Solutions to Homework Three
CS 191 :
Discrete Structures I
Feb. 20
, Winter 2007
Due Date: Mar. 6, Tuesday, before the class meeting.
Section 3.1, p123
In Exercises 912, draw the diagraph of the relation.
Exercise
(11)
The relation
R
= {(1,2), (2,3), (3,4), (4,1) } on
X
= {1, 2, 3, 4}.
Answer:
1
2
3
4
In Exercises 1316, write the relation as a set of ordered pairs.
Exercise
(14)
1
2
3
4
5
Answer:
{(1,1), (2,2), (3,3), (3,5), (4,3), (4,4), (5,5), (5,4)}.
1
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R
on the set {1, 2, 3, 4, 5} defined by the rule (
x
,
y
)
∈
R
if 3 divides
x
–
y
.
Exercise
(20)
List the elements of
R
1
.
Answer
:
{(1,1), (4,1), (2,2), (5,2), (3,3), (1,4), (4,4), (2,5), (5,5)}
Exercise
(27)
Is the relation of Exercise 25 reflexive, symmetric, antisymmetric, transitive, and/or a
partial order? Note, the relation
R
of Exercise 25 is defined on the set {1, 2, 3, 4, 5} by
the rule (
x
,
y
)
∈
R
if
x
+
y
≤
6..
Answer:
It is symmetric. It is not reflexive, not antisymmetric, not transitive, and not a partial order.
In Exercises 2934, determine whether each relation defined on the set of positive integers is
reflexive, symmetric, antisymmetric, transitive, and/or a partial order.
Exercise
(30)
(
x
,
y
)
∈
R
if
x
>
y
.
Answer:
It is not reflexive.
It is antisymmetric, transitive.
It is not a partial order.
Exercise
(34)
(
x
,
y
)
∈
R
if 3 divides
x
+ 2
y
.
Answer:
It is reflexive because
x
+ 2
x
= 3
x
is divisible by 3.
It is symmetric because if
x
+ 2
y
is divisible by 3, then
y
+ 2
x
= 2(
x
+ 2
y
) – 3
y
will be also
divisible by 3.
It is transitive because if
x
+ 2
y
is divisible by 3 and
y
+ 2
z
is divisible by 3, then
x
+ 2
z
= (
x
+ 2
y
) + (
y
+ 2
z
) 3
y
will be also divisible by 3.
It is not antisymmetric obviously and not a partial order.
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 Winter '06
 Shen
 Equivalence relation, Binary relation, Transitive relation, relation, 1 j, r1

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