Abstract In this paper, we evaluate b(Γ(Z n )). Our main result is, we give maximum value of b(Γ(Z n )) is 0.99999999796427626489236243072661, where n is any positive integer upto fiftieth million. Mathematics Subject Classification: 05C25, 05C35, 05C7

J Ravi Sankar

R Vasanthakumari

S Meena

S Sankeetha

CiteSeerX

International Journal of Algebra, Vol. 7, 2013, no. 5, 229 - 236
HIKARI Ltd, www.m-hikari.com
The Binding Number of a Zero Divisor Graph
S. Sankeetha and J. Ravi Sankar
Department of Mathematics, Saradha Gangadharan College
Puducherry, India - 605 004
sangeetha.sivarajp@gmail.com, ravisankar.maths@gmail.com
R. Vasanthakumari
Principal, Kasthurba College for Women, Villianur
Puducherry, India - 605 110
vasunthara1@gmail.com
S. Meena
Department of Mathematics, Government Arts College
Chidambaram, India - 608 104
meenasaravanan14@gmail.com
Copyright c© 2013 S. Sankeetha et al. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
Abstract
In this paper, we evaluate b(Γ(Zn)). Our main result is, we give
maximum value of b(Γ(Zn)) is 0.99999999796427626489236243072661,
where n is any positive integer upto fiftieth million.
Mathematics Subject Classification: 05C25, 05C35, 05C70
Keywords: Commutative ring, Zero divisor graph, Binding number
1 Introduction
Let R be a commutative ring and let Z(R) be its set of zero-divisors. We
associate a graph Γ(R) to R with vertices Z(R)∗ = Z(R) − {0}, the set of
230 S. Sankeetha, J. Ravi Sankar, R. Vasanthakumari, S. Meena
non-zero zero divisors of R and for distinct x, y ∈ Z(R)∗, the vertices x and
y are adjacent if and only if xy = 0. Throughout this paper, consider the
commutative ring R as Zn and zero divisor graph Γ(R) as Γ(Zn). The binding
number of Γ(Zn), denoted by b(Γ(Zn)) is defined by, Γ(Zn) = { |N(S)||S| , where
S ⊆ V (Γ(Zn)), S = φ, N(S) = V (Γ(Zn))} which satisfies the following condi-
tions; (i) N(S) ∪ S = V (Γ(Zn)) (ii) N(S) ∩ S = φ (iii) d(u) ≤ d(v) for u ∈ S
and v ∈ N(S) (iv) no two vertices in S are adjacent. For notation and graph
theory terminology, we generally follow [1, 2, 3, 5, 6].
2 Binding Number of a Zero Divisor Graph
Lemma 2.1 [4] A graph Γ(Zn) has a domination set iff Γ(Zn) is connected
and n is a composite number.
Theorem 2.2 For any prime p>2, then b(Γ(Z2p)) =
1
p−1 .
Proof: The vertex set of Γ(Z2p) is {2, 4, 6, ....2(p − 1), p}. Using theorem
(4.4) in [4], Γ(Z2p) is a star graph K1,p−1. Let S be a non-empty subset of
the vertex set V (Γ(Z2p)), then for any x ∈ S, such that d(x) < d(y), where
y ∈ V − S. Clearly, all the vertices are of minimum degree except p, then
S = {2, 4, 6, ....2(p − 1)}, that is |S| = p − 1 and the neighbourhood of the set
S = N(S) and |N(S)| = p − (p − 1) = 1. Hence, b(Γ(Z2p)) = |N(S)||S| = 1p−1 .
Theorem 2.3 For any prime p, b(Γ(Zp2)) =
1
p−2 .
Proof: The vertex set of Γ(Z2p) is {p, 2p, 3p, ....p(p−1)}. Any two vertices
in b(Γ(Zp2)) are adjacent. Clearly, b(Γ(Zp2)) is a complete graph namely Kp−1.
Let S be a non-empty maximun subset of b(Γ(Zp2)) then {p, 2p, 3p, ....p(p −
2)} ∈ S implies |S| = p − 2 and the neighbourhood of the set S contains only
one point {p(p − 1)} that is |N(S)| = 1. Clearly, b(Γ(Zp2)) = |N(S)||S| = 1p−2 .
Theorem 2.4 If p and q are distinct prime numbers with p < q, then
b(Γ(Zpq)) =
p−1
q−1 .
Proof: The proof is by the method of induction on p and q. The vertex
set of Γ(Zpq) is {p, 2p, 3p, ..., p(q − 1), q, 2q, 3q, ..., (p − 1)q}. Let S and N(S)
be the minimum degree set and the neighbourhood of S respectively.
Case(i): Let p = 2, q is any prime > 2.
Using theorem (2.1), b(Γ(Z2q)) =
1
q−1 =
p−1
q−1 .
Case(ii): Let p = 3, q is any prime > 3.
The vertex set of Γ(Z3q) is {3, 6, 9, ..., 3(q−1), q, 2q}. Let u = q and v = 2q
be two vertices in Γ(Z3q) with maximum degree then there exist any other vertex
w = q and w = 2q in Γ(Z3q) such that w is adjacent to both u and v. That is,
uw = vw = 0. But uv = 0. Therefore u and v are non-adjacent vertices. Then
The binding number of a zero divisor graph 231
the vertex set V (Γ(Z3q)) can be partitioned into two parts S and N(S) such that
S = {3, 6, 9, ...., 3(q−1)} and N(S) = {u, v} = {q, 2q}. Clearly |S| = q−1 and
|N(S)| = 2, then |V (Γ(Z3q))| = |S|+ |N(S)| = q−1+2 = q +1. Note that the
vertices in the set S have the smallest degree compared to the set N(S). Clearly,
any two vertices in S are non-adjacent. Moreover V (Γ(Z3q)) = S ∪N(S) and
S ∩ N(S) = φ and d(u) ≤ d(v) for all u ∈ S and v ∈ N(S).
Then, b(Γ(Z3p)) =
|N(S)|
|S| =
2
q−1 =
p−1
q−1 , where p = 3 and q > 3.
Case(iii): Let p < q.
The vertex set of Γ(Zpq) is {p, 2p, 3p, ..., p(q−1), q, 2q, 3q, ..., (p−1)q}. Using
the above cases, the vertex set V (Γ(Zpq)) can be partitioned into two parts S and
N(S) which implies that the vertex p, multiples of p are in S and q, multiples
of q are in N(S). Clearly, every vertices in S are non-adjacent which holds for
N(S). Then, |V (Γ(Zpq))| = |S|+ |N(S)| = p − 1 + q − 1 = p + q − 2. That is
S = {p, 2p, .., p(q − 1)} and N(S) = {q, 2q, .., (p − 1)q}. Clearly, d(u) < d(v)
where u ∈ S and v ∈ N(S). We note that, every vertex in S are adjacent to
all the vertices in N(S). Using all the above cases, b(Γ(Zpq)) =
|N(S)|
|S| =
p−1
q−1 .
Theorem 2.5 For any graph Γ(Z2n), where n > 2 is a positive integer then,
a) If n is even, b(Γ(Z2n)) =
2n−1−2n2
∑n−4
2
i=0 2
i−2
2
n
2
∑n−4
2
i=0 2
i+1
.
b) If n is odd, b(Γ(Z2n)) =
2
n−1
2 (2
n−1
2 −
∑n−3
2
i=0 2
i)−1
2
n−1
2
∑n−3
2
i=0 2
i
.
Proof: The vertex set of Γ(Z2n) is {2, 4, ., 2(2n−1 − 1)} and |V (Γ(Z2n))|
= 2n−1 − 1. The proof is by the method of induction on n.
Case(a): When n is even.
Subcase(i): Let n = 4. The vertex set of Γ(Z24) is {2, 4, 6, 8, 10, 12, 14}.
Let S be a vertex subset of V such that d(u) ≤ d(v), where u ∈ S and v ∈ N(S).
Let P be a set of all pendant vertices in Γ(Z24). Clearly, P = {2, 6, 10, 14} with
d(u) = 1, for all u ∈ P . It seems that P ⊆ S. Let v = 2n−1 = 24−1 = 8 and
w = 24−2 be any other vertex in Γ(Z24) then vw = 8×(24−2) = 112. Clearly,
24 must divides 112. Thus, the vertex v is adjacent to all vertices in Γ(Z24)
which implies v = 8 ∈ N(S). Let x = 4 and y = 12 be the remaining vertices
in V such that xv = yv = 0. That is, x, y and v are adjacent vertices. Clearly,
either x = 4 ∈ S or y = 12 ∈ S. Suppose, x, y ∈ S, we get a contradiction to
our definition that no two vertices in S are adjacent. Finally we conclude that
S = {2, 4, 6, 10, 14} or S = {2, 6, 10, 12, 14} and N(S) = {8, 12} or N(S) =
{4, 8}, respectively. That is |S| = 5 and |N(S)| = 2. Clearly, V (Γ(Z24)) =
S ∪ N(S) and S ∩ N(S) = φ. Since, degree of any vertex in S is less than or
equal to degree of any vertex in N(S) and |N(S)| = V (Γ(Z24))−|S| = 7−5 = 2.
Then, b(Γ(Z24)) =
|N(S)|
|S| =
2
5
=
24−1−1−2 42
∑ 4−4
2
i=0 2
i−1
=22+20
=
24−1−2 42
∑ 4−4
2
i=0 2
i−2
=24/2(20)+20
232 S. Sankeetha, J. Ravi Sankar, R. Vasanthakumari, S. Meena
=
2n−1−2n2
∑n−4
2
i=0 2
i−2
2
n
2
∑n−4
2
i=0 2
i+1
,where n = 4.
Subcase(ii): Let n = 6.
The vertex set of Γ(Z26) is {2, 4, 6, ...., 62}. That is |V (Γ(Z26))| = 31. Let S
be a vertex subset of V such that d(u) ≤ d(v), where u ∈ S and v ∈ N(S). Let
P be the set of all pendant vertices in Γ(Z26). Clearly, P = {2, 6, ..., (26 − 2)}
with d(u) = 1, for all u ∈ P . It seems that P ⊆ S. Using subcase (i), let
v = 2n−1 = 26−1 = 32 and w = 26 − 2 be any other vertex in Γ(Z26) such that
26 must divides vw = 32×(26−2) = 1984. Thus, the vertex v is adjacent to all
the vertices in Γ(Z26) which implies v = 32 ∈ N(S). Similarly, 24 and 3 × 24
are adjacent to all the vertices in Γ(Z26) except P , then {16, 48} ∈ N(S).
Let U be a vertex subset of V with U = {4, 12, 20, ..., (26 − 4)}. Clearly, no
two vertices in U is adjacent and every vertex in U are adjacent to {16, 32, 48}.
It seems that d(U) < d(N(S)) which implies that U ⊆ S.
Let W = V − (P ∪ U ∪ N(S)) = {8, 24, 40, 56} be a vertex subset of V .
Finally, the vertices in W make a complete subgraph, namely K4 and all the
vertices in W are adjacents to N(S). Using theorem (2.4), any one of the vertex
in W is in S. Otherwise, if any two vertices in W belongs to S, then we get a
contradiction that no two vertices are adjacent in S. Hence, |S| = |P |+ |U |+
any one vertex in W = 25 and |N(S)| = V (Γ(Z26))−|S| = 31−25 = 6. Then,
b(Γ(Z26)) =
|N(S)|
|S| =
6
25
=
26−1−1−2 62
∑ 6−4
2
i=0 2
i−1
24+23+20
=
26−1−2 62
∑ 6−4
2
i=0 2
i−2
2
6
2 23(21+20)+1
=
2n−1−2n2
∑n−4
2
i=0 2
i−2
2
n
2
∑n−4
2
i=0 2
i+1
, where n=6.
Subcase(iii): Let n > 6 is even.
The vertex set of Γ(Z2n) is {2, 4, ...., 2(2n−1 − 1)} and |V (Γ(Z2n))| = 2n−1−
1. Since P is a pendant vertex set with |P | = 2n−2. Using above cases, |S| =
2
n
2 (20 + ................ + 2
n
2
−1) + 20 = 2
n
2
∑n−4
2
i=0 2
i + 1, and |N(S)| = V (Γ(Z2n)) −
|S| = 2n−1 − 1 − 2n2 ∑
n−4
2
i=0 2
i − 1 = 2n−1 − 2n2 ∑
n−4
2
i=0 2
i − 2. Then,
b(Γ(Z2n)) =
|N(S)|
|S| =
2n−1−2n2
∑n−4
2
i=0 2
i−2
2
n
2
∑n−4
2
i=0 2
i+1
, where n is even.
Case(b): When n is odd.
Subcase(i): Let n = 3. The vertex set of Γ(Z23) is {2, 4, 6}. Let S be
a vertex subset of V and let P be the set of all pendant vertices in Γ(Z23).
Clearly, P = {2, 6} with d(u) = 1, for all u ∈ P . It seems that P ⊆ S. Let
v = 6 and w = 2 be any other vertex in Γ(Z23) then vw = 0. Thus, the
vertex v is adjacent to all the vertices in Γ(Z23) which implies v = 4 ∈ N(S).
Let x = 2 and y = 6 be the remaining vertices in V such that xv = yv = 0
and xy = 0. Finally we conclude that S = {2, 6} and N(S) = {4}. Hence,
b(Γ(Z23)) =
|N(S)|
|S| =
1
2
=
23−1−1−2 3−12
∑ 3−3
2
i=0 2
i
2(3−1)/2(20) =
2
n−1
2 (2
n−1
2 −
∑n−3
2
i=0 2
i)−1
2
n−1
2
∑n−3
2
i=0 2
i
.
The binding number of a zero divisor graph 233
Subcase(ii): Let n = 5.
The vertex set of Γ(Z25) is {2, 4, ..., 30}. Let P be the set of all pendant
vertices in Γ(Z25). Clearly, P = {2, 6, ..., 30} with d(u) = 1. It seems that
P ⊆ S. Let v = 16 and w = 2 be any other vertex in Γ(Z25) then vw =
32 = 0. Clearly, 25 must divides 32. Thus, the vertex v is adjacent to all
vertices in Γ(Z25) which implies v = 16 ∈ N(S). Let U be a vertex subset
of V with U = {4, 8, 12, 20, 24, 28}. Since, U has a induced subgraph K2,4.
Clearly, d(4) = d(12) = d(20) = d(28) < d(8) = d(24) implies that the vertices
8, 24 ∈ N(S) and the remaining vertices belongs to S. Therefore the set S =
{2, 4, 6, 10, 12, 14, 18, 20, 22, 26, 28, 30} with |S| = 12 and |N(S)| = 3. Then,
b(Γ(Z25)) =
|N(S)|
|S| =
3
12
=
25−1−1−2 5−12
∑ 5−3
2
i=0 2
i
22(21+20)
=
2
5−1
2 (2
5−1
2 −
∑ 5−3
2
i=0 2
i)−1
2(5−1)/2(21+20)
=
2
n−1
2 (2
n−1
2 −
∑n−3
2
i=0 2
i)−1
2
n−1
2
∑n−3
2
i=0 2
i
, where n = 5.
Subcase(iii): Let n > 5 is any odd number
The vertex set of Γ(Z2n) is {2, 4, ...., 2n−1, 2(2n−1 − 1)} and |V (Γ(Z2n))| =
2n−1 − 1. Using the above subcases, |S| = 2n−12 (20 + 21 + ................ + 2n−32 ) =
2
n−1
2
∑n−3
2
i=0 2
i and |N(S)| = V (Γ(Z2n)) − |S| = 2n−1 − 1 − 2n−12 ∑
n−3
2
i=0 2
i =
2
n−1
2 (2
n−1
2 − ∑
n−3
2
i=0 2
i) − 1. Then, b(Γ(Z2n)) = |N(S)||S| =
2
n−1
2 (2
n−1
2 −
∑n−3
2
i=0 2
i)−1
2
n−1
2
∑n−3
2
i=0 2
i
.
Theorem 2.6 If p > 4 is any prime, then (Γ(Z4p)) =
3
2(p−1) .
Proof: The proof is by the method of induction on p. Let P, S, N(S) be
the pendant set, minimum degree set, neighbourhood of S, respectivily.
Case(i): Let p = 5.
The vertex set of Γ(Z20) is {2, 4, .., 2(10 − 1), 5, 10, 15} with |V (Γ(Z20))|=
11. Clearly, the vertex v = 2p = 10 is adjacent to all the vertices in V (Γ(Z20))
except 5 and 15, then 10 ∈ N(S). Let x = 4 and y = 24, then 96 is not divisible
by 20 which implies x and y are non adjacent vertices. Then, the pendant set
P = {2, 6, 14, 18} with degree of any vertex in P is 1 and P ⊆ S.
Let U = {4, 8, 12, 16} be the vertex subset of V (Γ(Z20)). Clearly no two
vertices in U are adjacent. That is 20 does not divide 32(= 4 × 8). But, the
vertices in U are adjacent to the vertices 5, 10, and 15 with d(4) = d(8) =
d(12) = d(16) < d(5) = d(15). Clearly, U ⊆ S and the vertices 5, 15 ∈ N(S)
then N(S) = {5, 10, 15}. Clearly, |S| = |P | + |U | = 4 + 4 = 8 . Hence,
b(Γ(Z20)) =
|N(S)|
|S| =
3
8
= 3
2×5−2 =
3
2(p−1) , where p = 5.
Case(ii): Let p = 7
The vertex set of Γ(Z28) is {2, 4, .., 2(14 − 1), 7, 14, 21}.Clearly, the vertex
v = 2p = 14 is adjacent to all the vertices in V (Γ(Z28)) except 7 and 21, then
14 ∈ N(S). Let x = 6 and y = 18 then 108 is not divisible by 28 which implies x
and y are non adjacent vertices. Then, the pendant set P = {2, 6, 10, 18, 22, 26}
with degree of any vertex in P is 1 and P ⊆ S.
234 S. Sankeetha, J. Ravi Sankar, R. Vasanthakumari, S. Meena
Let U = {4, 8, 12, 16, 20, 24} be a vertex subset of V (Γ(Z20)). Clearly no two
vertices in U is adjacent. But, the vertices in U are adjacent to the vertices
7, 14, and 21. Clearly, U ⊆ S and the vertices 7, 21 ∈ N(S) then N(S) =
{7, 14, 21}. Then, |S| = |P | + |U | = 6 + 6 = 12. Hence,
b(Γ(Z42)) =
|N(S)|
|S| =
3
12
= 3
2×7−2 =
3
2(p−1) , where p = 7.
Case(iii): Let p > 7
The vertex set of Γ(Z4p) is {2, 4, ......., 2(2p − 1), p, 2p, 3p} with |V (Γ(Z4p))|
= 2p + 1. Since,the vertex v = 2p is adjacent to all the vertices in V (Γ(Z4p))
except p and 3p, then v = 2p ∈ N(S). Let P be the pendant vertex set and
using above cases, P = {2, 6, ..., 2(p − 2), 2(p + 2), ..., 2(2p − 1)}. Similarly,
Let U = {4, ..., 4(p − 1)}. Since, no two vertices in U are adjacent. But, the
vertices in U are adjacent to the vertices p, 2p and 3p. Clearly, U ⊆ S and the
vertices p, 3p ∈ N(S) then |N(S)| = 3. Hence, |S| = |V (Γ(Z4p))| − |N(S)| =
2p + 1 − 3 = 2(p − 1). Thus,
b(Γ(Z4p)) =
|N(S)|
|S| =
3
2×p−2 =
3
2(p−1) , where p is any prime > 4.
Theorem 2.7 In Γ(Z8p) , b(Γ(Z8p)) =
7
4(p−1) where p is any prime > 8.
Proof: Since, the vertex set of Γ(Z8p) is {2, ..., 2(4p− 1), p, 2p, .., 7p} with
|V (Γ(Z8p))| = 4p + 3. Using theorem (2.6), N(S) = {p, 2p, 3p, .., 7p} and
|N(S)| = 7. Hence, |S| = |V (Γ(Z8p))|− |N(S)| = 4p+3− 7 = 4(p− 1). Then,
b(Γ(Z8p)) =
|N(S)|
|S| =
7
4(p−1) , where p is any prime > 8.
Theorem 2.8 In Γ(Z2np) where p is any prime > 2
n and n is any positive
integer, then b(Γ(Z2np)) =
2n−1
2n−1(p−1) .
Proof: The vertex set of Γ(Z2np) is {2, .., 2(2n−1p − 1), p, 2p, ...., (2n − 1)p}
with |V (Γ(Z2np))| = 2n−1p+2n−1−1. Using theorems (2.6) and (2.7), N(S) =
{p, 2p, ......., (2n − 1)p} then |N(S)| = (2n − 1). Then, |S| = |V (Γ(Z2np))| −
|N(S)| = 2n−1p + 2n−1 − 1 − (2n − 1) = 2n−1(p − 1).
Hence, b(Γ(Z2np)) =
|N(S)|
|S| =
2n−1
2n−1(p−1) .
Theorem 2.9 For any prime p > 3, b(Γ(Z3n)) =
7
3n−1−8 .
Proof: The vertex set of Γ(Z3n) is {3, 6...., 3(3n−1 − 1)} and |V (Γ(Z3n))| =
3n−1 − 1.The proof is by the method of induction.
Case(i): Let n = 4.
The vertex set of Γ(Z81) is {3, 6, ..78} and |V (Γ(Z81))| = 26. Let S be
the vertex subset of V and N(S) be the neibourhood of S such that d(u) <
d(v) where u ∈ S and v ∈ N(S). Let x = 27 ,y = 54 and u = 3 then
ux = uy = 0. This implies that the vertices 27 and 54 are adjacent to all the
remaining vertices of Γ(Z81). Clearly. 27, 54 ∈ N(S). Consider another vertex
set X = {9, 18, 36, 45, 63, 72} which is the next maximum degree compared to
the vertices 27, 54. Let u = 18 and v = 72 then uv is divided by 81 that
is u and v are adjacent. Since, X has a subgraph K6 implies that any five
The binding number of a zero divisor graph 235
vertices ∈ N(S). Thus, N(S) = {9, 18, 27, 36, 45, 54, 63, 72}. Then, |S| =
|V (Γ(Z81))| − |N(S)| = 19. Hence, b(Γ(Z81)) = |N(S)||S| = 719 = 734−1−8 = 73n−1−8 .
Case(ii): Let n = 5.
The vertex set of Γ(Z243) is {3, 6, ..240} and |V (Γ(Z243))| = 80. Using
case(i), the vertex set X = {81, 162}. Since, the vertices in X has highest
degree then X ∈ N(S). The vertex set Y = {27, 54, 108, 135, 189, 216} is the
next maximum degree compared to the vertex set X. Let u = 27 and v = 216 in
Y then uv is divided by 243 that is u and v are adjacent. Using case(i), any five
vertices in Y belongs to N(S). Thus, N(S) = {27, 54, 81, 108, 135, 162, 189}.
Then |S| = |V (Γ(Z243))| − |N(S)| = 80 − 7 = 73.
Hence, b(Γ(Z243)) =
|N(S)|
|S| =
7
73
= 7
35−1−8 =
7
3n−1−8 .
Case(iii): Let n > 5.
In general, V (Γ(Z3n)) is {3, 6, .., 3(3n−1 − 1)} and |V (Γ(Z3n))| = 3n−1 − 1.
Clearly, N(S) = {1.3n−2, 2.3n−2, .., 7.3n−2} then |S| = |V (Γ(Z3n))| − |N(S)| =
3n−1 − 1 − 7 = 3n−1 − 8. Hence, b(Γ(Z243)) = |N(S)||S| = 73n−1−8 .
3 Main Result
The value of the binding number of Γ(Zn) for some positive integer n forms an
inequalities that Γ(Z2p) ≤ Γ(Z4p) ≤ Γ(Z8p) ≤ Γ(Zpq) where p and q are any
distinct primes with p < q and Γ(Z3n) ≤ Γ(Z2n) ≤ Γ(Z2np) ≤ Γ(Zpq) where n
is any positive integer ≥ 2. Using the above two inequalities, we conclude that
the maximum value of the binding number is Γ(Zpq). Since b(Γ(Zpq)) =
p−1
q−1 .
That is the numerator is greater when compared to the other prime number
with respect to the denominator. The last two twin prime numbers of fiftieth
million are p = 982451579 and q = 982451581. The maximum value of the
(Γ(Zn)) is 0.99999999796427626489236243072661 for some positive integer n
upto fiftieth million.
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Received: January, 2013


The Binding Number of a Zero Divisor Graph

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The Binding Number of a Zero Divisor Graph

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