lec1031 - Well Ordering Principle Every non-empty subset of...

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Well – Ordering Principle Every non-empty subset of Z + (the positive integers) contains a smallest element. Essentially the set Z + is well-ordered. Well, DUH!!! This sounds like a really useless principle. BUT, it can be used to show that mathematical induction is a valid technique. This is going to be an useful proof technique. Here is the basic idea behind mathematical induction: The goal of mathematical induction is to prove an open statement s(n) for all non-negative integers n, or all positive integers n. We can do this in the following manner: 1) Show that s(1), the base case, is true. 2) Show that s(k) s(k+1), for all positive integers k. We must show that proving the two above statements is equivalent to proving the open statement for all non-negative integers n. Now, let’s assume that these two conditions could hold while the statement s(n) is NOT true for all positive integers n. This means that we can create a set F = { t Z + | s(t) is false } that is non-empty. Since this set is a subset of Z + , it must contain a smallest element. Let that smallest element be w. (Thus s(w) is NOT true is our assumption.) We know that w 1 since that is the first requirement for an inductive proof to hold. Thus, we know that w > 1, which means that w-1 Z + .
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Z + , AND that s(w-1) is TRUE because we assumed that w was the SMALLEST value for which s(n) was NOT true. Combining that with the fact that s(k) s(k+1), for all positive integers k, then we can plug in k = w-1 to get: s(w-1) s((w-1)+1) or s(w-1) s(w), implying that s(w) is true. But, this contradicts our assumption that s(w) is false. Thus, we can conclude that proving the 2 given statements
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This document was uploaded on 06/09/2011.

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lec1031 - Well Ordering Principle Every non-empty subset of...

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