72 more on sets and logic

More on the Language of Sets
Continue from the previous Farmer Andy’s family

section with the given Farmer
Andy’s family photo. We define
the following sets Andy Beth Cathy Dan
U = {Farmer Andy’s family} = {a, b, c, d}
R = {member who is wearing something red} = {a, b, d}
G = {member who is wearing something green} = {a, c}
B = {member who is wearing something blue} = {b, c, d}
F = {member who is a girl} = {b, c}
M = {member who is a boy} = { a, d}


Each description is called an attribute.


Example A. Name another attribute in Farmer Andy’s family
photo and list the respective subset.


For example, H = {member who has black hair} = {a, c}.


For example, H = {member who has black hair} = {a, c}.
Note that H = G, i.e. different attributes may define the same set.

R = {wearing something red} = {a, b, d} Farmer Andy’s family

G = {wearing something green} = {a, c}
B = {wearing something blue} = {b, c, d}
F = {is a girl} = {b, c}
M = {is a boy} = {a, d} Andy Beth Cathy Dan

New attributes may be formed by linking other attributes using
the three connective words “not”, “and” and “or”,



the three connective words “not”, “and” and “or”, and each of
these words correspond to an operation of the corresponding
set(s) involved.



set(s) involved.
The Complementary Operation and “Not”



set(s) involved.
Given an universal set U and A U, the complement of A,
∩
denoted as AC, consists of elements that are not in A,



set(s) involved.
∩
denoted as AC, consists of elements that are not in A, that is,
AC = {x | x ϵ A}. (See diagram ) U
C
A
A
xϵ A

The complement of A



set(s) involved.
∩
AC
For example, using the sets above, A
RC = xϵ A

The complement of A



set(s) involved.
∩
AC
RC = {x | x is not “wearing something red”} xϵ A

The complement of A



set(s) involved.
∩
AC
RC = {x | x is not “wearing something red”} xϵ A
= {c}. The complement of A



The Intersection Operation and “And”



Given two set A and B, the intersection of A and B,
denoted as A ∩ B consists of elements that common to
both A and B,



both A and B, that is,
A ∩ B = {x | x is a member of A and also a member of B}
= {x | x ϵ A and x ϵ B }



= {x | x ϵ A and x ϵ B } (See diagram ).
U
A A∩B B

The intersection A ∩ B



For example, using the sets above,
U
R∩B=
A A∩B B




U
R ∩ B = {x | x is “wearing red” and is also
“wearing blue”} A A∩B B




U

= {x | x is “wearing both red and blue”}



U

= {x | x is “wearing both red and blue”}
= {b, d}. The intersection A ∩ B



The Intersection Operation and “Or”



Given two set A and B, the union of A and B,
denoted as A U B consists of elements that belong to
either A or B,



either A or B, that is,
A U B = {x | x is a member of A or a member of B, or both}
= {x | x ϵ A or x ϵ B }



= {x | x ϵ A or x ϵ B } (See diagram ).
U
A AUB B

The union A U B



U
GUF
= A AUB B

The union A U B



U
GUF
= {x | “x is wearing green” or “x is a girl”} A AUB B

The union A U B



U
GUF
= {x | “x is wearing green” or “x is a girl”} A AUB B

= {a, c} U {b, c}
= {a, b, c}. The union A U B


F = {Is a girl} = {b, c}
M = {Is a boy} = {a, d} Andy Beth Cathy Dan

Example B. Using the above, name and describe each of the
following sets in the form of {x | x is ###} then list its elements.
a. BC



a. BC
BC is the “complement of B”.



a. BC
BC = {x | x is not “wearing something blue”}



a. BC
BC = {x | x is not “wearing something blue”} =
{a}.



a. BC
{a}. ∩ F
b. R



a. BC
{a}. ∩ F
b. R
R ∩ F is the “intersection of R with F”.



a. BC
{a}. ∩ F
b. R
R ∩ F = {x | x is “wearing something red” and “is a girl”}



a. BC
{a}. ∩ F
b. R
R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.



a. BC
{a}. ∩ F
b. R
c. G U M



a. BC
{a}. ∩ F
b. R
c. G U M
G U M is the “union of G with M”.



a. BC
{a}. ∩ F
b. R
c. G U M
G U M is the “union of G with M”. G U M
= {x | x is “wearing something green” or “x is a boy”}.



a. BC
{a}. ∩ F
b. R
c. G U M
G U M is the “union of G with M”. G U M = {a, c, d}
= {x | x is “wearing something green” or “x is a boy”}.



Mutually Exclusive Sets and Basic Counting



Given two sets A and B, we say A and B
are mutually exclusive if A ∩ B = Φ = { },
i.e. A and B do not have any common items.



Given two sets A and B, we say A and B U
are mutually exclusive if A ∩ B = Φ = { }, A
B
Mutually exclusive sets



B
Using the sets above, we have that Mutually exclusive sets
BC = {not wearing anything blue} = {a}
is mutually exclusive from the set of F = {is a girl} = {b, c},



B
i.e. BC ∩ F = Φ.



B
i.e. BC ∩ F = Φ.
Any set B and its complement BC, by definition, are always
mutually exclusive, or that BC ∩ B = Φ.



B
i.e. BC ∩ F = Φ.
Any set B and its complement BC, by definition, are always
mutually exclusive, or that BC ∩ B = Φ. Nothing can possess an
attribute while not possessing the same attribute simultaneously.



For the following discussion, we assume all the sets are finite.



Recall o(B) is the number of elements in the set B.



The set U = {a, b, c, d} can be split into two non-overlapping sets,



R = {wearing red} = {a, b. d} with o(R) = 3, and
RC = {not wearing any red} = {c} with o(RC) = 1.




U
RC: R:
d
c b
a



Each member of Farmer Andy’s family is in precisely one of
these two sets. U
RC: R:
d
c b
a



these two sets. U
RC: R:
The Complementary Counting Principle d
b
Given U be the universal set and that A U, ∩ c
a
then o(A) + o(AC) = o(U),
or that o(A) = o(U) – o(AC).



these two sets. U
RC: R:
The Complementary Counting Principle d
b
Given U be the universal set and that A U, ∩ c
a
then o(A) + o(AC) = o(U),
or that o(A) = o(U) – o(AC). o(R) + o(RC) = o(U) = 4

Sometime it’s easier to count the complement of a set.
Example C. The Following is a break down of 100 audiences
at a showing of “Mary Porter”. Minors are teens and children.
How many in the audience are minors?
Adult Child
Teen
Male 5 17 18
Female 15 26 19

We may add up the numbers of the Adult Child
Teen
teens and children to obtain the
Male 5 17 18
answer
Female 15 26 19

Teen
Male 5 17 18
answer or we may observe that
Female 15 26 19
a non–minor is an adult.

Teen
Male 5 17 18
Female 15 26 19
a non–minor is an adult. Since there are 5 + 15 = 20 adults out
of the 100 viewers,

Teen
Male 5 17 18
Female 15 26 19
of the 100 viewers, we see that there are 100 – 20 = 80 minors.

Teen
Male 5 17 18
Female 15 26 19
Counting the Union of Two Sets

Teen
Male 5 17 18
Female 15 26 19
Recall that G = {wearing something green} = {a, c} U
so o(G) = 2,
G:

a c

Teen
Male 5 17 18
Female 15 26 19
so o(G) = 2, F = {is a girl} = {b, c} so o(F) = 2,
G: F:

a c b
d

Teen
Male 5 17 18
Female 15 26 19
G: F:
and G U F = {a, b, c} with o(G U F) = 3.
a c b
d

Teen
Male 5 17 18
Female 15 26 19
G: F:
But note that o(G U F) = 3 ≠ o(G) + o(F) = 4. a c b
d
G U F = {a, b, c}

Teen
Male 5 17 18
Female 15 26 19
G: F:
But note that o(G U F) = 3 ≠ o(G) + o(F) = 4. a c b
d
The sum o(G) + o(F) is higher because their
G U F = {a, b, c}
common element “c” is counted twice as

The Formula for Counting the Union (A U B)
Given two sets and A and B,
o(A U B) = o(A) + o(B) – o(A ∩ B)

o(A U B) = o(A) + o(B) – o(A ∩ B)
Example D. a. Given two squares each with an area of 10 cm 2,
find the total area they cover if we overlay them as shown.

10 cm2 10 cm2 5 cm2

A B AUB

o(A U B) = o(A) + o(B) – o(A ∩ B)

10 cm2 10 cm2 5 cm2

A B AUB

The area the joint–squares covered is
10 + 10 – (the overlapping portion)

o(A U B) = o(A) + o(B) – o(A ∩ B)

10 cm2 10 cm2 5 cm2

A B AUB

10 + 10 – (the overlapping portion) total area of
= 10 + 10 – 5 = 15 cm2, A U B = 15 cm
2

o(A U B) = o(A) + o(B) – o(A ∩ B)

10 cm2 10 cm2 5 cm2

A B AUB

10 + 10 – (the overlapping portion) total area of
= 10 + 10 – 5 = 15 cm2, A U B = 15 cm
2

or that + is 5 + 10 = 15 cm2

b. To graduate from a school, the student must pass the
algebra test and the biology test.

algebra test and the biology test. In a class, there are ten
students that have passed the algebra test.

students that have passed the algebra test. There are also ten
students that have passed the biology test.

students that have passed the biology test. In these groups,
five students passed both the algebra and the biology tests.

How many students have passed at least one required test?

Let A = {students who passed algebra},
B = {students who passed biology},

B = {students who passed biology}, then
A ∩ B = {students who passed both tests} and

A U B = {students who passed at least one test}.


o(A)=10 o(A ∩ B)=5 o(B)=10

Hence o(A U B) = o(A) + o(B) – o(A ∩ B) =

o(A)=10 o(A ∩ B)=5 o(B)=10

Hence o(A U B) = o(A) + o(B) – o(A ∩ B) = 10 + 10 – 5 = 15

o(A)=10 o(A ∩ B)=5 o(B)=10

or that there are 15 people
who passed at least one test.
o(A)=10 o(A ∩ B)=5 o(B)=10

o(A U B) = o(A) + o(B) – o(A ∩ B) = 15

or that there are 15 people
who passed at least one test.
o(A)=10 o(A ∩ B)=5 o(B)=10
We may use the counting
formula for the union to
deduce the order of A. o(A U B) = o(A) + o(B) – o(A ∩ B) = 15

Example E. As in example D, suppose there are 14 students
that have passed the biology test and in this group,
5 students passed both the algebra and the biology.
There are 20 students that have passed at least one test.
How may students have passed only algebra and need to pass
biology? How many students have passed the algebra test?

Again A = {passed algebra},
B = {passed biology},
A ∩ B = {passed both tests} and
A U B = {passed at least one test}.
We’ll use the diagram method.

Again A = {passed algebra}, A

Again A = {passed algebra}, A B

Imagine the 20 students of AUB are
total number of students:
standing in two overlapping squares o(A U B) = 20
as shown.

A ∩ B = {passed both tests} and 14
as shown. There are 14 students
inside B.

A ∩ B = {passed both tests} and 6 14
inside B. So there are 6 outside who have passed only algebra

A ∩ B = {passed both tests} and 6 5 14
inside B. So there are 6 outside who have passed only algebra
and need to pass biology and o(A) = 6 + 5 = 11.

Or we may solve this algebraically by setting,
o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5
in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),

o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5
we have that
20 = o(A) + 14 – 5, or
11 = o(A)

o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5
we have that
20 = o(A) + 14 – 5, or
11 = o(A)
Hence there are 11 students that passed algebra and out of
which 9 still have to pass the biology test.

o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5
we have that
20 = o(A) + 14 – 5, or
11 = o(A)
Finally we note that if A and B are mutually exclusive, then
o(A U B) = o(A) + o(B) since o(A ∩ B) = o(Φ) = 0.

o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5
we have that
20 = o(A) + 14 – 5, or
11 = o(A)
Finally we note that if A and B are mutually exclusive, then
o(A U B) = o(A) + o(B) since o(A ∩ B) = o(Φ) = 0.
It A and B does
not overlap A U B, then
A B Area of A U B
= Area(A) + Area(B)
or
o(A U B) = o(A) + o(B)
AUB

If I and J are two sets of intervals, then I U J is the set obtained
by gluing the two intervals together, and that I ∩ J is the
overlapping part of the two intervals.

Example E. Given intervals I, J , and K, perform the following
set operation. Give the answers in intervals and in inequalities.
–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3
0
K

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J
0
K

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J –3
K 0 so K U J is 0

K U J = {–3 < x} or (–3, ∞).

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J –3
K 0 so K U J is 0

K U J = {–3 < x} or (–3, ∞).
b. I ∩ K

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J –3
K 0 so K U J is 0

K U J = {–3 < x} or (–3, ∞).
b. I ∩ K
–4
I

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J –3
K 0 so K U J is 0

K U J = {–3 < x} or (–3, ∞).
b. I ∩ K
–4 –3 –1 K 1
0
I

–4 –1
I: 0 J = {x > –2} K= {–3 < x ≤ 1}

a. K U J
–3 –2 1 J –3
K 0 so K U J is 0

K U J = {–3 < x} or (–3, ∞).
b. I ∩ K
–4 –3 –1 K 1
0
I
The common or overlapping portion is shown.
I ∩ K = {–3 < x < –1} or (–3, –1).

Exercise A. From the photo of Farmer Andy’s family in their
new dresses, we define the following sets
U = { the family} Farmer Andy’s family in new dresses

R = {red}
G = {green}
B = {blue}
F = {female}
M = {male} Andy Beth Cathy Dan
T = {has triangular decoration} C = {has circular decoration}
Exercise A.
1.a. List each set defined above and pick out all the mutually
exclusive pairs by inspection.
List the elements of the following sets. Name and describe
each set in the form of {x | x is ###}.
b. FC c. MC d. CC e. TC f. GC g. UC
h. T U F i. F ∩ M j. B U C k. C U FC l. GC ∩ T k. TC ∩ CC

2. From Example C. we have the table as shown.
Let A = {Adult} ,T = {Teen}, Adult Child
Teen
C = {Child}, F = {Female},
Male 5 17 18
M = {Male},
Female 15 26 19
Which pair of sets are mutually exclusively?
Name and describe each of the following sets in the form of
{x | x is ###} and find its order.
b. TC c. CC d. MC e. FC f. AC
g. A U F h. T ∩ C i. F ∩ MC j. AC U C k. CC U FC
3. Given intervals I, J , and K, do the set operation. Draw.
Put the answer in the interval and the inequality notations.
–8 –2 0 J: x < –3½ K: –5 ≤ x < 2
I:
a. K U J b. K ∩ I c. I U J d. K U I
e. I ∩ J f. K ∩ J g. (K ∩ J) U I h. (K U J) ∩ I

4. Given following shapes and their areas, find the total area if
we glue them together as shown.

a.
10 cm2 20 cm2 5 cm2
A B AUB

b.
8 12 5
A B AUB
c. Each overlap is 1
A 20 A

B 20
C B
AUBUC
C 20

5. Given following shapes and their two areas and we glue them
together as shown, find the missing area.

a.
Area of A U B = 31
A 20 cm2 5 cm2
Area of A = ?
B

b.
8 B Area of A U B = 19
5
A Area of B = ?
c.
A A

B Area of
C B A U B U C = 45
C
Each overlap is 1 Area of each = ?
A =B=C

7. Farmer Andy produces nuts and sells them in three types of
bags. These are: peanuts only, cashews only, and
mixed peanuts–cashews. We bought many bags of each type.
Draw a diagram that reflects each of the following questions.
Answer the question if possible. If not, explain why not.
a. Out of the bags we bought, 30 contain peanut and
12 contain cashews only. How many bags we bought in total?
How many peanuts only bags did we buy?
How many mixed begs did we buy?
b. Out of the bags we bought, 30 contain peanut only and
12 contain cashews only. How many bags we bought in total?
c. Out of the bags we bought, 30 contain peanut, 12 contain
cashews and there are 6 of mixed type. How many bags did
we buy in total and how may of each type did we buy?
d. Out of the 40 bags we bought, 30 contain peanut and 12 are
of mixed type. How many bags of each type did we buy?

8. Everyone in an extended family of 100 people have either
genetic traits A or B, or both. Answer each of the following
questions with the given information. Define and draw each set.
a. There are 85 people who have trait A. Out of the people with
trait A , 27 people have both traits A an d B. How many people of
each type did we have?
b. There are 85 people who have trait A and there 27 people
with B. How many people of each type did we have?
c. Is it possible to have 85 people who don’t have trait A and
there 15 people with both traits A, B? What may we
conclude?
d. Is it possible to have 85 people who don’t have trait B and
there 16 people with both traits A, B? What may we
conclude?

72 more on sets and logic

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