
n
≤
16
}
.
Proof.
This is true. (The collection of sets is pairwise disjoint, their union is the set of integers between
1 and 16, and none of them is empty.)
±
Problem 7.
Let
A
,
B
, and
C
be sets. Then
A

(
B
∪
C
) = (
A

B
)
∩
(
A

C
)
.
Proof.
This is true, and is one of DeMorgan’s laws for sets.
±
Problem 8.
One way to read
P
=
⇒
Q
is “
P
is necessary for
Q
.”
Proof.
This is false. It should read “
P
is suﬃcient for
Q
” or “
Q
is necessary for
P
”.
±
Problem 9.
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 '11
 Smith
 Logic, Empty set, 2k, 4k, 16 g, J. Neukirch

Click to edit the document details