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COMPSCI 230: Discrete Mathematics for Computer Science January 16, 2019
Lecture 3
Lecturer: Debmalya Panigrahi Scribe: Erin Taylor
1 Overview
In this lecture, we will introduce an intuitive principle and illustrate how it can serve as a powerful technique in proving some non-obvious theorems.
2 Well-Ordering Principle
The well-ordering principle (WOP):
There is a smallest number in any set of positive integers.
While this may seem obvious, what if instead we had claimed there is a greatest number in any set of positive integers?
Example 1: Consider the set of even numbers. The smallest even number is 2, but there is no greatest even integer in this set. Our sets may not have finite size. Here are a few more examples:
Example 2: The set of positive odd integers: The smallest element in this set is 1. Example 3: The set of positive integers divisible by 3: The smallest element in this set is 3.
Example 4: The set of positive integers not divisible by 3: The smallest element in this set is 1.
We will consider a few theorems to demonstrate how we may use the WOP in a proof.
Theorem 1. For any rational number q we can express it as a ratio of two integers, x and y such that y 6 = 0 and x, y are coprime.
Recall:
Definition 1. Two integers x and y are coprime if gcd(x, y) = 1 , i.e. they share no common factors other than 1.
Theorem 1 seems apparent: any rational number can be represented as a ratio of two integers. Take one such ratio, and cancel common factors until you have a fraction in lowest terms. However, this intuition does not prove the statement. Let’s prove the statement formally.
Proof of Theorem 1. Assume the theorem is false, that is assume there exist rational numbers that cannot be expressed as the ratio of two integers with gcd 1. Let q be such a rational number. Without
loss of generality, assume q > 0. If q < 0, we can redefine q as −q. Consider all possible ratios q can be expressed as:
q =
x 1 y 1
x 2 y 2
where y 1 6 = 0, y 2 6 = 0,.... Furthermore, y 1 > 0, y 2 > 0,... and x 1 > 0, x 2 > 0,... since q > 0. We define a set of positive integers by taking the numerators of the above fractions:
S = {x 1 , x 2 ,... }
This set may be infinite, but it is a set of positive integers so we can apply the well-ordering principle. The WOP implies there exists a smallest integer in S. Call this integer x∗. Then, q = x
∗ y∗ for some y∗^ > 0. By our original assumption, gcd(x∗, y∗) ≥ 2 so x∗^ and y∗^ must share a factor greater than 1, call this factor p. We have:
q =
x∗ y∗^
x′^ · p y′^ · p
x′ y′
y′^6 = 0 because y∗^6 = 0. Thus, x′^ is a numerator of a ratio of two integers equal to q, so by the definition of S, it must be that x′^ ∈ S. However, x′^ < x∗^ because p > 1. This is a contradiction since we assumed x∗^ was the smallest element of S. Therefore, our original assumption was incorrect and the theorem holds.
Consider another theorem you are familiar with:
Theorem 2. Every positive integer can be written as the product of primes.
Proof. We will proceed with a proof by contradiction. Assume there exists some positive integer that cannot be written as the product of primes. Let S be the set of positive integers that do not have a prime factorization. By the WOP on S, there is a smallest positive integer that cannot be written as the product of primes. Call this integer n∗. Consider the following cases:
- n∗^ is prime. In this case, the prime factorization of n∗^ is n∗^ itself. This is a contradiction, as n∗^ ∈ S so by definition does not have a prime factorization.
- n∗^ is composite (not prime). Then n∗^ = a · b such that a 6 = n∗, a 6 = 1 and b 6 = n∗, b 6 = 1. This implies that a < n∗^ and b < n∗. Since n∗^ is the smallest element of S, a 6 ∈ S and b 6 ∈ S. Therefore, a, b have prime factorizations:
a = p 1 · · · pk
b = q 1 · · · q`
for prime numbers p 1 ,... , pk, q 1 ,... , q`. But,
n∗^ = a · b = p 1 · · · · · pk · q 1 · · · · · q`.
This is a prime factorization of n∗. This is a contradiction of our original assumption.
In all cases, we reached a contradiction. Thus, our original assumption was incorrect and every positive integer can be written as the product of primes.
Counter Example: Define the following set of rational numbers:
S = { 1 + x | x is any rational number s.t. 0 < x < 1 }
S has a lower bound of 1. For the claim to hold, we would need to show that every non-empty subset of S has a minimum element. In particular, this implies that S must have a minimum element. Let this be x∗^ = 1 + pq for some integers p, q where p, q > 0. Now define x′^ = 1 + (^) q+p 1.
0 < (^) q+p 1 < 1 because 0 < pq < 1. Additionally (^) q+p 1 is rational because pq is rational. Thus, x′^ ∈ S.
However, x′^ < x∗, which contradicts our assumption that x∗^ was the smallest element of S. Thus, S is not well-ordered and the claim is false.
An intuitive idea about the well-ordered property is as follows: You start with some number in the set and you consider smaller and smaller elements. Eventually, there will not be a smaller element left to consider. For the set of positive integers you may start at 5 and begin counting down to 1, where you stop. This idea captures the fact that there can be no infinite length decreasing sequence from a fixed starting point in a well-ordered set. Otherwise, there would always exist a smaller element in the set and there would be no minimum. Recall some examples of familiar infinite decreasing sequences:
Example 5: The negative integers: −1, −2, −3,...
Example 6: 1 2
x
x + 1
Although a well-ordered set does not have a decreasing sequence of infinite length, we will shortly see that a well-ordered set can have decreasing sequences of arbitrarily large finite length. Using the definition of well-ordered sets, we can prove the following observation:
Observation 7. A well-ordered set cannot contain a decreasing sequence of infinite length.
Proof. Let S be any well-ordered set, and for contradiction, suppose S contains a sequence A of decreasing numbers with infinite length. Thus, A is of the form (a 1 , a 2 , a 3 ,.. .) where a 1 > a 2 > a 3 > · · ·. Since each term of A is smaller than the preceding term and A has infinite length, A does not contain a minimum element. However, since A is a subset of S and S is well-ordered, the set A does have a minimum element. Since no set cannot simultaneously contain and not contain a minimum element, we conclude that S does not contain a decreasing sequence of infinite length.
However, it is possible for a well-ordered set to contain decreasing sequences that are arbitrarily long. More precisely, there are well well-ordered sets that satisfy the following property: for any positive integer k, there exists a decreasing sequence of length k in the set. (Note that the set of positive integers is such a set: simply take the sequence (k, k − 1,... , 2, 1).) Now let us consider the following subset of rational numbers:
S =
n n + 1
: n is a positive integer
Theorem 8. The set S, defined above, is well-ordered.
To prove Theorem 8, we will first prove a useful lemma that allows us to order the elements of S by simply looking at their numerators.
Lemma 9. If n 1 and n 2 are positive integers such that n 1 < n 2 , then (^) n 1 n+^11 < (^) nn 2 +^21.
Proof. Let n 1 and n 2 be positive integers such that n 1 < n 2. Adding n 1 n 2 to both sides gives us
n 1 n 2 + n 1 < n 1 n 2 + n 2.
Factoring n 1 from the left and n 2 from the right yields
n 1 (n 2 + 1 ) < n 2 (n 1 + 1 ).
Finally, dividing the left by (n 1 + 1 ) and the right by (n 2 + 1 ) yields
n 1 n 1 + 1
n 2 n 2 + 1
as desired.
(Note that starting with the conclusion, cross-multiplying, simplifying, and stating that all steps are reversible is an equally valid proof of Lemma 9). We now use Lemma 9, combined with the fact that the set of positive integers is well-ordered, to prove that S is well-ordered.
Proof of Theorem 8. To prove that S is well-ordered, we need to show that any subset of S contains a minimum element. Thus, let A be a subset of S, and define the set B as follows:
B =
k :
k k + 1
∈ A
In other words, B is the set of numerators that appear in the fractions of A. Since B is a subset of the set of positive integers (which is well-ordered), B contains a minimum element which we denote by b. Since b is the smallest numerator appearing in A, by Lemma 9 we have
b b + 1
a a + 1
where a is any integer larger than b. In particular, a can be any other numerator that appears in A, so this means (^) b+b 1 is the the minimum element of A.
Recall that we denote the set of positive integers by the symbol Z +, and this set is well-ordered. Now consider the the following set:
Z +^ + S =
n + s : n ∈ Z +, s ∈ S
In other words, this set is formed by adding all possible (n, s) pairs, where n is a positive integer and s is an element of S (as defined above). For example, 1 + 12 and 5 + 89 are elements of this set. Both Z +^ and S are well-ordered; we now show that their sum is also well-ordered.
Theorem 10. The set Z +^ + S is well-ordered.
Observation 11. For any positive integers n, k where n ≥ 2 , there exists a decreasing sequence of length k in the set Z +^ + S, such that every element of the sequence is less than n.
For example, if n = 3 and k = 5, we can use the sequence ( 2 +
Since a similar sequence can be exhibited for any value of k (which can be extremely large), we can conclude Z +^ + S contains decreasing sequences that are arbitrarily long. However, recall that Observation 7 states that a well-ordered set cannot contain a decreasing sequence of infinite length (which we showed earlier). These two observations illustrate the difference between “arbitrarily long” and “infinite.”
4 Summary
In this lecture, we considered the well-ordering principle and discussed well-ordered sets. We used the well-ordering principle to formally prove familiar theorems. We also showed that this principle can be used to show that various sets are well-ordered. Finally, we saw a well-ordered set that contains arbitrarily long decreasing sequences.
