please prove theorems 3.1 and 3.2 completely step by step. What makes a and b adjacent and what makes a and b not adjacent? then give examples for p is Odd prime and p = 2 for theorem 3.1 and theorem 3.2 respectively along with explanations and graphic images
Definition 2.11. [7] For a ring R, a simple undirected graph = (V, E) is said to be a graph I') (R) if all the nonzero elements of R as vertices, and two distinct vertices a and b are adjacent if and only if eithera-b=0o0rb-a = 0ora+ bis aunit. Definition 2.12. [7] For a ring R, a simple undirected graph = (V, E) is said to be a graph I'2( R) if all the nonzero elements of R as vertices, and two distinct vertices a and b are adjacent if and only if eithera-b =0 orb - a = 0 or a + b is a zero divisor (including zero). 3 The Prime Graph of PG (Z,,) In this section we investigate the chromatic number of PG (Z,,) for some particular values of n. Theorem 3.1. For the ring Z,, p+1 2 ' =2=0p, ifp=2. xPG\\(Z,) = if p is odd prime. Proof. Let a and b be any two elements of Z, {0}, then a and b are not adjacent if and only if pla+b. In PG, (Z, {0}) there are p 1 elements which contains two sets of non-adjacent ele- ments. So these non-adjacent elements can be colored with same Pz_' colors. Also, the vertex 0 is adjacent to all the vertices of PG, (Z,). So, the vertex 0 can be colored with a single color except the color assigned to the "5' vertices. Thus, the graph PG (Z,) can be properly colored with %' + 1 equals to }' colors. Therefore, xPG(Z,) = }' if pis an odd prime. In case when p = 2, Z, = {0, 1}. As the graph PG (Z,,) we are considering a simple graph. So, the graph PG (Z;) is a star graph and hence y PG (Z,) =2 = p. o Theorem 3.2. For the ring Z,,, xPG\\(Z,) = p(p; ]). if p is odd prime. =3=p+1, ifp=2. Coloring of Prime Graph PG\\ (1) and PGa(R) of a Ring 99 Proof. In PG\\(Z, {0}) there are p 1 elements which are divisible by p and all are adjacent to each other. So, these elements induces a complete subgraph k,_;. All the vertices of this clique can be colored with p 1 colors. The remaining p(p 1) elements which are not divisible by p, are adjacent to p | elements form two complete subgraphs having L'{\"%\" elements in each subgraph. So, the vertices in these two complete subgraphs can be properly colored with M colors except the colors assigned to the vertices of the clique k;, . Also, the vertex 0 is adjacent to all the other vertices of PG (Z,:). So, the vertex 0 can be colored with a single color except the colors assigned to the vertices of the cliques k- and k-1 . Thus, the graph PG\\ (Z,2) can be properly colored with (p 1) + M + 1 colors. Therefore, x PG1(Z,) = Pl if pis an odd prime. In case when p = 2, Zy = {0, 1,2,3}. The elements 0 and 2 are adjacent to every element in PG\\(Zs). So, the vertices 0 and 2 are colored by two different colors. Also, the elements | and 3 are adjacent to elements 0 and 2 but not adjacent to each other. So, the vertices 1 and 3 can be colored with a single color except the colors assigned to the vertices 0 and 2. Thus, the graph PG\\ (Z4) can be properly colored with 3 colors. Hence, x PG (Zs) =3 =p + 1. 0
