Chapter 5
Relations and Functions
relaof
This chapteris a continuation Chapter 1. In this chapter, the concept of
and a special class of relations called functions are discussed
tion is introduced
is on set-theoretic properties of functions and
some depth. The emphasis
in
their
immediate consequences.
Product of
Cartesian
K1
Sets
(a, b), where a e Aand b e B, is
pairs
ordered
all
of
set
the
Then
is
Bbe two sets.
Product Set of Aand B(in this order) and
or
Let Aand
Product
Cross
Cartesian Product, or
calledthe
Thus.
denoted by AxB.
B= ((a, b) | aeAandbe B}.
Ax
It is to be noted that the
AxB #BXA, ingeneral.
as the product set
same
the
not
is
B
x
A
product set
B XA; that is,
Because,
BXA = {(b, a) | be Band
ae A)
and (a, b) # (b, a) in general."
and B = (2, 3}, then
-1)
{1,0,
=
A
if
example,
For
(0,3), (-1, 2), (-1,3)}
2),
(0,
3),
(1,
{(1,2),
AxB=
(3,1), (3, 0), (3,-1)}.
-1),
(2,
0),
(2,
1l),
(2,
and B xA =
can have
Evidently, AXB# BxA.
A. Thus, we
=
B
when
even
It should be noted that A x B can be defined
dehned by
Poduct of a set A with itself. and this product is
and bE A).
AxA = {(a, b)|aE A
The product IAxAis also denotedIby42.
"If (a,b)
and(c,
if and
d) are ordered pairs, then (a, b) =(C, d)
193
d.
onlyif a =c, b=
the
yield
Example
elements,
non-empty
ordered
have
194
words:
otherIn have (a, and
5x
(i)
A(a1, As Ax For
idea
Theof 8 8For
For
From
()
IfA
b)with If
IfA the canbe
little
=
s
e
t
a
x
and We (2x,
with
Bx
example,
example,
example,
k-tuples
and
8
az,...az)
and
following
this
then
8 x5
every has m
ynote
=
thinking
C
sets
A A'
1
x
t
h
e
B
B
=
64
=
result,
are chosen
+y) Find
l-x=-2. (2x,
are
=
AxA =
Aj Cartesian
that
Aj ordered {(1,2,
elements.
one
of
A1
,
40 if
if
if XAg (a1,
finite
fniteresult: elementsand
=
= x
xwill
A2,
|
B
xA|
(b1,bz,...,b)
we
elements has A
A
A
inm
(6, and
= =
=
a2,
0),
XXA,
choices
these
X
...,A7,
sets
(1,1),(1,
sets,
note
(1,0,-1)
indicate pairs, (1,2, (1,0),
x 1) y
..., productof
5
ways;
X
n
+
|
A
j
in
=
with
elements
y)
then
|B| A| that
each,
ak), the
each
if -1), XA7={(a1,
B
X A7
|A|
(aj1,
=
(1,
(6,1) =
we
has that
A)
set aB 0),
where
(
2
,
of
k-fold
=
(
1
,
=m
if
this
(i)
-2,0),
(of
A
sets
X.
and
if and
njn2n3 a2,...,ak),
the
-2),
-1), have
×A |A|B| =|A|B|.
|Ax B|
if
has
n
and
a; can
A,
following XA7l
B
means a),
and (y-2,
(0,
E
has only (1,-2,-1),
C= a2, product
=
has
will
Aj
,
|
B
|
be
b
elements,
...
1),
...n/ n
only if
A
that can
if(b1,
=
2x
extended
=
8
{0, ,a7)
(0,
x
have
i=
elements,
|Aj|· elements, a,
n,
be
B|
,
Aj
-1},
0),
+
elements.
cases:
= b2, (0,2,
then
A
2x 1)
| 1,2, X
chosen
from
5
(0,-1),(1,
a;
x then
b;
=A2lAkl.
then
A2to
>X and
b3...br)
...,k.
has B
=
f
o
r
(x-1,y+
¬
X
0),
A
6
XB
5=then
any
A
A i
Aj
,
and
a
(0,
=
exactly
can
25
That has
xA|
IS
i
2,
1,2,...,
=1,2, That XA7 finiteelements,
X+y= x
be
1),(-1,0),
A
-1),
jnute a
X
is, n are
=
in
B
5.
is
2)
i
s
,
chosen
B |A|.
number
elements, k-tuples,
(0,
defined
Relations,
mn
...k}.
and
set
n
k. -2,0),(0,
1.
elements.ways.
with
and BxAwihayel
from (-1,-1))
These
of
as
then
and
...,
the sets. BxB
Accordngly,
A
AxBIo
-2,
in
Functions
A
Thus,we
s
e
t
For
m
-1)
r
nas
ways
wil
ofany
=
t
º
5.1. Cartesian Product of Sets
195
(ii) (y -2, 2x + 1) ==(r- 1,y + 2) if aand
only if y-2 =X-1 and 2x
+ 1 = y +2; that
-y-1=0. These yield x =2, y =3.
is,
Example 2 Let A =(1,3,5), B= (2,3}, and C=
(4,6}. Write down the
following:
X-y+l=0 and 2x
(1)Ax B
(2) B×A
(3) Bx C
(7) (Ax B) UC
(5) (AUB) x C
(4) Ax C
(6) AU(B×C)
(8) An (BxC)
(9) (Ax B) U
(Bx C) (10) (A XB)n (B x A) (11) (Ax
B)n(B xC)
By using the definition of the product of sets,
we find that
Ax B= ((1,2), (1,3), (3, 2), (3,3), (5, 2), (5,3),
BxA = {(2, 1), (2,3), (2, 5), (3, 1), (3,3), (3, 5)},
BxC= ((2,4), (2, 6), (3,4), (3, 6)),
AxC= ((1,4),(1,6), (3,4), (3, 6), (5,4), (5,6)),
(AUB) xC= (1,2, 3,5,} x (4, 6)
= ((1,4),(1,6), (2,4), (2,6),(3,4),(3,6), (5,4), (5, 6))
AU(BxC) =(1,3,5} U{(2,4), (2, 6),(3,4),(3, 6))
= {1,3,5, (2,4), (2, 6), (3,4), (3, 6)}
(A xB)U C= {(1,2), (1,3), (3, 2), (3,3), (5, 2),(5,3), 4, 6}
An(Bx C) = 0.
(Ax B) U(Bx C) = {(1,2),(1, 3),(3, 2),(3, 3), (5, 2), (5, 3)
(2,4), (2, 6),(3,4), (3, 6)),
(Ax B) n (Bx A) = {(3,3)}
(Ax B) n(Bx C) =0.
Example 3 Suppose A, B, CsZxZ with A=((x, y) |y =5x- 1}, B=((x,y) | y=6x),
C={(x,y) |3x -y =-1}.
Find : (i) AnB, (i) BoC, (ii) ÄU, (iv) BUC.
(i) We note that
(x, y) E ANB
(x, y) E A and (x, y)e B
y=5x - landy = 6x
5x- l =y= 6x
x= -l, y= -6.
5. Relations and Functions
196
Thus. ANB= ((-I,-6)).
(i) We note that
(x.y) E BnC
(x, y) ¬ Band (x, y) EC
y=6x, and 3I-y=-7
6x = 3x +7
3r =7
i.e., x=7/3.
which is not possible, because xe Z.
Thus,
BOC= ).
(ii) We note that U =AnCso that
Now,
(x, y) EAnCe (y) ¬Aand (x,y) e C
y= 5x- 1and 3x- y= -7
5x 1=y=3x+7
x= 4, y = 19.
Thus,
AnC= ((4,19)).
(iv) We note that B UC= BOC. It has been seen in (ii) above that Bn C= 0. Therefore,
BU= BOC=ZxZ (universal set).
Example 4 For any set A, prove that
AxO=
×A= ).
º Suppose A x ¢ # . Then, A x Õ has at least one element (a, b) in it such that a E A
and b ¬ . Now, b ¬ means that is not the nullset. This is a contradiction. Therefore.
Ax¢ = . Similarly, x A= 0.
Example 5 Let A, B,C,D be non-empty sets. Prove that (Ax B) (C x D) if and only if
AcC and BCD. What happens to this resuit if any of the sets is empty?
º First, suppose that (A x B) C(C x D). Take any a e Aand any b e B, then (a, b) e (A x B).
Since (A x B) C(Cx D), it follows that (a, b) e (C x D). Therefore, a e Cand b ¬D. This
proves that A C and BCD.
Conversely, suppose that ACCandBCD. Take any (x, y) e AxB. Then x¬ Aand ye B.
Since ACCand BCD, it follows that x¬ C and y e D. Therefore, (x, y) E CxD. This proves
that (A B) C(Cx D). The required result is thus proved.
Next, suppOse Ais empty, and take B =(1,2), C = (2,4,5), D = (1,3). We find that
AZB=B= . Also, ) cCxD. Therefore, here., AxBCCxD. But B D. Thus, it A
empty, then (A x B) E (Cx D) does not imply that AC C and BCD.
Chapter 6
Relations - II
This chapter is a continuation of Chapter 5. In this chapter, the topic of re
lations is discussed further. Special types of relations called equivalence re
lations and partial orders are dealth-with in some detail. The matrix repre
sentation of relations and the pictorial representation of relations (known as
digraphs) are used toillustrate the concepts and results.
6.1 Zero-one Matrices and Directed graphs
Consider the sets A = aj, a2, ., am)and B = (b1,b2,... , ba) of orders mand n respectively.
Then Ax Bconsists of all ordered pairs of the form (a;, b;), 1 sism, 1 sjsn, which are
mn in number. Let R be a relation from A
to B so that R is a subset of A x B.
Now, letus put m;; = (a;, bi) and assign the values 1or 0 to mjaccording to the following
Rule:
(a, b;) eR
|0 if (a, b,) &
R.
1
mij =
if
Xn matrix formed by these m;;'s is called the matrir of the relation R, or the relation
matrix for R, and is denoted by Me or MR)." Since M(R) contains only 0 and 1as its elements,
M(R) is also called the Zero-one matrix tor R.
It is to be noted that the rows of Mp correspond to the elements of Aand the columns to
those of B.
When B = A, the matrix Me becomes an nxn matrix whose elements are mij = (a;,j)
with
mij
|1 if (a, a) e R
0 if (a, a) &
R.
,O example, consider the sets A = (0. 1.2 and B = (p, al and the relation R
from Ato 5
defined by
R= {(0, P), (1,9). (2, p).
This matrix is used as a Lool for the cOnputer recognition and analysis of relations and is an exanple of a data
Structre.
259
260
6.
Here, A = (a1, a2, a3}) = {0, 1, 2) and B = (bË, b2) = (p, q). We note that
m12
(aj, bj) =(0, p) = 1, because (0, p) eR
(aj, b2) = (0, q) = 0, because (0, g) R
m21
m22
(a2, bi) (1, p) = 0,
(az,bz) = (1,q) = 1,
m31
(a,, bË) = (2, p) = 1,
m32
(as, ba) = (2, q) = 0.
M11
Relations - I|
Accordingly, the matrix of the relation R is
|1 0
M(R) =MR =my=|0
As another Example, consider the set A = (1,2,3, 4) and a relation R defined on Aby
R= ((1,2), (1,3), (2,4), (3, 2)}. Thus, here, A = {aj, az, a3, a4) = B where aj = 1, ag = 2,
a3 = 3, a4 = 4.
Accordingly, mËj = (aj, aj) = (i,), i= 1,2,3, 4; j= 1,2, 3,4, and we find that
m1| = (1, 1) = 0,
m12 = (1,2) = 1,
m13 = (1,3) = 1,
m21 = (2, 1) = 0,
m31 = (3, 1) = 0,
m41 = (4, 1) = 0,
because (1,1) ¢R
because (1,2)ER
m14 = (1,4) = 0,
m22 = (2, 2) = 0, m23 = (2, 3) = 0,
m32 = (3, 2) = 1, m33 = 3, 3) = 0,
m42 = (4,2) = 0, m43 = (4,3) = 0,
m24 = (2,4) = 1,
m34 = (3, 4) = 0,
m44 = (4,4) = 0.
Thus, the matrix of R is
0
M¡ =(my]=
0
1 1 0]
00
1
1 0
0
0
0 0
Remarks
For any relation Rfrom a finite set Ato afinite set B, thefollowing results are o0
(1) MR is the zero matrix if and only if R = .
(2) Every element of MR is 1if and only if R= Ax B.
Digraph ofa Relation
Let Rbe a relation on afinite set A. Then Rcan be represented
below:
pictorially as
described
61.
Zero-one Matrices and Directed graphs
Draw asmall circle or a bullet for each
261
element of Aand label the circle
VRYPondingelement of A. These circles (bullets):
are called
(bullet) with the
a vertex x to a
from
edge,
or
an
vertices
yif and
called
vertex
nodes.
only if (x, y) ¬R. The Draw an arrow,
rpresentation of Ris called a
directed graph or digraph of R.
resulting pictorial
ta relation is pictorially
repreSented by a digraph, a vertex from which an edge leons
acalled the origin or the SOurce for that edge, and a
vertex where an edge ends is
terminusfor that edge. Avertex which is
a
called the
neither source nor a
of any edge is
isolated vertex. An edge for which the sourCe and
terminus
called
terminus are one and the same vertex
called aloop. The number of edges (arrows)
is
at a vertex is called the
terminating
hat vertex and the number of
of
edges (arrows) leaving a vertex is called the
in-degree
vertex.
of
out-degree that
For example, consider the set A
c,
(c). (d. a), (d, c)} defined on A. The={a,b, d)ofand the relation R= {(a, b), (b, b), (b, d), (c, b),
digraph this relation is as shown below:
Figure 6.1I
ODserve that, since (b. b) e R. there is a loop at the vertex b. We also see that the in
degrees of the vertices a, b, c, d are 1,3, 1,2 respectively. Further, the
out-degrees of a, b, c, d
are 1,2,2,2 respectively.
As
another example, consider the set A ={(1,2, 3,4, 5} and the relation
R= {(1, 1), (1, 2), (1,4), (3, 2)}
defined on A. The digraph of this relation is as shown below:
Figure 6.2
6.
262
We observe that the above digraph has a loop at the vertex 1.
Relations - II
Further, the vertex 5 is
an isolated vertex; no edge leaves this vertex and no edge terminates at this vertex. We also
note that the in-degrees of the vertices 1,2,3,4 are l,2,0,1 and their out-degrees are 2o
respectively.
Example 1 Let A = {1.2} and B= (p, q, r, s} and let the relation Rfrom Ato Bbe defined by
R= {(1,q),. (1, r), (2, p), (2, q), (2, s).
Write down the matrix ofR.
We first consider all elementsof Ax Band make the following observation:
(1,p) ¢ R, (1,q) ¬R, (1,) eR, (1, s) ¢ R
(2, p) e R, (2.q) ¬ R, (2,r) ¢ R, (2, s) ¬ R
In view of these and by using the definition of the matrix of a relation, we find that, for the
given R,
1 1
MR
1 0
Example 2 Let A = {1,2,3,4} and let R be the relation on A defined by xRy if and only if
y= 2x.
(a) Write down R as a set ofordered pairs.
(b) Draw the digraph of R.
(c) Determine the in-degrees and out-degrees of the vertices in the digraph.
º (a) We observe that for x, y ¬ A, (x, y) E R if and only ify = 2x.
R= {(1,2), (2,4)).
(b) The digraph of R is as shown below:
3
ligure 6.3
Thus,
6.1. Zero-one Matrices and Directed graphs
(c) Fromthe above digraph, we note that 3is
2,4. the in-degrees and the
263
an isolated
and that for the
out-degrees are as shown invertex
the following Table. vertices 1,
Vertex
12 4
11
In-degree
Out-degree
Bxample 3 Let A = {1,2,3, 4} and let Rbe the relation on A
defined by xRy if and only if "x
divides y'", written xly.
(a) Write down R as aset of ordered pairs.
(b) Draw the digraph of R.
(c) Determine the in-degrees and
out-degrees of the vertices in the digraph.
º (a) We observe that
1|1, 1|2, 1|3, 1|4, 2|2, 2|4,3|3,4|4
Hence
R= {(1, 1), (1,2), (1, 3), (1,4), (2,2), (2, 4), (3,3), (4, 4)}.
(b) The digraph of R is as shown below.
Figure 6.4
(c) By examining the digraph we note that, for the vertices 1,2,3,4, the in-degrees andthe
out-degrees are as shown inthe following Table.
4
Vertex
In-degree
Out-degree
4
2
6.
264
Relations - I
ifand onty i. :
Example 4 Let A = {1,2,3, 4, 6} andR be a relation on A defined by aRb
matrix and draw its digraph.
amultiple of b. Represent the relation R as a
From the definition of the given R, we note that
R={(1, 1), (2, 1), (2, 2),(3, 1), (3, 3), (4, 1), (4, 2), (4, 4), (6, 1), (6, 2), (6, 3), (6,
6))
the matrix of R is
By examining the elements of R, we find that
1 0 0 0 0]
|1 1 0 0 0
MR =|1 01 0 0
1 1 0 1 0
|1 1 1 0 1
The diagraph of R is as shown below:
Figure 6.5
a set B as described by the following
Example 5 Determine the relation R from aset A to
matrix:
1
0. 1
1
MR =
1 0
1
|1 0 0|
|A| =
mnatrix. Therefore, JA| = 4 and |B = 3. If
We note that the given MR is a 4 x3
observing the elements of MR, we find that
by
then
b3},
{b1,b2,
=
B
and
a4}
a3,
{aj,az,
(aj, bi) E R, (aj,b2) R, (aj, b3) E R
(az, bË)E R,(ay, b2) E R, (a2, b3) R
(a3, b) R, (a3, b) R, (a3, b3)E R
(a4, bj) ER, (a4, b2) ¢ R, (a4, b3) ¢ R
Thus, R = {(a,,b1), (aj, b3), (a2, b1),(ag, b2),(a3, b3), (a4,bi)}.
6.1.
Zero-one Matrices and Directed graphs
265
Example 6 Let A ={u, v, x, y, z} and R be a relation on A whose matrix is as given below.
Determine Rand also drawthe associated digraph.
0 1 1 0 0]
1 0 10 0
M(R) =|1 1 0 0 1
1 0 0 0 1
0 01 10
By examining the elements of the given M(R), we find that*
R= ((u, v), (u, x), (v, u), (", x), (*, u), (x, v), (x, z), (y, u), (y, z), (<, x), (z. y)
The digraph of this relation is as shown below:
Figure 6.6
Example7 Find the relation represented by the digraph given below.
Also, write down its
matrix.
Figure 6.7
we note that the relation Rrepre
vertices,
four
has
which
digraph
yCKamining the given
ented by it is defined on the set A = (1,2, 3,4) and is given by
(4, ),(4, 4)).
R= ((1,2), (1,4), (2, 2), (2, 3),
ol ¤.
a.
enents of AInaybe desienated as
4), du, a for the purpose of
writing the clements
6.
266
The matrix of R is
Relations - I1
[0 1 0
0
MR =
1 1
00 0 o
|1
0 0
Example 8 For A = (a, b,c, d, e, fl, the digraph in Figure below represents a relation P
A. Determine R as well as its associated relation matrix.
b
Figure 6.8
º By examining the given digraph,we find that
R= ((a, b), (b, e), (d, b),(d,c), (e, f).
Also, the matrix ofR is given by
[0 1
0 00 0
0 0 0 0
0
0
0
0
0
0
0
0
1 0
0
0
M(R)= 0 1 1 0 0 0
0
1
|0 0 0 0 0
0
Example 9 Let A = {a, b, c, d}and R be arelation on A that has the matrix
MR =
[1 0 0
|0 1 0 0
1
1
1 0
|0
10
Construct the digraph of Rand list the in-degrees and out-degrees of all vertices.
º By examining the entries in the given matrix, we find that the given relation Rhas ut
following representation as a set of ordered pairs:
R= ((a, a, (b, b),(c, a), (c, b), (c, c), (d, b), (d, d)}.
6.1. Zero-one. Matrices and Directed graphs
267
The digraph of this relation is as shown below:
d
a
Figure 6.9
The in-degrees and out-degrees of the vertices are
shown in the following Table.
Vertex
d
In-degree
Out-degree
3
1
1
1
3
2
Exercises
I. Let A ={a, b, c} and B = {0, 1}, and R = ((a, 0), (b,0), (c, 1)} be a relation from A to B.
Write
down the matrix of this relation.
d Let A ={1,2,3,4) andR be the relation on Adefined by (a, b) ¬ R if and only if a < b. Write
own R as a set of ordered pairs. Also write down the matrix of this relation.
Determine the relation Rfrom a set Ato a set Bas represented by the following matrix:
|1 0 0
MR = 0 1 0
0
1
1
0
0
1,2,3,4} and R = {(1, 1),(1, 2), (2, 1), (2, 2), (3, 4), (4, 1)} be arelation on A. Write down
the digraph of R.
5. Let A ={a, b, c, d, e, f) and Rbe the relation on Adefined by R = ((a, b), (a, d), (6, c), (b, e),
(d,b), (d, e), (e, c)), (e, f), (f, d)). Draw the digraph of R.
6. If R {x, y) | x > y} is a relation defined on the set A = (1,2, 3, 4), write downthe matrix and
he digraph of R.
268
6. Relations -I
7. Find the relation Rdetermined by each of the digraphs given below. Also, write
of the relation.
()
(ii)
down the
matrIK
a
1
(a)
(b)
Figure 6.10
8. Let A= {2,4,5,7), and let R be the
relation on A having the matrix
|1 1 0 11
MR =
0
1 1
00
1 1
|1 0 0
0
Construct the digraph of R.
9. Find the relation R on the set A and write down its
matrix of R is
|1
digraph, given that A =(a, b, c, d,e) and the
1 0 0 0]
0 0
1 1 0
MR =0 0 0 1 1
0 1 10 0
0 0 0
0
Answers
1. Mp =1
2. R= {(0, 1), (1,2),(1,3), (1,4), (2,2), (2,3),
(2,4),(3, 3), (3,4), (4,4))
|| 1 1
Mp =
1
1
0
0
|0
0 0
1 1
1|
61.
Zero-one. Matrices and
Directed graphs
269
3. IfA =(aj, az, as), B =(b1,b2, b3, ba), then R=
((a,,b). (a2, b2),(a,, ba) ,(a,, b,))
5.
4
Figure 6.12
Figure 6.11
0
6. M =
|1
0 0 01
0
0
1
0 0
1 1
0
0
Figure 6.13
7.
() R= {(1,2), (2,3), (2,4), (3, 2), (3,3), (3,4), (4,4)}
MR =
|0 1
0 0
0
0
1
1
0
1
1
1
0
0 0 1
(i) R= ((a, c), (a, e), (c,e), (c, d), (d, f), (e,b), (e, d))
fo
0 1 01 0)
00 0 0 0 0
0 0 0 1 1 0
MR =
|0 0 0 00 1
0 1 0 1 0 0
Jo 0 0 0 0 0
63.Properties of
Relations
285
Answers
1
RoS = {(1,3), (1,4), (3, 1), (4, 1)},
SoR= ((2, 1),(2,3), (3,2). (3,4), (4, 1), (4, 3)}
{(1,
RoR= 1),(1,2), (1,3),(1,4),(3,2)}, SoS
=(3,3),(3, 1), (3,4)}
2.
R = ((1, 1),(1,2), (1,3), (2,3),
(2,4), (3,4)}
R = R= ((1,2), (1, 3),
(1,4),
(2,3), (2,4), (3,4)}
2
R2
(a)
(b)
Figure 6.28
1
0
1
1
0
1
0
0
1
1
1
0
3. M(Rosy =
0 0
0.5
In this
|1 1 1
Properties of Relations
Section we consider someimportant properties of relations defined on a set.
Reflexive
Relation
A
R on a set A is said to be reflexive (or said to have the reflexive
relation
property) if
(a, a) e R. for
allae A.
In other words, a relation Ron a set Ais refiexive whenever every element a of Ais related
to itself by R(i.e., aRa, for all a EA).
It follows that Ris not reflexive if there iS Some a EAsuch that (a, a) R.
For examnle. the relation "is less than or cqual to" is a reflexive relation on the set ta1
real numbers. Because, a = a for every real number a.
It is obvious that the relations "is less than" and "is greater than" are not retlexive on the
Set of all real numbers.
6.
286
Relations - II
As another example, we observe that if A = (1,2, 3, 4), then the relation R= {(1,1). (2,2),
4) ¢ R.
(3,3)} is not reflexive. Because, 4 E A but (4,
The following results are easy to see:
diagonal.
(1) The matrix of areflexive relation must have 1's on its main
cycle of lenoth 1
(2) Atevery vertex of the digraph of a reflexive relation there must be a
(3) On a set A, the relation AA defined by
A = {(a,a) |a e A}
is reflexive.* Furthermore, AA is a subset of every reflexive relation on A. The matrix of
AA contains 1's on the main diagonal and 0's in all other positions.
Irreflexive Relation
a relation
A relation on aset A is said to be irreflexive if (a, a) R for any a ¬ A. That is,
Ris irreflexive if no element of Ais related to itself by R.
of all
For example, the relations is less than" and is greater than'are irreflexive on the set
real numbers.
It is to be noted that an irreflexive relation is not the same as a
non-reflexive relation.
R=
A relation can be neither reflexive nor ireflexive. For example, consider the relation
R
{(1, 1),(1,2)} defined on the set A = {1,2,3}. This relation is not reflexive because (2,2)
and (3,3) ¢ R. The relation is not irreflexive because (1, 1) ¬ R.
The following results are obvious:
(1) The matrix of an irreflexive relation must have O's on its main diagonal.
(2) The digraph of an irreflexive relation has no cycle of length 1at any vertex.
Symmetric Relation
A relation R on a set is said to be symmetric (or said to have the symmetric propery) if
(b, a) e R whenever (a, b)¬R for all a, b e A.
It follows that R isnot symmetric if there exist a, b e A such that (a, b) ER but (b, a) EK.
A relation which is not symnmetric is called an asymmetric relation.
For example, if A= {1,2, 3)and RË = ((1,1), (1, 2), (2, 1)) and R, = {(1,2),(2, 1), (1,5)}
relations on A,then RË is symmetric but Rz is asymmetric; because (1, 3) e R, but (3, 1)*
Jt is evident that for the matrix MR =[m;il of a symmetric relation the following property
holds:
If mËj = ] then mj = 1, and if mij = 0then mji = 0.
"The relation A is called the equality relation on Aor the diagonal line of A XA.
Properties of
63.
Relations
that
Thismeans
a
287
the matrix MR Of a symmetric rela ion R is such that the (i, jh element of
th
tothe (j, i)" element of
equal
Mpis
*
symmetriCmatriy
MR. In other worc s, the matrix of a
symmetric relation is
symmetric relation, if there is ai edge from vertex ato
a vertex b, then
a
Inthe digraph of
vertic
two
S are connected by an edge, they must
fronm bto a; this means that if
thereisan edge
in both directions. Because of this, in a digraph of asymmetric relation,
alwaysbe connected
without arrows
the arrows are understood both ways. The digraph of
the edges are shown
symmetricrelation is called the graph of the relation and an edge connecting two vertices
a
nihis always a bidirected edge; it is denoted by {a, b}. Two vertices a and b of a graph which
vertices.
are connected by an edge are called adjacent
For example, consider the relation
R= {(1,2), (2, 1), (1, 3), (3, 1)}
on the set A = {1,2, 3}. Evidently, this relation is a symmetric
relation, and its graph is as
shown below:
Figure 6.29
vertices, 1 and 3 are
dle above graph, 1and 2 are adjacent
adjacent vertices , but 2 and 3
are not adjacent vertices.
Antisymmetric Relation
have the antisymmetric
Arelation R on a set A is said to be antisymmetric (or said to
property) if whenever (a, b) e Rand (6, a) ¬Rthena= b.
R and
Asuch that (a, b) E
¬
b
a,
exist
there
It
that R is not antisymmetric if
e
6,a) Rbut a # b.
numbers is an
of all real
follows
to" on the set
the relation is less than or equal
a = b).
antisymmetric relations
(because if a <b and b <a, then
relation
and
symmetric)
It
asymmetric (i.e., not
antisymmetric. Arelation can
and
are not one be emphasized that
symmetric
and the same. A relation can be both
be
For
exampl
e
,
antisymmetric
should
neiForther symmetric nor antisymmetric.
= ((1,2),(2,
1), (2, 2)) and R2
). (2, 3)}).
example, let A =(1,2,3} and RË = {(1,
of A.
A',the transpose
=
A
equivalently
"A square
sym1netric if ajj = dji or
bec
to
said
is
Flaj]
=
matrixA
288
6.
We check that R1 is both symmetric and antisymmetric, and Rz is neither
SVMmetic.
Relations - I|
symmetric nor anti-
The following resultsare obvious:
(1) If MR =[m;]is the matrix of an antisymmetric relation, then, for i # i we
mËj = 0 or m = 0.
have either
(2) In the digraph of an antisymmetric relation, for two different vertices a and h a
cannot be a bidirectionaledge between aand b.
Transitive Relation
Arelation R on a set A is said to be transitive (or said to have the transitive pronert) i:
whenever (a, b) e Rand (b, c) ¬ R then (a, c) ER, for all a, b, c e A.
It follows that R is not transitive if there exist a, b, c¬ A such that (a, b) E R and (b, c) e R
but (a, c) ¢ R.
For example, the relations "is less than or equal to" and is greater than or equal to are
transitive relations on the set of all real numbers. Because, if a s b and b s c then a s c. and
if a> band b > c then a > c, for allreal numbers a, b, c.
As another example, if we consider the set A= (1,2,3) and the relations R= (1, 1),(1,2),
(2,3), (1,3), (3, 1), (3, 2)} and Ry = {(1,2), (2, 3), (1,3), (3, 1)} on A, then RË is transitive but&:
is not transitive.
The following result is easy to prove:
A relation R on a set A istransitive if and only if its matrix M¹ = [ml has the following
property:
If mk = land mkj = 1, then m;;=1.
Example1 Let A= {1, 2, 3). Determine the nature of the following relations on A:
) R ={1, 2), (2, 1), (1, 3), (3, 1)) () Rz = ((1, 1), (2, 2), (3,3), (2, 3))
(iii) R3 = {(1, 1), (2, 2),(3,3)}
(v) Rs = {(1,1), (2,3), (3, 3)}
(iv) R4 = {(1, 1), (2, 2), (3,3), (2,3),(3, 2))
(vi) R6 = (2,3), (3, 4), (2,4)}
(vii) R, = {(1,3), (3, 2)}.
º By examining all ordered pairs present in the relations given, we find that:
k, is symmetricand irreflexive, but neither reflexive nor transitive.
Ra is reflexive and transitive, but not symmetric.
R, and R4 are both reflexive and sym1metric.
Rs is neither rellexive nor symmetric.
Ro is transitive and irreflexive, but not symmetric.
Rq is irreflexive, but neither transitive nor symmetric.
Propertiesof
63.
Reclations
289
Examplè2 Let A =(1, 2, 3, 4). Determine the nature of the following relations on A.
()
) R ={(1, 1),(1,
2),(2,,I1), (2,2), (3, 3), (3,4),(4,3), (4,4))
(2).R=((1, 2).(1,3).(3, 1),(1, 1),(3, 3),(3, 2),(1,4), (4,2), (3,4)
(3) R, represented by the following digraph:
Figure 6.30
By examining all ordered pairs present in Rand R2, we find that:
(1) RË is reflexive, symmetric and transitive, and
(2) R, is transitive.
By examining the edges in the digraph in Figure 6.30, we find that the relation Rg is both
asymmetric and antisymmetric.
Eample 3 Find the nature of the relations represented by the following matrices:
[0 1 1 01|
(a) |1 1 0 0
|1 0 1 1
[1
(b)
0
1 0]
1
0 1
1 0 1 0
0 1 0 1
[0 0 1 1]
|0 0 1 0
(c) 0 0 0 1
1 0 0 0
(a) Here, the given matrix is symmetric (that is, aji = aij for i,j = 1,2,3). Therefore, the
Corresponding relation is symmetric.
Therefore, the
(b) Here., the
and is symmetric.
diagonal
matrix
has
1's
on
its
main
given
symmetric.
ng matrix is not symmetric. Therefore, the correspondng relation is ot
(cy cor
Here,respondi
the given
relation is reflexive and
. Further,
isymmet
ndicatesricthat
the relation is not antisymmelric.
and
the presence of | in the (1,4)th
(4, I)h positions of the
matrix
64.EquivalenceRelations
299
5
Figure 6.33
20. Hint: Prove (i) and disprove (ii)-(iv) through counterexamples.
6.4 Equivalence Relations
ArelationR on a set A is said to be an equivalence relation on A if (i)
R is reflexive, (i) R is
symmetric, and (11i) R is transitive, on A.
Atrivial example of an equivalence relation is the relation is equal to on the set of all
real
numbers, R. An example of a relation which is not an equivalence relation is the relation is
less than" on R.
Example 1 Let A (1,2,3,4} and
R= {(1,1), (1, 2),(2, 1), (2, 2), (3,4), (4,3), (3,3),(4,4)}
be a relation on A. Verify that R is an
equivalence relation.
We have to show that R is reflexive, symmetric and transitive.
First, we note that all of (1,1),, (2, 2),(3,3),(4, 4) belong to R. That is, (a, a) e Rfor all
a¬ A. Therefore.Ris a
Next, we note the
reflexive relation.
following:
(1,2), (2, 1) e R and (3,4), (4, 3) ¬ R.
That is, if
a, be A. Therefore, Ris a symmetric relation.
(a, b) ¬ Rthen (b, a) ERfor
e
whenever
Lastly, we note that
(1,2), (2, 1), (1, 1) E R, (2, 1), (1, 2), (2, 2) ¬ R,
(4,3),(3,4), (4, 4)e R.
That is, if whenever (a, b) ERand (b, c) ER then (a, c) e R, for a, b, c¬ A. Therefore, Ris a
transitive relation.
Accordingly,
Ris an equivalence relation.
300
6. Relations -I|
Example 2 Let A= (1,2,3,4), and
R= {(1,1), (1, 2), (2, ),(2, 2), (3, 1), (3, 3), (1, 3), (4, 1), (4,4))
be a relation on A. Is R an equivalence relation?
Here, we have to check whether or not R is reflexive, symmetric and transitive,
By examining the elements of R, we note the following:
(i) (a,a) ER for every of ae R, Therefore R is reflexive.
(ii) (4, 1) ¬R, but (1,4) ¢ R. Therefore, R is not symmetric.
Since R is not symmetric, R is not an equivalence relation. (We need not check Riur
transitivity).
Example 3 Let A ={1,2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12}. On this set define the relation Rhy
(x, y) ¬ Rif and only if x - yisa multiple of 5. Verify that R is an equivalence relation.
º For any x ¬ A, we have x-x = 0 which is a multiple of 5(because 0 = 5 -0). Therefore
(x, x) ER and soR is reflexive.
For any x, y E A, if (x, y) ER then x-y=5k for some integer k. Consequently, y-X= 5(-4)
so that (, x) ¬ R. Therefore, R is symmetric.
For any x, y, z E A, if (x, y) E R and (y, z) ER then x-y= 5kË and y -z=5k, for some
integers kË and k2. Consequently.
x-2= (*-y) +(y-)=5kË + 5k2 =5(kj + k2)
from which it follows that (x, z) R. Therefore, R is transitive.
Thus, R is an equivalence relation.
Example 4 IfA = Aj UAz U A3, where A = {1,2}, A, = (2,3,4} and A3 = {5}, defne
Is Kan
the relation R on Aby xRy if any only ifx and y are in the same set A,, i =1,2,3.
equivalence relation?
Therefore, K
We note that xRx for every x in A; because x and x belong to the same A;.
reflexive.
Further, if x, y E A;then y, xE A;for all x, yinA. Therefore, R issymmetric.
and(2,3)eR
Lastly, we observethat (1, 2) ER (because 1 and 2 are in the same set, Aj) inthesame
R(because 1 and 3 are not
(because 2 and 3 are in the same set, A2) but (1,3)
set). Hence R is not transitive.
Accordingly, R is not an equivalence relation.
Example 5 Arelation R on a set A =(a, b, c) is represented by the following matrix:
|1 0
MR = 0
1
0
0
Determine wlhether R is an equivalence relation.
1
Relations
301
the clcmcnts of MR. We find that
NCha(a,c)e)Rbut (c, a)
R=((a, c). (a,
(6, b), (c, c)).Ris not an
R. Therefore, Ris not symmetric. c),Accordingly,
aNake
relation.
symmetric can also be seen by the fact that
the matrix MR 0S not
not
RIS
That
symmetric.)
Aample6 7The.digraph ofa relation Ron the set A= {1,2, 3} is as given below.
Determine
Risanequivalence relation.
Figure 6.34
ºBy examining the digraph, we note that the given relation is symmetric and transitive but not
rtexive; observe that (3,3) R. Therefore, Ris not an equivalence relation.
Example 7 Let S be the set ofall non-zero integers, and A = S xS. On A, define the relation
Rby (a, b)R(c, d) if and only if ad= bc. Show that R is an equivalence relation.
R is reflexive on
FrIst, we note that (a, a)R(a, a), because aa = aa for any a eS. Therefore,
A.
d)R(a, b).
Next, suppose (a, b)R(C.d), Then ad = bc and therefore cb =da. Hence (c,
Accordingly, Ris symmetric on A.
= de, which
suppose that (a, b)R(c, d) and (C, d)R(e, f). Then ad = bc and cf
Last
l
y
.
af =be. Hence (a, b)R(e,f). Accordingly, Ris transitive on A.
yield
Ihis proves the required result.
Exampl & For afixed integer n > 1, prove that the relation "congruent nodulo n" is
an
tuivalence relation on the set of all integers, Z.
ora.beZ, if a - bis a
[written symbolically as
some &¬Z.
equivalently, a -b= knfor
we say that "a is Congruentto b modulo n"
of n, or ,
We have to
a = b (mod n),
laus denote this relationmultiple
means
aRb
that
by RsSo
equjvalence relation.
Iit, we note that
-a = 0is
for every a E Z, a
lherclore, Ris reflexive.
n; that is, a
a muliple of
prove that ¤
a(iod n), or
314
6.Relalions -
Answers
5. No
7. [] = (4k +i| keZ}
8. {(1,4), (2,5), (3, 6), (4,7), (5, 8), (6,9)}
10. No
11. (a),(c): Yes;
13. (1,2), (3), (4})
(b), (d): No
14. R= ((1, 1), (1,2), (2, 1), (2, 2), (3,3), (3,4), (4,3), (4,4), (5, 5)}.
15. ((1, 1),(1,3), (1, 5),(3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(2,2), (2, 6), (2,4), (4, 2), (4,4), (4, 6), (6,2), (6,4), (6, 6)}.
17.
(ii) [(2, 14)] = {(2, 14)}, [(-3,-9)] = {(-3,-9), (-1, -3), (4,
[(4,8)]= {(-2, -4), (1, 2), (3, 6). (4, 8)}.
12)),
(i) A= [(-4,-20)] U[(-3, -9)]U[(-2,-4)] V[(-1,-11)] U[(2,
18. {(1,6, 11,16}, {2,7, 12, 17}, (3,8, 13, 18}, (4,9, 14, 19), (5,
14)].
10, 15, 20}.
19. RoS need not be an equivalence relation.
23. 52.
6.5
24. A= ([1] U[4]U[2]
Partial Orders
A relation R on set A is said to be a partial ordering relation or a
(i) Ris relexive, (i) Ris antisymmetric, and
partial order on Aif
(iii) Ris transitive, on A.
A set A with a partial order R defined on it is called a partially ordered set or an ordereu
set or aposet, and isdenoted by the pair (A, R).
The most familiar partial order is the relation less than or equal to", denoted by s, on U
set Z of all integers. (Because, this relation is reflexive, antisymmetric and transitive). Thus,
(2, s) is a poset.
The relation is greater than or equal to", denoted by >, is also a partial order on Z;thatis
(Z, >) is also a poset.
a|b)torall
The "divisibility relation" on the set Z+
divides b(denoted by
defined
a
by
a,be Z+ is a partial order on Z*; see Example 6,
Section 6.3.
The subset relation"
defined on the power set of a set S is a partial order on S;se¢
Example 7, Section 6.3. Thus, for any set S,
(P(S) S) is a poset.
because,these
The relations "is less
than"
and
"is
partial
orders
on
Z;
are not reflexiye.
greater than"
are not
65.PartialOrders
315
The relation "congruent modulo n"
defined on the set of all integers Z is
order; becausethis relation is not antisymmetric.*
also not a partial
Total Order
..phe apartial order on a set A. Then R is
called a total order * on A if
either xRy oryRx. In this case, the poset (A, R) is called atotally ordered set.* for all x. y¬ A.
Eor example, the partial order relation "less than or
equal to" is a total order on the set R.
Docase. for any x,ye R, we have x <y or y < x. Thus, (R, s) is a totally
ordered set.
If we consider the divisibility relation on the setA =
{1,2,4,8}, this relation is a total order
am A. The same relation is not a total order on the set A =
{1,2,4,6, 8} although it is a partial
order on A. (Observe that neither 4 divides 6 nor 6 divides 4).
The subset relation is also not a total order on the power set of an
arbitrary set S although it
is a partialorder; because for any two subsets S 1 and S2 of S, neither S1 S S2 nor S2 CS can
be true. (For example, if S = {1,2,3), SË ={1,2) and S2 = {1,3), then S S S2 and S S
but S1 S2 and S2 S).
From the definition of a total order and the examples given above it is clear that every total
order is a partial order, but not every partial order is a total order.
Hasse Diagrams
Since a partial order is a relation on a set, we can think of the digraph of a partial order
if the set is finite. Since a partial order is reflexive, at every vertex in the digraph of a partial
order there would be acycle of length 1.In view of this, while drawing the digraph of apartial
order, we need not exhibit such cycles explicitly; they willbe automatically understood (by
convention).
Ii, in the digraph of apartial order, there is an edge from a vertex a toa vertex b and there
1S an edge from the vertex b to a vertexc. then there should be an edge from a toc (because of
As such. we need not exhibit an edge from atoc explicitly; it will be automatically
transitivity). (by convention).
understood
lo simplify the format of the digraph of a partial order, we represent the vertices by dots
(bullets) and draw the digraph in such a way that all edges point upward. With this convention,
We need not
put arrowS in the edges.
The digraph of a partial order drawn by adopting the conventions indicated in the above
paragraphs is1 called aposet digram or the Hasse diagramfor the partial order.
|Example
Let A =(1,2,3,4), and
R={(0,1),(1, 2), (2, 2), (2, 4), (I, 3), (3, 3), (3,4), (1,4), (4,4)}
Verify that Riss apartial onder on A, Also, write down the Husse diagrm for R. b(nod n)
a=
6.4. Note tlhat
he definition of the relation COngruent modulo n lrom IExanple &, Section =2 (uodl 3) but 2 #8.
"Recall
8
and b =u
2 8 (nod 3) and
(mod ) do always imply thata= b, 'or example,
"A totaly ordered set is also called alinewly ordered set or achain.
"Iotal order is also notcalled alinear order.