Discrrete mathematics for computer science 03well ordering

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 nonneg

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

• Every nonempty set of

nonnegative integers has a least element

Well Ordering Principle

 Every nonempty set of nonnegative integers has a least element

arguing that a fraction can be

reduced to “lowest terms”

integer is nonempty

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)

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)

One way to do it

• Grab under the

biggest pancake

and bring it to the

top

• 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

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

Summing powers of 2

• Thm: 1+2+22+23+…+2n =2n+1-1

• E.g 1+2+22 = 1+2+4 = 7 = 23-1

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

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

Summing powers of 2

• But if

1+2+2 2 +2 3 +…+2m ≠2m+1-1

• then subtracting 2m from both sides:

1+2+2 2 +2 3 +…+2m-1 ≠2m+1-1-2m

= 2 m-1

(since 2m+2m = 2m+1)

• But then P(m-1) is also false, contradiction.

Summing powers of 2

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

A Notational Note

• Learn to avoid ellipses …!

P(n) ≡ 2ι

ι =0

ν

∑ = 2ν+1 −1 Τηεορεµ : (∀ ν ) Π ( ν )

1+½+¼+⅛ 1+½+¼+⅛+…

A geometric “proof”

2i

i=0

ν

ι =0

ν

2 = 1

ι − ν

ι =0

ν

∑

1

1

11+½ 1+½+¼