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MIT6_042JF10_lec03

MIT6_042JF10_lec03 - "mcs-ftl 2010/9/8 0:40 page 43#49 3...

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3 “mcs-ftl” — 2010/9/8 — 0:40 — page 43 — #49 Induction Now that you understand the basics of how to prove that a proposition is true, it is time to equip you with the most powerful methods we have for establishing truth: the Well Ordering Principle, the Induction Rule, and Strong Induction. These methods are especially useful when you need to prove that a predicate is true for all natural numbers. Although the three methods look and feel different, it turns out that they are equivalent in the sense that a proof using any one of the methods can be automat- ically reformatted so that it becomes a proof using any of the other methods. The choice of which method to use is up to you and typically depends on whichever seems to be the easiest or most natural for the problem at hand. 3.1 The Well Ordering Principle Every nonempty set of nonnegative integers has a smallest element. This statement is known as The Well Ordering Principle . Do you believe it? Seems sort of obvious, right? But notice how tight it is: it requires a nonempty set—it’s false for the empty set which has no smallest element because it has no elements at all! And it requires a set of nonnegative integers—it’s false for the set of negative integers and also false for some sets of nonnegative rationals —for example, the set of positive rationals. So, the Well Ordering Principle captures something special about the nonnegative integers. 3.1.1 Well Ordering Proofs While the Well Ordering Principle may seem obvious, it’s hard to see offhand why it is useful. But in fact, it provides one of the most important proof rules in discrete mathematics. In fact, looking back, we took the Well Ordering Principle for granted in proving that p 2 is irrational. That proof assumed that for any positive integers m and n , the fraction m=n can be written in lowest terms , that is, in the form m 0 =n 0 where m 0 and n 0 are positive integers with no common factors. How do we know this is always possible?

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44 “mcs-ftl” — 2010/9/8 — 0:40 — page 44 — #50 Chapter 3 Induction Suppose to the contrary 1 that there were m; n 2 Z C such that the fraction m=n cannot be written in lowest terms. Now let C be the set of positive integers that are numerators of such fractions. Then m 2 C , so C is nonempty. Therefore, by Well Ordering, there must be a smallest integer, m 0 2 C . So by definition of C , there is an integer n 0 > 0 such that m 0 the fraction cannot be written in lowest terms. n 0 This means that m 0 and n 0 must have a common factor, p > 1 . But m 0 =p m 0 ; D n 0 =p n 0 so any way of expressing the left hand fraction in lowest terms would also work for m 0 =n 0 , which implies m 0 =p the fraction cannot be in written in lowest terms either. n 0 =p So by definition of C , the numerator, m 0 =p , is in C . But m 0 =p < m 0 , which contradicts the fact that m 0 is the smallest element of C .
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