3-regular (cubic) graphs with a bridge

[math8]

How can you prove that a cubic graph with a bridge cannot be 3-edge colored?

I guess one could try a proof by contradiction, so we assume a 3 edge coloring is possible for such a graph. But then I am not sure in which direction to continue. I have tried to draw such graphs, and clearly, they can't be 3 edge-colored. But a more formal proof would be helpful.
Or maybe a proof by contrapositive, so let say we have a cubic graph with a 3-edge coloring. How do we show this graph is bridgeless?

 

[mmwave]

I would approach this problem by first defining the terms and concepts involved. A cubic graph is a graph where every vertex has exactly three edges connected to it. A bridge is an edge in a graph that, if removed, would disconnect the graph into two or more components. Edge coloring is the assignment of colors to the edges of a graph such that no two adjacent edges have the same color.

Next, I would consider the properties of a cubic graph with a bridge. Since a bridge is an edge that, when removed, disconnects the graph, it follows that a cubic graph with a bridge must have at least two components. This means that there must be at least two vertices of degree one in the graph. However, in a 3-edge coloring, each vertex must have exactly three edges of different colors connected to it. Therefore, it is impossible for a vertex of degree one to have three different colored edges connected to it, as it would require at least four different colors.

This leads us to the conclusion that a cubic graph with a bridge cannot be 3-edge colored. If it were possible, it would require at least four colors, contradicting the definition of a 3-edge coloring. This is a proof by contradiction.

Alternatively, we could use a proof by contrapositive. We start with the assumption that a cubic graph can be 3-edge colored. This means that each vertex has exactly three edges of different colors connected to it. But if a graph has a bridge, it must have at least two components, and as we have established, at least one of these components must have a vertex of degree one. This vertex cannot have three different colored edges connected to it, as it would require at least four colors. Therefore, a cubic graph with a 3-edge coloring must be bridgeless.

In conclusion, both proof techniques lead us to the same result: a cubic graph with a bridge cannot be 3-edge colored. This is because a bridge would require at least two components, and at least one of these components would have a vertex of degree one, which is incompatible with a 3-edge coloring.