If k vertices has kC2 edges, show that (k+1) vertices has (k+1) C 2 edges
To show that (k+1) vertices have (k+1) C 2 edges, we need to prove that the number of edges in a complete graph with (k+1) vertices is equal to (k+1) C 2.
First, let's consider a complete graph with k vertices. We are given that this graph has kC2 edges. The C symbol represents the combination formula, which calculates the number of ways to choose 2 items from a set of k items. In other words, kC2 = k*(k-1)/2.
Now, let's add a new vertex to the existing graph, resulting in a complete graph with (k+1) vertices. When we add this new vertex, it needs to be connected to every other vertex in the graph to maintain completeness.
Since the new vertex needs to be connected to each of the k existing vertices, it will have k edges. Additionally, the k existing vertices will have one additional edge each, connecting them to the new vertex. Therefore, we have k+1 edges from the new vertex and k additional edges from the existing vertices, resulting in a total of k+(k+1) edges.
Simplifying this expression, we find that the total number of edges in the complete graph with (k+1) vertices is 2k + 1.
Now, let's calculate the value of (k+1) C 2. Using the combination formula, (k+1) C 2 = (k+1)*(k+1-1)/2 = (k+1)*k/2.
To show that (k+1) vertices have (k+1) C 2 edges, we need to prove that 2k + 1 = (k+1)*k/2.
Let's simplify the right-hand side of the equation:
(k+1)*k/2 = (k^2 + k)/2
Multiplying both sides by 2:
(k+1)*k = 2k^2 + 2k
Expanding the product on the right-hand side:
k^2 + k = 2k^2 + 2k
Rearranging the terms:
k^2 + k - 2k^2 - 2k = 0
Combining like terms:
-k^2 - k = 0
Factoring out -k:
-k(k + 1) = 0
Simplifying the equation:
k(k + 1) = 0
This equation holds true for k = 0 or k = -1. However, since we are dealing with the number of vertices and edges, the value of k cannot be negative or zero. So, we exclude these values.
Hence, we have shown that (k+1) vertices have (k+1) C 2 edges, given that k vertices have kC2 edges.