⇐ Proofs by Contradiction | Proof by Cases ⇒ |

Claim:There exists a triangle-free graph that is 4-colorable, but not 3-colorable.

Proof:Consider the following graph and its 4-coloring:By checking every triple of vertices, the reader can verify that the graph is triangle-free. We leave the proof that the graph is not 3-colorable as an exercise. QED

Claim:There exist irrational numbersxandysuch thatxis rational.^{y}

Proof:Letz= √2^{ √2 }. Ifzis rational thenzis our desired number withx= √2 andy= √2.Now, suppose that

zis irrational. Then, letx=zandy= √2.In this case,x= (√2^{y}^{ √2 })^{ √2 }= √2^{ ( √2 ⋅ √2 ) }= √2^{2}= 2.xis again rational.^{y}In either case, whether

zis rational or irrational, we've shown the existence of irrational numbersxandysuch thatxis rational.^{y}QED

The proof in Example 2 is definitely non-constructive. The proof establishes the claim but does not tell us if it is √2

⇐ Proofs by Contradiction | Proof by Cases ⇒ |

Last Modified: 12 Feb 2009 12:55:13 EST by Richard Chang to Spring 2009 CMSC 203 Homepage