n6 - CS 70 Spring 2008 Discrete Mathematics for CS David...

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Discrete Mathematics for CS Spring 2008 David Wagner Note 6 Well Ordering Principle How can the induction axiom fail to be true? Recall that the axiom says the following: [ P ( 0 ) ( n . P ( n ) = P ( n + 1 ))] = ⇒ ∀ n . P ( n ) . What would it take for n N . P ( n ) to be false, i.e., for ¬ ( n N . P ( n )) to be true? This would mean that ( n )( P ( n ) is false ) . In other words, at least one of the propositions P ( 0 ) , P ( 1 ) , . . . , P ( n 1 ) , P ( n ) must be false. Let m be the smallest integer for which P ( m ) is false. In other words, P ( m 1 ) is true and P ( m ) is false. But this directly contradicts the fact that P ( m 1 ) = P ( m ) . It may seem as though we just proved the induction axiom. Not quite. Instead, what we have actually done is to show that the induction axiom follows from another axiom, which was used implicitly in defining “the smallest m for which P ( m ) is false.” Well ordering principle: If S N and S n = /0, then S has a smallest element. We assumed something when defining m that is usually taken for granted: that we can actually find a smallest element of the set. This cannot always be accomplished, and to see why consider the set { x R : 0 < x < 1 } . Whatever number is claimed to be the smallest of the set, we can always find a smaller one. This shows that the real real numbers are not well ordered: if we consider a set S R and S n = /0, then S may or may not have a smallest element. In other words, every non-empty set of natural numbers has a smallest element, but not every non-empty set of real numbers has a smallest element. Again, the well ordering principle may seem obvious but it should not be taken for granted. It is only because the natural numbers (and any subset of the natural numbers) are well ordered that we can find a smallest element. Not only does the principle underlie the induction axioms, but it also has direct uses in its own right. Here is a simple example. Round robin tournament: Suppose that, in a round robin tournament, we have a set of n players { p 1 , p 2 ,..., p n } such that p 1 beats p 2 , p 2 beats p 3 , . . . , and p n beats p 1 . This is called a cycle in the tournament: Claim: If there exists a cycle in a tournament, then there exists a cycle of length 3. Proof
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This note was uploaded on 04/09/2008 for the course CS 70 taught by Professor Papadimitrou during the Spring '08 term at University of California, Berkeley.

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n6 - CS 70 Spring 2008 Discrete Mathematics for CS David...

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