⇐ Direct Proofs | Proofs by Contradiction ⇒ |

Claim:Letnbe an integer. If 5n+ 4 is even, thennis even.

Proof:We will prove this claim indirectly. So, suppose thatnisodd. Then, there must be an integerksuch thatn= 2k+ 1. So, we have5Thus, 5n+ 4 = 5 (2k+ 1) + 4 = 10k+ 9 = 2 (5k+ 4) + 1n+ 4 = 2m+ 1 wherem= 5k+ 4. Therefore, 5n+ 4 is an odd integer.QED

Claim:If a graphGhas an Euler circuit, then every vertex inGhas even degree.

Proof:Suppose thatGhas a vertexvwith odd degree. Letπbe any circuit inG. Ifπdoes not visitv, thenπis not an Euler circuit since it misses all edges of the edges incident onv. (Sincevhas odd degree, there is at least one edge incident onv.) Letkbe the number of times thatvappears on the circuitπ. Each time thatπvisitsv, it must entervusing an edge and leavevusing a different edge. If any of these edges are repeated, thenπis not an Euler circuit. Thus,πuses exactly 2kedges incident onv. Since 2kis an even number andvhas odd degree,πmust miss at least one edge incident onv. Thus,πis not an Euler circuit andGdoes not have any Euler circuits.QED

Using an indirect proof allows us to assume the existence of a
vertex *v* with odd degree. We can then use this "fact" to
conclude that none of the circuits in *G* are Euler circuits.

Note: the proof above assumes that *v* does not have any
self-loops (that is, an edge from *v* to itself). As discussed
in class, self-loops are inconsequential since a graph has an
Euler circuit if and only if the graph with all self-loops removed has an
Euler circuit.

⇐ Direct Proofs | Proofs by Contradiction ⇒ |

Last Modified: 30 Sep 2010 15:33:30 EDT by Richard Chang to Fall 2010 CMSC 203 Homepage