Chapter  4, Relations
Section  4.2  Relations
Summary
 Relation: Suppose and are sets. Then a set is called a relation from to .
 Suppose is a relation from to , then:
 Domain of relation is a set:
.  Range of relation is a set:
.  The inverse of is the relation from to :
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 Domain of relation is a set:
 Composition: Suppose is a relation from to and is a relation from to . Then the composition of
and is the relation from to defined as follows:
.  If is a relation from to , is a relation from to , and is a relation from to . Then:
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Soln1
(a) Domain is the set of living persons who are parents of some living people and Range is the set of all living persons who have a living parent.
(b) Domain is the set of all real numbers, . Range is the set of all positive numbers, .
Soln2
(a) Domain is the set of all living persons who are brother of some living person. Range is the set of all living persons who have a living brother.
(b) Domain : . Note that if domain lies in then
will be negative. Thus there will be no such that is negative.
Range: .
Soln3
(a) is a relation from to . is a relation from to . Thus:
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(b) is a relation from to . is a relation from to . Thus:
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Soln4
(a) .
(b)
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Soln5
(a)
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(b)
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Soln6
(a)
and are both subsets of . Let is an arbitrary element of . Then,
iff .
iff .
Since is arbitrary, .
(b)
Suppose , then . Since , it follows that , or .
(c)
Clearly and are both relations from to . Let be an arbitrary element of .
Suppose . Thus there must exist an element such that and . Since , there must exist a n element such that and . Now since and , it follows that . Thus we have and , thus it follows that . Since is arbitrary, it follows that .
(d)
Clearly is a relation from to . Thus is a relation from to . Also, it can be noted that is also a relation from to . Suppose is an arbitrary element of to .
()Suppose . It follows that . Thus there exist an element such that and . Since , it follows that . Similarly, since , it follows that . Now, we have and , it follows . Now since is arbitrary, it follows that .
()Similarly, suppose . Thus there exists an element such that and . Thus it follows that and . Now since and , it follows that . And since , it follows that . Since is arbitrary, we can conclude that .
Thus from both directions, we can conclude that .
Soln7
Suppose is the enemy of and is the enemy of , then it is given that is the friend of . Thus we have if and , then . Now since and , it follows . Thus the given statement says: if , then , or Enemy of an enemy is a friend. Since is arbitrary, it follows that .
Soln8
(a)
Suppose . Thus there exist a pair such that . Since , it follows that . Thus if , then . Since is arbitrary, we can conclude that .
(b)
Suppose . Suppose . Thus there must exist an element such that . Thus . Since , it follows that . Thus there must exist an element such that . Now since and , it follows that . Thus . Thus we have, if , then . Since is arbitrary, we can conclude that .
Thus from part (a) and this proof we can conclude that .
(c)

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Suppose . Thus there must exist a pair . Since , there must exist an element such that and . Now since , it follows that . Since is arbitrary, we can conclude that . 
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Suppose . Suppose . Thus there must exist an element such that . Thus . Now since , it follows that . Thus there must exist an element such that . Thus we have and , it follows that . Thus . Thus we have, if , then . Since is arbitrary, we can conclude that .
Since and , we can conclude that .
Soln9
(a). True.
Suppose . Then and . Thus . Thus we can say that, if , then . Since is arbitrary, we can conclude that .
(b) True.
Suppose . Suppose . Thus . Since , it follows . Now since , it follows . Thus if , then . Since is arbitrary, we can conclude that .
(c) True.
Suppose .
iff .
iff .
iff .
iff .
Since is arbitrary, we can conclude that .
Soln10
We will prove it by contrapositive.
()Suppose . Thus there must exist some element and . Since , there must exist an element such that . Similarly, since , there must exist an element such that . Thus we have and , it follows that . Thus . Hence by contrapositive we can conclude that if , then .
()Suppose . Thus there exist atleast one pair . Thus there exist an element such that and . Thus it follows , or . Hence by contrapositive, we can conclude that if , then .
Soln11
(a)
Clearly , , and are relations from \to . Suppose is an arbitrary element in . Thus and . It follows that and and . Thus , which is equivalent to saying that . Now since and , there exist an element such that and . Thus it follows that . Sicne is arbitrary, we can conclude that .
(b)
The theorem and proof both not correct. In the proof, incorrect statement is:
Since and .
The problem is, there might exist some other element such that and . Thus even if , it might be a case that may still is in .
(c)
As noted in the above part b, we can easily find such element:
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Soln12
(a) True.
Suppose . Suppose . Thus there exists an element such that and . Since , it follows that . Now since and , it follows . Thus if , then . Since is arbitrary, we can conclude .
(b) True.
Suppose . Thus there exists an element such that and . It follows that and . Since and , . Similarly, since and , . Thus .
(c) False.
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Thus, .
(d)
Suppose . Thus,
iff, such that, .
iff, .
iff, .
iff, .
Since, is arbitrary, we can conclude that .