A full 3ary tree with 24 internal vertices must have
V
=(
3
)(
2
4
)+1=7
3 vertices.
Since 24 are internal, there
must bel=7
32
4=4
9 leaves.
_________________________________________________________________
8. (10 pts.)
Suppose that A is the set consisting of all real
valued functions with domain consisting of the interval [1,1] .
Let R be the relation on the set A defined as follows:
R = { (f,g)
(
∃
C)( C
ε
and f(0)  g(0) = C) }
Prove that R is an equivalence relation on the set A.
Proof.
We must establish that R is reflexive, symmetric, and
transitive to show that it is an equivalence relation.
To see that R is reflexive, suppose that f is an element of
the set A.
Then f:[1,1]
→
is a function. Since we have that
f(0)  f(0) = 0 and 0
ε
, (f,f)
ε
R.
Since f was arbitrary, it
follows that (
∀
f)( f
ε
A
→
(f,f)
ε
R ).
Thus R is reflexive.
To show that R is symmetric, pretend that f and g are
members of A with (f,g)
ε
R.
It follows from the definition of
the relation that there is some real number C for which
f(0)  g(0) = C.
Now C
ε
implies C
ε
.
Since we have
g(0)  f(0) = C, it follows that (g,f)
ε
R.
Consequently, we
have shown that R is symmetric.
Finally, to verify that R is transitive, let f, g, and h be
elements of A with (f,g)
ε
R and (g,h)
ε
R.
It follows from the
definition of the relation R that there are real numbers, say C
1
and C
2
, such that f(0)  g(0) = C
1
and g(0)  h(0) = C
2
.
Since C
1
+C
2
ε
when C
1
and C
2
are in
, and we have that
f(0)  h(0) = f(0)  g(0) + g(0)  h(0) = C
1
2
, (f,h)
ε
R.
Consequently, R is transitive.//
Question:
How many equivalence classes does this relation have?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 STAFF
 Graph Theory, Equivalence relation, Transitive relation, Tree traversal

Click to edit the document details