§20 Smallest Counterexample
Tom Lewis
Fall Term 2010
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
1 / 14
Outline
1
The method of smallest counterexample
2
The well-ordering principle
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
2 / 14
The method of smallest counterexample
A joke with a purpose
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
3 / 14
The method of smallest counterexample
A joke with a purpose
Theorem
All of the natural numbers are interesting.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
3 / 14
The method of smallest counterexample
A step-stone
We proved this theorem in the previous section.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
4 / 14
The method of smallest counterexample
A step-stone
We proved this theorem in the previous section.
Theorem
An integer cannot be both even and odd.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
4 / 14
The method of smallest counterexample
Another stone
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
5 / 14
The method of smallest counterexample
Another stone
Theorem
Every natural number is either even or odd.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
5 / 14
The method of smallest counterexample
Our destination
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
6 / 14
The method of smallest counterexample
Our destination
Theorem
Every integer is either even or odd.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
6 / 14
The method of smallest counterexample
The method of the smallest counterexample
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
We proceed by contradiction. The negation of the theorem’s
statement is: ∃x ≥ 0 such that Px is false.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
We proceed by contradiction. The negation of the theorem’s
statement is: ∃x ≥ 0 such that Px is false.
Consider the smallest i ≥ 0 such that Pi is false. Call it x. We will
call Px the smallest counterexample.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
We proceed by contradiction. The negation of the theorem’s
statement is: ∃x ≥ 0 such that Px is false.
Consider the smallest i ≥ 0 such that Pi is false. Call it x. We will
call Px the smallest counterexample.
Show that P0 is true. This establishes the base case or basis step. We
may conclude that x ≥ 1. This is important!
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
We proceed by contradiction. The negation of the theorem’s
statement is: ∃x ≥ 0 such that Px is false.
Consider the smallest i ≥ 0 such that Pi is false. Call it x. We will
call Px the smallest counterexample.
Show that P0 is true. This establishes the base case or basis step. We
may conclude that x ≥ 1. This is important!
We now know that “Px is false” and “Px−1 is true”. Show that this
leads to a contradiction.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
The method of the smallest counterexample
You wish to prove a theorem of the form: ∀i ≥ 0, Pi is true, where
each Pi is a statement.
We proceed by contradiction. The negation of the theorem’s
statement is: ∃x ≥ 0 such that Px is false.
Consider the smallest i ≥ 0 such that Pi is false. Call it x. We will
call Px the smallest counterexample.
Show that P0 is true. This establishes the base case or basis step. We
may conclude that x ≥ 1. This is important!
We now know that “Px is false” and “Px−1 is true”. Show that this
leads to a contradiction.
The last step is often the difficult step. It may require Herculean
ingenuity.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
7 / 14
The method of smallest counterexample
Theorem
Let n be a positive integer. The sum of the first n odd natural numbers is
n2 . In other words
1 + 3 + 5 + · · · (2k − 1) = k 2
Tom Lewis ()
§20 Smallest Counterexample
for all k ≥ 1.
Fall Term 2010
8 / 14
The well-ordering principle
The well-ordering principle
Every non-empty set of natural numbers contains a least element.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
9 / 14
The well-ordering principle
The well-ordering principle
Every non-empty set of natural numbers contains a least element.
Some things to note
You cannot alter any of the elements of this theorem without endangering
the result. Why do neither of the following sets have a least element?
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
9 / 14
The well-ordering principle
The well-ordering principle
Every non-empty set of natural numbers contains a least element.
Some things to note
You cannot alter any of the elements of this theorem without endangering
the result. Why do neither of the following sets have a least element?
Let A = {z ∈ N : z is even and odd}.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
9 / 14
The well-ordering principle
The well-ordering principle
Every non-empty set of natural numbers contains a least element.
Some things to note
You cannot alter any of the elements of this theorem without endangering
the result. Why do neither of the following sets have a least element?
Let A = {z ∈ N : z is even and odd}.
Let B = {r ∈ Q : r > 0}.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
9 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
We suppose that this is not true, for the sake of a contradiction.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
We suppose that this is not true, for the sake of a contradiction.
Let X = {n ∈ N : Pn is false}. Our assumption is that X 6= ∅.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
We suppose that this is not true, for the sake of a contradiction.
Let X = {n ∈ N : Pn is false}. Our assumption is that X 6= ∅.
Show that 0 ∈
/ X . This is the basis step.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
We suppose that this is not true, for the sake of a contradiction.
Let X = {n ∈ N : Pn is false}. Our assumption is that X 6= ∅.
Show that 0 ∈
/ X . This is the basis step.
By WOP, let x be the least element of X .
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Proofs using the Well-Ordering Principle
Let Pn be a statement about the nth natural number.
You want to prove the theorem: ∀n ∈ N, Pn is true.
We suppose that this is not true, for the sake of a contradiction.
Let X = {n ∈ N : Pn is false}. Our assumption is that X 6= ∅.
Show that 0 ∈
/ X . This is the basis step.
By WOP, let x be the least element of X .
Show that “Px is false” and “Px−1 is true” leads to a contradiction.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
10 / 14
The well-ordering principle
Problem
Use the Well-Ordering Principle to show that
∀n ≥ N,
Tom Lewis ()
1 + 21 + 22 + · · · + 2n = 2n+1 − 1
§20 Smallest Counterexample
Fall Term 2010
11 / 14
The well-ordering principle
Problem
Use the Well-Ordering Principle to show that
∀n ≥ 5,
Tom Lewis ()
2n > n 2 .
§20 Smallest Counterexample
Fall Term 2010
12 / 14
The well-ordering principle
Definition (The Fibonacci numbers)
The Fibonacci numbers are the list of integers F0 , F1 , F2 , . . . defined
recursively according to the rule: F0 = 1, F1 = 1 and, for each n ≥ 2,
Fn = Fn−1 + Fn−2 .
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
13 / 14
The well-ordering principle
Definition (The Fibonacci numbers)
The Fibonacci numbers are the list of integers F0 , F1 , F2 , . . . defined
recursively according to the rule: F0 = 1, F1 = 1 and, for each n ≥ 2,
Fn = Fn−1 + Fn−2 .
Problem
Find F7 .
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
13 / 14
The well-ordering principle
A stronger form of proof by WOP
So far we have been using the following fact: if x is the least element of
X , then
Px is false and Px−1 is true.
In fact, we can assert something much stronger. If x is the least element
of X , then
Px is false and P0 , P1 , P2 , · · · , Px−1 are true.
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
14 / 14
The well-ordering principle
A stronger form of proof by WOP
So far we have been using the following fact: if x is the least element of
X , then
Px is false and Px−1 is true.
In fact, we can assert something much stronger. If x is the least element
of X , then
Px is false and P0 , P1 , P2 , · · · , Px−1 are true.
Theorem
For n ∈ N, Fn ≤ (1.7)n .
Tom Lewis ()
§20 Smallest Counterexample
Fall Term 2010
14 / 14