# 83 The Basics of Sets

*You will need: **Coins **(Material Card 1), A–Blocks (Material Cards 2A–2E)*

Have you ever collected anything? If so, what?

Name some things people might collect, including anything people in your group collect.

Suppose someone collected 25 coins from the years 1964 –1969.

a. What different types of coins might be in their collection?

b. Name one way you might sort the coins into groups.

c. Can you think of a different way to sort them into groups? Explain.

d. How do you think a child would sort them?

Let’s assume these are the 25 coins that were collected:

1966 penny, 1967 nickel, 1966 quarter, 1967 penny, 1965 penny, 1966 half dollar, 1967 quarter, 1965 dime, 1967 dime, 1968 quarter, 1964 dime, 1966 nickel, 1965 nickel, 1967 half dollar, 1966 dime, 1964 nickel, 1969 quarter, 1969 half dollar, 1965 half dollar, 1968 penny, 1968 dime, 1964 quarter, 1965 quarter, 1969 dime, 1968 nickel

To simplify writing each coin out, let’s abbreviate 1966 penny by (6P), and 1967 nickel by (7N), etc. So in our collection, we have the following:

(6P, 7N, 6Q, 7P, 5P, 6H, 7Q, 5D, 7D, 8Q, 4D, 6N, 5N, 7H, 6D, 4N, 9Q, 9H, 5H, 8P, 8D, 4Q, 5Q, 9D text{ and } 8N)

A physical model for these coins is found on Material Card 1. If you haven’t already done so, cut out a set of coins from this Material Card and use them to do several of the following exercises.

Sort the coins into five groups; pennies, nickels, dimes, quarters and half dollars. Using the abbreviations on each coin, list the coins that go into each group (such as (6P, 7N, 6Q, 7P, 5P), etc. each coin must be listed with **a number ***and*** a letter**. You can’t just write a letter or a number, like 6 or 5 or (P). Each coin is made up of one of each! It’s like writing the initials of a first and last name instead of the full name.)

a. pennies: ___ (did you list 4 pennies?)

b. dimes: ___ (did you list 6 dimes?)

c. half dollars: ___ (did you list 4 of them?)

d. Now sort the coins into groups by year. How many groups are there? List two different groups and which coins go into each group. Remember to list each coin individually each of the 25 coins has a digit followed by a letter.

The collection of coins we have been working with is called a ** set**. A

**is simply a collection of objects. That is the definition of a set (without the word “simply”).**

*set*Define ** set**: A

**is**

*set*In exercises (e text{ and } f), you grouped the ** elements **or

**of the set into**

*members**different*

**. The**

*subsets***of a set are the objects in the set. In this case, the**

*elements***of the set are the 25 different coins.**

*elements*Define ** elements**: The

**of a set are**

*elements*A ** subset** is also a set. When we talk about

**, it is in reference to another set. For example, the set of pennies in the collection above is a**

*subsets***of the collection of coins.**

*subset*The following is a formal definition of ** subset**:

A set ** B** is a

**of a set**

*subset***if every element of set**

*C***is an element of set**

*B***.**

*C**To show that one set is not a subset of another set, there would have to be an element in the first set that is *

**not**

**in the second set.**There is one thing about sets that confuses a lot of people. By definition, we say that a set is a collection of objects. But in reality, a set can have no elements in it. A set containing no elements is called the ** empty set **or

**. Look at the definition of a subset very carefully. Is the null set is a subset of every set?**

*null**set*I hope this was your conclusion: *The null set is a subset of every set.*

Define ** null set **: ___

Note that in my definition of a subset on the previous page, I used the letters B and C for my sets. I did not have to use these letters. They are just dummy variables much like when we use (x, y text{ and } n) in algebra!! Here is another perfectly fine way to define subset: A set ** R** is a

**of a set**

*subset***if every element of set**

*Q***is an element of set**

*R***.**

*Q*Use different letters than just used to define ** subset**:

We usually use a lot of abbreviations in math so that it isn’t so cumbersome writing everything out. Many times, we pick a capital letter to represent a set or subset and *define* the set by that letter. I will define some sets now: The set of 25 coins we’ve been working with will be defined as **C**. The subsets of pennies, nickels, dimes, quarters and half dollars will be denoted by **P**, **N**, **D**, **Q** and **H**, respectively. Notice I gave them meaningful letters so it is easy to remember which letter stands for which subset.

One way to describe a set is to list its objects, if possible. (Later, we will discuss why this is not always practical or possible.) When we list the objects, we do so *inside of braces using commas to separate the elements*. The order in which the elements are listed is irrelevant. Using the listing method, one way to write our coin collection, set **C**, is

( mathbf{C} = { 6P, 7N, 6Q, 7P, 5P, 6H, 7Q, 5D, 7D, 8Q, 4D, 6N, 5N, 7H, 6D, 4N, 9Q, 9H, 5H, 8P, 8D, 4Q, 5Q, 9D, 8N })

Let **S** be the subset of coins from 1964, **V** from 1965, **W** from 1966, **X** from 1967, **Y** from 1968, **Z** from 1969 and **T **from 1970. Are **P**, **N**, **D**, **Q**, **H**, **S**, **V**, **W**, **X**, **Y**, **Z** and **T** all subsets of **C**? ___ Using the correct notation, write out these 12 sets using the listing method by listing the elements in each set. *The only elements in C are those listed above this exercise. Note that 6P is one element in P because 6P is a particular penny in C but 6 is not an element and P is not an element. *The difference between how you write the answers here and how you did it in exercise 4 is that you are using proper notation with the braces, and commas in between the elements.

N = ___ | W = ___ |

Q = ___ | Y = ___ |

S = ___ | T = ___ |

The way to express that an element, like 4D, is in set **C** is with the symbol which means “is an element of”. We write: 4D (read “4D is an element of **C**“).

Let **A** represent the one dollar coins in our set **C**. Notice that set **A** is empty. We can show that **A** contains no elements by writing **A = {**(varnothing)**}**. The null or empty set is also written (varnothing). So, we can also write **A** = (varnothing). In fact, **{ }** = (varnothing) always. It is correct to write the null or empty set either way. But note that it is ** not correct** to write the null set like this: {(varnothing)}. That would be a set that

**contain at least one element the null set! See how this can be a bit confusing?**

*does*Now to some questions about subsets…

a. Is there *any* element in **N** that is not in **D**? ___ If so, name *one*: ___

b. From your work in part a, answer this question: Is **N** a subset of **D**? ___

c. Is there *any* element in **A** that is not in **C**? ___ If so, name *one*: ___

d. From your work in part a, answer this question: Is **A** a subset of **C** ? ___

The symbol (subseteq) means “is a subset of” and so the way to express that **P** is a **subset** of **C** is by writing . Since **N** is also a subset of **C**, we can write **N** (subseteq) **C**. If the first set has **fewer** elements than the second set, then the first set is a **proper subset** of the second set. In this case, the symbol (subseteq) may be used in place of (subseteq). So, in the case of the coins, it is also correct to write P (subset) C and N (subset) C. A first set is **not** a subset of a second set if there is *at least one element in the first set *that is *not in the second set*. Think again about why the null set has to be a subset of every set. For it not to be a subset, there would have to be *at least one element in it *that wasn’t in the second set. But, the null set is empty!

**P** is not a subset of **N**. To show that **P** is not a subset of **N**, we must name at least one element in **P** that is not in **N**. Name an element in **P** that is not in **N**: ___

Think about and discuss the subtle differences between the following statements: (6P in C), ({6P} subset C), (D subseteq D) and (P subset C) are correct whereas (6P subset C), ({6P} in C), (D subset D) and (P in C) are **not** correct!

Decide which of the following statements are true and which ones are false:

a. _______ (varnothing subset A) | b. _______ (varnothing subseteq varnothing) | c. _______ (S subseteq C) |

d. _______ (C subseteq C) | e. _______ (B subset B). | f. _______ (7P in P) |

g. _______ (T subseteq varnothing) | h. _______ (9D in Z). | i. _______ (5H in N) |

Make up three true statements and three false statements using the newly defined subset and element symbols. Use the null set in at least one true and one false statement.

True Statement |
False Statement |

Something else to consider when working with sets is how many elements are in a given set. If we count the number of elements in a set, we call that number the **cardinality** or **cardinal number** of that set. To denote the cardinality of set **C**, we write (n)(**C**) = 25. Think of the *n* as the *number **of objects in the set*. Do you remember any functional notation from algebra like (f)(x) = 2x + 3 implies (f)(5) = 2(5) + 3 = 10 + 3 = 13 – that is, when you plug 5 into the function (f), out pops 13? Well, when working with the cardinality of a set, it is the same kind of thing and we use similar notation. In this case, when you plug **C** into the function (n)*,* out pops the number 25!!! When you are asked about the cardinality of a set, the answer is a number. We say that two sets are **equivalent **if they each have the *same number of elements* which is to say they have the

*same cardinality*. We use the symbol, ~, to denote equivalence. For instance, since

**P**and

**H**each contain four elements, we can write

**P**~

**H**. Two sets are

**equal**if they have the

**.**

*exact same elements*Compute the following, using the information from exercise 9.

a. n(**P**) = ____ b. n(**N**) = ____ c. n(**Q**) = ____ d. n(**D**) = ____

e. n(**Z**) = ____ f. n(**H**) = ____ g. n(**Y**) = ____ h. n(**T**) = ____

i. n(**S**) = ____ j. n(**W**) = ____ k. n(**X**) = ____ l. n(**V**) = ____

m. n(**C**) = ____ n. n(**A**) = ____ (**A **represents the one dollar coins in our set **C**)

o. Consider the subsets of **C** we have defined so far. Using the equivalence symbol, express which pairs of subsets are equivalent to each other.

What if you were asked which coins were pennies and ** also** from the year 1965? In set

**C**, there is only one coin that satisfies both criteria and that is (5P). When trying to find the elements that are common to two sets, in this case,

**P**(pennies) and

**V**(coins from 1965), we use the word

**intersection**. A formal definition of

**intersection**follows:

The **intersection** of two sets, **A** and **B**, written A (cap) B, is the set of elements which are found in both **A** and **B**.

Using the correct notation, we could now write P (cap) V = {5P}

Complete the following, using correct notation:

a. = ____ X (cap) Y | b. = ____ H (cap) Z | c. = ___ N (cap) S |

d. = ____ Q (cap) W | e. = ____ D (cap) P |

Two sets are ** disjoint** if their intersection is empty. For instance,

**D**and

**Q**are disjoint because (D cap Q = varnothing). Name three other pairs of sets that are disjoint.

Define ** disjoint** sets: ___

What if you were asked which coins were either pennies **or** from the year 1965? List which coins would satisfy one or both of these conditions:

Did you get eight different coins this time? ___ Any coin having a 5 or P or both would do it. You just found the ** union** of the two sets

**P**and

**V**. A formal definition of union follows:

The ** union** of two sets,

**A**and

**B**, written (A cup B), is the set of elements that are

*either*in

**A**

*or*

**B**

*or*

**both**.

Using correct notation, we write P (cup = {6P, 7P, 5P, 5D, 5N, 5H, 8P, 5Q}) Remember that the order in which the elements are written is irrelevant! *IMPORTANT:** If an element is in both sets, it is only written down once in the union.*

Complete the following, using correct notation:

a. N (cup) S = ____ |

b. Q (cup) W = ____ |

c. D (cup) P = ____ |

d. X (cup) Y = _____ |

Write the definition for ** union** and then the definition for

**. Use different letters than A and B.**

*intersection*Make up one new problem involving the coins using the intersection symbol and one new problem using the union symbol. Write the solutions to the problems.

Make up one true and one false statement involving the coins using the intersection and union symbols. State which statement is true and which is false.

Put aside the coin set for now. Let’s work with another set. When working with sets, we need to be clear about what we are talking about. Here is another definition for you.

Mathematicians often refer to the set under consideration as the ** universe **or

**and usually use the letter**

*universal set,***U**to represent the universe.

In our coin problem, the ** universe **wasn’t specified. It could have been the

*25*coins in the collection, all of the coins made between 1960 and 1970 or maybe all the coins in the world. Without knowing the universe, there are some things we couldn’t answer in set theory. Now that we have that out of the way, here are two more important definitions.

The ** complement **of a set

**A**, written

**A**or (bar{A}) or

^{c}**A’**

**,**is the set of elements in the universal set that are not in

**A**. We read

**A**as “

^{c}**A**

**complement”**or “

**not**

*A”.**The*For example, if

**universe must be specified in order to compute the complement.****U**={1, 2, 3, 4, 5} and

**H**= {3, 4}, then H’ = {1, 2, 5).

The ** difference **of two sets

**A**and

**B**, written

**A –**–

*B (or A***B) ,**is the set of elements of

**A**that are

**in**

*not also***B**. (Some people list the elements of

**A**and then cross off any that are in

**B**to get the answer for

**A – B**.

**So**

**if**

**F**= {1, 2, 3 ,4} and

**G**= {1, 3, 5, 7}

*,*then to find

**F – G**, list the elements of

**F**and cross off any that are in G: {1,2,3,4}. The answer is: {2, 4)

Try the following problems in which the universe is the first 9 counting numbers. If U represents the universe, we have U = {1, 2, 3, 4, *5, *6, 7, 8, 9}* *Three subsets of U are defined as follows: **A **= {1, 2, 3, 4, 5} **B **= {2, 4, 6, 8} **C **= {3, 5, 7}

a. A (cap) B = _____ |
b. A (cap) C = ____ |

c. B (cap) C = _____ |
d. A (cup) B = ____ |

e. A (cup) C = _____ |
f. B (cup) C = ____ |

g. A^{c} = _____ |
h. B^{c }= ____ |

i. C^{c} = _____ |
j. A – B = _____ |

k. B – A = ____ |
1. A – C = _____ |

m. C – A = ____ |
n. B – C = _____ |

o. C – B = _____ |

*Did you notice that the order matters for difference, but not for union or intersection? *

Continuation of exercise 23 where **U** = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The subsets are:

**A**= {1, 2, 3, 4, 5} **B** = {2, 4, 6, 8} **C** = {3, 5, 7}

Try these “more involved” problems. These will take more than one step to do. You might need scratch paper.

p. (A (cup) C) – B = ____ |
q. (A – C)^{c} = ____ |

r. (A (cap) B) (cup) C = ____ |
s. (A (cup) C) (cap) B = ____ |

t. (A (cup) C)^{c} = ____ |
u. A^{c }(cap) B^{c} = ____ |

v. (bar{A cap B}) = ____ | w. (bar{A cup B}) = ____ |

x. A(cup) (^{c }B^{c }(cap) C^{c}) = _____ |

Answer true or false for the following:

y. (varnothing) (subset) B ___ |
z. (A – C) ~ (C – B) ___ |

aa. 5 (in) (C– (A (cup) B)) ___ |
bb. n(A (cup) B) = 9 ___ |

cc. n(A) + n(B) = 9 ___ |
dd. (A (cup) B)^{c} = A^{c }(cup) B^{c} ___ |

ee. (A (cup) B)^{c }= A^{c }(cap) B^{c} ___ |
ff. A – B = A – (A (cup) B) ___ |

gg. A and B are disjoint ___ |
hh. B and C are disjoint ___ |

ii. U – A = A^{c} ___ |
jj. {3, 4, 5, 6, 8} (subset) {4, 5, 6, 7, 8, 9} __ |

Make sure you can work through all of the problems above perfectly on your own. If you have trouble understanding any of them, check with your classmates or ask your teacher if something still needs clarifying!!! Then, make up some more problems of your own as instructed below before going on.

Make up four of your own “more involved problems” and show the solutions. Use the same sets (**U**, **A**, **B** and **C**) as in exercise 23.

1. ___ |

2. ___ |

3. ___ |

4. ___ |

Now it’s time for you to make up some original problems on your own and to provide the solutions. First, decide on a universal set, U, and define three subsets, D, E and F. Then, make up two original and nontrivial (thoughtful and involving two or more steps) problems using combinations of union, complement, intersection and difference

Remember when I mentioned that the listing method wasn’t always practical or possible? Consider a universal set that contained all of the whole numbers. These could not be listed. Another way of expressing this set would be by description. You could say **U** = (the whole numbers) This is a very formal way of writing this same set **U** = {x|x is a whole number} which is read “**U** is the set of all x such that x is a whole number”. You may have seen this notation in algebra. We won’t be so formal here. Another way to get around the listing problem is to use three consecutive dots which means “and so on”. We would write **U** = {0,1,2,3,…}. Here is a way to express all the even numbers between 13 and 509: {14,16,18,…,508}. You could list them out (if you didn’t have a life!), but you can see the advantage of using three dots!!!! Using the three dots doesn’t always work either. Consider the set of all real numbers. They cannot be listed. You would have to write something like this: {all real numbers) or {x|x is a real number}.

Use correct set notation to express the following sets in one or more ways:

a. the letters of the alphabet

b. the even integers between and including 124 and 698

c. the whole numbers strictly in between 100 and 1000

d. the names of the past presidents of the United States up until 1995

e. the names of the presidents of the U.S. from 1981 – 1995

It’s now time get out your **A-blocks**. They can be found on Material Cards 2A – 2E. The white cards are value label cards (which you won’t use until Exercise Set 3), and the colored objects are the actual A-blocks. There are 24 objects in each set of A-blocks.

We will use the following abbreviations when referring to each of the 24 elements.

S = small |
L = large |
Y = yellow |
R = red |
B = blue |
G = green |
Q = square |
T = triangle |
C = circle |

We’ll use three letters to specify each object in this order – size, color, shape. The small blue square would be denoted by SBQ and the large red circle would be denoted by LRC. These abbreviations are on the objects that you cut out.

Let’s define our universe of A-blocks by **A**.

Play with the A-blocks a little bit. How do you think children might play with them?

Arrange the A-blocks into some piles of subsets. What kinds of piles did you make and how many were in each pile?

Divide them up differently. This time, what kinds of piles did you make and how many were in each pile?

a. Select a shape and take all the blocks of that shape and put them on a blank piece of paper. Close your eyes and have a friend remove one of objects. Look at the remaining pieces and determine which piece was removed. Think about how you figured it out. How do you think a child would figure it out?

b. Repeat part (a), but this time select a color and put all the blocks of that color on a blank sheet of paper.

c. Repeat part (a), but this time select a size and put all the blocks of that size on a blank sheet of paper.

d. Which was the fastest for you to figure out — shape, color or size? Why do you think that was so?

e. **Extra Credit:** Work this activity with 2 young children. One should be very young (you decide) and the other several years older. Document your findings and share on the Forum along with a picture or video. Compare how they figured out the missing piece to how you did. Be sure to do the activity yourself first.

In this exercise, parts a through f are not related to each other. Start with the whole set of A-Blocks for each part and then proceed to follow the directions each time.

a. Make subsets of the blocks, dividing them by shape (disregard size)

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets of shapes. State which shape you picked: ____________

Using correct set notation, and abbreviations, list its elements

b. Make subsets of the blocks, dividing them by color.

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets you made. State which color you picked: _____________

Using correct set notation, and abbreviations, list its elements

c. Make subsets of the blocks, dividing them by size.

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets you made. State which size you picked: _____________

Using correct set notation, and abbreviations, list its elements

d. Make subsets of the blocks, dividing them by color shape.

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets you made. State which color and shape you picked: _____

Using correct set notation, and abbreviations, list its elements

e. Make subsets of the blocks, dividing them by color and size.

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets you made. State which color and size you picked: ______

Using correct set notation, and abbreviations, list its elements

f. Make subsets of the blocks, dividing them by size and shape.

How many subsets are there? _____ How many blocks are in each subset? _____

Pick one of the subsets you made. State which size and shape you picked: _____

Using correct set notation, and abbreviations, list its elements

A particular **shape**, **color** or **size** is called a **value**. Let each of the values of A-blocks be defined in terms of subsets **S**, **L**,**R**, **B**, **G**, **Y**, **Q**, **T**, **C** will define small, large, red, blue, green, yellow, square, triangle and circle, respectively.

a. Have a friend think of two values (like blue and square, for instance). Then have your friend put all the blocks having either of these values on a blank paper, but intentionally leaving one out. Try to figure out which block is missing. What do you have to figure out first?

b. Which values were chosen? ____ Which piece was missing?

Do this exercise two more times, with other values. Try it with all the blocks hidden one time and with all of them in clear sight another time.

c. Is it easier if all the blocks, including the missing block, are in plain view somewhere or if the missing block as well as all the rest are hidden?

d. Try this game by having your friend choose three values. Is this easier or harder? Try it more than once. List the values chosen and the piece missing for one of the games.

e. When making a subset that has either of two values as in part *a*, you are forming the ____ of those two sets. Here’s a hint, *it isn’t* complement. Using correct set notation, write the set formed by the union of the two values chosen in part a. _____ (cup) _____ = _______

Choose a value (a particular shape, color or size). Put all the blocks with that value in one pile. Pick a different value and put all the blocks with that value in a separate pile.

a. Which values were chosen? ____ |

b. Were there any pieces that needed to go in both piles? ____ If so, which pieces? |

c. Suppose the values chosen were red and blue. Were there any blocks that needed to go in both piles? ____ If so, which ones? |

d. Suppose yellow and large were chosen. Were there any blocks that needed to go in both piles? ____ If so, which ones? ____ |

e. Elements belonging in both piles must be both yellow AND large. When making a subset where the objects must have both values, you are forming the ____ of those two sets, Using set notation, we write Y (cap) L =____ |

Put your A-blocks away for now. Don’t lose them!! Before we continue working with the A-blocks, we’ll need to learn about and work with Venn diagrams. We will then come back and do more work with our A-blocks.

Let’s consider a situation where a mother, Mary, has three children –Alicia, Brent and Carlos. Mary has red hair. The question we will attempt to answer is this: What possibilities exist regarding which, if any of her children, have red hair? List the possibilities you came up with:

Let’s work through the above problem. First of all, let’s break the problem down a little bit. It’s possible none of her children have red hair, only one has red hair, two out of three have red hair or all three of them ended up as redheads. Within each of these four situations, let’s write down each of the possibilities. We will list each possibility as a set, using the childrens’ names as the possible elements.

Situation 1: None of the children are redheads: { } |

Situation 2: One child is a redhead. This gives us the three possibilities listed below: {Alicia} {Brent} {Carlos} |

Situation 3: Two out of three have red hair. You can also think of this as only one child not having red hair. In any case, that gives us the three possibilities below: {Brent, Carlos} {Alicia, Carlos} {Alicia, Brent} |

Situation 4: All three children have red hair: {Alicia, Brent, Carlos} |

We came up with eight different possibilities.

The above problem really amounts to a set theory question. Given a set M, where M = {A, B ,C}, list all possible subsets of M. First of all, remember that the null set is a subset of every set and the entire set is a subset of itself (it’s just not a proper subset). You did remember those two points, didn’t you?

Let’s begin by listing the possibilities according to how many elements might be in the subsets either no elements, one element, two elements or three elements. For each of these situations, list all possibilities:

No elements: { } | Two elements: {A, B}, {A, C}, {B, C} |

One element: {A}, {B}, {C} | Three elements: {A, B, C} |

Here is a list of all possible subsets: { }, {A}, {B}, {C}, {A, B}, {A, C}, {B, C}, {A, B, C}

WOW! Did you see how we came up with the same eight possibilities? This set theory stuff is just amazing, isn’t it?

List all possible subsets for each of the sets given.

a. (varnothing ): ___ |

b. {P}: ___ |

c. {G, F}: ___ |

d. {X, Y, Z}: ___ |

e. {1, 2, 3, 4}: ___ |

Using the information you gained from exercise 34, answer the following question.

How many subsets are there for a set containing the given number of elements?

a. no elements: ___ |

b. one element: ___ |

c. two elements: ___ |

d. three elements: ___ |

e. four elements: ___ |

f. five elements: ___ |

g. n elements (use a formula): ___ |

There is one more operation we will define on sets. So far, you’ve worked with union, intersection, difference and complement. Now we will learn what it means to take the Cartesian product of two sets, A and B.

The **Cartesian product** of set **A** with set **B**, which is written **A** (times) **B** and is read as ‘**A** cross **B**” is the set of all possible ordered pairs (a,b), where a (in) **A** and b (in) **B**.

If you are finding the Cartesian product, the answer is a set that contains ordered pairs. The order matters! The first element must come from the set written to the left of the (times) and the second element must come from the set written to the right of the (times).

If A = {x, y, z} and B = {a, b}, find A (times) B and B (times) A.

###### Solution

A (times) B = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}

B (times) A = {(a, x), (b, x), (a, y), (b, y), (a, z), (b, z)}

Notice there were 3 elements in one set and two in the other. The Cartesian product has six elements in it, six ordered pairs.

If E = {1, 2} and F = {2,3}, find E (times) F, F (times) E, E (times) E, and F (times) F

###### Solution

E (times) F= {(1, 2), (1.3), (2, 2), (2, 3)} | E (times) E = {(1, 1), (1, 2), (2, 1), (2, 2)} |

F (times) E = {(2, 1), (3, 1), (2, 2), (3, 2)} | F (times) F = {(2, 2), (2, 3), (3, 2), (3, 3)} |

Notice there were two elements in one set and two in the other. Each Cartesian product has four elements in it – four ordered pairs.

Find {4, 5} (times) {7, 2, x, #}.

###### Solution

{4, 5} (times) {7, 2, x, #} = {(4, 7), (4, 2), (4, x), (4, #), (5, 7), (5, 2), (5, x), (5, #)}

Notice there were two elements in one set and four in the other. The Cartesian product has eight elements in it – eight ordered pairs.

In Example I, one set contained 3 elements and the other contained 2 elements, The Cartesian product contained 6 elements – 6 ordered pairs. In Example 2, one set contained 2 elements and the other contained 2 elements, The Cartesian product contained 4 elements, 4 ordered pairs. In Example 3, one set contained 2 elements and the other contained 4 elements. The Cartesian product contained 8 elements , 8 ordered pairs.

If a set **B** contains 5 elements (in other words, n(**B**) = 5) and a set **C** contains 7 elements (or n(**C**) = 7), then how many elements (where each element is an ordered pair) are in the Cartesian product **B** (times) **C**?

If a set **B** contains 5 elements (in other words, n(**B**) = 5) and a set **C** contains 7 elements (or n(**C**) = 7), then how many elements (where each element is an ordered pair) are in the Cartesian product **C **(times) **B**?

In general, for any two sets, F and G, is n(F (times) G) = n(G (times) F)?

Assume a set B contains b elements (in other words, n(B) = b) and a set C contains c elements (or n(C) = c). Use this information to compute the following, where you are asked how many elements (where each element is an ordered pair) are in a given Cartesian product.

a. n(B (times) C) = ____ | b. n(B (times) B) = ____ | c. n(C (times) C) = ____ |

In general, if you take any two sets, A and B, is A (times) B = B (times) A? ____ If you answered yes, provide an example of two different sets, A and B, and show that A (times) B = B (times) A. If you answered no, provide an example of two different sets, A and B, and show that A (times) B ≠ B (times) A.

This one is trickier: Find {(4, 3), 5) (times) {(3, 3),{4, 7, 5, 2}, 1}

Okay, take a breath…you can do this. Look carefully at the first set; it has two elements, the first element happens to be the ordered pair (4, 3) and the second element is the number

5. Look carefully at the second set; it has three elements, the first shown happens to be the ordered pair (3,3), the second happens to be set containing four elements, and the third is the element 1, So the first set contains 2 elements and the second set contains 3 elements. First, think about how many elements are in the Cartesian product. Did you get the idea to multiply 2 by 3? I hope so! There are six elements in the Cartesian product. Each of those elements is an ordered pair, and some of these ordered pairs will look kind of weird. See if you can understand the solution written below.

###### Solution

{(4, 3), 5} (times) {(3, 3), {4, 7, 5, 2}, l} = {((4, 3), (3, 3)), ((4, 3), {4, 7, 5, 2}), ((4, 3), 1), (5, (3 , 3)), (5, {4, 7, 5, 2}), (5, 1)}

Look at this solution carefully. There are six ordered pairs in the solution. The first three ordered pairs have (4, 3) as its first coordinate and the second three ordered pairs have 5 as its first coordinate. Then (3, 3) is the second coordinate for the first and fourth ordered pair, {4, 7, 5, 2} is the second coordinate for the second and fifth ordered pair and 1 is the second coordinate for the third and sixth coordinate. If you can follow this one, you should do great on the next exercise.

Write the Cartesian product. Each answer is a set containing ordered pairs. Before doing each problem, think about how many ordered pairs will be in the answer. Make sure you write your answer using correct notation, put the ordered pairs in a set. There is a comma between each coordinate in each ordered pair and there is a comma between each element in the set. Use braces { } around the set. One of the answers is the null set!

a. {3, 4} (times) {2, 6} = ____ |

b. {6, 7, 8, 9} (times) {5} = ____ |

c. {r, s, t} (times) { } = ____ |

d. {a} (times) {a} = ____ |

e. {x, y} (times) {x, y} = ____ |

f. {1, 3, 5} (times) {1, 3, 5} = ____ |

g. {(9, 4), C)} (times) {D, {a, b, c}} = ___ |

h. ({{5, 6, 7, 8, 9}} times {g, {4, 3}}) = ___ |