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MATH 135
Fall 2005
Assignment 2 Solutions
HandIn Problems
Exercise 164:
Use the Contrapositive Proof Method to prove that
(
S
∩
T
=
∅
) AND (
S
∪
T
=
T
) =
⇒
S
=
∅
.
Solution:
We want to prove the contrapositive of the statement. That is
(
S
6
=
∅
) =
⇒
(
S
∩
T
6
=
∅
) OR (
S
∪
T
6
=
T
)
.
(We have used the fact that NOT(
P
AND
Q
) is equivalent to (NOT
P
) OR (NOT
Q
).)
It then makes sense to use Proof Method 1.56 (for proving statements of the form
P
=
⇒
(
Q
OR
R
)).) So we now
want to prove the statement
((
S
6
=
∅
) AND (
S
∩
T
=
∅
) =
⇒
(
S
∪
T
6
=
T
)
.
Proving this will be equivalent to proving our original statement.
Therefore, we assume that
S
6
=
∅
and also that
S
∩
T
=
∅
.
Because
S
6
=
∅
then there is an
x
∈
S
. It follows that
x /
∈
T
(since
S
∩
T
=
∅
). Because
x
∈
S
∪
T
, then
S
∪
T
6
=
T
.
We have shown
(
S
6
=
∅
) AND NOT (
S
∩
T
6
=
∅
) =
⇒
(
S
∪
T
6
=
T
)
.
This is equivalent to the statement
(
S
6
=
∅
) =
⇒
(
S
∩
T
6
=
∅
) OR (
S
∪
T
6
=
T
)
.
Thus by the Contrapositive Law we have proven the original statement.
Exercise 168:
Prove or give a counterexample to each statement.
(
S
∩
T
)
∪
U
=
S
∩
(
T
∪
U
), for any sets
S
,
T
, and
U
.
Solution 1:
A speciﬁc counterexample would be if
S
=
{
1
}
,
T
=
{
1
,
3
}
and
U
=
{
2
}
.
Then, (
S
∩
T
)
∪
U
= (
{
1
} ∩ {
1
,
3
}
)
∪ {
2
}
=
{
1
} ∪ {
2
}
=
{
1
,
2
}
.
Also,
S
∩
(
T
∪
U
) =
{
1
} ∩
(
{
1
,
3
} ∪ {
2
}
) =
{
1
} ∩ {
1
,
2
,
3
}
=
{
1
}
.
Thus, in this example (
S
∩
T
)
∪
U
6
=
S
∩
(
T
∪
U
).
Solution 2:
The statement is false. To see this notice that for any set
A
,
A
∩ ∅
=
∅
and
A
∪ ∅
=
A
.
Let
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This homework help was uploaded on 04/18/2008 for the course LINEAR ALG MATH135 taught by Professor Vanderburgh during the Spring '08 term at Waterloo.
 Spring '08
 Vanderburgh

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