Chapter 1

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

This is not a well-defined set.

This is a well-defined set.

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

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

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

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

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

$n(P)=0$

$n(A)=26$

finite

infinite

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

neither

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

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

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

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

Any of the following combinations of three members would work:

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

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

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

512

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

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}.

There are multiple possible solutions. Each set must contain two players, but both players cannot be the same, otherwise the two sets would be equal, not equivalent. For example, {Maria, Shantelle} and {Angie, Maria}.

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 $L\subset <!--\; \subset \; -->U$.

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.

The universal set is the set of integers. Draw a rectangle and label it with $U=\text{Integers}$. Next, draw a circle in the rectangle and label with Natural numbers.

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

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

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

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

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

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

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

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

33

113

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

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

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

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

127

50

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

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

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

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

40

0

27

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

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

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

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

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

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

The left side of the equation is:

The right side of the equation is given by:

set

cardinality

not a well-defined set

12

equivalent, but not equal

finite

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

subset

To be a subset of a set, every member of the subset must also be a member of the set. To be a proper subset, there must be at least one member of the set that is not also in the subset.

empty

true

$2^{10}=1024$

equivalent

equal

relationship

universal

disjoint or non-overlapping

complement

disjoint

intersection

union

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

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

$A$

$B$

empty

$n(A\cup <!--\; \cup \; -->B)=n(A)+n(B)-<!--\; -\; -->n(A\cap <!--\; \cap \; -->B)$

overlap

central

intersection of all three sets, $A\cap <!--\; \cap \; -->B\cap <!--\; \cap \; -->C$

parentheses, complement

equation, true