Total Mean Cordiality ofk c n + 2K 2

Transcription

1 Palestine Journal of Mathematics Vol () (15), 1 8 Palestine Polytechnic University-PPU 15 Total Mean Cordiality ofk c n + K R Ponraj and S Sathish Narayanan Communicated by Ayman Badawi MSC 1 Classifications: 5C78 Keywords and phrases: Path, Cycle, Wheel, Complete graph, Complete bipartite graph Abstract A Total Mean Cordial labeling of a graph G = (V,E) is a mapping f : V(G) {, 1, } such that f(xy) = where x,y V(G), xy G, and the total number of f(x)+f(y), 1 and are balanced That is ev f (i) ev f (j) 1, i,j {, 1, } where ev f (x) denotes the total number of vertices and edges labeled with x (x =, 1, ) If there exists a total mean cordial labeling on a graph G, we will call G is Total Mean Cordial In this paper, it is shown that K c n + K is Total Mean Cordial iff n = 1 or or or 6 or 8 1 Introduction By a graph we mean a finite unoriented graph without loops and multiple edges A general reference for graph theoretic ideas can be seen in [] A vertex labeling of a graph G is an assignmentf of labels to the vertices ofgthat induces for each uv E(G) a label depending on the vertex labels f(u) and f(v) The vertex and edge set of a graph G are denoted by V(G) and E(G) so that the order and size of G are respectively V(G) and E(G) Let G 1 and G be two graphs with vertex setsv 1 and V and edge sets E 1 and E respectively Then their join G 1 +G is the graph whose vertex set is V 1 V and edge set is E 1 E {uv : u V 1 and v V } The notion of Total Mean cordial labeling was introduced and studied by Ponraj, Ramasamy and Sathish Narayanan [] Let f be a function from V(G) {, 1, } For each edge uv, assign the label f(u)+f(v) f is called a Total Mean Cordial labeling if ev f (i) ev f (j) 1 where ev f (x) denote the total number of vertices and edges labeled with x (x =, 1, ) A graph with a Total Mean Cordial labeling is called Total Mean Cordial graph In this paper, we investigate the Total Mean Cordial labeling behaviour of K c n + K Let x be any real number Then the symbol x stands for the largest integer less than or equal to x and x stands for the smallest integer greater than or equal to x Main result Theorem 1 K c n + K is Total Mean Cordial if and only if n = 1 or or or 6 or 8 Proof Let V(K c n + K ) = {u,v,x,y,u i : 1 i n} and E(K c n + K ) = {uv,xy,} {uu i,vu i,xu i,yu i : 1 i n} It is clear that V(G) + E(G) = 5n+6 Let l denotes the number of zeros to be used in u i (1 i n) and that for the label, we use r Suppose f is a Total Mean Cordial labeling of K c n + K Case 1 n (mod ) Let n = t Then V(G) + E(G) = 15t + 6 Here ev f () = ev f (1) = ev f () = 5t + Consider the set S = {u,v,x,y} and the label Here there are five possible cases * All the four vertices of S are labeled by * Any three of them are labeled by * Any two of them are labeled with [This may be adjacent vertices or two non adjacent vertices] * Only one vertex is labeled by * None of them received the label Now we discuss all the cases given above Subcase 1 f(u)=f(v) = f(x) = f(y)=

2 R Ponraj and S Sathish Narayanan Here, ev f () t, a contradiction Subcase f(u), f(v), f(x), f(y) In this case ev f () t, a contradiction Subcase Any three of them are labeled with zero With out loss of generality assume that f(u) = f(v) = f(x) = and f(y) So the vertices in S contributes only zeros We should utilize the remaining 5t zeros for both vertices and edges In this case, if u i is labeled with, apart from this label, each vertex u i contributes zeros So we havel+l = 5t Thereforel = 5t This is possible only whent (mod ) Now consider the label Suppose f(y) = then r + r + 1 = 5t+ This implies r = 5t+1, a contradiction since t (mod ) Suppose f(y) = 1 Then r = 5t+ But l + r > t, a contradiction Subcase Any two vertices from S are labeled by zero First we assume that any two adjacent vertices in u, v, x, y are labeled with zero Without loss of generality we assume that f(u) = f(v) = and f(x), f(y) At present we have used zeros In this case each u i contributes two edges with label zero So,l+ l+ = 5t+ That is l = 5t 1 Such a positive integer l exists only if t (mod ) Suppose t (mod ) Consider the label If f(x) = f(y) = Then r = 5t 1 It is clear that l + r t This is true only when t Since t (mod ), t 1 If t =, the following figure 1 shows that K6 c + K is Total Mean Cordial Figure 1 Suppose f(x) =, f(y) = 1 In this case r + r + = 5t + That is r = 5t Since t (mod ), such a postive integerr does not exists Iff(x) = f(y) = 1 thenr+ r = 5t+ Hence r = 5t+ Then l+r > t, a contradiction Now consider the case, zero is labeled with any two non adjcent vertices from the set S Without loss of generality assume that f(u) = and f(y)= Here, l+l+ = 5t+ Thereforel = 5t This is true only if t (mod ) If f(v) = f(y) = then in this case r+r + = 5t+ This implies r = 5t Here l +r > t, a contradiction Suppose f(v) = 1, f(y) = Here r + r + 1 = 5t + Then r = 5t+1 Since t (mod ), r can not be a positive integer Assume that f(v) = f(y) = 1 In this case r+ r = 5t+ This implies r = 5t+ Since t (mod ), r is not a positive integer Subcase 5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u) = In this case each vertex u i contributes one edge with label zero Hence l+l+1 = 5t+ That is l = 5t+1 Since l is an positive integer, t 1 (mod ) Now assume f(x) = f(y) = f(v) = Then r + r + = 5t+ Therefore r = 5t Since t 1 (mod ), r can not be a positive integer Suppose f(x) = f(y) = f(v) = 1 In this case, r+r = 5t+ Hence r = 5t+ Since t 1 (mod ), r can not be a positive integer If f(x) = f(y)= 1 and f(v) = Then r+ r+ 1 = 5t+ That isr = 5t+1 Again a contradiction sincel+r > t Forf(x) = 1 andf(y) = f(v) =, we haver+ r+ = 5t+ Hence r = 5t, a contradiction since t 1 (mod ) If f(v) = 1 and f(x) = f(y) = then r + r + = 5t + That is r = 5t 1 It follows that l + r > t, a contradiction Suppose f(v) = f(x) = 1 and f(y)= In this case r+r+ = 5t+ Hence r = 5t a contradiction since t 1 (mod ) Case n 1(mod )

3 Total Mean Cordiality of K c n + K Let n = t+1 Then V(G) + E(G) = 15t+11 Here we have three possibilities a ev f () = ev f () = 5t+, ev f (1) = 5t+ or b ev f () = ev f (1) = 5t+, ev f () = 5t+ or c ev f (1) = ev f () = 5t+, ev f () = 5t+ Suppose ev f () = ev f () = 5t+, ev f (1) = 5t+ Subcase a1 f(u)=f(v) = f(x) = f(y)= Here, ev f () t+1, a contradiction Subcase a f(u), f(v), f(x), f(y) In this case ev f () t+1, a contradiction Subcase a Any three vertices of S are labeled with zero With out loss of generality assume thatf(u) = f(v) = f(x) = and f(y) Here, if a vertex u i is labeled with zero then it contributes three edges with label zero So we have l + l + = 5t+ Thereforel = 5t Such a positive integerlexisits only whent (mod ) Now suppose f(y) = A vertex u i with label contributes one edge with label So r + r + 1 = 5t + That is r = 5t+ Since t (mod ), r can not be a positive integer If f(y) = 1 then in this case also a vertex u i with label contributes one edge with label Therefore r + r = 5t+ That is r = 5t+ But l+r > t+1, a contradiction Subcase a Any two vertices of S are labeled with zero First we assume that f(u) = f(v) = and f(x), f(y) In this case, if a vertex u i is labeled by zero then it gives two edges with label zero Therefore l+l+ = 5t+ That is l = 5t+1 Since l is a positive integer, t 1 (mod ) Consider the vertices x and y Suppose these two vertices are labeled with 1 If a vertex u i is labeled by then each u i contributes two edges with labele It follows that r + r = 5t + Hence r = 5t+ But l + r > t + 1, a contradiction Suppose the vertices x and y are labeled by Here a vertex u i with label contributes two edges with label Then r + r + = 5t + Therefore, r = 5t+1 We know that l + r t + 1 This is true only when t = 1 The Total Mean Cordial labeling of K c + K is given in figure Suppose f(x) = 1 and f(y) = Then each vertex u i with a Figure label contributes two edges with label This impliesr+ r+ = 5t+ and hence r = 5t+ Since t 1 (mod ), such a positive integer r does not exists Suppose f(u) = f(x) = Here l + l + = 5t + Therefore l = 5t+ It follows that t (mod ) Now assume f(y) = f(v) = Thenr+r+ = 5t+ That isr = 5t+ Butl+r > t+1, a contradiction If f(y) = f(v) = 1 then r+ r = 5t+ Hence r = 5t+, a contradiction since t (mod ) Forf(y)=1 andf(v) =, we haver+ r+ 1 = 5t+ That isr = 5t+, again a contradiction since t (mod ) Subcase a5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u) = In this case l + l + 1 = 5t + and hence l = 5t+ Since l is a positive integer, t 1 (mod ) If f(v) = f(x) = f(y) = 1 then r+r = 5t+ This impliesr = 5t+ For the values oft,r could not be an integer Iff(v) = 1 and f(x) = f(y) = then r + r + = 5t + Therefore r = 5t+1 Here, l + r > t + 1, a contradiction If f(v) = and f(x) = f(y) = 1 then r + r + 1 = 5t + This implies r = 5t+ Here also l+r > t+1, a contradiction Suppose f(v) = f(x) = 1 and f(y) = Here r + r + = 5t + and hence r = 5t+ This is impossible since t 1 (mod ) For f(v) = f(x) = f(y)=, we have r+r+ = 5t+ Thereforer = 5t But t 1(mod ), a contradiction

4 R Ponraj and S Sathish Narayanan Consider the case ev f () = ev f (1) = 5t+ and ev f () = 5t+ Subcase b1 f(u)=f(v) = f(x) = f(y)= Here, ev f () t+1, a contradiction Subcase b f(u), f(v), f(x), f(y) In this case ev f () t+1, a contradiction Subcase b Any three vertices of S are labeled with zero With out loss of generality assume that f(u) = f(v) = f(x) = and f(y) In this case l+l+ = 5t+ That isl = 5t Sincel is a positive integer,t (mod ) Supposef(y)= then r + r + 1 = 5t+ Therefore r = 5t+ But l + r t+1 This is true only if t = A Total Mean Cordial labeling of K1 c + K is given in figure For f(y) = 1, r+r = 5t+ Figure Therefore r = 5t+ This is a contradiction to t (mod ) Subcase b Any two vertices of S are labeled with zero First we assume that f(u) = f(v) = and f(x), f(y) Then l+l+ = 5t+ This implies l = 5t+1 It follows that t 1(mod ) Now assume that f(x) = f(y) = In this case r + r + = 5t+ Then r = 5t, a contradiction to t 1 (mod ) If f(x) = f(y) = then r+ r = 5t+ and hence r = 5t+ This is impossible since t 1 (mod ) For f(x) = 1 and f(y) =, we haver+r+ = 5t+ Thenr = 5t+1 Butl+r t+1 This is true only when t 1 Since t 1 (mod ), t For t = 1, the Total Mean Cordial labeling of K c + K is given in figure Suppose f(u) = f(x) = and f(v), f(y) Then l+l+ = 5t+ 1 Figure and therefore l = 5t+ It follows that t (mod ) Consider the vertices v and y Suppose these two vertices are labeled by then r + r + = 5t + So we have r = 5t+1 Since t (mod ), such a positive integer does not exists Suppose v and y are labeled by 1 In this case r+r = 5t+ and hence r = 5t+ This is impossible since t (mod ) If f(v) = 1 and f(y)=then r+r+1 = 5t+ That is r = 5t+ But l+r > t+1 a contradiction Subcase b5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u)= In this case l+l+ 1 = 5t+ Thus l = 5t+ This is true only ift 1(mod ) Supposef(v) = f(x) = f(y)= Herer+ r+ = 5t+ Then r = 5t 1 But l + r > t + 1, a contradiction If f(v) = f(x) = f(y) = 1 then r + r = 5t + and hence r = 5t+ In this case l + r > t + 1, a contradiction Suppose f(v) = and f(x) = f(y) = 1 Here r+r+1 = 5t+ Thus r = 5t+ This is impossible since t 1 (mod ) For f(v) = f(y) = and f(x) = 1, we have r + r + = 5t + This implies r = 5t, a contradiction to t 1 (mod ) If f(v) = 1 and f(x) = f(y) = then r+r+ = 5t+ Hencer = 5t Again a contradiction tot 1(mod ) Forf(v) = f(x) = 1 and f(y)=, r+ r+ = 5t+ Therefore r = 5t+1 But l+r > t+1, a contradiction Suppose ev f (1) = ev f () = 5t+ and ev f () = 5t+

5 Total Mean Cordiality of K c n + K 5 Subcase c1 f(u) = f(v) = f(x) = f(y)= Here, ev f () t+1, a contradiction Subcase c f(u), f(v), f(x), f(y) In this case ev f () t+1, a contradiction Subcase c Any three vertices of S are labeled with zero With out loss of generality assume that f(u) = f(v) = f(x) = and f(y) Here l + l + = 5t + Thus l = 5t 1 This implies t 1 (mod ) Now assume f(y) = Then r+r+ 1 = 5t+ Hence r = 5t+ But l+r > t+1, a contradiction Forf(y)=, we have r+r = 5t+ This implies 5t+ This is a contradiction to t 1 (mod ) Subcase c Any two vertices of S are labeled with zero Assume f(u) = f(v) = and f(x), f(y) In this case l+l+ = 5t+ Therefore l = 5t Such a positive integer l exists only if t is a multiple of Suppose f(x) = f(y) = thenr+ r+ = 5t+ Thusr = 5t+1 This shows thattis not a multiple of, a contradiction If f(x) = f(y) = 1 then r + r = 5t + Therefore r = 5t+ Since t (mod ), such a positive integer r does not exists Suppose f(x) = 1 and f(y) = then r + r + = 5t+ Therefore r = 5t+, a contradiction since t (mod ) Consider the case f(u) = f(x) = and f(v) f(y) Here l+ l+ = 5t+ Thus l = 5t+1 It follows that t 1 (mod ) Suppose f(v) = f(y) = then r + r + = 5t+ Hence r = 5t+ Since t 1 (mod ), r could not be a positive integer If f(v) = f(y) = 1 then r + r = 5t+ Therefore r = 5t+ But l +r > t+1, a contradiction For f(v) = 1 and f(y) =, we have r+r + 1 = 5t+ and hence r = 5t+, a contradiction since t 1 (mod ) Subcase c5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u)= Here, l+l+1 = 5t+ and hence l = 5t+ It follows that t (mod ) Now assume f(v) = f(x) = f(y) = Then r+ r+ = 5t+ This implies r = 5t But l+r t+1 This is true only when t = Kc 1 + K with a Total Mean Cordial labeling is given in figure 5 Suppose f(v) = f(x) = f(y) = 1 In this case Figure 5 r + r = 5t + Therefore r = 5t+ But l + r > t + 1, a contradiction If f(v) = and f(x) = f(y)=1then r+ r+ 1 = 5t+ That is r = 5t+ Such a positive integer r does not exists since t (mod ) Assume f(v) = f(y)= and f(x) = 1 Here r+r+ = 5t+ Then r = 5t+1 This is a contradiction to t (mod ) If f(x) = f(y) = and f(v) = 1 then r + r+ = 5t+ and hence r = 5t+1 Here also a contradiction arises since t (mod ) Further if f(v) = f(x) = 1 and f(y) = then r + r + = 5t+ Therefore r = 5t+ But l+r > t+1, a contradiction Case n (mod ) Let n = t + Then V(G) + E(G) = 15t + 16 In this case we have three possibilities a ev f () = ev f (1) = 5t+5, ev f () = 5t+6 or b ev f () = ev f () = 5t+5, ev f (1) = 5t+6 or c ev f (1) = ev f () = 5t+5, ev f () = 5t+6 Suppose ev f () = ev f (1) = 5t+5, ev f () = 5t+6 Subcase a1 f(u)=f(v) = f(x) = f(y)= Here, ev f () t+, a contradiction Subcase a f(u), f(v), f(x), f(y) In this case ev f () t+, a contradiction Subcase a Any three vertices of S are labeled with zero With out loss of generality assume thatf(u)= f(v)= f(x) = andf(y) Thenl+l+ = 5t + 5 That is l = 5t+1 It follows that t (mod ) Suppose f(y) = In this case r+r+1 = 5t+6 This implies r = 5t+5 Here, l+r > t+, a contradiction For f(y) = 1, we have r+r = 5t+6 and hence r = 5t+6 This is impossible since t (mod )

6 6 R Ponraj and S Sathish Narayanan Subcase a Any two vertices of S are labeled with zero Assume f(u) = f(v) = and f(x), f(y) Here, l + l + = 5t + 5 This implies l = 5t+ It follows that t (mod ) Now consider the vertices x and y Suppose these two vertices are labeled by then r + r + = 5t + 6 Hence r = 5t+, a contradiction to the nature of t If f(x) = f(y) = 1 then r + r = 5t + 6 Therefore r = 5t+6 Here also a contradiction arises to the values of t Now we consider the case that f(x) = 1 and f(y) = In this case r + r + = 5t+6 That is r = 5t+, a contradiction to t (mod ) Now we consider the case f(u) = f(x) = and f(v), f(y) In this case l + l + = 5t+5 and hence l = 5t+ This shows that t should be a multiple of If f(v) = f(y) = Then r+r+ = 5t+6 Hencer = 5t+, a contradiction to the values oft Assumef(v) = f(y) = 1 Here r+r = 5t+6 and then r = 5t+6 But l+r > t+, a contradiction Supposef(v) = 1, f(y) = then r+r + 1 = 5t+6 That is r = 5t+5 Such a positive integer r does not exists since t (mod ) Subcase a5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u) = Then l + l + 1 = 5t + 5 and therefore l = 5t+ This is possible only when t is a multiple of If f(v) = f(x) = f(y) = then r + r + = 5t + 6 Therefore r = 5t+ But l + r > t +, a contradiction Suppose f(v) = f(x) = f(y) = 1 then r + r = 5t+6 and hence r = 5t+6 Here also l+r > t+, a contradiction For the case f(v) = and f(x) = f(y) = 1, we have r + r + 1 = 5t + 6 and therefore r = 5t+5 This is impossible since t (mod ) Assume f(v) = f(y) = and f(x) = 1 Herer+ r+ = 5t+6 Thereforer = 5t+ Here also a contradiction to the nature oft Iff(v) = 1 andf(x) = f(y)= thenr+r+ = 5t+6 Hencer = 5t+, a contradiction to t (mod ) Assume f(v) = f(x) = 1 and f(y) = Here r+r + = 5t+6 That is r = 5t+ But l+r > t+, a contradiction Assume ev f () = ev f () = 5t+5, ev f (1) = 5t+6 Subcase b1 f(u)=f(v) = f(x) = f(y)= Here, ev f () t+, a contradiction Subcase b f(u), f(v), f(x), f(y) In this case ev f () t+, a contradiction Subcase b Any three vertices of S are labeled with zero With out loss of generality assume thatf(u)= f(v)= f(x) = andf(y) Thenl+l+ = 5t + 5 This implies l = 5t+1 This is true only if t (mod ) Suppose f(y) = then r+r+1 = 5t+5and therefore r = 5t+, a contradiction to the values of t For f(y) = 1, we have, r+r = 5t+5 and hence r = 5t+5 But l+r > t+, a contradiction Subcase b Any two vertices of S are labeled with zero Assume f(u) = f(v) = and f(x), f(y) Then l + l + = 5t + 5 and therefore l = 5t+ It follows that t (mod ) Now consider the vertices x and y Suppose both of this two vertices are Simultaneously labeled by Here r + r + = 5t + 5 Therefore r = 5t+ Clearly the value of l+r should be less than or equal to t+ This should be true only if t But we discussed earlier that t is a multiple of and t is a positive integer Hence t and t 1 For t =, we display a Total Mean Cordial labeling of K c 8 + K in figure 6 Assume f(x) = f(y) = 1 Then r + r = 5t + 5 and therefore r = 5t+5 But l + r > t +, a contradiction Suppose f(u) = f(x) = and f(v), f(y), then l+ l+ = 5 5t+ t + 5 This implies l = It follows that t is a multiple of Iff(v) = f(v) = thenr+r+ = 5t+5 and hnecer = 5t+ Nowl+r should not exceed t+ This is possible only if t This implies t = A Total Mean Cordial labeling of K c+ K is given in figure 7 For f(v) = f(y) = 1 we have r + r = 5t+5 Hence r = 5t+5 This is a contradiction to the values of t Assume f(v) = 1 and f(y) = In this case r+ r+1 = 5t+5 and therefore r = 5t+ This is also a contradiction to the nature of t Subcase b5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u) = Here l + l + 1 = 5t + 5 That is l = 5t+ This implies t is a multiple of Now assume f(v) = f(x) = f(y) = Then r + r + = 5t + 5 and hence r = 5t+1 Since t is a multiple of, r is not an integer, a contradiction If f(v) = f(x) = f(y) = 1 then r + r = 5t + 5 This implies r = 5t+5 For the same reason as discussed above, we have a contradiction For f(v) = and f(x) = f(y) = 1, we have r + r + 1 = 5t + 5 Therefore 5t+ Now l + r > t +, a contradiction If f(x) = 1 and f(v) = f(y) = then r + r + = 5t + 5 That is r = 5t+ Here also the value of l + r exceeds t +, a contradiction Consider the case when f(v) = 1 and f(x) = f(y) = In this case r+r + = 5t+5 Therefore r = 5t+ Again l +r > t+, a contradiction For f(v) = f(x) = 1 and f(y) = we have r+ r+ = 5t+5 Hence r = 5t+ Such a positive

7 Total Mean Cordiality of K c n + K 7 Figure 6 Figure 7 integer r does not exists since t is a multiple of Consider the case ev f () = 5t+6 and ev f () = ev f (1) = 5t+5 Subcase c1 f(u) = f(v) = f(x) = f(y)= Here, ev f () t+, a contradiction Subcase c f(u), f(v), f(x), f(y) In this case ev f () t+, a contradiction Subcase c Any three vertices of S are labeled with zero With out loss of generality assume thatf(u)=f(v) = f(x) = andf(y) Herel+l+ = 5t + 6 That is l = 5t+ It follows that t (mod ) Now consider the vertex y Suppose f(y) = then r+r+ 1 = 5t+5 and hence r = 5t+ But l+r > t+, a contradiction For f(v) = 1 we have r+r = 5t+5 therefore r = 5t+5, a contradiction to the values of t Subcase c Any two vertices of S are labeled with zero Assume f(u) = f(v) = and f(x), f(y) In this case l + l + = 5t + 6 This implies l = 5t+ It follows that t is a multiple of Assume f(x) = f(y) = Then r+r+ = 5t+5 That isr = 5t+ Such a positive integerrdoes not exists sincet (mod ) If f(x) = f(y) = 1 then r + r = 5t+5 Hence r = 5t+5 Suppose f(x) = 1, f(y) = In this case r + r + = 5t+5 and hence r = 5t+ Now l +r t+ This is possible only when t = A Total Mean Cordial labeling of K c + K is given in figure 8 Consider the case 1 Figure 8

8 8 R Ponraj and S Sathish Narayanan f(u) = f(x) = andf(v),f(y) In this casel+l+ = 5t+6 This impliesl = 5t+ It follows that t 1 (mod ) Next consider the vertices v and y If possible f(v) = f(y) = then r+r+ = 5t+5 and hence r = 5t+ This is impossible since t 1 (mod ) Suppose f(v) = f(y) = 1 Here r+r = 5t+5 Therefore r = 5t+5, a contradiction to the values of t For f(v) = 1 and f(y) = we have r+r+1 = 5t+5 Then r = 5t+ But l+r > t+, a contradiction Subcase c5 Only one vertex from the set S is labeled by zero Without loss of generality assume that f(u) = Then l + l + 1 = 5t + 6 Hence l = 5t+5 This implies t 1 (mod ) Suppose f(v) = f(x) = f(y) = then r + r + = 5t + 5 and hence r = 5t+1 But l + r > t +, a contradiction If f(v) = f(x) = f(y) = 1 then r+ r = 5t+5 Thereforer = 5t+5 Here alsol+r > t+, a contradiction Forf(v) = and f(x) = f(y)=1 we haver+ r+ 1 = 5t+5 Then r = 5t+, a contradiction tot 1(mod ) Assume f(v) = f(y) = and f(x) = 1 In this case r + r + = 5t + 5 Hence r = 5t+ This is impossible since t 1 (mod ) Consider the case f(v) = 1 and f(x) = f(y) = Here r + r + = 5t+ Then r = 5t+ Here also a contradiction to the values of t When f(v) = f(x) = 1 and f(y) =, we have r + r + = 5t + 5 and hence r = 5t+ But l+r > t+, a contradiction References [1] J A Gallian, A Dynamic survey of graph labeling, The Electronic Journal of Combinatorics, 16 (1) # Ds6 [] F Harary, Graph theory, Addision wesley, New Delhi (1969) [] R Ponraj, A M S Ramasamy and S Sathish Narayanan, Total Mean Cordial labeling of graphs, International JMath Combin, (1) Author information R Ponraj and S Sathish Narayanan, Department of Mathematics, Sri Paramakalyani College, Alwarkurichi- 67 1, India ÔÓÒÖ Ñ Ø Ñ ÐºÓÑ Ø ÖÚ Ñ ÐºÓÑ Ê Ú ÔØ ÂÙÒ ½ ½¼½ º ÇØÓ Ö ¾ ¾¼½ º