COT3100
Spring’2001
Practice problems for the test#2
Relations.
1.
Let
A
={1, 2, 3},
B
={
w
,
x
,
y
,
z
}, and
C
= {4, 5, 6}. Define the relation
R
⊆
A
×
B
,
S
⊆
B
×
C
, and
T
⊆
B
×
C
, where
R
= {(1,
w
), (3,
w
), (2,
x
), (1,
y
)},
S
={(
w
, 5), (
x
, 6), (
y
, 4),
(
y
, 6)}, and
T
={(
w
, 4), (
w
, 5), (
y
, 5)}.
a)
Determine
R
(
S
∪
T
) and (
R
S
)
∪
(
R
T
).
S
∪
T
= {(
w
, 4), (
w
, 5), (
x
, 6), (
y
, 4), (
y
, 5), (
y
, 6)},
R
(
S
∪
T
) = {(1, 4), (1, 5), (1, 6),
(2, 6), (3, 4), (3, 5)}
R
S
= {(1, 4), (1, 5), (1, 6), (2, 6), (3, 5)},
R
T
= {(1, 4), (1, 5), (3, 4), (3, 5)}
(
R
S
)
∪
(
R
T
) = {(1, 4), (1, 5), (1, 6), (2, 6), (3, 4), (3, 5)}.
b)
Determine
R
(
S
∩
T
) and (
R
S
)
∩
(
R
T
).
S
∩
T
={(
w
, 5)};
R
(
S
∩
T
) ={(1, 5), (3, 5)};
(
R
S
)
∩
(
R
T
) = {(1, 4), (1, 5), (3, 5)}.
2.
Consider the relation
T
over real numbers:
T
= {(
a
,
b
) 
a
2
+
b
2
=1}. Define all
properties of this relation (reflexive, irreflexive, symmetric, antisymmetric, and
transitive).
Not reflexive (for example,
a
=0.5, (
a
,
a
)
∉
T
).
Not irreflexive (
a
=1/
√
2, (
a
,
a
)
∈
T
)
Symmetric (for any real
a
,
b
, (
a
,
b
)
∈
T
→
(
b
,
a
)
∈
T
, because
a
2
+
b
2
=1
→
b
2
+
a
2
=1).
Not antisymmetric (counterexample: (0, 1)
∈
T
and (1, 0)
∈
T
)
Not transitive (counterexample: (0, 1)
∈
T
, (1, 0)
∈
T
, but (0, 0)
∉
T
)
3.
Let
R
⊆
A
×
A
be a symmetric relation. Prove that
t
(
R
) is symmetric.
Proof.
Assume
R
⊆
A
×
A
is symmetric. To prove that
t
(
R
) is symmetric, we need to show
that if (
a
,
b
)
∈
t
(
R
) then (
b
,
a
)
∈
t
(
R
). So, pick arbitrary (
a
,
b
)
∈
t
(
R
) (1). By the definition
of transitive closure, (1) implies that there is a path from
a
to
b
in
R
. By the definition of
a path it means, that there are some elements
a
0
=
a
,
a
1
,
a
2
, … ,
a
n
=
b
∈
A
such that (
a
0
,
a
1
),
(
a
1
,
a
2
), … (
a
n

1
,
a
n
)
∈
R
. But since
R
is symmetric by assumption, all inverted pairs belong
to
R
as well, i. e. (
a
n
,
a
n

1
), …, (
a
2
,
a
1
), (
a
1
,
a
0
)
∈
R
. It signifies that
R
contains the path
from
a
n
=
b
to
a
0
=
a
. By the definition of the transitive closure it implies that (
b
,
a
)
∈
t
(
R
).
4.
Prove or disprove the following propositions about arbitrary relations on
A
:
a)
R
1
,
R
2
symmetric
→
R
1
∪
R
2
is symmetric.
Proof.
Assume
R
1
and
R
2
are symmetric. To prove that
R
1
∪
R
2
is symmetric we need
to show that if (
a
,
b
)
∈
R
1
∪
R
2
, then (
b
,
a
)
∈
R
1
∪
R
2
. So, take arbitrary (
a
,
b
)
∈
R
1
∪
R
2
. It
implies that either (
a
,
b
)
∈
R
1
or (
a
,
b
)
∈
R
2
. In the first case (
a
,
b
)
∈
R
1
implies (
b
,
a
)
∈
R
1
by symmetric property of
R
1
and (
b
,
a
)
∈
R
1
∪
R
2
by the union property. In the
second case, (
a
,
b
)
∈
R
2
implies (
b
,
a
)
∈
R
2
by symmetric property of
R
2
, and then (
b
,
a
)
∈
R
1
∪
R
2
by the union property. So, in both cases we showed that (
b
,
a
)
∈
R
1
∪
R
2
.is
implied.
b)
R
1
,
R
2
transitive
→
R
1
∩
R
2
is transitive.
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Proof.
Assume
R
1
and
R
2
are transitive. To prove that
R
1
∩
R
2
is transitive, we need to
show that if (
a
,
b
)
∈
R
1
∩
R
2
and (
b
,
c
)
∈
R
1
∩
R
2
then (
a
,
c
)
∈
R
1
∩
R
2
. So, assume that
(
a
,
b
)
∈
R
1
∩
R
2
(1) and (
b
,
c
)
∈
R
1
∩
R
2
(2).
By the intersection definition we can imply from (1) that (
a
,
b
)
∈
R
1
and (
a
,
b
)
∈
R
2
.
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 Spring '09
 Equivalence relation, Binary relation, Transitive relation, Symmetric relation, ∈R

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