# Ch3m252

```Chapter Three
Relations
3.1. Relations.
Relation is an important fundamental concept in the theory of
sets. Let recall that if x , y  R . Then x  y means that y  x is a nonnegative numbers. If we have an ordered pair (x , y ) of real numbers,
then we can determine that x  y , we check if y  x is non-negative.
Hence the x  y describe a property of the order pair (x , y ) . We call
it R  of R 2 defined as follows:
R   (x , y )  R 2 x  y   (x , y )  R 2 y  x is non-negative
The set R  determines the relationship between x and y relative to the
property less than or equal. Hence x  y if and only if (x , y )  R  .
Generalizing this we get the concept of relation.
Definition 3.1. A relation R between a set A and a set B is a subset of
A  B , that is, R is a relation between A and B if and only if R  A  B .
We simply write (a, b )  R or aRb for a  A , and b  B . R is not relation
between A and B if and only if R  A  B , that is (a, b )  R or a R b if
and only if R  A  B . A relation R on a set A is a subset of
A 2  A  A . We note that if A is a set with m elements and B is a set
with n elements, then there are 2mn relations from A to B .
Example 3.1. Let A  1, 2 and B  3, 4 . Then
A  B  (1,3),(1, 4),(2,3),(2, 4)
Then R  (2,3),(2, 4  A  B is a relation between the A and B . Hence
(2,3)  R (or 2R 3) and (2, 4)  R (or 2R 4) .
Example 3.2. Let A  1, 2 . Then A 2  (1,1),(1, 2),(2,1),(2, 2) . Since A 2
has 4 elements, (A 2 ) has 24  16 subsets, thus 16 relations on A  1, 2 .
Here are several relations on A .
R1  (1,1),(2, 2)
R 2  (1,1),(1, 2)
R 3  (1,1),(1, 2),(2, 2)
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R 4  (1, 2)
R 5  (2,1),(2, 2)
Example 3.3. Let A  1, 2,3 and B  a,b  . Then
R1  (1, a),(1, b ),(2, a),(2, b )
R 2  (3, a),(3, b )
R 3  (1, a),(1, b ),(3, a)
are relations from A to B . It is clear that 1R1a, and 1R 3a , but 1 R 2 a .
Definition 3.2. Let A be a set , and I A  (a, a ) a  A   A 2 . Then I A is
called the identity relation.
Example 3.5. Let A  1, 2,3 . Then I A  (1,1),(2, 2),(3,3)  A 2 is the
identity relation on A .
Definition 3.3. Every relation R from a set A to a set B has an inverse
relation R 1 from B to A which is defined by
R 1  (b , a ) (a, b )  R , a  A , b  B 
Example 3.6. Let A  1, 2,3 and B  a,b  . Then
R1  (1, a),(1, b ),(2, a),(2, b ) , and R11  (a,1),(b ,1),(a, 2), (b , 2)
R 2  (3, a),(3, b ) , and R 21  (a,3),(b ,3)
R3  (1, a),(1, b ),(3, a) , and R 31  (a,1),(b ,1),(a,3)
Then R11 , R 21 and R 31 are inverse relations from B to A .
3.2. Properties of relations
We continue our study of relations by looking to various
properties of relations.
Definition 3.4. A relation R on a set A is called reflexive if aRa , that
is (a, a )  R for all a  A . Hence every element is related to itself.
We note that the identity relation I A  (a, a ) a  A   A 2 is
reflexive.
Example 3.7. Let A  1, 2,3 . Then R1  (1,1),(1, 2),(2, 2),(2,3),(3,3) is a
reflexive relation, but R 2  (1,1),(1, 2) is not a reflexive relation.
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Example 3.8. Let R be the relation on the real numbers R defined by
the sentence x  y for all x , y  R . Then R is not reflexive, for x  x for
any real number x .
Example 3.9. Let  be a family of sets, and let R be the relation on
 defined by the sentence A is a subset of B . Then R is a reflexive
relation on  , for every set is a subset of itself.
Definition 3.5. A relation R on a set A is called symmetric if
aRb implies that bRa , that is (a , b )  R implies (b , a )  R for all a , b  A .
Hence R is symmetric if a is related to b , then b is also related to a .
We note that R is a symmetric relation on A if and only if R  R 1 , for
if (a, b )  R , then (b , a)  R 1 .
Example 3.10. Let A  1, 2,3, 4 . Then R1  (1,3),(2, 4),(3,1),(4, 2) is a
symmetric relation, but R 2  (1,3),(2, 4),(3,1) is not a symmetric
relation, for (2, 4)  R 2 , but (4, 2)  R 2 .
Example 3.11. Let R be the relation on the natural numbers N which
is defined by x divides y . Then R is not a symmetric relation, since 2
divides 4 , but 4 does not divides 2 , that is (2, 4)  R but (4, 2)  R .
Definition 3.6. A relation R on a set A is called transitive if aRb and
bRc implies that aRc , that is (a , b )  R and (b , c )  R implies (a, c )  R
for all a, b , c  A . Hence R is transitive if a is related to b , and b is
related to c , then a is also related to c .
Example 3.12. Let R be the relation on the real numbers R defined by
the sentence x  y for all x , y  R . Then R is
reflexive, for
a  b , and b  c , then a  c .
Example 3.13. Let  be a family of sets, and let R be the relation on
 defined by the sentence A is a subset of B . Then R is a transitive
relation on  , for if A  B and B  C , then A  C .
Example 3.14. Let A  1, 2,3, 4 . Then R1  (1,3),(3, 4),(1, 4) is a
transitive relation, but R 2  (1,3),(3, 4) is not a transitive relation, for
(1, 4)  R 2 .
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Definition 3.7. A relation R on a set A is called anti-symmetric if
aRb and bRa imply a  b , that is (a , b )  R and (b , a )  R imply a  b , for
all a , b  A . Hence R is anti-symmetric if a is related to b , and b is also
related to a , then a  b . We note that if a  b , then possibly a is related
to b or possibly b is related to , but never both.
Example 3.15. Let A  1, 2,3, 4 . Then R1  (1,1),(1,3),(3, 4),(4, 4) is an
anti-symmetric relation, but R 2  (1,3),(3,1),(3, 4) is not anti-symmetric,
since (1,3)  R 2 and (3,1)  R 2 , but 1  3 .
Example 3.16. Let  be a family of sets, and let R be the relation on
 defined by the sentence A is a subset of B . Then R is an antisymmetric relation on  , for if A  B and B  A , then A  B .
Definition 3.8. A relation R on a set A is called irreflexive if a R a ,
that is (a, a )  R for a  A . Hence every element is not related to itself.
Definition 3.9. A relation R on a set A is called asymmetric if
aRb imply b R a , that is (a , b )  R imply (b , a )  R , for all a , b  A . Hence
R is asymmetric if a is related to b imply b is not related to a .
Example 3.17. Let A  1, 2,3 . Then the relation
R1  (1,1),(1, 2),(2,1),(2,3)
is neither reflexive, irreflexive, symmetric, asymmetric,
symmetric, nor transitive.
Let consider the relation
anti-
R 2  (1,1),(2, 2),(1, 2),(3,3)
Then R 2 is reflexive, anti-symmetric, and transitive, but not symmetric.
Definition 3.10. A relation R on a set A is called an equivalence
relation if
1. R is reflexive, that is aRa for all a  A .
2. R is symmetric, that is if aRb implies bRa ,for all a , b  A .
3. R is transitive, that is if aRb and bRc , then aRc for all a, b , c  A .
If R is an equivalence relation on a set A , and aRb , then we say
that a is equivalent to b .
Example 3.18. An important example of an equivalence relation is that
of equality, for any elements in any set, we have
1. a  a
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2. a  b  b  a .
3. a  b and b  c  a  c .
Example 3.19. Let A  1, 2,3, 4,5,6 and consider the relation
R  (1,1),(2, 2),(3,3),(4, 4),(5,5),(6,6),(2,3),(3, 2)
Then R is an equivalence relation on A .
Example 3.20. Let n be a fixed positive integer in Z . Define the
relation  n on Z by for all x , y  Z ,
[x n y ]  [n (x  y )]
[x n y ]  [x  y  kn , for some k Z]
[x n y ]  [x  y  kn , for some k Z]
Sometimes, we write
[x  y (mod n )]  [n (x  y )]
[x  y (mod n )]  [x  y  kn , for some k  Z]
[x  y (mod n )]  [x  y  kn , for some k  Z]
Now, we show that  n is an equivalence relation on Z .
1. For all x  Z , we have x  x  0  0n , for 0Z . Hence, for all
x  Z , we get x  n x . Thus,  n is a reflexive relation.
2. Let x , y  Z , and suppose that x n y . Then n (x  y ) , and hence
there exist k  Z such that x  y  kn . Thus y  x  (k )n , and so
n ( y  x ) , that is y  n x . Hence,  n is a symmetric relation.
3. Let x , y , z  Z , and suppose that x n y and y n z . Then
n (x  y ) and n ( y  z ) , and hence, there exist k , l  Z such that
x  y  kn , and y  z  ln .Thus x  z  (k  l )n , and so n (x  z ) ,
for (k  l )  Z . Hence,  n is a transitive relation.
Therefore,  n is an equivalence relation on Z .
Definition 3.11. Let R be an equivalence relation on a set A , and let
a  A . Then the set
a   a  b  A bRa
is called the equivalence class determine by a with respect to R . The
set of all equivalence classes is called A modulo R and denoted by
A .
R
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Definition 3.12. Let A be a set and P be a collection of nonempty
subsets of A . Then P is called a partition of A if the following
properties satisfied:
1. for all X ,Y  P , either X Y or X Y   .
2. A 
X .
X P
Example 3.21. Let A  1, 2,3, 4,5,6 . Let X  1 , Y  2, 4,6 , and
X Y Z   , and A  X Y Z . So
Z  3,5 . Then,
P  X ,Y , Z  .
Theorem 3.1. Let R be an equivalence relation on the set . Then
1. for all a  A , a   ,
2. if b  a , then a  b , where, a , b  A ,
3. for all a , b  A , either a  b or a b   ,
4. A 
a , that is, A is the union of all equivalence classes with
aA
respect to R .
5. The set P  a a  A  is a partition of A .
Proof.
1. Let a  A . Since R is reflexive, we have aRa Hence,
a  a  [a ] and so a   .
2. Let b  A such that b  a . Then bRa and by symmetric
property of R , we have aRb . Now, if x  a , then xRa . But
aRb and R is transitive for R is an equivalence relation. Then
xRb , and hence x b . Therefore, a  b . Similarly, x b ,
then xRb . But bRa and R is transitive for R is an
equivalence relation. Then xRa , and hence x  a . Therefore,
b  a . Thus a  b for all a , b  R .
3. Let a , b  R and a b   . Then there exist x  a b and
hence, x  a and x b . Thus xRa and xRb . Since R is
symmetric, we have bRx , and by transitivity, we have bRa ,
and b  a . Hence, by (2), a  b .
aA.
4. Let a  A . Then a  a  a . Thus A  a . Also,
aA
Hence, A 
aA
aA
a .
aA
Example 3.22. Let n be a fixed positive integer in Z . Define the
relation  n on Z by for all x , y  Z ,
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[x n y ]  [n (x  y )]
As in Example 3.20. Let Zn  a a  Z . By Theorem 3.1, we have Z n is
a partition of Z . Then Zn  0,1, 2,..., n  1 , where,
0  0  kn  kn k  Z
1  1  kn k  Z
2  2  kn k  Z
n  1  (n  1)  kn k  Z
Suppose that n  5 . Then
0  5k k  Z  ..., 10, 5, 0,5,10,... 
 10  5  5  10 
1  1  6k k  Z  ..., 9, 4,1, 6,11,... 
 9  4  6  11 
2  2  kn k  Z  ..., 8, 3, 2, 7,12,... 
 8  3  7  12 
3  3  kn k  Z  ..., 7, 2,3,8,13,... 
 7  2  8  13 
4  3  kn k  Z  ..., 6, 1, 4,9,14,... 
 6  1  9  14 
Hence, Z5  0,1, 2,3, 4 .Similarly, we find that Z8  0,1, 2,3, 4,5, 6, 7 .
Let
The
A  1, 2,3, 4 .
R  (1,1),(2, 2),(3,3)(4, 4),(4, 2),(2, 4) . Then R is an equivalence relation
on A . The distinct equivalence relations are
Example
3.23.
a  [a ]  x  A (a, x )  R 
1  [1]  x  A (1, x )  R   1
2  [2]  x  A (2, x )  R   2, 4  4
3  [3]  x  A (3, x )  R   3
Example 3.24. Let A  0,1, 2,3, 4 .Then
R  (0,0),(1,1),(2, 2),(3,3),(4, 4),(0, 4),(1,3),(4,0),(3,1)
is an equivalence relation, and the distinct equivalence classes are:
a  [a ]  x  A (a, x )  R 
0  [0]  x  A (1, x )  R   0, 4  4
1  [1]  x  A (1, x )  R   1,3  3
2  [2]  x  A (2, x )  R   2
Example 3.25. Consider the set of all real numbers , and the relation


R  (x , y )  R  R x 2  y 2  0
1. We want to show that R is reflexive. Let x  R . Then 0  x 2  x 2 . Thus
(x , x )  R , that is xRx . Therefore, R is reflexive.
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2. Now, we show that R is symmetric. Let x , y  R such that (x , y )  R .
Then x 2  y 2  0 , and this implies that y 2  x 2  0 . Hence ( y , x )  R ,
that is yRx . Therefore, R is symmetric.
3. Finally, we show that R is transitive. Let x , y , z  R such that
(x , y )  R and ( y , z )  R . Thus x 2  y 2  0 and y 2  z 2  0 .So
x 2  y 2  y 2  z 2  0 , and x 2  z 2  0 . Thus (x , z )  R , that is xRz .
Therefore, R is transitive.
Therefore, R is reflexive, symmetric and transitive, and hence, R is an
equivalence relation. The equivalence classes has the form
x  [x ]   y  R yRx 
 y  R ( y , x )  R 

 y  R y

 y R y 2 x 2  0
2
x2

  y  R y  x 
  x , x 
For any x  R . If x  4 , then
4  [4]   y  R ( y , 4)  R 
  y  R y  4
 4, 4
Similarly, 3  [3]  3,3 .
Example 3.26. Let A be any set. Then the identity relation
I A  (a , a ) a  A   A 2
is an equivalence relation and the equivalence classes are
a  [a ]  b  A a I A b 
 b  A (a , b )  I A 
 b  A a  b 
 a
for any a  A . If a , b  A such that a  b , then a  b  , that is a  b and the
two classes are distinct.
If A  1, 2,3, 4 , then the equivalence classes for the identity relation I A
are 1 , 2 , 3 , and 4 .
3.3. Domain and range of relations
Let R be a relation between a set A and a set B . Then R  A  B .
The domain of D R of the relation R is the set
D R  a  A (a, b )  R , b  B   A
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The range Rang R of the relation R is the set
Rang R  b  B (a , b )  R , a  A   B
Example 3.27. Let A  1, 2,3, 4 , and B  5,6,7 . Consider the relation
R  (2,5),(4,5),(4,7) between the set A and the set B . Then the domain of R
is DR  2, 4 and the range of R is Rang R  5,7 .
Also, R 1  (5, 2),(5, 4),(7, 4) is a relation between B and A .
domain of R
1
is D R 1  5,7 , and the range of R
Example 3.28. Let consider the relation

 (x , y ) x , y  R,9 y
1
The
is Rang R 1  2, 4 .


R  (x , y ) x , y  R, 4x 2  9 y 2  36  R  R
2
 36  4x 2

4 
  ( x , y ) x , y  R, y 2  4  x 2 
9 


4 
  ( x , y ) x , y  R, y   4  x 2 
9 


4

  ( x , y ) x , y  R, 4  x 2  0 
9



4 
  ( x , y ) x , y  R, 4  x 2 
9 



 ( x , y ) x , y  R, x 2  9
 (x , y ) x , y  R, x  [ 3,3]
Thus D R  [3,3] . Similarly,

 (x , y ) x , y  R, 4x


R  (x , y ) x , y  R, 4x 2  9 y 2  36  R  R
2
 36  9 y 2

9 
  ( x , y ) x , y  R, x 2  9  y 2 
4 


9 
  ( x , y ) x , y  R, x   9  y 2 
4 


9

 (x , y ) x , y  R,9  y 2  0 
4



9 
 (x , y ) x , y  R,9  y 2 
4 



 ( x , y ) x , y  R, y 2  4
 (x , y ) x , y  R, y  [ 2, 2]
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Thus, Rang R  [2, 2] . Furthermore,


R 1  (x , y ) x , y  R,9x 2  4 y 2  36
Hence D R 1  [2, 2] , and Rang R 1  [3,3] .
Example 3.29. Let R be the relation defined on the natural numbers N such that
R  (x , y ) x , y  N,3x  y  15
(x , y ) x , y  N, y
 15  3x 
 (1,12), (2,9), (3, 6), (4,3)
R 1  (12,1),(9, 2),(6,3),(3, 4)
Note that x  5, y  0 is a solution for 3x  y  15 in general, but (5,3)  R for
0 N.. Hence, DR  1, 2,3, 4  Rang R 1 , and Rang R  12,9,6,3  DR 1 .
3.4. Partially and totally ordered sets
Definition 3.13. Let R be a relation on a set A . Then R is called a partial
order relation, if it satisfies the following conditions:
1. R is reflexive, that is aRa for all a  A .
2. R is anti-symmetric, that is if aRb and bRa , then a  b for all
a,b  A .
3. R is transitive, that is if aRb and bRc , then aRc for all a, b , c  A .
If a relation R on a set A defines a partial order on A , then aRb is
denoted by a b which reads a precedes b , and we write
(A , R )  (A , ) to denote the set A and the partial ordering R  , and we
call A is a partially ordered set (poset).
Example 3.30. Let consider the set R . Define the relation on R ,
R   (x , y ) x , y  R, x  y  . We know that  is reflexive, anti-symmetric,
and transitive. So R  is a partial order relation on R , that is (A , R  ) .
Example 3.31. Let  be a family of sets. Define the relation on the set  ,
R   (A , B ) A , B , A  B  . We know that  is reflexive, anti-symmetric,
and transitive. Thus R  is a partial order relation on  , that is (A , R  ) .
Example 3.32. Let consider the set of all real numbers R . Define the relation
on R , R   (x , y ) x , y  R, x  y  . We know that &lt; is not reflexive. Thus
R  is not a partial order relation on R .
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Example 3.33. Let A  1, 2,3 , and the relation R  (1,1),(2, 2),(3,3),(1, 2 on
A . Thus R is reflexive, anti-symmetric, and transitive, that is R  is a partial
order relation on A .
Example 3.34. Let A  l l is a line in the plane . Define
R   (l1 , l 2 ) l1 , l 2  A , l1  l 2 , that is l1 is perpendicular to l 2   A  A
Then R  is not a partial order on A , since any line can not be a perpendicular
to itself, that is (l , l )  R  , and hence R  is not reflexive.
Definition 3.14. (i) Let R be a partially ordered relation on a set A . Elements
a , b  A are said to be comparable if and only if either aRb or bRa .
Otherwise, a and b are non-comparable.
(ii) A partially ordered relation R on a set A is called a totally order
relation, if either aRb or bRa for any a , b  A .
Example 3.35. A partially ordered relation on any set A of real numbers is
totally ordered set.
a
Example 3.36. Let A  1, 2,3, 4,5,6 . Define the relation
R  (x , y ) x , y  A , x divieds y 

 (x , y ) x , y  A , x y

 (1,1), (1, 2), (1,3), (1, 4), (1,5), (1, 6), (2, 2), (2, 4), (2, 6), (3,3), (3, 6), (4, 4), (5,5), (6, 6)
Then R is a partial order relation , but not a total order, since (3,5)  R and
(5,3)  R .
Solved Problems
1. Let A  1, 2,3, 4 , and B  1,3,5 . Define a relation R between A and
B by R  (x , y ) x  A , y  B ; x  y  . Write the relation R .
Solution. R  (x , y ) x  A , y  B ; x  y   (1,3), (1,5), (2,3), (2,5), (3,5) .
2. Let A  2,3, 4,5 , and B  3,6,7,10 . Define a relation R between A


and B by R  (x , y ) x  A , y  B ; x y . Write the relation R , R 1 ,and
find the domain and range R .
R  (x , y ) x  A , y  B ; x y  (2, 6), (2,10), (3,3), (3, 6), (5,10) .
Solution.

Also R
1

 (6, 2),(10, 2),((3,3),(6,3),(10,5) . Then DR  2,3,5  Rang R 1
and Rang R  3,6,10  D R 1 .
3. Answer the following True (T) , or false (F):
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