Dr. Ameria Eldosoky
Discrete mathematics
1
Relations and functions
2
Relations
• Relation
Let A and B be nonempty sets. A relation R from A to B
is a subset of A × B. If R ⊆ A× B and (a, b) ∈ R, we
say that a is related to b by R, write a R b, if a is not
related to b by R, we write a b.
If A and B are equal. We say that If R ⊆ A× A is a
related on A, instead of a relation from A to A.
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Relations
• Example 1
Let A ={1, 2, 3} and B={r, s}. Then we define R={(1, r),
(2, s), (3, r)} is a relation from A to B.
• Example 2
Let A and B be sets of real numbers. We define the
following relation R(equals) from A to B
Y
O
X
4
Relations
• Example 3
Let A={1, 2, 3, 4}. Define the following relation R (less
than) on A
a R b if and only if a<b
Then R= { (1,2) (1,3) (1,4) (2, 3) (2,4) (3,4) }
• Example 4
Let A=Z+, the set of all positive integers. Define the
following relation R on A:
a R b if and only if a divides b
Then 4 R 12, 6 R 30, but 5 7 .
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Relations
• Example 5
Let A be the set of all people in the world. We define the
following relation R on A: a R b if any only if there is a sequence
a0, a1, a2, …,an of people such that a0=a, an=b and ai-1 knows ai,
i=1,2,…,n (n will depend on a and b)
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Relations
• Example 6
Let A=R, the set of real numbers. We define the
following relation R on A
x R y if and only if x and y satisfy the equation
x2/4+y2/9=1
(0,3)
Y
O
(2,0)
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X
Relations
• Sets arising from relations
Let R ⊆ A× B be a relation from A to B.
a) The domain of R
Denoted by Dom(R), is the set of all first elements in
the pairs that make up R. (Dom(R) ⊆ A )
b) The range of R
Denoted by Ran(R), is the set of all second elements in
the pairs that make up R. (Ran(R) ⊆ B )
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Relations
• R-relative set of x
If R is a relation from A to B and x ∈ A, we define R(x),
the R-relative set of x, to be the set of all y in B with the
property that x is R-related to y.
R(x) = {y ∈ B | x R y}
If A1 ⊆A, then R(A1), the R-relative set of A1, is the set
of all y in B with the property that x is R-related to y for
some x in A1.
R(A1) = {y ∈ B | x R y for some x in A1}
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Relations
• Example 13
Let A={a, b, c, d} and let R={(a,a), (a,b), (b,c),
(c,a), (d,c), (c,b)}. Then
R(a) = {a, b} R(b)={c}
If A1={c, d}, Then
R(A1)= {a, b, c}
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Relations
• Theorem 1
Let R be a relation from A to B, and let A1 and A2 be
subsets of A, then
(a) If A1 ⊆ A2, then R(A1) ⊆ R(A2)
(b) R(A1 U A2 ) = R(A1) U R(A2)
(c) R (A1 ∩ A2 ) ⊆ R(A1 ) ∩ R(A2)
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Relations
• (a) If A1 ⊆ A2, then R(A1) ⊆ R(A2)
Proof:
If y ∈ R(A1), then x R y for some x in A1, since
A1 ⊆ A2, x ∈A2, thus y ∈ R(A2) and therefore
R(A1) ⊆ R(A2)
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Relations
• (b) R(A1 U A2 ) = R(A1) U R(A2)
Proof :
(1) R(A1 U A2 ) ⊆ R(A1) U R(A2)
If y ∈ R(A1 U A2 ), then x R y for some x in A1 U A2, then x in
A1 or A2. If x in A1 , y ∈ R(A1); if x in A2 . y ∈ R(A2). In either
cases, y ∈ R(A1) U R(A2) , therefore
R(A1 U A2 ) ⊆ R(A1) U R(A2)
(2) R(A1) U R(A2) ⊆ R(A1 U A2 )
A1 ⊆ A1 U A2 , then R(A1) ⊆ R(A1 U A2 )
A2 ⊆ A1 U A2 , then R(A2) ⊆ R(A1 U A2 )
Thus R(A1) U R(A2) ⊆ R(A1 U A2 )
Therefore, we obtain R(A1 U A2 ) = R(A1) U R(A2)
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Relations
• (c) R (A1 ∩ A2) ⊆ R(A1) ∩ R(A2)
Proof:
If y ∈ R (A1 ∩ A2 ) , then x R y for some x in A1 ∩ A2,
since x is in both A1 and A2 , it follows that y is in both
R(A1) and R(A2) ; that is, y ∈ R(A1 ) ∩ R(A2).
Therefore
R (A1 ∩ A2) ⊆ R(A1) ∩ R(A2)
Note: R(A1) ∩ R(A2) ⊆ R (A1 ∩ A2)
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Relations
• Theorem 2
Let R and S be relations from A to B, If R(a) =S(a) for
all a in A, then R=S
Proof:
If a R b, then b ∈ R(a). Therefore, b ∈ S(a) and a S b. A
completely similar argument shows that, if a S b, then a
R b. Thus R=S.
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Properties of Relations
• Reflexive & Irreflexive
A relation R on a set A is reflexive if (a, a)∈ R for all a
∈A. A relation R on a set A is irreflexive if a R a for all
a∈ A
1 * ... * 


*
1
...
*



... 


 * ... 1 
 0 * ... * 


*
0
...
*



... 


 * ... 0 
Dom(R)
=Ran(R)
=A
reflexive
irreflexive
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Properties of Relations
• Theorem 2
Let R be a relation on a set A, then
(a) Reflexivity of R means that a∈R(a) for all a in A.
(b) Symmetry of R means that a∈R(b) iff b∈R(a).
(c) Transitivity of R means that if b∈R(a) and c∈R(b),
then c∈R(a).
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Properties of Relations
• Example 1
(a) Δ = {(a,a) | a ∈A}
Reflexive
(b) R = {(a,b)∈A×A | a ≠ b}
Irreflexive
(c) A={1,2,3} and R={(1,1), (1,2)}
not Reflexive ((2,2) not in R), not Irreflexive ((1,1) in R)
(d) A is a nonempty set. R=Ø ⊆ A×A, the empty
relation.
not Reflexive, Irreflexive
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Properties of Relations
• Symmetric, Asymmetric & Antisymmetric
Symmetric:
if a R b, then b R a
Asymmetric:
if a R b, then b R a (must be irreflexive)
Antisymmetric: if a ≠ b, a R b or b R a
( if a R b and b R a, then a=b )
not Symmetric: some a and b with a R b, but b R a
not Asymmetric: some a and b with a R b, but b R a
not Antisymmetric :
some a and b with a ≠ b, but both a R b and b R a
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Properties of Relations
• Example 2
Let A=Z, the set of integers, and let
R={ (a,b) ∈A×A | a < b }
Is R symmetric, asymmetric, or antisymmetric?
Solution:
not symmetric: if a<b then b<a is not true.
asymmetric: if a<b, then b<a must be true
Antisymmetric: if a ≠ b, then a<b or b<a must be true
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Properties of Relations
• Example 4
Let A={1,2,3,4} and let
R={(1,2), (2,2), (3,4), (4,1)}
R is not symmetric, since (1,2) ∈ R, but (2,1) ∈ R.
R is not asymmetric, since (2,2) ∈ R
R is antisymmeric, since if a ≠ b, either (a, b) ∈ R or (b,a)
∈R
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Properties of Relations
• Example 5
Let A=Z+, the set of positive integers, and let
R={(a,b) ∈A×A | a divides b }
Is R symmetric, asymmetric, or antisymmetric?
Solution:
not symmetric: 3 | 9 but 9 | 3
not asymmetric: 2 | 2
antisymmetric: if a | b and b | a, then a=b (Section 1.4)
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Properties of Relations
• Example 9
Let A=Z+, and R={(a,b) ∈A×A | a divides b }
Is R transitive?
Solution:
Suppose that a R b and b R c, so that a | b and b | c. It
then does follow that a | c Thus R is transitive.
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Properties of Relations
• Example 10
Let A ={1,2,3,4} and let R={(1,2),(1,3),(4,2)}
Is R transitive?
Solution:
Since there are no elements a, b and c in A such that a R
b and b R c, but a R c, R is transitive.
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Operations on Relations
• Let R and S be relations from a set A to a set B
R and S are subsets of A×B. We can use set
operations on the relations R and S
Relations: (three representations)
1. the set of ordered pairs (finite or infinite)
2. digraph (finite)
3. matrix (finite)
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Operations on Relations
• Complementary relation

a R b if and only if a R b
• The intersection R ∩ S
a (R ∩ S) b means that a R b and a S b
• The union R U S
a (R U S) b means that a R b or a S b
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Operations on Relations
• Inverse
Let R be a relation from A to B, the relation R-1 is a
relation from B to A (reverse order from R) denoted by
b R-1 a if and only if a R b
Note: (R-1) -1=R
Dom(R-1) = Ran (R)
Ran(R-1) = Dom(R)
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Operations on Relations
• Example 1
Let A={1,2,3,4} and B= {a,b,c}. We Let
R={(1,a), (1,b), (2,b),(2,c),(3,b),(4,a)}
and S= {(1,b),(2,c),(3,b),(4,b)}

Compute: (a) R (b) R ∩ S (c) R U S (d) R-1
Solution:
A×B = { (1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b),
(3,c), (4,a), (4,b), (4,c) }

R  A B  R
 { (1,c), (2,a), (3,a), (3,c), (4,b), (4,c) }
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Operations on Relations
• Example 1
R∩S = { (1,b), (3,b), (2,c) }
R∪S = { (1,a), (1,b), (2,b), (2,c), (3,b), (4,a), (4,b) }
R-1= { (a,1), (b,1), (b,2), (c,2), (b,3), (a,4) }
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Operations on Relations
• Example 2
Let A=R. Let R be the relation ≤ on A and let S be ≥
Then the complement of R is the relation >, since a ≤ b
means a>b
Similarly, the complement of S is the relation <.
On the other hand, R-1=S, since for any number a and b
a R-1 b if and only if b R a if and only if b ≤ a
if and only if a ≥ b if and only if a S b
R ∩ S : “=”
R U S : A×A
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Operations on Relations
• Theorem 1
Suppose that R and S are relations from A to B
(a) If R ⊆ S, then R-1 ⊆ S-1.
(b) If R ⊆ S, then S ⊆ R
(c) (R∩S) -1 =R -1 ∩ S -1 and (R U S) -1 = R -1 U S -1
(d) (R∩S) = R U S
and R U S = R ∩ S
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Operations on Relations
• Theorem 2
Let R and S be relations on a set A
(a) If R is reflexive, so is R-1
(b) If R and S are reflexive, then so are R∩S and R U S
(c) R is reflexive if and only if R is irrflexive
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