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300 Mathematical Induction
Find a formula for a n and prove that your formula is correct. (Hint: These
numbers are related to the Fibonacci numbers.)

6.5. Closures Again

In Chapter 4 we promised to give an alternative treatment of transitive clo-
sures of relations using mathematical induction. In this section we fulfill this
promise.
Recall that if R is a relation on a set A, then the transitive closure of R is the
smallest relation S on A such that R ⊆ S and S is transitive. In this section we’ll
find this relation S by starting with R and then adding only those ordered pairs
that must be added if we want to end up with a transitive relation. We begin
with a sketchy description of how we’ll do this, motivated by the examples in
Section 4.5. Then we’ll use recursion and induction to make this sketchy idea
precise and prove that it works.
The examples in Section 4.5 suggested that if a 0 , a 1 ,..., a n is a list of
elements of A such that (a 0 , a 1 ) ∈ R, (a 1 , a 2 ) ∈ R,..., (a n−1 , a n ) ∈ R, then
to create a transitive relation S extending R we must have (a 0 , a n ) ∈ S. Let’s
rephrase this idea in terms of composition of relations. Because (a 0 , a 1 ) ∈ R
and (a 1 , a 2 ) ∈ R, by the definition of composition, (a 0 , a 2 ) ∈ R ◦ R. Similarly,
from (a 0 , a 2 ) ∈ R ◦ R and (a 2 , a 3 ) ∈ R it follows that (a 0 , a 3 ) ∈ R ◦ (R ◦ R). It
3
isnaturaltocallthislastrelation R .NotethatbyTheorem4.2.5,compositionof
relations is associative, so there is no ambiguity if we leave the parentheses out
3
3
3
of the definition of R and write R = R ◦ R ◦ R. Thus, we have (a 0 , a 3 ) ∈ R ,
3
and because (a 3 , a 4 ) ∈ R, it follows that (a 0 , a 4 ) ∈ R ◦ R = R ◦ R ◦ R ◦ R =
4
R . Continuing in this way we’ll eventually reach the conclusion that (a 0 , a n ) ∈
n
R = R ◦ R ◦ ··· ◦ R, where there are nR’s in the last composition. We’ll show
that the ordered pairs that must be added to R to create a transitive relation are
n
the elements of R for every positive integer n.
The use of an ellipsis in the last paragraph suggests that it might be best to
n
define R by recursion. Here’s the precise definition:
1
R = R;
n
for every n ≥ 1, R n+1 = R ◦ R.
Before using this definition to construct the transitive closure of R,weprove a
lemma about it. Of course, the proof will be done by induction!
m
n
Lemma 6.5.1. For all positive integers m and n, R m+n = R ◦ R .
Proof. We let m be an arbitrary positive integer and then proceed by induction
on n.