CSci/Math2112 Assignment 6 Due July 3, 2015
(20) 1.
For the following, if the statement is true, prove it. If the statement is false, disprove it.
(a) (BoP 9 #4) For every natural number n , the integer n 2 + 17 n + 17 is prime.
Solution: This statement is false.
Disprove.
Consider n = 1, a natural number. Then n 2
Since 5 and 7 are both integers, this shows that n 2 + 17
+ 17 n n + 17 = 1 + 17 + 17 = 35 = 5 · 7.
+ 17 is a composite number, i.e. is not prime.
(b) (BoP 9 #8) If A , B , and C are sets, then A − ( B ∪ C ) = ( A − B ) ∪ ( A − C ).
Solution: This statement is false.
Disprove.
Consider the sets A = { 0 , 1 } , B = { 0 , 2 } , and C = { 1 , 2 } . Then
A − ( B ∪ C ) = { 0 , 1 } − { 0 , 1 , 2 } = ∅ while
( A − B ) ∪ ( A − C ) = { 1 } ∪ { 0 } = { 0 , 1 } .
Since { 0 , 1 } 6 = ∅ , we have A − ( B ∪ C ) = ( A − B ) ∪ ( A − C ).
(c) (BoP 9 #16) If A and B are finite sets, then | A ∪ B | = | A | + | B | .
Solution: This statement is false.
Disprove.
Consider the sets A = { 0 , 1 } and B = { 0 , 2 } . Then
| A ∪ B | = |{ 0 , 1 , 2 }| = 3 and
| A | + | B | = 2 + 2 = 4 .
Thus | A ∪ B | 6 = | A | + | B | .
(d) (BoP 9 #18) If a, b, c ∈
N
, then at least one of a − b , a + c , and b − c is even.
Solution: This statement is true.
Proof.
Let a , b , and c be random natural numbers. By the Pigeon-hole Principle (buckets are the parities, items are a, b, c ), at least two of these have to have the same parity. Note that if two natural numbers x and y have the same parity, then x + y and x − y is even, while if they have opposite parity, then their sum and difference is odd (if this is not immediately clear, you should check this).
Thus if a and b have same parity, then a − b is even. If a and c have same parity then a + c is even. And if b and c have same parity, then b − c is even.
Thus in all cases at least one of a − b , a + c , and b − c is even.
Alternatively, the statement can also be proven in the following way:
Proof.
Let a, b, c ∈
N and assume that all of a − b , a + c , and b − c are odd. Then since the difference of two odd integers is even, we have that ( a + c ) − ( a − b ) = b + c is even. Since the sum of two integers with opposite parity is odd, we have that ( b + c ) + ( b − c ) = 2 b is odd. Since b is a natural number, and thus an integer, we also have by definition that 2 b is even, a contradiction.
Thus the assumption that all of a − b , a + c , and b − c are odd is false, i.e. at least one of a − b , a + c , and b − c is even.

2. Consider a row of n squares. You want to fill this row with round tokens (taking up one square) and dominoes (taking up two adjacent squares). Let a n be the number of ways to do this. The goal of this project is to find (and prove) an explicit formula for a n
. As an example, a
2 strip with either two tokens or a domino (see picture below).
= 2 since we can fill the
(5)
(10)
(6)
(a) Find a n for n = 1 , 2 , 3 , 4 , 5 , 6. Put your values into a table with columns n and a n
.
Solution: n a n
1 1
2 2
3 3
4 5
5 8
6 13
(b) What is the recursive pattern for a n
? Prove that your recursion is true.
Solution: The recursive pattern is a n
= a n − 1
+ a n − 2 with starting values a
1
= 1 and a
2
= 2.
Proof.
We have that a
1 the round token.
= 1 since a strip of length 1, i.e. a single square, can only be tiled with
We have that a
2 round tokens.
= 2 since a strip of length 2 can be tiled with either a single domino, or with two
Now consider a strip of length n ≥ 3 and look at the right-most piece. This piece could either be a round token or a domino.
In the case that this piece is a round token, it will take up one square, thus there remains a strip of length n − 1 to fill. There are a n − 1 ways to do this.
In the case that the last piece is a domino, it will take up two squares, thus there remains a strip of length n − 2 to fill. There are a n − 2 ways to do this.
Thus in total there are dominoes.
a n − 1
+ a n − 2 ways to tile a strip of length n with round tokens and
(c) Enter the sequence of numbers you got in part (a) into The On-Line Encyclopedia of Integer
Sequences (oeis.org) and look at the first entry (which should be sequence A000045).
i. What is the sequence a n called?
Solution: The sequence is called the Fibonacci Numbers.
ii. Look at some of the comments. What is something interesting you have learned about this sequence?
iii. Check out the section called “Formula”. Find the explicit formula for a n and state it.
Solution: If F
0
, F
1
, F
2
, F
3
, . . .
is the Fibonacci sequence, then according to the OEIS we have
F n
=
1+
√
5
2 n
−
√
5
1 −
√
5
2 n
.
Note though that the Fibonacci sequence has F
1
F n − 1
+ F n − 2
. Thus the sequence a n
= 1 and F
2
= 1 with the recursion F n
= is the Fibonacci sequence shifted by one index, i.e.
a n
= F n +1
. Thus a n
=
1+
√
5
2 n +1
−
√
5
1 −
√
5
2 n +1
.

(15) (d) Prove the explicit formula for a n you have found in the OEIS using induction.
Solution:
Proof (by induction).
Let f ( n ) =
1+
√
5
2 to show by induction that a n
= f ( n ).
Base cases: For n = 1 we have a
1
= 1 while n +1
−
√
5
1 −
2
√
5 n +1 for all natural numbers n . We need f (1) =
1+
√
5
2
2
−
√
5
1 −
√
5
2
2
=
(6 + 2
√
5) − (6 − 2
4 5
√
5)
=
4
4
√
5
√
5
= 1 .
Thus a
1
= f (1).
For n = 2 we have a
2
= 2 while f (2) =
1+
√
5
2
3
−
√
5
1 −
√
5
2
3
=
(16 + 8
√
5) − (16 − 8
8 5
√
5)
=
16
√
5
8
√
5
= 2 .
Thus a
2
= f (2).
Induction step: Assume that a k − 1
= f ( k − 1) and a k − 2
= f ( k − 2) for some natural number k ≥ 3.
Then a k
= a k − 1
+ a k − 2
= f ( k − 1) + f ( k − 2)
=
=
=
=
=
1+
√
5
2
1+
√
5
2
1+
√
5
2
1+
√
5
2
1+
√
5
2 k
−
√
5
1 −
√
5
2 k
+
1+
√
5
2 k − 1
−
√
5
1 −
√
5
2 k − 1 k
− k − 1 k − 1 k +1
−
√
5
1 −
√
5
2 k k +1 k − 1
= f ( k )
− 1 −
√
5
2 k − 1
1+
√
5
2
4
+ 1 −
√
5
(1+
√
5)
2
−
√
5
1 −
√
5
2
1 −
√
5
2 k − 1 k − 1
1 −
√
5
2
(1 −
√
5)
2
4
+ 1
1 −
√
5
2
+
√
5
1+
√
5
2
Thus by induction, we have that a n
= f ( n ) for all natural numbers n .
(e) Who did you work with?