Mathematics for Computer Science
MIT 6.042J/18.062J
The Well Ordering
Principle
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.1
Well Ordering principle
Every nonempty set of
nonnegative integers
has a
least element.
Familiar? Now you mention it, Yes.
Obvious? Yes.
Trivial?
Yes. But watch out:
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.2
Well Ordering principle
Every nonempty set of
nonnegative integers
rationals
has a
least element.
NO!
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.3
Well Ordering principle
Every nonempty set of
nonnegative integers
has a
least element.
NO!
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.4
2
proof used Well Ordering
m
Proof: …suppose 2 
n
…can always find such m, n>0
without common factors…
why always ?
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.5
Proof using Well Ordering
WOP: find smallest m s.t.
m If m, n had a
2 .
n
common factor, c, then
m / c

2=
n / c 
Feb. 25, 2008
and m/c < m
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.6
Proof using Well Ordering
WOP: find smallest m s.t.
m
2 .
n
This contradiction implies
m, n have no common factors.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.7
Prime Products
Thm: Every integer > 1 is a
product of primes.
Proof: (by contradiction) Suppose
{nonproducts} is nonempty. By WOP,
there is a least m > 1 that is a
nonproduct. This m is not prime
(else is a product of 1 prime)
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.8
Prime Products
Thm: Every integer > 1 is a
product of primes.
…So m = j·k for integers j,k
where m > j,k > 1. Now j,k < m
so both are prime products:
j = p1·p2··· p94 k = q1·q2···q213
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.9
Prime Products
Thm: Every integer > 1 is a
product of primes.
…now
m = j·k = p1·p2···p94·q1·q2···q213
is prime product, contradiction.
So {counterexamples} =
Feb. 25, 2008
. QED
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.10
Well Ordering Principle Proofs
To prove n
. P(n) using WOP:
• define set of counterexamples
C ::= {n
| not P(n)}
• assume C is not empty.
By WOP, have minimum element m C.
• Reach a contradiction somehow …
usually by finding c C with c < m.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.11
Well Ordered Postage
available stamps:
5¢
Thm: Get any amount
3¢
8¢
Prove by WOP. Suppose not.
Let m be least counterexample:
if m > n
Feb. 25, 2008
8, can get n¢.
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.12
Well Ordered Postage
m
8:
m
9:
m
10:
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.13
Well Ordered Postage
So m 11. Now m m-3 8
so can get m-3¢. But
+
m-3¢
Feb. 25, 2008
= m¢
3¢ contradiction!
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.14
Well-founded
Partial Orders
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.15
Well Ordering Principle
WOP equivalent to:
has no
infinite decreasing sequence:
< nk <
Feb. 25, 2008
< n 2 < n1 < n0
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.16
Well-founded Partial Orders
generalize: a partial order, <,
is well founded iff it has no
infinite <-decreasing sequence
< ak <
Feb. 25, 2008
< a2 < a1 < a0
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.18
Well-founded Partial Orders
well founded partial order may
not have a minimum element:
on nonempty finite sets.
size 1 sets are minimal,
no minimum.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.19
minimal not minimum
{0,1,2} {0,1,8}
{0,1}
{1,8} {2,8}
{0} {1} {2}
Feb. 25, 2008
{8}
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.20
Well-founded Partial Orders
lemma: partial order, <, is
well-founded iff every
nonempty subset has a
minimal element.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.21
Well-founded Partial Orders
proof by contradiction:
Say
S µ domain(<)
has no minimal element.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.22
Well-founded Partial Orders
s1 S not minimal, so have s2 S,
s2 < s 1
s2 not minimal, so have s3 S,
< s 3 < s2 < s1
Continue this way
to get <-decreasing sequence!
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.23
Lexicographic order on
2
Define:
(a1,b1) lex (a2,b2) ::=
a1 a2 or (a1=a2 and b1 b2)
Fact: lex is a well-founded
total order on 2.
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.24
Ackermann’s Function
2
A: 

A(m,n) = 2n if m=0 or n 1,
A(m,n) = A(m-1,A(m,n-1))
if m > 0 and n > 1.
problem:
A(m,n) calls A(m-1,big)
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.28
Ackermann’s Function
A: 

A(m,n) = 2n if m=0 or n 1,
A(m,n) = A(m-1,A(m,n-1))
if m > 0 and n > 1.
do the recursive calls
terminate?
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.29
Ackermann’s Function
A:  
A(m,n) = 2n if m=0 or n 1,
A(m,n) = A(m-1,A(m,n-1))
if m > 0 and n > 1.
YES: because
(m,n) lex(m-1,big)
so no inf. seq. of recursive calls
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.30
Team Problems
Problems
13
Feb. 25, 2008
Copyright © Albert R. Meyer, 2008. All rights reserved.
lec 4M.31