CS237. Practice Problems Set 1: Sets, Basic Counting.
Not Graded. Please complete by Thursday Jan 28, 2011.
January 27, 2011
Reading.
Schaum’s Chapter 1, 2.1. L&L Counting I.
Exercise 1.
Let
A, B
⊆
U
. Prove the following. (Hint, use proofs by contradiction.):
•
A
⊆
B
iff
A
∩
B
c
=
∅
.
Solution: In (
→
) direction we can use the proof by contradiction. Let us assume that
A
⊆
B
and
A
∩
B
c
6
=
∅
.
From the latter assumption, it follows that there exists an element
x
∈
A
that is not in
B
. But, if
x
∈
A
then
x
∈
B
because of
A
⊆
B
. Contradiction. For the other direction, we again use proof by contradiction. Assume
that
A
⊆
B
is false. Then we must have some
x
∈
A
, so that
x /
∈
B
. But then,
A
∩
B
c
is not empty as assumed.
Contradiction.
•
A
⊆
B
iff
A
c
∪
B
=
U
.
Solution:
The (
→
) direction: By contradiction. Let us assume that
A
⊆
B
and
A
c
∪
B
6
=
U
. From the latter
assumption, it follows that there exists an element
x
∈
U
that is not in
A
c
and not in
B
. That means that
x
∈
A
.
But, if
x
∈
A
then
x
∈
B
because of
A
⊆
B
. Contradiction. For the other direction, by contradiction: Assume
that
A
⊆
B
is false, so that for some
x
∈
A
,
x /
∈
B
. But then, we must have
x
∈
U
so
A
c
∪
B
=
U
implies that
either
x
∈
A
c
which is not the case, or
x
∈
B
which is a contradiction.
•
A
⊆
B
iff
B
c
⊆
A
c
.
Solution:
The (
→
) direction: By contradiction. Let us assume that
A
⊆
B
and
B
c
*
A
c
. From the latter
assumption, it follows that there exists an element
x /
∈
B
that is not in
A
c
, i.e. it is in
A
. But, if
x
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 Spring '11
 Goldberg
 Logic, Product Rule, Mathematical Induction, Grammatical person, Wile E. Coyote and Road Runner

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