An IBL Introduction to Discrete Mathematics

Proof.

Let \(C_n\) be a cycle with \(n\) vertices. Since \(C_n\) is a cycle, it's vertices can be labeled \(v_0, v_1, \ldots, v_{n-1}\) so that the only edges in \(C_n\) are of the form \(v_iv_{i+1}\text{,}\) with addition performed modulo \(n\text{.}\) Furthermore, every such edge exists in \(C_n\text{.}\) Since there are \(n\) such edges, the number of edges is equal to the number of vertices.

Proof.

Direct Proof Definition 1.3.17 Means No edge is incident with two nodes in \(K_n\text{.}\) and also No edge is incident with two nodes in \(K_m\text{.}\) Definition 1.3.3 implies a node is incident to zero or two edges in a cycle. Because no edge is incident with two nodes in \(K_n\)/\(K_m\text{,}\) Each edge must be incident with some node in \(K_n\) and some node in \(K_m\text{.}\) For any path In \(K_{n,m}\text{:}\) if the ith edge starts in kn and ends in km, the i+1th edge starts in km and ends in kn if the ith edge starts in km and ends in kn, the i+1th edge starts in kn and ends in km It follows: For a path starting in n: the nth edge where n is even, ends in n. and the nth edge where n is odd, ends in m. A cycle contains a path starting and ending at the same node. which means it must start and stop in the same n/m. From above for a path to start and stop in the same n/m it must be have an even number of edges. Therefor: A cycle contained in \(K_{n,m}\) must be even. Which Proves: \(K_{n,m}\) contains no odd cycles.

Proof.

The definition of a \(K_n\) graph is that all vertices are adjacent to every other vertice. The degree of all vertices in the graph is n, if an edge were removed, a vertex would still be adjacent to n-1 vertices in the graph. To make the graph disconnected, n-n edges would have to be removed from a single vertex.

Proof.

We will show that any two vertices of\(K_n\) have a path between them. Let \(u\) and \(v\) be vertices of \(K_n\text{.}\) Then, by the definition of \(K_n\text{,}\) \(u\) and \(v\) are adjacent, so \(uv\) is a path between them. Since \(u\) and \(v\) are arbitrary, this suffices to show that \(K_n\) is connected.

Proof.

Let \(K_{n,m}\) be a complete bipartite. And let X,Y be vertices in \(K_{n,m}\text{.}\)

I will prove that \(K_{n,m}\) is connected for all n,m. This means there exists a path between X,Y.

There are two cases:

Case 1: X,Y are in different sets.

Given the definition of a \(K_{n,m}\) graph, there exisits a path XY.

Case 2: X,Y are in the same set.

Let Z be a vertex that is in a different set than X,Y. From case 1, Z is both adjacent to X and Y. Thus the path XZY exists and X,Y are connected.

Proof.

Suppose G is a disconnected graph that has been formed by removing some number of edges in \(K_{n,m}\) where \(n \leq m\text{.}\) To construct G, select \(x\) elements from \(n\) and \(y\) elements from \(m\text{.}\)

Beginning with \(K_{n,m}\) (in this case \(n=3\) and \(m=4\) we select two vertices to be in \(x\) (green) and two vertices to be in \(y\) (red).

Delete every connection from the elements in \(x\) to the elements of \(m\) that are not part of \(y\)

Delete the connections from the vertices of \(x\) to the vertices in \(m\) that are not in \(y\text{.}\)

Now delete every connection from the elements in \(y\) to the elements of \(n\) that are not part of \(x\)

Delete the connections from the vertices of \(y\) to the vertices in \(n\) that are not in \(x\text{.}\)

This results in a graph that is disconnected.

The graph is not disconnected.

Since \(m-y\) edges were deleted from each element of \(x\) and \(n-x\) edges were deleted from each element of \(y\text{,}\) an equation can be constructed for the total number of edges deleted from the graph.

Our desire is to find the smallest number of edge deletions needed to disconnect \(K_{n,m}\text{.}\) To find this, we begin by searching the equation for an absolute minimum by taking \(\frac{\partial r}{\partial x}\) and \(\frac{\partial r}{\partial y}\) and attempting to find some values of \(x\) and \(y\) that can simultaneously set both of these to zero.

It is immediately apparent that any combination of \(x=0, x=n, y=0, y=m\) will set both the expressions to zero. However, these combinations are not valid inputs, as it would not result in a disconnected graph. Since this is the case, the minimum value of \(r\) must be achieved by some pair of boundary values for x and y.

Suppose \(x=1\) and \(y=0\text{:}\) then \(r=1 \cdot (m-0)+0 \cdot (n-1)=m\)

Suppose \(x=0\) and \(y=1\text{:}\) then \(r=0 \cdot (m-1)+1 \cdot (n-0)=n\)

Suppose \(x=n\) and \(y=0\text{:}\) then \(r=n \cdot (m-0)+0 \cdot (n-n)=nm\)

Suppose \(x=0\) and \(y=m\text{:}\) then \(r=0 \cdot (m-m)+m \cdot (n-0)=mn\)

As we traverse along any boundary between these corner values, we can see that the output increases monotonically as we hold \(x\) or \(y\) constant at a value of one and increase the other. Therefore, since \(n \leq m\text{,}\) the minimum number of edge deletions is \(n\) and \(K_{n,m}\) is r-connected with a value of \(n\)