Answer: To show that the graph G3 = (V, E ∪ E2) has a chromatic number at most 100, we need to prove that there exists a proper coloring of G3 using at most 100 colors.
Given that x(G) = x(G2) = 10, it means that both G and G2 can be properly colored using 10 colors. Let's consider a proper coloring of G with 10 colors, where each vertex in V is assigned a color from {1, 2, ..., 10}. Similarly, we have a proper coloring of G2 with the same 10 colors.
Now, let's construct a proper coloring for G3 using these 10 colors. We'll start by assigning the same color to each vertex in G and G2, maintaining the proper coloring of each individual graph. Since G and G2 have the same vertex set, every vertex in V has been assigned a color.
Next, we consider the edges in E2. For each edge (u, v) in E2, if the colors assigned to u and v in G2 are different, we select a new color from the remaining 90 colors and assign it to either u or v in G3 (whichever vertex already has a color assigned). Since (u, v) is an edge in E2, it means that u and v are adjacent in G2. Assigning a new color to one of them does not violate the proper coloring of G2, as the new color is distinct from the colors already assigned to their neighbors.
By extending this process for all edges in E2, we can ensure that all vertices in V are properly colored in G3. Since we started with a proper coloring of G, and only introduced new colors for the vertices connected by edges in E2, the resulting coloring is proper for G3.
Since we used at most 10 colors for G and introduced at most 90 new colors for the edges in E2, the total number of colors used in G3 is at most 100. Therefore, the chromatic number of G3 is at most 100.
Explanation: