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CSci/Math2112
(20)
Assignment 6
Due July 3, 2015
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 n2 + 17n + 17 is prime.
Solution: This statement is false.
Disprove. Consider n = 1, a natural number. Then n2 + 17n + 17 = 1 + 17 + 17 = 35 = 5 · 7.
Since 5 and 7 are both integers, this shows that n2 + 17n + 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) 6= (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) = 2b is odd. Since b is a
natural number, and thus an integer, we also have by definition that 2b 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 an be the number of ways to do this. The goal of this
project is to find (and prove) an explicit formula for an . As an example, a2 = 2 since we can fill the
strip with either two tokens or a domino (see picture below).
(5)
(a) Find an for n = 1, 2, 3, 4, 5, 6. Put your values into a table with columns n and an .
Solution:
n
1
2
3
4
5
6
(10)
an
1
2
3
5
8
13
(b) What is the recursive pattern for an ? Prove that your recursion is true.
Solution: The recursive pattern is an = an−1 + an−2 with starting values a1 = 1 and a2 = 2.
Proof. We have that a1 = 1 since a strip of length 1, i.e. a single square, can only be tiled with
the round token.
We have that a2 = 2 since a strip of length 2 can be tiled with either a single domino, or with two
round tokens.
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 an−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 an−2 ways to do this.
Thus in total there are an−1 + an−2 ways to tile a strip of length n with round tokens and
dominoes.
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(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 an 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 an and state it.
Solution: If F0 , F1 , F2 , F3 , . . . is the Fibonacci sequence, then according to the OEIS we have
√ n √ n
1+ 5
− 1−2 5
2
√
Fn =
.
5
Note though that the Fibonacci sequence has F1 = 1 and F2 = 1 with the recursion Fn =
Fn−1 + Fn−2 . Thus the sequence an is the Fibonacci sequence shifted by one index, i.e.
an = Fn+1 . Thus
√ n+1 √ n+1
1+ 5
− 1−2 5
2
√
an =
.
5
(15)
(d) Prove the explicit formula for an you have found in the OEIS using induction.
Solution:
Proof (by induction). Let f (n) =
√ n+1
− 1−2 5
√
for all natural numbers n. We need
5
√ n+1
1+ 5
2
to show by induction that an = f (n).
Base cases: For n = 1 we have a1 = 1 while
√ 2 √ 2
√
√
√
1+ 5
− 1−2 5
2
(6 + 2 5) − (6 − 2 5)
4 5
√
√
f (1) =
=
= √ = 1.
5
4 5
4 5
Thus a1 = f (1).
For n = 2 we have a2 = 2 while
√ 3 √ 3
√
√
√
1+ 5
− 1−2 5
2
(16 + 8 5) − (16 − 8 5)
16 5
√
√
f (2) =
=
= √ = 2.
5
8 5
8 5
Thus a2 = f (2).
Induction step: Assume that ak−1 = f (k − 1) and ak−2 = f (k − 2) for some natural number k ≥ 3.
Then
ak = ak−1 + ak−2
= f (k − 1) + f (k − 2)
√ k−1 √ k−1
√ k √ k
1+ 5
1+ 5
− 1−2 5
− 1−2 5
2
2
√
√
+
=
5
5
√ k √ k √ k−1 √ k−1
1+ 5
− 1−2 5 + 1+2 5
− 1−2 5
2
√
=
5
√ k−1 √
√ k−1 √
1+ 5
1+ 5
1− 5
+ 1 − 1−2 5
+1
2
2
2
√
=
5
√ k−1 √ 2 √ k−1 √ 2 (1+
(1− 5)
5)
1+ 5
− 1−2 5
2
4
4
√
=
5
√ k+1 √ k+1
1+ 5
− 1−2 5
2
√
=
= f (k)
5
Thus by induction, we have that an = f (n) for all natural numbers n.
(e) Who did you work with?