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Proof by the Well ordering principle

Proof by the Well ordering principle. Well Ordering Principle. Every nonempty set of nonnegative integers has a least element. Well Ordering Principle. Every nonempty set of nonnegative integers has a least element. Well Ordering Principle.

quentin-briggs
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Proof by the Well ordering principle

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  1. Proof by theWell ordering principle

  2. Well Ordering Principle Every nonempty set of nonnegative integers has a least element

  3. Well Ordering Principle Every nonempty set of nonnegative integers has a least element

  4. Well Ordering Principle Every nonempty set of nonnegative integers has a least element

  5. Well Ordering Principle • Every nonempty set of nonnegative integers has a least element • We actually used this already when arguing that a fraction can be reduced to “lowest terms” • The set of factors of a positive integer is nonempty

  6. To prove P(n) for every nonnegative n: • Let C = {n: P(n) is false} (the set of “counterexamples”) • Assume C is nonempty in order to derive a contradiction • Let m be the smallest element of C • Derive a contradiction (perhaps by finding a smaller member of C)

  7. A Proof Using WOP Given a stack of pancakes, make a nice stack with the smallest on top, then the next smallest, …, and the biggest on the bottom By using only one operation: Grabbing a wad off the top and flipping it! Theorem: n pancakes can be sorted using 2n-3 flips (n≥2)

  8. One way to do it • Flip the entire stack over • Repeat, ignoring the bottom pancake Grab under the biggest pancake and bring it to the top

  9. Why does this take 2n-3 flips? For n≥2, let P(n) := “n pancakes can be sorted using 2n-3 flips” Suppose this is false for some n Let C = {n: P(n) is false} C has a least element by WOP. Call it m. So m pancakes cannot be sorted using 2m-3 flips and m is the smallest number for which that is the case

  10. Why does this take 2n-3 flips? m≠2 since one flip sorts 2 pancakes But if m>2 then it takes 2 flips to get the biggest pancake on the bottom … and 2(m-1)-3 to sort the rest since P(m-1) is true (since m-1 < m) … for a total of 2(m-1)-3+2 = 2m-3, contradicting the assumption that P(m) is false

  11. Summing powers of 2 • Thm: 1+2+22+23+…+2n=2n+1-1 • E.g. 1+2+22 = 1+2+4 = 7 = 23-1

  12. Summing powers of 2 • Thm: For every n≥0, 1+2+22+23+…+2n=2n+1-1 • E.g. 1+2+22 = 1+2+4 = 7 = 23-1 • Let P(n) be the statement 1+2+22+23+…+2n= 2n+1-1

  13. Summing powers of 2 • Let C = {n: P(n) is false} = {n: 1+2+22+23+…+2n≠2n+1-1}. • Then C is nonempty by hypothesis. • Then C has a minimal element m by WOP. • m cannot be 0 since P(0) is true: 1=20=20+1-1 • So m > 0

  14. Summing powers of 2 • But if 1+2+22+23+…+2m≠2m+1-1 • then subtracting 2mfrom both sides: 1+2+22+23+…+2m-1 ≠2m+1-1-2m = 2m-1 (since 2m+2m = 2m+1) • But then P(m-1) is also false, contradiction.

  15. Summing powers of 2 • Where did we use the fact that P(0) is true, so m > 0?

  16. A Notational Note Learn to avoid ellipses …!

  17. A geometric “proof” 1 1 1/2 1/2 1+½+¼+⅛ 1+½+¼+⅛+… 1 1+½ 1+½+¼

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