Proof by the
Well ordering principle
4/9/2015
Well Ordering Principle
• Every nonempty set of nonnegative
integers has a least element
4/9/2015
Well Ordering Principle
• Every nonempty set of nonnegative
integers has a least element
4/9/2015
Well Ordering Principle
• Every nonempty set of nonnegative
integers has a least element
4/9/2015
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
4/9/2015
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)
4/9/2015
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)
4/9/2015
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
4/9/2015
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
4/9/2015
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
4/9/2015
4/9/2015
Summing powers of 2
• Thm: 1+2+22+23+…+2n =2n+1-1
• E.g. 1+2+22 = 1+2+4 = 7 = 23-1
4/9/2015
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
4/9/2015
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
4/9/2015
Summing powers of 2
• But if
1+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.
4/9/2015
Summing powers of 2
• Where did we use the fact that P(0) is true,
so m > 0?
4/9/2015
A Notational Note
• Learn to avoid ellipses …!
n
P(n)   2  2
i
n1
1
i0
Theorem: (n)P(n)
4/9/2015
A geometric “proof”
n
2
i
2
n1
1 
i0
n
2
n1
 1 2 
i
i0
1
n
1
in
2  n  2
2
i0
4/9/2015
1
1/2
1+½+¼+⅛+…
1+½+¼+⅛
1
1+½+¼
+½
1/2