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Game Math Chapter 1 Solutions
Game Math Chapter 1 Solutions
Frank Luna
www.gameinstitute.com
Sunday, December 12, 2004
1. Determine which of the following are sets:
a. F = { p | p is a prime }
b. M = (1, 2, 3, 4, … )
c. R = {}
d. Q = { { x , y } | x is an integer, y is an integer, x > y }
1a. F = { p | p is a prime} is read as F is the set of all p such that p is a prime number, and
yes it is a set.
1b. M = (1, 2, 3, 4, …) is not a set because the parenthesis notation in this example has
no particular set interpretation.
1c. R = {} is a set; in particular, it is the empty set (or null set) ∅ .
x ∈ ] , x > y } is the set of all sets with two integer elements such
that one element is greater than the other element. This set consists of all subsets of ]
with exactly two elements.
,
2. Translate the following sets from English into set builder notation:
a. A is the set that contains the names of all electronic components.
b. L is a set that contains all integers that are not less than the square root of five.
c. W is a set that contains all other sets.
d. Z is a set that contains everything that is not contained in the universal set.
2a. A = { x | x is the name of an electronics component}
2b. L = { x | x ∈ ] ,
x
≥
5
}
2c. W = { x | x is a set, x ≠ }
x ∈ } = ∅ . Remark: the universal set contains everything. So the
complement of everything is nothing.
'
Frank Luna
Page 1
www.gameinstitute.com
1d. Q = {{ x , y } |
2d. Z = { x |
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Game Math Chapter 1 Solutions
3. Translate the following sets from set builder notation to English:
= ∈ \
b. D = { s | s is a string of letters }
c. P S = { x |
Gyy y
|
,
0
}
x ⊂ }
S
d. {
}
Nxx x
= ∈ ]]
|
+
,
−
= ∈ \ reads as, G is the set of all y such that y is an element of the real
numbers \ and y is greater than zero.
|
,
0
}
3b. D = { s
s | is a string of letters} reads, D is the set of all s such that s is a string of
letters.
= | reads, P is the set of all x such that x is a proper subset of S . The ⊆
symbol is used to denote a subset and ⊂ is sometimes used to denote a proper subset.
Recall from the readings that every set is also a subset of itself (i.e.,
x
x
⊂
S
B ⊆ ). Sometimes
we want to refer to a subset A of a set B such that
A ≠ . To do this we define a proper
B
subset . A is a proper set of B , denoted as
A ⊂ , if and only if A is a subset of B and A is
B
⊂ , . Now, the author did not discuss the proper
subset symbol so maybe the author meant { }
A
B
⇔
A
⊆
B
A
≠
B
= | , which would say, P is the
set of all x such that x is a subset of S . In other words, P is the set of all subsets of S .
P S
x
x
⊆
S
Remark: The ⇔ symbol is the logical equivalence operator. It is used to denote that two
statements mean the same thing. It can be translated to English as “if and only if.”
= ∈ ]] . We translate the comma as and . So, N is the set of all x
such that x is a positive integer and x is a negative integer. This is the empty set since
there is no integer that is both negative and positive.
|
+
,
−
}
4. Prove that if A and B are sets, and A = B , then A is a subset of B , and B is a subset of A .
Let’s first recall the definition of a subset: A ⊆ if and only if for every x , if x ∈ then
x ∈ . Now, let x ∈ . From the hypothesis we have A = , so x ∈ . Hence A ⊆ .
Let x ∈ . From the hypothesis we have A = , so x ∈ . Hence B ⊆ .
Frank Luna
Page 2
www.gameinstitute.com
a. {
3a. {
Gyy y
3c. { }
P S
B
not B . In symbols,
3d. {
Nxx x
736239497.007.png
Game Math Chapter 1 Solutions
5. Prove that the intersection operator is distributive with respect to the union operator, or
draw a Venn diagram to illustrate this.
∩∪=∩∪∩. Let
us look at some Venn diagrams first to convince ourselves that this is true.
ABC AB AC
Figure 1: Left) ( )
∪ is shown in red; that is, ( )
∪ is all the elements in B or in C . Right) the
intersection ( )
A BC
∩∪ is shown in blue; that is, ( )
∩∪ is all the elements in A and in ( )
B C
∪ .
Figure 2: Left) ( )
∩ is shown in red; that is, ( )
∩ is all the elements in A and in B . Middle)
∩ is shown in blue; that is, ( )
∩ is all the elements in A and in C . Right) ( ) ( )
∩∪∩ is
shown in green; that is, ( ) ( )
AB AC
∩∪∩ is all the elements in ( )
∩ or ( )
AC
So from the Venn diagrams we see that graphically, the elements of ( )
∩∪ are the
∩∪∩. So we have a good intuitive feeling that the
relationship is true. However, a Venn diagram is not a formal proof. The usual proof
that ( ) ( ) ( )
AB AC
∩∪=∩∪∩ requires a small theorem of logic, namely that the
logical and operator is distributive with respect to the logical or operator. Since we
haven’t discussed any logic, we will instead prove this using a membership table , which
is similar to a truth table. The idea is as follows. An element is either in or out of a set
(these are the only possible cases). Because we only have three sets A , B , and C in this
problem, we can take a brute force approach and look at every possible way an arbitrary
element can be in these sets. If, for every possible situation, ( )
ABC
∩∪ and
Frank Luna
Page 3
www.gameinstitute.com
In symbols, we want to prove this relationship: ( ) ( )( )
BC
BC
A BC
A B
A B
( )
AC
AC
AB AC
A B
∩ .
ABC
same elements of ( ) ( )
ABC AB AC
736239497.008.png 736239497.009.png 736239497.001.png 736239497.002.png 736239497.003.png 736239497.004.png
Game Math Chapter 1 Solutions
∩∪∩ have the same membership value, then we conclude that they are
equivalent. We use the notation I to denote that an arbitrary element is in a set and O to
denote that an arbitrary element is outside a set. When studying the following
membership table it may help to recall the following:
Let x be an arbitrary element.
∈∪⇔∈ ∈.
• x A B
∈∩⇔∈ ∈.
• x A C
x A and x B
∈∩⇔∈ ∈.
• ( )
x A and x C
∈∩ ∪ ⇔∈ ∈ ∪ .
• ( ) ( ) ( ) ( )
B C
x A and x B C
( )
x AB AC x ABrx AC
Remark: Note how the union operator translates to an or statement and the intersection
operator translates to an and statement. This comes straight from the definitions of these
operators. Think about it and review the definitions if you do not see this.
A
B
C ( )
∪ ( )
A B
∩ ( )
AC
∩
∩∪ ( )( )
AB AC
∩∪∩
I
I
I
I
I
I
I
I
I
I
O
I
I
O
I
I
I
O
I
I
O
I
I
I
I
O
O
O
O
O
O
O
O
I
I
I
O
O
O
O
O
I
O
I
O
O
O
O
O
O
I
I
O
O
O
O
O
O
O
O
O
O
O
O
∩∪∩ always have the
same membership value, we conclude that they are equivalent; that is,
( ) ( ) ( )
ABC
∩∪ and ( ) ( )
AB AC
ABC AB AC
6. Calculate the Cartesian products of the following sets (list the ordered pairs explicitly
if the resulting set is finite):
a. A = { 1, 2, 3 }, B = { a, b, c }
b. A = { !, @, # }, B = { alpha, beta, gamma, delta }
c. A = { x | x is a real number }, B = { y | y is a letter of the alphabet }
d. A =
2 \]
Frank Luna
Page 4
www.gameinstitute.com
( ) ( )
AB AC
• x BC xBrxC
x A
∈∩∪∩⇔∈∩ ∈∩.
( )
BC
ABC
Since for every possible situation, ( )
∩∪=∩∪∩ .
736239497.005.png
Game Math Chapter 1 Solutions
A × contains every
combination of ordered pairs such that the first component is an element of A and the
second component is an element of B .
AB xy x Ay B
×= ∈ ∈ . In other words, the set
{
, |
,
}
B
A
×
B
=
{ } { } ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )
2
3
×
a
,
b
,
c
=
{
1
a
,
1
b
,
1
c
,
2
a
,
2
b
,
2
c
,
3
a
,
3
b
,
3
c
}
6b. { } { } =
A
( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( )( )
×
B
=
!
@,
#
×
α ,
,
,
γ
δ
{
δ
!
α
,
!
β
,
!
γ
,
!
δ
,
@,
α
,
@,
β
,
@,
γ
,
@,
δ
,
#
,
α
,
#
,
β
,
#
,
γ
,
#
,
6c. =
A
× B
{ x | x is a real number}×{ y | y is a letter in the alphabet} =
{( x, y ) | x is a real number, y is a letter in the alphabet}
6d.
A
=×=××=
\]\\] \ \ ]
2
{
( )
x y z x
,, |
∈ ∈ ∈
,
y
,
z
}
7. Determine whether the following mappings are functions:
a. 1 →2
3 →4
5 → 2
4 →1
b. 3 →2
1 →1
5 → 2
1 →3
7a. For reference, let us recall the definition of a function as stated in the text: “A
function f is a mapping from a set A to a set B such that with each element in set A , the
function associates exactly one item from set B .”
To be clear, in this problem, A = {1, 3, 5, 4} and B = {2, 4, 1}. Here f defines the
following mappings:
1 →
2
3 →
2
4
5 →
1
4 →
Each element in A is associated with exactly one element from set B . This agrees with
the definition, so it is a function.
Frank Luna
Page 5
www.gameinstitute.com
6a. Recall ( )
1

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