for 1/30/10
Mathematics Computer Science
MIT 6.042J/18.062J
The Well Ordering
Principle
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Albert R Meyer
February. 8, 2010
Lec 2M.1
Well Ordering principle
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:
Albert R Meyer
Every nonempty set of
nonnegative integers
has a
least element.
NO!
NO!
February. 8, 2010
Lec 2M.3
Well Ordering Principle Proofs
To prove n N. P(n) using WOP:
define set of counterexamples
C ::= n 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
Albert R Meyer
February. 8, 2010
Lec 2M.2
Well Ordering principle
Every nonempty set of
nonnegative integers
rationals
has a
least element.
Albert R Meyer
February. 8, 2010
Lec 2M.11
Albert R Meyer
February. 8, 2010
Lec 2M.4
Well Ordered Postage
available stamps:
5
3
Thm: Get any amount n 8
Prove by WOP. Suppose not.
Let be m least counterexample:
if m > n 8, can get n.
Albert R Meyer
February. 8, 2010
Lec 2M.12
1
1/30/10
Well Ordered Postage
Well Ordered Postage
So m 11. Now m > m-3 8
so can get m-3. But
m > 8:
m > 9:
m > 10:
February. 8, 2010
Lec 2M.13
Geometric sums
1 + r + r + r ++ r =
3
n
r
n +1
1 + r + r2 + r3 + + rm1 =
1
February. 8, 2010
1 + r + r2 + r3 + + rm1 =
r 1
r 1
r 1
February. 8, 2010
Albert R Meyer
Lec 2M.14
Geometric sums
Proof by WOP. Let m be
smallest n with . But = for
n = 0, so m > 0, and m
Albert R Meyer
3 contradiction!
m-3
Albert R Meyer
2
= m
+
add rm to both sides
rm 1
r 1
LHS = 1 + r + r2 + r3 + + rm1 + rm
m
r m +1 1
rm 1 rm +1 rm
RHS =
+r
=
r 1
r 1
r 1
so = at m, contradicting :
there is no counterexample.
Lec 2M.15
Albert R Meyer
February. 8, 2010
Lec 2M.16
Team Problems
Problems
1 3
Albert R Meyer
February. 8, 2010
Lec 2M.17
2
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6.042J / 18.062J Mathematics for Computer Science
Spring 2010
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