Chapter 1

Set notation lists the elements inside curly braces, separated by commas. Your list may include different tools. Sets are usually designated with capital letters.

$T=\left\{\mathrm{w}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{h},\mathrm{s}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{w}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{r},\mathrm{h}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r},\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{r}\mathrm{s}\right\}$

One possible solution: $T=\{\text{wrench, screwdriver, hammer, plyers}\}$.

A well-defined set clearly communicates whether the element is a member of the set or not. “A collection of medium-sized potatoes” is not well defined. What makes a potato medium-sized?

This is not a well-defined set.

A well-defined set clearly communicates whether the element is a member of the set or not. You can determine which musical artists were members of the group when it formed in 1995. $B=\left\{\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}.\mathrm{i}.\mathrm{a}\mathrm{m},\mathrm{a}\mathrm{p}\mathrm{l}.\mathrm{d}\mathrm{e}.\mathrm{a}\mathrm{p},\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{o}\right\}$

This is a well-defined set.

When there are no elements that satisfy the set conditions, you represent the set with either version of the empty set notation: a zero with a slash through it or empty curly braces.

$\mathrm{\varnothing <!--\; \varnothing \; -->}\mathrm{o}\mathrm{r}\{\}$

$\mathrm{\varnothing <!--\; \varnothing \; -->}\text{ or { } }$

Be sure to include zero because the requirements include “greater than or equal to zero.” You stop before 10 because 10 is not a single-digit number.

$D=\left\{0,1,2,3,4,5,6,7,8,9\right\}$

$D=\{0,1,2,\dots <!--\; \dots \; -->,9\}$

This set is infinite. You represent large sets by giving enough numbers to define the pattern, and then using three dots, the ellipsis, to show that the set continues. When no number follows the ellipsis, you know the set continues infinitely.

$M=\left\{1,3,5,\cdots <!--\; \cdots \; -->\right\}$

$M=\{1,3,5,\dots <!--\; \dots \; -->\}$

The roster method lists the elements of the set. Set builder notation uses a letter to represent a sample member of the set, a vertical bar, and then a description of a sample member. In this case, because it is not practical to list all types of cars, it is much easier to use set builder notation.

$C=\left\{c|c\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{c}\mathrm{a}\mathrm{r}\right\}$

$C=\{c|c\text{is a car}\}$

The roster method lists the elements of the set. Set builder notation uses a letter to represent a sample member of the set, a vertical bar, and then a description of a sample member. In this case, because it is not practical to list all musical instruments, it is much easier to use set builder notation.

$I=\left\{i|i\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{m}\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\right\}$

$I=\{i|i\text{is a musical instrument}\}$

The cardinal value is the number of elements in the set. There are no prime numbers less than 2.

$n\left(\mathrm{P}\right)=0$

$n(P)=0$

The cardinal value is the number of elements in the set. There are 26 lowercase letters of the English alphabet.

$n\left(\mathrm{P}\right)=26$

$n(A)=26$

There are four elements in set B, so this set is finite.

finite

There is no largest or smallest real number, so this set is infinite.

infinite

Equal sets have the same elements, without regard to order. These sets have the same elements, although in a different order.

Set $B$ is equal to set $A$, $B=A$

Equal sets have the same elements, without regard to order. These sets are not equal because only one has the element f. Equivalent sets have the same number of elements. One set has four elements, while the other set has five elements, so they are not equivalent. These sets are neither equal nor equivalent.

neither

Equal sets have the same elements, without regard to order. Equivalent sets have the same number of elements. These sets are equivalent but not equal because they have the same number of elements (four), but not the same elements. One has a b, while the other has a c.

Set $B$ is equivalent to set $C$, $B\sim <!--\; \sim \; -->C$

Step 1: Begin with the set itself, as every set is a subset of itself. The original set has two elements, so its cardinality is 2.

$\{\text{head, tails}\}$

Step 2: Next, list all the proper subsets containing one less element than the original set. That is, list the sets with one element.

$\{\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d}\mathrm{s}\}$

$\{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{s}\}$

Step 3: Continue this process, reducing the number of elements in the subset by one each step. On this step, list the subsets with zero elements, or the empty set.

$\{\}$

$\{\text{heads, tails}\};$$\{\text{heads}\},\{\text{tails}\};$ and $\mathrm{\varnothing <!--\; \varnothing \; -->}$

A set is a proper subset of L if every member of the subset is a member of L and the subset does not contain an element that is in L. The proper subsets of L with one member include the following.

$\{\mathrm{A}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{u}\mathrm{n}\mathrm{o}\}$

$\{\mathrm{Z}\mathrm{a}\mathrm{p}\mathrm{d}\mathrm{o}\mathrm{s}\}$

$\{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\}$

$\{\mathrm{M}\mathrm{e}\mathrm{w}\mathrm{t}\mathrm{w}\mathrm{o}\}$

A set with one member could contain any one of the following:

$\{\text{Articuno}\}\text{, }\{\text{Zapdos}\}\text{, }\{\text{Moltres}\}\text{, or }\{\text{Mewtwo}\}$.

A set is a proper subset of L if every member of the subset is a member of L and the subset does not contain an element that is in L. The proper subsets of L with three members will include three elements and leave out one.

$\{\mathrm{A}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{u}\mathrm{n}\mathrm{o},\mathrm{Z}\mathrm{a}\mathrm{p}\mathrm{d}\mathrm{o}\mathrm{s},\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\}$

$\{\mathrm{A}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{u}\mathrm{n}\mathrm{o},\mathrm{Z}\mathrm{a}\mathrm{p}\mathrm{d}\mathrm{o}\mathrm{s},\mathrm{M}\mathrm{e}\mathrm{w}\mathrm{t}\mathrm{w}\mathrm{o}\}$

$\{\mathrm{A}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{u}\mathrm{n}\mathrm{o},\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s},\mathrm{M}\mathrm{e}\mathrm{w}\mathrm{t}\mathrm{w}\mathrm{o}\}$

$\{\mathrm{Z}\mathrm{a}\mathrm{p}\mathrm{d}\mathrm{o}\mathrm{s},\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s},\mathrm{M}\mathrm{e}\mathrm{w}\mathrm{t}\mathrm{w}\mathrm{o}\}$

Any of the following combinations of three members would work:

$\{\text{Articuno, Zapdos, Mewtwo}\}$,$\{\text{Articuno, Moltres, Mewtwo}\}$, or $\{\text{Zapdos, Moltres, Mewtwo}\}$.

A set is a proper subset of L if every member of the subset is a member of L and the subset does not contain an element that is in L. The only proper subset of L with no members is the empty set, represented by $\{\}$ or $\mathrm{\varnothing <!--\; \varnothing \; -->}$.

The empty set is represented as $\{\}$ or $\mathrm{\varnothing <!--\; \varnothing \; -->}$.

The set of even numbers given in this question is a subset of the set of natural numbers. The set of even numbers is a proper subset because there are members of the natural numbers (1 and 3 for example) that are not even numbers. The notation for “the even numbers are a proper subset of the natural numbers”:

$E\subset <!--\; \subset \; -->\mathbb{N}$

Note that there are also negative even numbers that were not a part of this question.

$E\subset <!--\; \subset \; -->\mathbb{N}$

Use the formula for the number of subsets of set A with n elements: $2^{n(A)}$.

Set T has nine elements.

$2^{n(T)}=2^9=512$

512

Set builder notation uses a letter to represent a sample member of the set, a vertical bar, and then a description of a sample member.

$\{m|m=5n\text{ where }n\in <!--\; \in \; -->\mathbb{N}\}$

$\{m|m=5n\text{ where }n\in <!--\; \in \; -->\mathbb{N}\}$

Serena also ordered a fish sandwich and chicken nuggets, because for the two sets to be equal they must contain the exact same items: {fish sandwich, chicken nuggets} = {fish sandwich, chicken nuggets}.

Equal sets have the same elements. For the choices to be equal, they needed to order the same food. Serena ordered a fish sandwich and chicken nuggets.

L is a subset of U if every element of L is a member of U. Because all lions are cats, every member of the set of Lions is a member of the set of Cats.

$L\subset <!--\; \subset \; -->C$

The set of lions is a subset of the universal set of cats. In other words, the Venn diagram depicts the relationship that all lions are cats. This is expressed symbolically as $L\subset <!--\; \subset \; -->U$.

Sets are disjoint if they have no members in common. No member of the set of Eagles is a member of the set of Canaries. No member of the set of Canaries is a member of the set of Eagles. These sets are disjoint.

The set of eagles and the set of canaries are two disjoint subsets of the universal set of all birds. No eagle is a canary, and no canary is an eagle.

$A^{\prime }$ represents the complement of set A. The complement of set A is the set of all elements in the universal set U that are not in set A. List the elements of U that are not in A.

$\{\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e},\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{n},\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{g}\mathrm{o},\mathrm{v}\mathrm{i}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{t}\}$

$A^{\prime }=\left\{\text{orange, green, indigo, violet}\right\}$

$A^{\prime }$ represents the complement of set A. The complement of set A is the set of all elements in the universal set U that are not in set A. List the elements of U that are not in A. Because A includes the cats that are not lions, the complement will include the cats that are lions.

$A^{\prime }=\{c|c\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{o}\mathrm{n}\}$

$A^{\prime }=\{c\in <!--\; \in \; -->U|c\text{ is a lion}\}$ or $A^{\prime }=\{c\in <!--\; \in \; -->U|c\notin <!--\; \notin \; -->A\}$

The members that two sets have in common are included in the intersection of the sets. Only element a is in both sets.

$A\cap <!--\; \cap \; -->B=\left\{a\right\}$

$A\cap <!--\; \cap \; --><!--\; \; -->B=\{a\}$

The members that two sets have in common are included in the intersection of the sets. The sets have no elements in common, so their intersection is the empty set. You can represent the empty set by either the zero with a slash or empty curly braces.

$A\cap <!--\; \cap \; -->B=\mathrm{\varnothing <!--\; \varnothing \; -->}$ or $A\cap <!--\; \cap \; -->B=\{\}$

$A\cap <!--\; \cap \; -->B=\{\}$

The members that two sets have in common are included in the intersection of the sets. The two sets have the vowels in common.

$A\cap <!--\; \cap \; -->B=\left\{a,e,i,o,u\right\}$

$A\cap <!--\; \cap \; -->B=B=\{a,e,i,o,u\}$

The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

$A\cup <!--\; \cup \; -->B=\left\{a,d,h,p,s,y\right\}$

$A\cup <!--\; \cup \; -->B=\{a,d,h,p,s,y\}$

The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

$A\cup <!--\; \cup \; -->B=\left\{\text{red, yellow, blue, orange, green, purple}\right\}$

$A\cup <!--\; \cup \; -->B=\{\text{red, yellow, blue, orange, green, purple}}$

The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice. The vowels in set B are a subset of set A, so the union is equal to set A.

$A\cup <!--\; \cup \; -->B=\left\{a,b,c,d,...,z\right\}$

$A\cup <!--\; \cup \; -->B=\{a,b,c,\dots <!--\; \dots \; -->,z\}=A$.

33

Use the formula: $n\left(A\cup <!--\; \cup \; -->B\right)=n(A)+n(B)-<!--\; -\; -->n\left(A\cap <!--\; \cap \; -->B\right)$

$n\left(A\cup <!--\; \cup \; -->B\right)=n(A)+n(B)-<!--\; -\; -->n\left(A\cap <!--\; \cap \; -->B\right)=23+17-<!--\; -\; -->7=33$

113

Use the formula: $n\left(A\cup <!--\; \cup \; -->B\right)=n(A)+n(B)-<!--\; -\; -->n\left(A\cap <!--\; \cap \; -->B\right)$

$n\left(A\cup <!--\; \cup \; -->B\right)=n(A)+n(B)-<!--\; -\; -->n\left(A\cap <!--\; \cap \; -->B\right)=35+78-<!--\; -\; -->0=113$

To find A or B, find $A\cup <!--\; \cup \; -->B$. The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

$A\mathrm{o}\mathrm{r}B=A\cup <!--\; \cup \; -->B=\left\{h,a,p,y,w,e,s,o,m\right\}$

$A$ or $B=A\cup <!--\; \cup \; -->B=\{h,a,p,y,w,e,s,o,m\}.$

To find A and C, find $A\cap <!--\; \cap \; -->C$. The intersection of two sets includes the elements that are in both sets.

$A\mathrm{a}\mathrm{n}\mathrm{d}C=A\cap <!--\; \cap \; -->C=\left\{a,h\right\}$

$A$ and $C=A\cap <!--\; \cap \; -->C=\{a,h\}.$

To find B or C, find $B\cup <!--\; \cup \; -->C$. The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

$B\mathrm{o}\mathrm{r}C=B\cup <!--\; \cup \; -->C=\left\{a,w,e,s,o,m,t,h\right\}$

$B$ or $C=B\cup <!--\; \cup \; -->C=\{a,w,e,s,o,m,t,h\}.$

To find (A and C) and B, you need to follow the order of operations and evaluate the parentheses first.

First, evaluate inside the parentheses. Find $A\text{ and }C=A\cap <!--\; \cap \; -->C$. The intersection of two sets includes the elements that are in both sets.

$A\mathrm{a}\mathrm{n}\mathrm{d}C=A\cap <!--\; \cap \; -->C=\{h,a,p,y\}\cap <!--\; \cap \; -->\left\{m,a,t,h\right\}=\left\{a,h\right\}$

To find (A and C) and $B=\left(A\cap <!--\; \cap \; -->C\right)\cap <!--\; \cap \; -->B$, find $\left\{a,h\right\}\cap <!--\; \cap \; -->B$.

$\left\{a,h\right\}\cap <!--\; \cap \; -->\left\{a,w,e,s,o,m\right\}=\left\{a\right\}$

($A$ and $C$) and $B=(A\cap <!--\; \cap \; -->C)\cap <!--\; \cap \; -->B=\{a,h\}\cap <!--\; \cap \; -->\{a,w,e,s,o,m\}=\{a\}.$

127

Total number of guests = 150

n(neither soup nor salad) = 23

To find the number who had soup, salad, or both, subtract 23 from 150.

$150-<!--\; -\; -->23=127$

50

Total number of guests = 150

n(neither soup nor salad) = 23

To find the number who had soup, salad, or both, subtract 23 from 150.

$150-<!--\; -\; -->23=127$

This represents the number of elements in the union of the sets of Soup and Salad eaters.

n(Soup $\cup <!--\; \cup \; -->$ Salad) = 127

n (Soup) = 92

n (Salad) = 85

Use the formula: $n\left(A\cup <!--\; \cup \; -->B\right)=n(A)+n(B)-<!--\; -\; -->n\left(A\cap <!--\; \cap \; -->B\right)$

$A\cap <!--\; \cap \; -->B=\{3,5,7\}$.

The members that two sets have in common are included in the intersection of the sets. Look for the elements where the circles overlap.

$A\cap <!--\; \cap \; -->B=\{3,5,7\}$

$A\cup <!--\; \cup \; -->B=\{1,2,3,5,7,9\}$.

The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice. List the elements that are inside the two circles.

$A\cup <!--\; \cup \; -->B=\{1,2,3,5,7,9\}$

$A\cap <!--\; \cap \; -->B^{\prime }=\{1,9\}$.

First, find $B^{\prime }$, the complement of B. The complement of set B is the set of all elements in the universal set U that are not in set B. List the elements of U that are not in B.

$B^{\prime }=\left\{0,1,4,6,8,9\right\}$

Now find

$A\cap <!--\; \cap \; -->B^{\prime }=\left\{1,3,5,7,9\right\}\cap <!--\; \cap \; -->\left\{0,1,4,6,8,9\right\}$

The intersection of two sets are the elements that are in both sets.

$A\cap <!--\; \cap \; -->B^{\prime }=\left\{1,9\right\}$

$n(A\cap <!--\; \cap \; -->B^{\prime })=2$.

First, find $B^{\prime }$, the complement of B. The complement of set B is the set of all elements in the universal set U that are not in set B. List the elements of U that are not in B.

$B^{\prime }=\left\{0,1,4,6,8,9\right\}$

Now find

$A\cap <!--\; \cap \; -->B^{\prime }=\left\{1,3,5,7,9\right\}\cap <!--\; \cap \; -->\left\{0,1,4,6,8,9\right\}$

The intersection of two sets are the elements that are in both sets.

$A\cap <!--\; \cap \; -->B^{\prime }=\left\{1,9\right\}$

$n\left(A\cap <!--\; \cap \; -->B^{\prime }\right)$ is the number of elements in the set, which is two.

To find A or B, find $A\cup <!--\; \cup \; -->B$. The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

Because the two sets are disjoint, the number of elements in the union of the two sets is the sum of the number of elements in the two sets.

$23+17=40$

40

To find A or B, find $A\cap <!--\; \cap \; -->B$. The intersection of two sets includes the elements they have in common. They have no elements in common, so the number of elements in the intersection is zero.

0

$A^{\prime }$ is the complement of A. The complement of set A is the set of all elements in the universal set U that are not in set A. The number of elements that are not in set A:

$17+10=27$

27

$n(B)=n(A{B}^{+})+n(A{B}^{-<!--\; -\; -->})+n(B^{+})+n(B^{-<!--\; -\; -->})=14$

Count every blood type that has any type of B as part of the blood type.

$n(B)=n(A{B}^{+})+n(A{B}^{-<!--\; -\; -->})+n(B^{+})+n(B^{-<!--\; -\; -->})=3+1+8+2=14$

$n(B^{\mathrm{\prime <!--\; \prime \; -->}})=n(U)-<!--\; -\; -->n(B)=86$

You could count every type that does not have a B as part of the blood type. However, if you did the previous exercise, you already know the complement of B.

$n\left(B^{\prime }\right)=n\left(U\right)-<!--\; -\; -->n(B)=100-<!--\; -\; -->14=86$

$n(B\cup <!--\; \cup \; -->R{h}^{+})=87$

Count the number of people who are in either the B circle or the RH⁺ circles.

$1+3+2+8+36+37=87$

Draw a rectangle to represent the universal set. Label it with “U = Conference Attendees = 50.”

Inside the rectangle, draw three overlapping circles. Label one circle Soup, one Sandwich, and one Salad. In the Soup circle, write 4 in the part that does not overlap any other circle, write 2 in the part that only overlaps the Sandwich circle, write 8 in the part that overlaps all three circles, and write 3 in the part that overlaps only the Salad circle. In the Sandwich circle, write 15 in the part that does not overlap any other circle and write 10 in the part that only overlaps the Salad circle. In the Salad circle, write 8 in the part that does not overlap any other circle. Write 0 outside all the circles.

$A\cap <!--\; \cap \; -->\left(B\cap <!--\; \cap \; -->C\right)$

First, evaluate the operation inside the parentheses.

$B\cap <!--\; \cap \; -->C=\left\{0,2,4,6,8,10,12\right\}\cap <!--\; \cap \; -->\left\{0,3,6,9,12\right\}$

The intersection of two sets includes the elements that are common to both sets.

$B\cap <!--\; \cap \; -->C=\left\{0,6,12\right\}$

$A\cap <!--\; \cap \; -->\left(B\cap <!--\; \cap \; -->C\right)=\left\{0,1,2,3,4,5,6\right\}\cap <!--\; \cap \; -->\left\{0,6,12\right\}$

$A\cap <!--\; \cap \; -->\left(B\cap <!--\; \cap \; -->C\right)=\left\{0,6\right\}$

$A\cap <!--\; \cap \; -->(B\cap <!--\; \cap \; -->C)=\{0,1,2,3,4,5,6\}\cap <!--\; \cap \; -->\{0,6,12\}=\{0,6\}$

$\left(A\cap <!--\; \cap \; -->B\right)\cup <!--\; \cup \; -->\left(A\cap <!--\; \cap \; -->C\right)$

Evaluate the expression in the left parentheses.

$A\cap <!--\; \cap \; -->B=\left\{0,1,2,3,4,5,6\right\}\cap <!--\; \cap \; -->\left\{0,2,4,6,8,10,12\right\}$

The intersection of two sets includes the elements that are common to both sets.

$A\cap <!--\; \cap \; -->B=\left\{0,2,4,6\right\}$

Evaluate the expression in the right parentheses.

$A\cap <!--\; \cap \; -->C=\left\{0,1,2,3,4,5,6\right\}\cap <!--\; \cap \; -->\left\{0,3,6,9,12\right\}$

$A\cap <!--\; \cap \; -->C=\left\{0,3,6\right\}$

Now, you can find $\left(A\cap <!--\; \cap \; -->B\right)\cup <!--\; \cup \; -->\left(A\cap <!--\; \cap \; -->C\right)$.

$\left(A\cap <!--\; \cap \; -->B\right)\cup <!--\; \cup \; -->\left(A\cap <!--\; \cap \; -->C\right)=\left\{0,2,4,6\right\}\cup <!--\; \cup \; -->\left\{0,3,6\right\}$

$\left(A\cap <!--\; \cap \; -->B\right)\cup <!--\; \cup \; -->\left(A\cap <!--\; \cap \; -->C\right)=\left\{0,2,3,4,6\right\}$

$(A\cap <!--\; \cap \; -->B)\cup <!--\; \cup \; -->(B\cap <!--\; \cap \; -->C)=\{0,2,4,6\}\cup <!--\; \cup \; -->\{0,3,6\}=\{0,2,3,4,6\}$

$\left(A\cup <!--\; \cup \; -->C^{\prime }\right)\cap <!--\; \cap \; -->\left(B\cup <!--\; \cup \; -->C^{\prime }\right)$

$C^{\prime }$, the complement of C, includes the elements of U that are not in C.

$C=\{0,3,6,9,12\}$ and $U=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$

$C^{\prime }=\{1,2,4,5,7,8,10,11\}$

Evaluate the expression in the left parentheses.

$A\cup <!--\; \cup \; -->C^{\prime }=\left\{0,1,2,3,4,5,6\right\}\cap <!--\; \cap \; -->\left\{1,2,4,5,7,8,10,11\right\}$

The intersection of two sets includes the elements that are common to both sets.

$A\cup <!--\; \cup \; -->C^{\prime }=\left\{0,1,2,3,4,5,6,7,8,10,11\right\}$

Evaluate the expression in the right parentheses.

$B\cup <!--\; \cup \; -->C^{\prime }=\left\{0,2,4,6,8,10,12\right\}\cap <!--\; \cap \; -->\left\{1,2,4,5,7,8,10,11\right\}$

$B\cup <!--\; \cup \; -->C^{\prime }=\left\{0,1,2,4,5,6,7,8,10,11,12\right\}$

Now, you can find $\left(A\cup <!--\; \cup \; -->C^{\prime }\right)\cap <!--\; \cap \; -->\left(A\cup <!--\; \cup \; -->C^{\prime }\right)$.

$\left(A\cup <!--\; \cup \; -->C^{\prime }\right)\cap <!--\; \cap \; -->\left(B\cup <!--\; \cup \; -->C^{\prime }\right)=\left\{0,1,2,3,4,5,6,7,8,10,11\right\}\cap <!--\; \cap \; -->\left\{0,1,2,4,5,6,7,8,10,11,12\right\}$

$\left(A\cup <!--\; \cup \; -->C^{\prime }\right)\cap <!--\; \cap \; -->\left(B\cup <!--\; \cup \; -->C^{\prime }\right)=\left\{0,1,2,4,5,6,7,8,10,11\right\}$

$(A\cup <!--\; \cup \; -->C^{\mathrm{\prime <!--\; \prime \; -->}})\cap <!--\; \cap \; -->(B\cup <!--\; \cup \; -->C^{\mathrm{\prime <!--\; \prime \; -->}})=\{0,1,2,3,4,5,6,7,8,10,11\}\cap <!--\; \cap \; -->\{0,1,2,4,5,6,7,8,10,11,12\}=\{0,1,2,4,5,6,7,8,10,11\}$

First, draw the left side of the equation.

For A ∩ B, draw a rectangle to represent the universal set U. Label the rectangle “U.” Inside the rectangle, draw overlapping circles to represent two sets. Label one circle “A” and the other circle “B.” Shade the overlap.

Now, draw the complement of the intersection. Draw the same Venn diagram, but shade everything except the overlap to represent ${\left(A\cap <!--\; \cap \; -->B\right)}^{\mathrm{\prime <!--\; \prime \; -->}}$.

Now, draw the right side of the equation: $A^{\prime }\cup <!--\; \cup \; -->B^{\prime }$.

First, draw $A^{\prime }$. Draw a rectangle to represent the universal set U. Label the rectangle “U.” Inside the rectangle, draw a circle and label the circle “A.” Shade everything outside the circle.

To the right of the rectangle, draw the ∪ symbol.

Now, draw $B^{\prime }$. Draw a rectangle to represent the universal set U. Label the rectangle “U.” Inside the rectangle, draw a circle and label the circle “B.” Shade everything outside the circle.

To the right of the second rectangle, draw the equal symbol: =.

For $A^{\prime }\cup <!--\; \cup \; -->B^{\prime }$, draw a rectangle to represent the universal set U. Label the rectangle “U.” Inside the rectangle, draw overlapping circles to represent two sets. Label one circle “A” and the other circle “B.” Shade everything except for the overlap.

The left side of the equation is:

The right side of the equation is given by:

Alternate answer: cardinal value

cardinality

A well-defined set clearly communicates whether the element is a member of the set or not. This is not well defined because not everyone’s list would be the same.

not a well-defined set

A well-defined set clearly communicates whether the element is a member of the set or not. You can list the names of the twelve people who had walked on the moon as of March 12, 2021.

M = {Neil Armstrong, Edwin “Buzz” Aldrin, Charles “Pete” Conrad, Alan Bean, Alan Shepard Jr., Edgar Mitchell, David Scott, James Irwin, John Young, Charles Duke, Eugene Cernan, Harrison Schmitt}

12

Equal sets have the same elements, without regard to order. Equivalent sets have the same number of elements. These sets are equivalent but not equal because they have the same number of elements, but they have different elements.

equivalent, but not equal

While there are a great number of butterflies in the world, it is a finite number. This is a finite set.

finite

The roster method lists the individual members of the set inside curly braces, separated by commas. Once you have established the pattern, you can use the ellipsis to avoid needing to list them all. Be sure to list the final member of the set (Z) to avoid giving the appearance of an infinite set.

$A=\left\{A,B,C,D,\cdots <!--\; \cdots \; -->,Z\right\}$

Set builder notation uses a letter to represent a sample member of the set, a vertical bar, and then a description of a sample member.

$A=\left\{a|a\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{E}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{h}\mathrm{a}\mathrm{l}\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{b}\mathrm{e}\mathrm{t}\right\}$

Roster method: $\{\text{A, B, C, }\dots <!--\; \dots \; -->\text{, Z}\},$ and set builder notation: $\{x|x\text{ is a capital letter of the English alphabet}\}$

True. A is a subset of itself because every element of A is contained in A.

true

Use the formula for the number of subsets of set A with n elements: $2^{n(A)}$.

Set A has ten elements.

$2^{n(A)}=2^{10}=1024$

$2^{10}=1024$

Equivalent sets have the same number of elements. Because $n\left(A\right)=n\left(B\right)$, the sets are equivalent.

equivalent

equal

Equal sets have the same elements. For each set, there can be no element that is not an element of the other set. Sets A and B are equal because every member of A is in B and every member of B is in A.

To help you remember, think of the intersection of two streets. The intersection is the pavement that the two streets have in common.

intersection

To help you remember, think of the union of the United States. A union joins things together.

union

$A\cup <!--\; \cup \; -->B$

It is very helpful that the union symbol $(\cup <!--\; \cup \; -->)$ looks somewhat like a U.

The intersection of two sets includes the elements they have in common. Because A is a subset of B, all the elements of A are also in B. There can be no other elements in B that are in A. Thus, $A\cap <!--\; \cap \; -->B=A$.

$A$

The union of two sets includes all the members of the first set and all the members of the second set. Do not list duplicates twice.

Because A is a subset of B, all the elements of A are also in B. There can be no elements in A that are not in B. Thus, $A\cup <!--\; \cup \; -->B=B$.

$B$

Disjoint sets have no elements in common. The intersection of two sets includes the elements they have in common. Because set A and set B are disjoint, their intersection is the empty set.

empty

overlap

By beginning with the region involving the most overlap, you start with the regions involving the most sets and the most complexity. Work toward the simpler parts of the diagram.