1. Introduction
All graphs considered in this paper are finite, simple, and undirected. Let  and  be the vertex set and edge set of a graph G, respectively. The open neighborhood  of a vertex v in G is the set of all vertices that are adjacent to v, the closed neighborhood is the set , and the set of edges incident with v is  The degree of a vertex v is the number of vertices in . The maximum degree among all vertices of G is denoted by . The union of simple graphs G and H is the graph  with vertex set  and edge set . A star of order  is the complete bipartite graph . The center of the star is the vertex of maximum degree.
A leaf of G is a vertex with degree one and a support vertex is a vertex adjacent to a leaf. For a vertex subset , we denote by  the subgraph induced by S. A subdivision of an edge  is obtained by removing the edge , adding a new vertex w, and adding edges  and . Throughout this paper, when an edge  is subdivided, the subdivision vertex for e is denoted by . For a set F of edges in a graph G, we use  to denote the graph obtained from G by subdividing every edge in  Note that  for every two different edges 
A set 
 is a 
paired-dominating setof 
G, PD-set for short, if each vertex in 
 has at least one neighbor in 
S and 
 contains a perfect matching. The minimum cardinality of a PD-set of 
G is called the 
paired-domination number of 
G and is denoted by 
. Let 
S be a PD-set of 
G with a perfect matching 
M. Then, two vertices 
u and 
v are called 
partners (or paired) in
S if the edge 
. Paired domination in graphs was first studied in [
1] and has been studied since then by several authors (for example, see [
2,
3,
4,
5,
6]). The literature on the subject of paired domination has been detailed in the recent book chapter [
7].
As good models of many practical problems, graphs sometimes have to be changed to adapt the changes in reality. Thus, we must pay attention to the change of graph parameters under graph modifications, such as deletion of vertices, deletion or addition of edges, and subdivision of edges. For example, Kok and Mynhardt [
8] introduced the reinforcement number, which is the minimum number of edges which must be added to 
G in order to decrease the domination number of 
G. Fink et al. [
9] introduced the bondage number of a graph, which is the minimum number of edges in which removal increases the domination number. For the subdivision of edges, Velammal [
10], in his thesis, introduced the domination subdivision number which is the minimum number of edges that must be subdivided (where each edge can be subdivided at most once) in order to increase the domination number. The study of this kind of problems has been extended to other domination parameters (see, for instance [
11,
12,
13,
14,
15,
16,
17,
18]).
In this paper, we are interested in studying the paired-domination subdivision number introduced by Favaron et al. in [
19]. In order to increase the paired-domination number of 
G, the minimum number of edges that must be subdivided (where each edge in 
G can be subdivided no more than once) is called the 
paired-domination subdivision number and is denoted by 
. We note that the subdivision of the unique edge of a path of order 2 does not increase the paired-domination number. Thus, we always assume that all graphs involved have a component of order at least 3. The minimum cardinality of a set 
 such that 
 is called an 
-set. The paired-domination subdivision number has been studied by several authors (see, for instance [
20,
21]).
Let 
G be a connected graph of order at least 3. Favaron et al. [
19] posed the following question: Is it true that, for any edge 
? A negative answer to this question was given by Egawa et al. [
22]. However, they approved the question in the affirmative if the following additional condition is added: each edge 
 satisfies 
. We can further specify that, if 
 for some edge 
, then the difference 
 can be arbitrary large. To see this, consider the connected graph 
 obtained from 
 disjoint 
 by adding a new vertex attached to one vertex of each 
 Now, for two leaves 
x and 
y of 
, one can easily see that 
, while 
 and 
.
Let 
 denote the subdivided star obtained from a star 
 of order 
t by subdividing all edges of 
. Let 
 be obtained from 
t copies of 
 by adding a new vertex 
x and joining 
x to the central vertices of subdivided stars, 
 be obtained from 
 copies of 
 by adding a new vertex 
y and joining 
y to the central vertices of subdivided stars and adding a pendant edge 
, and let 
 be the union 
. Note that 
. It is not hard to verify that 
 and 
, where the graph 
 for 
 is illustrated in 
Figure 1. Hence, the difference of 
 can be arbitrary large for some edge 
. Thus, an interesting problem is to find good bounds on 
 in terms of 
 and 
 if 
.
In this paper, we provide an upper bound for 
 for any 
 in terms of 
 and 
, the proof of which will be given in 
Section 3. More precisely, we mainly show the following.
Theorem 1. Let G be an isolated-free graph different from  Then, for every , Furthermore, this bound is sharp.
 We close this section by recalling three useful results.
Proposition 1 ([
19])
. For any connected graph G of order at least three and any graph  formed from G by subdividing an edge , . Proposition 2 ([
22])
. Let G be a graph with no isolated vertex. Then, for every edge , either  or . Proposition 3 ([
22])
. Let G be a connected graph of order at least three, and let  satisfy . Then, .   2. Preliminary Results
In this section, we give some preliminary results useful for the proof of Theorem 1. We begin by extending the result of Proposition 3 to disconnected graphs different from  and having no isolated vertices.
Proposition 4. Let G be an isolated-free graph different from . If  for some edge  then .
 Proof.  Let F be an -set and observe that . We shall show that . Assume that  and let P be a -set. If  or , and they are not partners in P, then, clearly, P is a PD-set of , so . Hence, we assume that . First, let  be two partners in P. Since P is a -set, we may assume that . Let . If y has a neighbor , then the set  (in which x and y are partners with  and  respectively) is a PD-set of ; thus, .
Hence, we can assume that  Then, the set  (in which x and  are partners) is a PD-set of , and the result follows as above. Finally, let . Without loss of generality, assume that . If y has a neighbor in P other than x, then P is a PD-set of  and the result follows as above. Now, if x is the unique neighbor of y in  then, by considering a vertex , one can see that the set  (in which y and  are partners) is a PD-set of ; thus, . In either case,  implying that , which completes the proof.    □
 Lemma 1. Let G be an isolated-free graph different from . If  for some edge  then .
 Proof.  Assume that  and let  and . We denote by  the graph formed from  by subdividing the three edges  and adding three new vertices , respectively. In addition, we denote by  the graph formed from G by subdividing the two edges  and adding two new vertices , respectively, and we denote by  the graph formed from G by subdividing only the edge  and adding a new vertex . Let P be a -set. If , then P is a PD-set of , so . Hence, assume that , and let, without loss of generality, x be the partner of  in P.
Assume first that . If x has a neighbor w in  such that , then  is a PD-set of , and, as before, we have . Thus, we can assume that all neighbors of x in  belong to P. Then, clearly,  is a PD-set of , and, as before, . Assume now that . Then, we have . If x has a neighbor w in  such that , then  is a PD-set of , and, as above, one can easily see that . Finally, if all neighbors of x in  belongs to P, then  is a PD-set of ; thus, . Therefore, .    □
 As an immediate consequence of Proposition 4 and Lemma 1, we, therefore, have the following result.
Corollary 1. Let G be an isolated-free graph different from . If  for some edge  then .
 Lemma 2. For any isolated-free graph G different from , let F be a set of edges of G in which subdivision increases , , and let  be the graph formed from  by subdividing the edges in  If P is a -set such that  or , then  Proof.  According to Proposition 4, we may assume that  (otherwise, the result is straightforward from this proposition). If , then, clearly, P is a PD-set of ; thus, . Hence, assume that . Since, by assumption, , we may assume, without loss of generality, that x and  are partners in P. If all neighbors of x in  belong to P, then  is a PD-set of ; thus, . Now, if x has a neighbor w in , then  is a PD-set of , and, as before, , which completes the proof.    □
 Lemma 3. Let G be an isolated-free graph different from , and let F be an -set. If  such that  and , then  Proof.  If , then by Corollary 1, the assertion is trivial. So, in the following, we may assume that . Since  and , let  and  be the neighbors of x and  respectively, such that . Let  be the graph formed from  by subdividing the edges in . We denote by P a -set. According to Lemma 2, we may assume that  and  (otherwise, the result is straightforward from this lemma). Without loss of generality, assume that x is the partner of .
First, let . If x has a neighbor w in  such that , then, clearly,  is a PD-set of  which is obtained from G by subdividing the edges of  It follows that . Hence, we assume that all neighbors of x in  belong to P. In this case,  is a PD-set of , and, as before, we obtain .
Assume now that . Therefore,  (to paired-dominates ). If x has a neighbor w in  such that , then, clearly,  is a PD-set of  which is obtained from G by subdividing the edges of , and as before one can see that . Hence, we can assume that all neighbors of x in  belong to P. In this case,  is a PD-set of ; thus,  In either case,     □
 Lemma 4. Let G be an isolated-free graph different from . If  such that x or y is a support vertex, then  Proof.  If , then the result follows from Proposition 4. Hence, we assume that . Without loss of generality, let x be a support vertex, and let  be a pendant edge. Suppose that F is an -set. We denote by  the graph formed from G by subdividing the edges in F. In addition, we denote by  the graph formed from G by subdividing the edges in , and we denote by  the graph formed from  by subdividing the edges in . Let P be a -set. By Lemma 2, we may assume that  and  (otherwise, the result is straightforward from this lemma).
First, let x be the partner of  in P. Then, we must have . If , then the set  in which x and  are partners, is a PD-set of  yielding . Hence, assume that  and let w be a neighbor of y in  Then, the set  in which x and y are partners with  and  respectively, is a PD-set of  As before, we get .
Now, assume that y is the partner of  in P. Clearly,  (to paired-dominates ). If y has a neighbor w in  such that , then  is a PD-set of ; thus, . Now, if all neighbors of y in  belong to P, then  is a PD-set of , and, as before, we obtain . In either case,     □
 Before going further, we give some notation and definitions. For a vertex , the set of isolated vertices in the subgraph induced by  is denoted by . We also denote by  the set of pairs  of non-adjacent vertices in G. Moreover, for a pair , let  in other words,  is the set of edges incident with x in which end vertices are neighbors of  In addition, we consider two functions  and  on  as follows.
- (a)
  defined by 
 if neither 
 nor 
 is independent, and
          
          otherwise. Note that 
 if and only if 
x or 
y is a support vertex.
- (b)
  defined by 
 if there exits an sd
-set 
M such that 
 and 
, and
          
          otherwise.
Lemma 5. Let G be an isolated-free graph different from , and let .
- 1. 
 If neither  nor  is independent, then - 2. 
 If F is an -set such that ,  and  is not independent, then - 3. 
 If , then 
 Proof.  According to Proposition 4, we may assume that . Note that since G is isolated-free and different from  We now show items of the lemma one by one.
        
If there is an 
-set 
F such that 
 and 
, then 
, and, by Lemma 3, we have
            
Hence, we may assume that, for every 
-set 
F, 
 or 
. In that case, it is clear that 
. Now, let 
F be an 
-set with, without loss of generality, 
. Since, by assumption, neither 
 nor 
 is independent, let 
 be two adjacent vertices of 
 likewise 
 two adjacent vertices of 
 In addition, consider the graph 
 formed from 
 by subdividing the edges in 
 and all edges in 
 and let 
P be a 
-set. If 
 or 
, then, by Lemma 2, we have 
. Hence, we may assume that 
 and 
. We claim that 
 is the unique subdivision vertex adjacent to 
x belonging to 
P. Suppose, to the contrary, that 
 is a subdivision vertex adjacent to 
x such that 
. If 
x is the partner of 
, then the set 
 in which 
x and 
z are partners, is a PD-set of 
 which is obtained from 
 by subdividing all edges in 
 and the edges of 
 It follows that 
; thus,
            
Hence, we can now assume that 
y is the partner of 
 If all neighbors of 
y in 
 are in 
P, then 
 is a PD-set of 
 which is obtained from 
G by subdividing all edges in 
 and the edges of 
 It follows that 
; thus,
            
If 
y has a neighbor 
w in 
 with 
, then 
 is a PD-set of 
 (defined before) and the desired result follows as before. Thus, 
 is indeed the unique subdivision vertex adjacent to 
x that belongs to 
P. We now claim that 
 is the unique subdivision vertex adjacent to 
y belonging to 
P. Suppose, to the contrary, that 
 is a subdivision vertex adjacent to 
y such that 
. If 
y is the partner of 
, then the set 
 in which 
y and 
z are partners, is a PD-set of 
 which is obtained from 
 by subdividing all edges in 
 and the edges of 
 It follows that 
; thus,
            
Therefore, we may now suppose that 
x is the partner of 
 If all neighbors of 
x in 
 are in 
P, then 
 is a PD-set of 
 (defined before), and the desired result follows. If 
x has a neighbor 
w in 
 with 
, then 
 is a PD-set of 
, and the desired result follows as before. Thus, 
 is indeed the unique subdivision vertex adjacent to 
y that belongs to 
P. Moreover, to paired-dominate vertices 
 and 
, we may assume that 
. In this case, 
 (if 
y is the partner of 
) or 
 (if 
x is the partner of 
) is a PD-set of 
 which is obtained from 
G by subdividing the edges of 
, implying that 
. Therefore,
            
We first note that, since 
, we have 
. In addition, since, by assumption, 
 and 
 is not independent, let 
, and let 
 be two adjacent vertices of 
. We denote by 
 the graph formed from 
 by subdividing the edges of 
 and all edges in 
, and we denote by 
 the graph formed from 
 by further subdividing the edge 
 Now, let 
P be a 
-set, and let 
 be the set of all subdivided edges of 
, except 
, such that their subdivision vertices belong to 
P. We denote by 
 the graph formed from 
G by subdividing the edges of 
. It is easy to check that, if 
 or 
, then, by Lemma 2,
            
Hence, we may assume that 
 and 
. As in the proof of Item 1, we can see that 
 is the unique subdivision vertex adjacent to 
x and 
y that belongs to 
P. To paired-dominate vertex 
, we may assume that 
. Now, if 
x and 
 are partners in 
P and 
 is a subdivision vertex adjacent to 
x, then 
 is a PD-set of 
 which is formed from 
G by subdividing the edges in 
, as well as the edges of 
. It follows that 
. However, if 
y and 
 are partners in 
P, then, clearly, 
; thus, 
 is a PD-set of 
, so 
. In either case,
            
Let 
. If there is an 
-set 
F such that 
 and 
, then 
, and, by Lemma 3, we have
            
Hence, we can assume that, for every 
-set 
F, 
 or 
. Clearly, in this case, 
. Now, let 
F be an 
-set, such that, without loss of generality, 
. We denote by 
 the graph formed from 
 by subdividing the edges of 
. Let 
P be a 
-set, and let 
 be the set of all subdivided edges of 
 except 
 in which subdivision vertices belong to 
 If 
 or 
, then, by Lemma 2, 
. Hence, we assume that 
 and 
. As in the proof of Item 1, we can see that 
 is the unique subdivision vertex adjacent to 
x and 
y that belongs to 
P. Then, clearly, 
 (to paired-dominate either 
 or 
, thus, 
 is a PD-set of the graph 
 which is obtained from 
G by subdividing only the edges of 
 Consequently, 
; hence,
            
The proof is completed.
 □
   3. Proof of Theorem 1
In this section, we prove Theorem 1.
Proof of Theorem 1. We start by noting that since G is isolated-free and different from  Now, let  If  then, by Proposition 4. Hence, we assume that . By Lemma 5-(3), we can assume that  for, otherwise, the result is obviously valid. If , then x or y is support vertex; thus, the result follows by Lemma 4. Now, we consider two cases.
Case 1.. 
Without loss of generality, assume that  where . Clearly, in this case v has degree two. Let z be the neighbor of v different from x. Moreover, let F be an -set such that  is minimized. We can assume without loss of generality that  (otherwise, the result follows from Lemma 3). Let . Consider the graph  and let P be a -set, and let  be a subset of F in which subdivision vertices are in P. According to Lemma 2, we can assume that  and that . As in the proof of item 1 of Lemma 5, one can see that  is the unique subdivision vertex adjacent to x and y that belongs to P.
First, let  and . If , then ; thus, the set  is a PD-set of the graph  which is obtained from G by subdividing the edges of  It follows that ; thus,  Now, if , then  is the partner of v in P; thus, the set  in which v and x are partners, is a PD-set of the graph  defined before, which leads to . Now, let  and . Then, we must have ; thus, the set  is a PD-set of  which is obtained from  by subdividing the edges in , yielding  as above.
Case 2.. 
Assume that there exists some -set F satisfying  and . By Lemma 3, the result follows. Hence, we assume that, for every -set F, either  or . By Lemma 5-(3 and 1), we may assume that  and either  or  is independent. Let z be a vertex in  such that  or . Moreover, let F be an -set. We denote by  the graph formed  by subdividing the edges of . Note that since either  or , the number of subdivided edges is at most . Let P be a -set. Among all edges of G that have been subdivided resulting in the graph , let  be the set of those in which subdivision vertices are in P. If  or , then, clearly, the result follows from Lemma 2. Hence we may assume that  and . In addition, we assume, without loss of generality, that  and .
By the similar method to the proof of Lemma 5-(1), we may assume that no subdivision vertex adjacent to x or y other than  belongs to P. Since , we have . First, let . Then,  and it has as a partner a subdivision vertex, say . In this case, one can easily see that the set  in which y and z are partners, is a PD-set of the graph  which is obtained from  by subdividing all edges in  It follows that . Now, let . Then, P contains a subdivision vertex  that may have as a partner either z or  If  then let , and if , then let  Regardless the situation that occurs,  is a PD-set of the graph  which is obtained from  by subdividing all edges in ; thus,  again. This completes the proof.    □