Relations

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

Math 320
Homework 7 - Relations

4.1

Relations

4.10
Given that A, B, and C are sets, prove that
(A ∪ B) − (A ∩ B ∩ C) = [A − (B ∩ C)] ∪ [B − (A ∩ C)]

Figure 1: Venn Diagram of Conjecture, 4-10
Suffice it to prove that:
1) (A ∪ B) − (A ∩ B ∩ C) ⊆ [A − (B ∩ C)] ∪ [B − (A ∩ C)], and
2) [A − (B ∩ C)] ∪ [B − (A ∩ C)] ⊆ (A ∪ B) − (A ∩ B ∩ C)
(A ∪ B) − (A ∩ B ∩ C) ⊆ [A − (B ∩ C)] ∪ [B − (A ∩ C)]
Proof: x ∈ (A ∪ B) − (A ∩ B ∩ C), means that x is in A or in B, but not in the intersection of A, B, and C.
If x is in A, then it must not be in both B and C, in other words A − (B ∩ C).
If x is in B, then it must not be in both A and C, in other words B − (A ∩ C).
Because x is in either A or B, then it lies in the union of A − (B ∩ C) and B − (A ∩ C).
Therefore, (A ∪ B) − (A ∩ B ∩ C) ⊆ [A − (B ∩ C)] ∪ [B − (A ∩ C)].
[A − (B ∩ C)] ∪ [B − (A ∩ C)] ⊆ (A ∪ B) − (A ∩ B ∩ C)
Proof: For x in [A − (B ∩ C)] ∪ [B − (A ∩ C)] ⊆ (A ∪ B) − (A ∩ B ∩ C)
x is either in [A − (B ∩ C)] or in [B − (A ∩ C)].
If x is in [A − (B ∩ C)], then this means it is in A, but not in both B and C.
If x is in [B − (A ∩ C)], then it is in B, but not both A and C.
In other words, x is in A or B, but it is not in A, B, and C.
Therefore, [A − (B ∩ C)] ∪ [B − (A ∩ C)] ⊆ (A ∪ B) − (A ∩ B ∩ C).
Because the left is a subset of the right, and the right is a subset of the left, then the two sets are equivalent
and the equality is true.

Math 320
Homework 7 - Relations

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

4.11
a) Let S and T be relations. Prove that if S ⊆ T and x is a member of the domain of S, then S[x] ⊆ T [x].
Proof :
Suppose S ⊆ T and x is a member of the domain of S.
Then there exists y such that y ∈ S[x].
If y ∈ S[x], then y ∈ T [x] because S ⊆ T .
Therefore, S[x] ⊆ T [x].
b) Let A and B be nonempty sets. Prove that
(A × B)[x] = B if x ∈ A and (A × B)[x] = ∅ if x 6∈ A
Proof :
(A × B)[x] means that (x, y) ∈ (A × B).
(A × B) means that for every x ∈ A and every y ∈ B, (x, y) ∈ (A × B).
Therefore, if x ∈ A, then the ordered pair for every y in B will be in (A × B).
Thus, the set (A × B)[x] for every x ∈ A is B, and (A × B)[x] ⊆ B.
Take any y ∈ B. Then if x ∈ A, the ordered pair (x, y) ∈ (A × B).
Then y ∈ (A × B)[x].
Therefore, B ⊆ (A × B)[x].
Thus, if x ∈ A, (A × B)[x] = B.
Now, if x 6∈ A, then no ordered pairs (x, y) will be in A × B.
Therefore, (A × B)[x] ⊆ ∅.
Because ∅ ⊆ (A × B)[x].
Therefore, ∅ = (A × B)[x]

4.13
Let N denote the set of all natural numbers. Let R = {(a, b) ∈ N × N : a divides b}. List five members of
R[7], and list five members of R[14]. For which n ∈ N is it true that R[n] = N?
(7, 7), (7, 14), (7, 21), (7, 28), (7, 35) ∈ R[7]
(14, 14), (14, 28), (14, 42), (14, 56), (14, 70) ∈ R[14]
If R[n] = n, then (n, 1), (n, 2), · · · ∈ R[n] which means that n = 1.

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

Math 320
Homework 7 - Relations

4.2

Cartesian Graphs and Directed Graphs

4.19
Let A = {1, 2, 3, 4, 5} be the set of vertices of the accompanying directed graph. Draw the Cartesian Graph
of the relation on A associated with this directed graph.
2

3

1
4

5

Figure 2: Directed Graph of the relation on A

Figure 3: Cartesian Graph of the relation on A

4.20
Let A = {1, 2, 3, 4, 5} as in Exercise 19. Draw the directed graph associated with each of the following
relations on A.

3

2
1

5

4

Figure 4: 4.20a) R = {(a, b) ∈ A × A : a divides b}

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

Math 320
Homework 7 - Relations

4

2

1

3

5

Figure 5: 4.20b) U = {(a, b) ∈ A × A : a 6= b}

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

Math 320
Homework 7 - Relations

3

4

1

2

5

Figure 6: 4.20c) E = {(a, b) ∈ A × A : a + b is even}
5

2

3

1
4

Figure 7: 4.20d) O = {(a, b) ∈ A × A : a + b is odd}

4.25
Let R be a relation such that R−1 ⊆ R. Must R by symmetric? Prove your answer.
A relation R such that R−1 ⊆ R must be symmetric.
Proof: Let S be the set on which R is defined and x and y be elements in S such that (x, y) is in R.
Then (y, x) is in R−1 and,
(y, x) is in R because R−1 ⊆ R
Then for any x, y in S such that (x, y) in R, (y, x) in R.
Therefore, R is symmetric.

Math 320
Homework 7 - Relations

4.3

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014

Equivalence relations

4.34
Let R be the set of all real numbers and consider the following subsets of R × R. Which of these is i) Reflexive
on R ii) Symmetric? iii) Transitive?
a) R1 = {(x, y) ∈ R × R : xy = 0}
Reflexivity: Not reflexive for (1, 1).
Symmetry: Not symmetric for (1, 0).
Transitivity: Not transitive for (1, 0) and (1, 1)
b) R2 = {(x, y) ∈ R × R : |x − y| < 5}
Reflexivity: Reflexive.
• Proof:
Let x ∈ R.
x−x=0<5
Therefore, (x, x) ∈ R.
Symmetry: Symmetric.
• Proof:
Let x, y ∈ R such that (x, y) ∈ R
Then, |x − y| < 5 which means that −5 < x − y < 5.
Therefore, 5 > y − x > −5 or |y − x| < 5.
Thus, (y, x) ∈ R.
Transitivity: Not transitive for (−5, 0) and (0, 5).
c) R3 = {(x, y) ∈ R × R : xy 6= 0}
Reflexivity: Not reflexive for (0, 0).
Symmetry: Symmetric.
• Proof:
Let x, y ∈ R such that (x, y) ∈ R
Then, xy 6= 0 means yx 6= 0
Therefore, (y, x) ∈ R.
Transitivity: Transitive.
• Proof:
Let x, y, z ∈ R such that (x, y), (y, z) ∈ R
Then, xy 6= 0 and yz 6= 0.
xy 2 x 6= 0 =⇒ xz 6= 0
Therefore, (x, z) ∈ R.

Math 320
Homework 7 - Relations

d) R4 = {(x, y) ∈ R × R : x ≥ y}
Reflexivity: Reflexive.
• Proof:
Let x ∈ R.
x≥x
Therefore, (x, x) ∈ R.
Symmetry: Not symmetric for (1, 0).
Transitivity: Transitive.
• Proof:
Let x, y, z ∈ R such that (x, y), (y, z) ∈ R
Then, x ≥ y and y ≥ z.
x ≥ y ≥ z =⇒ x ≥ z
Therefore, (x, z) ∈ R.
e) R5 = {(x, y) ∈ R × R : x2 + y 2 = 1}
Reflexivity: Not reflexive for (1, 1).
Symmetry: Symmetric.
• Proof:
Let x, y ∈ R such that (x, y) ∈ R
Then, x2 + y 2 = 1. Also, y 2 + x2 = 1
Therefore, (y, x) ∈ R.
Transitivity: Not transitive for (1, 0), (0, 1).

Joshua Cook
Dr. Katherine Stevenson
Lindsay Justice
October 15, 2014