JDEP 183H

Homework Assignment 1A – Solution

Sets, Relations, and Functions

Assigned: September 5, 2006

Due: September 19, 2006 (12:30 PM)

(Homework 5 minutes late will not be accepted)

Set Basics

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

(a)

(b)

(c) where is the set of all positive numbers

(d) where is the set of all positive numbers

(e) where is the set of all positive numbers

Solution

(a) { 1, ½, 1/3, ¼}

(b) When n = 0, ;

n = 1, ;

n = 2, ;

n = 3, ;

n = 4, .

{ 0, 0, 2, 6, 12 }

(c) When n is even, is 1, and thus . When n is odd, is -1, and thus . Thus, = {1, 3}.

(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 - 2 | t Î Z }

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

(a) A = B

(b) A = C

(c) B = 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 - 2 | t Î Z } = { …, -11, -8, -5, -2, 1, 4, 7, … }

(a) A = B. This is true.

(b) A = C. This is true.

(c) B = C. This is true.

(d) D = E. This is false.

(e) D = F. This is true.

(f) E = F. This is false.

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 (14 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 }

(c) = { ε, ab, ba }

(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} }.

Binary Relations

4. (18 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.”

(c) A is the set of integers; if and only if is odd.

(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.

(f) A is the set of natural numbers, N; if and only if is an integer.

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.

(c) Not Reflexive, since, for example, 1 + 1 is even. So, (1,1) is not part of the relation.

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

Not Transitive. For example, suppose a = 10, b = 7, c = 4. Both (a,b) and (b,c) are members since which is odd, and , which is odd. However, (a, c) is not a member of the relation. Why? This is because which is even. Thus, the relation is not transitive.

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.

(f) Reflexive, as since is always equal to 1, an integer

Not symmetric. For example, let and a = 2 and b = 1. Then is 2, an integer. But because is ½, or 0.5, not an integer.

Transitive. We want to show that if then . If , then we have = j, and = k, where j and k are integers. So, a = bj, and b = ck. Substituting for b, we have a = ckj. Then, we have = kj. Since j and k are both integers, then the product kj is also an integer. Thus, . Therefore, if , then .

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

5. (8 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)

(c) 5 members: (2,2), (2,4), (4,6), (4,4), (4,8)

(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).

Functions

6. (15 points) Determine whether each of these functions is onto and one-to-one (where Z is the set of integers, and R is the set of real numbers). Explain or show an example to support your answers. (Hint: For (e)-(g), think carefully about the domain, target, and range involved.)

(a) f: Z → Z ,

(b) f: Z → Z ,

(c) f: R → R ,

(d) f: R → R ,

(e) f: Z → Z ,

(f) f: R → R ,

(g) f: R → Z ,

Solution:

First, remember that a function from a set A to a set B is a binary relation f from A to B with the property that for every , there is exactly one such that . That means that every single element in the domain must have a link to one of the elements in the target. Also, for a function to be onto, then all the elements in the target, B, must have a link from the domain.

Problems	One-to-one?	Onto?
(a) f: Z → Z ,	Yes. Since this function maps from an integer to another integer (smaller by 1), all elements in the domain set will have a unique corresponding element in the target. For example, { …, (-2,-1), (-1, 0), (0, -1), (1,0), (2,1), (3,2), … }	Yes. All elements in the target will have a corresponding element in the domain.
(b) f: Z → Z ,	No. For example, and . In this case, there are two elements in the domain set that map to the same element in the target set.	No. The target set is the set of integers. However, the function only maps to positive integers since is always positive. As a result, the negative integers of the target set do not have corresponding elements in the domain.
(c) f: R → R ,	No. For example, and . In this case, there are two elements in the domain set that map to the same element in the target set.	No. The target set is the set of real numbers. However, the function only maps to real numbers smaller than 7 since is always positive. As a result, the real numbers greater than 7 in the target set do not have corresponding elements in the domain.
(d) f: R → R ,	Yes. Each real number will be mapped to a unique real number. In fact, the function resembles a graph like this:	Yes. The graph shows that goes to infinity and also to negative infinity.
(e) f: Z → Z ,	Yes. Each integer will be mapped to itself. (Note that the ‘ceiling’ function actually does not do anything when the domain is the set of integers.)	Yes. Each integer will be mapped to itself and all integers in the target set will have corresponding elements in the domain. (Note that the ‘ceiling’ function actually does not do anything when the domain is the set of integers.)
(f) f: R → R ,	No. Now we consider the set of real numbers. So, it is possible to have two elements from the domain set to map to the same element in the target set. For example, and . Thus this function is no longer one-to-one.	No. Now we consider the set of real numbers in the target set. Since we know that the ceiling function always return an integer, that means many real numbers in the target set will not have corresponding elements in the domain. For example, an element such as 6.7 in the target set does not have a corresponding element in the domain. Thus this function is no longer onto.
(g) f: R → Z ,	No. Now we consider the set of real numbers. So, it is possible to have two elements from the domain set to map to the same element in the target set. For example, and . Thus this function is no longer one-to-one.	Yes. See, here, the target is the set of integers!!!! So, all the elements of the target set will have corresponding elements in the domain!

7. (15 points) We define functions mapping R to R (where R is the set of real numbers) as follows: , , . Find

(a)

(b)

(c)

(d)

(e)

Solution:

(a) = .

(b) = .

(c) = .

(d) = .

(e) = .

8. (15 points) Find the inverses of the following functions mapping R to R (where R is the set of real numbers):

(a)

(b)

(c)

(d)

(e) where is a constant.

Solution:

(a) We have . Let , and . Solving for , we have

Thus, is the inverse of .

(b) We have . Let , and . Solving for , we have

Thus, is the inverse of .

(c) We have . This function is not one-to-one. For example, when x = 2, ; when x = 0, . Thus it is possible for two different values of x to map to the same value in the target. Thus, there is no inverse for this function.

(d) We have . Since k(1) = -1/6 or 1/6. This is not a function. Thus there is no inverse.

(e) We have where is a constant. Let , and . Solving for , we have

Thus, is the inverse of .