CS173 Discrete Mathematical Structures
Spring 2007
Homework #8 Solution
Due Sun 03/12/07, 8AM.
(35 points)
1) Give a recursive definition to each of the followings: (13 points, 2+2+3+3+3)
a)
2
3 2
n
a n n
= + +
,
1, 2, 3,
n
=
L
2 2
1 1
6, ( 1) 3( 1) 2 3 2 2 4 2 4
n n
a a n n n n n a n
+
= = + + + + = + + + + = + +
b)
2
log ( !)
n
a n
=
,
1, 2, 3,
n
=
L
1 2 1 2 2
log 1! 0, log (( 1)!) log ( 1)
n n
a a n a n
+
= = = + = + +
c)
2 ( 2)
n n
n
a
= + −
,
1, 2, 3,
n
=
L
1
1 2 1
1
, is even
0, 8,
4 , is odd
n
n
n
a n
a a a
a n
−
+
−
= = =
d) The set of positive integers congruent to 3 modulo 5.
3 , and if , then 5
S x S x S
∈ ∈ + ∈
e)
{( , ) | , , and is odd}
S a b a Z b Z a b
+ +
= ∈ ∈ +
(1,2) ,(2,1) , and if ( , ) , then ( 1, 1) , ( 2, ) , and ( , 2)
S S a b S a b S a b S a b S
∈ ∈ ∈ + + ∈ + ∈ + ∈
2) Suppose sequence
{ }
n
a
is defined as
0 1 2 1 2 3
1, 2, 3,
k k k k
a a a a a a a
− − −
= = = = + +
, for
3
k
≥
.
a) We want to find a bound for
each term of the sequence. (10 points, 6+2+2)
1.
Prove by
strong induction
that
2
n
n
a
≤
for all integers
0
n
≥
.
Base case:
0
0
1 2 1
a
= ≤ =
. Verified.
Inductive hypothesis: Assume this bound holds for all
k
a
, where1
k n
≤ ≤
.
Inductive step: (Prove
1
1
2
n
n
a
+
+
≤is true)
According to the definition of the series,
1 1 2
n n n n
a a a a
+ − −
= + + .
According to IH,
1 2
1 1 2
2 2 2
n n n
n n n n
a a a a
− −
+ − −
= + + ≤ + +
1 2 1 2 2 1
1
2 2 2 1 2 2 2 2 2
n n n n n n
n
a
− − − −
+
≤ + + ≤ + + + + + +
L
(Notice that this is a geometric
series)
1 2 2 1 1 1
1
1 2 2 2 2 2 2 1 2
n n n n n
n
a
− − + +
+
≤ + + + + + + = − ≤
L
Q.E.D.
6 points, 1 for the base case, 2 for the IH, 3 for the inductive step.
2. What is the difference between a strong induction and a normal mathematical induction?
The difference is in the inductive hypothesis. In mathematical induction, we only need to
assume
( )
P k
is true, and then show the implication
( ) ( 1)
P k P k
→ +
is true for every positive
integer k. In strong induction, however, the hypothesis is
(1) (2) ( )
P P P k
∧ ∧ ∧
L
is true, and
then show that
(1) (2) ( ) ( 1)
P P P k P k
∧ ∧ ∧ → +
L
is true for every positive integer k.
Apparently, the latter has a stronger hypothesis.
