Prove that if R is a reflexive relation on a nonempty set A, then
R
⊆
R
n
for every positive integer n. //
Remark:
Evidently we
need to do a proof by induction since the n
th
composition power
of a relation is defined recursively.
Proof.
The basis step is true since R
1
= R implies R
⊆
R
1
for
any relation R on a nonempty set A.
To deal with the induction
step, it suffices to show that if the relation R is reflexive,
then (
∀
n
ε
+
)( R
⊆
R
n
→
R
⊆
R
n+1
).
To this end, pretend n is an
arbitrary positive integer, R is reflexive, and R
⊆
R
n
.
Let
(a,b)
ε
R be arbitrary.
Then (a,b)
ε
R and the induction
hypothesis, R
⊆
R
n
, implies (a,b)
ε
R
n
.
R reflexive and a
ε
A
→
(a,a)
ε
R.
The definition of the composition R
n
R=R
n+1
,
(a,a)
ε
R, and (a,b)
ε
R
n
implies (a,b)
ε
R
n+1
.
Thus we have
shown that R
⊆
R
n
→
R
⊆
R
n+1
for an arbitrary n
ε
+
.
So we have
verified  ugh  that (
∀
n
ε
+
)( R
⊆
R
n
→
R
⊆
R
n+1
).
Since we
have verified the hypotheses of the induction axiom, modus ponens
wraps things up: If R is reflexive, (
∀
n
ε
+
)( R
⊆
R
n
).//
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_________________________________________________________________
7. (15 pts.)
(a) How many vertices does a tree with 37 edges
have?
V=
E
+
1
=
3
8
(b)
What is the maximum number of leaves that a binary tree of
height 6 can have?
If l denotes the number of leaves, then l
≤
2
6
= 64.
The maximum occurs when the tree is complete, i.e., all the
leaves are at the same level.
(c)
If a full 3ary tree has 24 internal vertices, how many
leaves does it have?
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 Spring '08
 STAFF
 Graph Theory, Equivalence relation, Transitive relation, Tree traversal

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