CSCE 235 HW 4 Solution

CSCE 235

Homework Assignment 4 (with Programming) – Solution

Assigned: February 6, 2009

Due: 12:30 p.m. February 16, 2009

Set Basics

1. (10 points) List all members of the following sets.

(a)

(b)

(d) where is the set of all positive numbers

(e) where is the set of all positive numbers

Solution

(a)

(b) When n = 0, ;

n = 1, ;

n = 2, ;

n = 3, ;

n = 4, .

{ 0, 0, 6, 24, 60 }

(d) The valid n values are 2, 4, 6, 8, and 10. So, we have {1/4, 1/16, 1/36. 1/64, 1/100 }.

(e) The valid n values are 2, 3, 5, 7, 11, and 13. So we have { 2, 3, 5, 7, 11, 13 }.

2. (10 points) Consider the following six subsets of Z. (Definition: Z = the set of all integers = ( …, -3, -2, -1, 0, 1, 2, 3, …}.)

A = { 2m + 1 | m Î Z }

B = { 2n + 3 | n Î Z }

C = { 2p -3 | p Î Z }

D = { 3r + 1 | r Î Z }

E = { 3s + 2 | s Î Z }

F = { 3t - 4 | t Î Z }

Which of the following statements are true and which are false? Show your reasons.

(a) A = B

(b) A = C

(d) D = E

(e) D = F

(f) E = F

Solution

First, let us enumerate the members of each set.

A = { 2m + 1 | m Î Z } = { …, -5, -3, -1, 1, 3, 5, 7, …}

B = { 2n + 3 | n Î Z } = { …, -3, -1, 1, 3, 5, 7, 9, …}

C = { 2p - 3 | p Î Z } = { …, -9, -7, -5, -3, -1, 1, 3, …}

D = { 3r + 1 | r Î Z } = { …, -8, -5, -2, 1, 4, 7, 10, … }

E = { 3s + 2 | s Î Z } = { …, -7, -4, -1, 2, 5, 8, 11, … }

F = { 3t - 4 | t Î Z } = { …, -13, -10, -7, -4, -1, 2, 5, … }

( …, -3, -2, -1, 0, 1, 2, 3, …}.)

(a) A = B. This is true.

(b) A = C. This is true.

(d) D = E. This is false.

(e) D = F. This is true.

(f) E = F. This is true.

Note that there are several ways to approach this question. The above is basically enumerating the members of each set and then just comparing the sets. Another way is as follows:

(a) Consider x in A. Then x = 2m + 1. Or, x = 2m - 2 + 2 + 1 = 2(m – 1) + 2 + 1 = 2(m – 1) + 3. Since m Î Z, m-1 Î Z too. And thus, we can rewrite the equation as x = 2n + 1 for n Î Z, or A = { 2n + 3 | n Î Z }. And now, A = B.

The proofs for (b)-(f) are similar to the above.

Set Operations

3. Application of Knowledge (21 points) Consider an alphabet Σ = { a, b }. It has only two letters. A word (or w) is any finite string of letters that you can form out of the alphabet. For example, abba is a word formed using only the letters from the alphabet Σ. The collection of all words is called a language, and is denoted as . For example, = { ε, a, aa, ab, ba, bb, aaa, aab, abb, bbb, bba, baa, …}. And is an infinite set! The symbol ε denotes the empty word or null word.

Now, suppose Σ = { a, b }; A = {a, b, aa, bb, aaa, bbb }; B = { : length(w) ≥ 2}; and C = { : length(w) ≤ 2}. (Hint: Don’t forget the empty word ε, which has a length of zero!) List all members of the following:

(a)

(b)

(c)

(d)

(e)

(f)

(g) power set of Σ

Solution

First, we list B = { ab, aa, ba, bb, aaa, aab, abb, bbb, bba, baa, …}.

Then, we list C = { ε, a, b, aa, ab, ba, bb }.

(a) = { a, b, bb, aa }

(b) = { aaa, bbb }

(d) = { ε, ab, ba, aaa, bbb }

(e) = { aa, bb, aaa, bbb }

(f) = { aa, ab, ba, bb }

(h) The set Σ = { a , b }. So the power set of Σ is { {}, {a}, {b}, {a, b} }.

4. (15 points) True or False (if false, show an example; if true, explain):

(a) for all A, B, and C.

(b) If A = B, then is .

(d) whenever .

(e) for all A and B.

Solution

(a) False. For example, suppose A = {1}, B = {1,2}, and C = {4}. Then = = {1}, and = ={1, 4}. So, the two sets are different.

(b) True. If A = B is true, then . Thus, we have . According to the identity laws, the intersection of a set and its complement is an empty set.

(d) True. According to the identity laws, . So, we have . Since , !

(e) False. Suppose U = {1,2,3,4,5}, A ={1,2}, and B = {2,3}. So = {3,4,5} and = {1,4,5}. And given the two sets, we know that = {4, 5}. Now, = {2}, and = {1, 3, 4, 5}. Therefore for all A and B.

5. (9 points) Prove the following formally, without using Venn diagrams:

(a) (5 points) for any sets of A, B, and C. (Hint: Use the fact .)

(b) (4 points) If and , then .

Solution

(a) Using the fact , Using the set distributive law, we have . Using the De Morgan’s law, we have . And with this, we can use the same fact again: . Thus, we have shown that for any sets of A, B, and C.

(b) To show “if and , then ”, consider an element x in A, . If , then x is also a member of B, . Similarly, if , then x is also a member of C, . So, if and , then and . Thus, . Therefore, if and , then .

Binary Relations

6. (20 points) Determine whether or not each of the binary relations defined on the given sets A are reflexive, symmetric, or transitive. If a relation has a certain property, prove this is so; otherwise, provide a counterexample to show that it does not. Determine also whether a relation is equivalence.

(a) A = { 3, 4 }; = { (3, 3), (4, 4)}

(b) A is the set of all humans; if and only if where means “a is the parent of b.”

(d) A is the set of cross products of integers, Z×Z; if and only if .

(e) A is the set of all humans; if and only if where means “a and b are both fans of the same musician.” Assume that a human may be a fan of more than one musician.

Solution

(a) Reflexive, since both (3,3) and (4,4) are part of the relation.

Symmetric, since we cannot find a counter example such as “(3, 4) and no (4, 3).”

Transitive, since we cannot find a counter example such as “(3, 4) & (4, 3) and no (4, 4).”

The relation is reflexive, symmetric, and transitive. It is an equivalence.

(b) Not Reflexive, since a person cannot be a parent of himself or herself. does not exist.

Not Symmetric, since if , then does not exist. If a person a is a parent of another person b, then that person b cannot be a parent of the person a.

Not Transitive, since if , and , does not exist. If a person a is a parent of another person b, and that person b is a parent of yet another person c, then a is the grandparent of c. a cannot be the parent of c.

Not Equivalence, since the relation does not satisfy all the above three properties.

Symmetric. If , then that means is even. By the commutative law of mathematics, =. So, if is even, then is also even. So, if , then we are guaranteed to have .

Transitive. If is even, and is also even, then we know that all a, b, and c are even. Thus, ac must also be even. As a result, we can conclude that if and , then .

Not Equivalence, since the relation does not satisfy all the above three properties.

(d) Reflexive, since . This is because is always true.

Symmetric, since is always true. This is because always yields the same result as .

Transitive. Suppose we have such that we know and . If we add c – m to , we have . Since , we can transform the equation to . Then we can reduce the equation to . Thus, we have an element . Since for every pair , is also in . The relation is transitive.

Equivalence. The relation satisfies all three of the above properties.

(e) A is the set of all humans; if and only if where means “a and b are both fans of the same musician.” Assume that a human may be a fan of more than one musician.

Reflexive, since is always part of the relation. If I am a fan of Eric Clapton, then I am also a fan of Eric Clapton.

Symmetric, since if , then too.

Not transitive. For example, suppose there are three fans: Adam, Belle, and Charlie. Adam is a fan of Britney Spears; Belle is a fan of Britney Spears and YoYo Ma. Charlie is a fan of YoYo Ma. So, we have , and ; but we do not have ! Adam and Charlie do not have a common musician that they are fans of.

Not Equivalence, since the relation does not satisfy all the above three properties.

7. (24 points) Given the following relations:

Solution

First, let us look at the four relations.

. Since a sum of two numbers is odd only when one of the number is odd and the other is even, we know that consists of (1, 2), (1,4), (3,6), and so on.

. Similarly, a sum of two numbers is even only when (a) both numbers are even or (b) both numbers are odd. So, we know that consists of (1,1), (2,2), (3,1), (5,9), and so on.

. Since a product of two numbers is odd only when both numbers are odd, we know that consists of (1,1), (3,5), (9,17), and so on.

. For a product of two numbers to be even, one of the numbers must be even. So, we know that consists of (1,4), (2,3), (10,2), and so on.

List 5 members for each of the following sets; if not possible, explain.

(a) This is an empty set. requires one of the numbers to be even; requires both to be even. So, the intersection is empty.

(b) 5 members: (1,4), (2,3), (10,2), (3,6), (2,2)

(d) 5 members: (1,4), (2,3), (10,2), (9,17), (1,1)

(e) Since and do not intersect, = . So, 5 members: (1,2), (1,4), (3,6), (5, 6), (6, 5)

(f) As we see from the answer to part (c), we know that and intersect. So, out of the members for , we need to filter out members that also satisfy . To satisfy , basically the two numbers have to be even or both have to be odd. The latter case cannot exist within . So, we only have to make sure that we filter out all members that have both numbers even. So, 5 members: (1,4), (2,3), (3,2), (5,6), (2,9).

(g) This is exclusive or. requires one even number and one odd number. requires at least one even number. And since is a subset of . So, to be exclusive, we consider for only the members that have two even numbers. 5 members: (2,2), (4,2), (10,2), (2,6), (4,4).

(h) This is exclusive or. requires either both even numbers or both odd numbers. requires both numbers to be odd. So, once again, we have a scenario where is a subset of . To be exclusive, we only consider the members that have two even numbers. 5 members: (2,2), (4,2), (10,2), (2,6), (4,4).

Equivalence Relations and Partitions

8. (6 points) Let A be the set of all students in the CSCE235 class. For , define when a is taller than or as tall as b. Is an equivalence relation? Why?

Note that is a binary relation on A×A such that

{(a,b) : a is as taller than or as tall as b }.

Now, we have to show that whether is reflexive, symmetric and transitive to prove whether an equivalence relation. First, every student is as tall as him or herself. So, the relation is reflexive. Second, if (a,b), then it is possible for (b,a) to not be true. That is, if person a is taller than b, then b cannot be taller than a. Thus, we now know that the relation is not symmetric. And thus, is not an equivalence relation.

9. (15 points) A partition of a set A is a collection of disjoint nonempty subsets of A whose union is A. These disjoint sets are called cells (or blocks). Suppose that A has the following cells: m, n, o, and p.

(a) What is ?

(b) What is ?

(d) What is ?

(e) What is ?

(a) since all cells are disjoint.

(b) since all cells are disjoint.

(d) .

(e) . This is because the intersection of the two cells is an empty set and the complement of an empty set is the set itself.

Programming

10. (50 points) Toolbox. Implement a relation-based toolbox to determine whether a binary relation is reflexive, symmetric, or transitive.