Proof by theWell ordering principle

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Well Ordering Principle

• Every nonempty set of nonnegative integers has a least element

04/19/23

Well Ordering Principle

• Every nonempty set of nonnegative integers has a least element

04/19/23

Well Ordering Principle

• Every nonempty set of nonnegative integers has a least element

04/19/23

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

04/19/23

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)

04/19/23

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)

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One way to do it• Grab under the

biggest pancake and bring it to the top

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• Flip the entire stack over

• Repeat, ignoring the bottom pancake

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

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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

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Summing powers of 2

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

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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

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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

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Summing powers of 2

• But if1+2+22+23+…+2m ≠2m+1-1

• then subtracting 2m from 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.

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Summing powers of 2

• Where did we use the fact that P(0) is true, so m > 0?

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A Notational Note

• Learn to avoid ellipses …!

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P(n)≡ 2i

i=0

n

∑ =2n+1 −1

Theorem: (∀n)P(n)

1+½+¼+⅛1+½+¼+⅛+…

A geometric “proof”

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2i

i=0

n

∑ =2n+1 −1≡

2n+1 =1+ 2i

i=0

n

∑ ≡

2 =12n + 2i−n

i=0

n

∑1

1

1/2 1/2

11+½1+½+¼