Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Discrete Mathematics
Relations
12.5. Let A, B and C denote sets.
• A ∪ B = B ∪ A:
Proof of A ∪ B ⊆ B ∪ A.
W Let x ∈ A ∪ B. Then x ∈ A and
x ∈ B. We have
(x
∈
A)
(x ∈ B). This is logically equivalent
W
to (x ∈ B) (x ∈ A). Thus x ∈ B ∪ A. Since this holds for
each element in A ∪ B, this implies A ∪ B ⊆ B ∪ A.
Proof of B ∪ A ⊆WA ∪ B. Likewise if x ∈ B and x ∈ A.
We haveW(x ∈ B) (x ∈ A). This is logically equivalent to
(x ∈ A) (x ∈ B). Thus x ∈ A ∪ B. Since this holds for each
element in B ∪ A, this implies B ∪ A ⊆ A ∪ B.
• A ∪ (B ∪ C) = (A ∪ B) ∪ C.
Proof of A ∪ (B ∪ C) ⊆W(A ∪ B) ∪ C. Let x ∈ A ∪ (B ∪ C).
This
that x ∈ A (x ∈ B ∪ C). This means that (x ∈
W means W
A) (x ∈ B x ∈ C). Because
W
W
W
W
(x ∈ A) (x ∈ B x ∈ C) = (x ∈ A x ∈ B) (x ∈ C)
we have x ∈ A ∪ B or x ∈ C, hence x ∈ (A ∪ B) ∪ C. Therefore
A ∪ (B ∪ C) ⊆ (A ∪ B) ∪ C.
Proof of (A∪B)∪C
⊆W
A∪(B ∪C). Likewise,Wif y ∈ (A∪B)∪C,
W
W
then (y ∈ A y ∈ B) (y ∈ C) = (y ∈ A) (y ∈ B y ∈ C)
implies y ∈ A ∪ (B ∪ C). Therefore (A ∪ B) ∪ C ⊆ A ∪ (B ∪ C).
• A∪∅=A
Proof of A ∪ ∅ ⊆ A. Let x ∈ A ∪ ∅. This means that x ∈ A or
x ∈ ∅. But x ∈ ∅ is impossable, because there is no element in
∅. So x has to be in A. Therefore A ∪ ∅ ⊆ A.
Proof of A ⊆ A ∪ ∅. Let x ∈ A, then this certainly implies
x ∈ A or x ∈ ∅, so x ∈ A ∪ ∅. Therefore A ⊆ A ∪ ∅.
• A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Proof of A∩(B∪C) ⊆ (A∩B)∪(A∩C). Let x ∈ A∩(B∪C) then
x ∈VA and x ∈W(B ∪ C). By the definitionVthis meansWthat (x ∈
A) ((x ∈ B) (x ∈ C)). Since (xV∈ A) ((xW∈ B) (x V
∈ C))
is logically equivalent to ((x ∈ A) (x ∈ B)) ((x ∈ B) (x ∈
C)), we have x ∈ (A ∩ B) ∪ (A ∩ C). Therefore A ∩ (B ∪ C) ⊆
(A ∩ B) ∪ (A ∩ C).
Proof of (A∩B)∪(A∩C)
Let x V
∈ (A∩B)∪(A∩
V ⊆ A∩(B ∪C).
W
C). Because
((x
∈
A)
(x
∈
B))
((x
∈
B)
(x ∈ C)) and
V
W
(x ∈ A) ((x ∈ B) (x ∈ C)) are logically equivalent we obtain x ∈ A∩(B ∪C). Therefore (A∩B)∪(A∩C) ⊆ A∩(B ∪C).
14.2. Express the relations on {1, 2, 3, 4} in words.
a. (a, b) ∈ R if b = a + 1.
b. (a, b) ∈ R if a ≤ b.
c. (a, b) ∈ R if a + b = 6.
d. (a, b) ∈ R if a|b
14.3. For each of the following relations on {1, 2, 3, 4}, please
determine whether the relation is reflexive, symmetric, and/or transitive.
a. reflexive, symmetric, transitive,
b. not reflexive, not symmetric, not transitive,
c. not reflexive, not symmetric, transitive,
d. not reflexive, symmetric, not transitive.
e. reflexive, symmetric, transitive.
14.4. For each of the following relations on the set of human beings, please determine whether the relation is reflexive, symmetric,
and/or transitive.
a. reflexive, symmetric, transitive,
b. not reflexive, not symmetric, not transitive,
c. reflexive, symmetric, transitive,
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d. reflexive, symmetric, not transitive.
e. not reflexive, symmetric, not transitive.
f. not reflexive, not symmetric, transitive.
14.6.
a. R = {(x, y) ∈ Z : |x − y| ≤ 2}
b. R is reflexive.
(We have to show (x, x) ∈ R for each x ∈ Z)
Proof. Let x ∈ Z, then |x − x| = 0 ≤ 2. Therefore (x, x) ∈ R.
d. R symmetric.
(We have to show: If (x, y) ∈ R, then (y, x) ∈ R)
Proof. Assume (x, y) ∈ R, then |x − y| ≤ 2. Since
y − x = −x + y = −(x − y)
we obtain |y − x| = | − (x − y)| = |x − y| ≤ 2. Therefore
(y, x) ∈ R.
f. R is not transitive.
Counterexample: We have (1, 3), (3, 5) ∈ R, because |1 − 3| =
2 ≤ 2 as well as |3 − 5| = 2 ≤ 2, but (1, 5) 6∈ R, because
|1 − 5| = 4 6≤ 2.
15.3. Which of the following are equivalence relations?
a. Yes. This relation is reflexive, symmetric, and transitive.
b. No. This relation is not reflexive, because (1, 1) 6∈ R.
c. No. This relation is not symmetric, because (1, 5) ∈ R, but
(5, 1) 6∈ R.
d. No. This relation is not symmetric, because (1, 5) ∈ R, but
(5, 1) 6∈ R.
e. Yes. This relation is reflexive, symmetric, and transitive.
f. No. This relation is not reflexive, because (4, 4) 6∈ R.
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1. Let R be the relation on the set of integers Z defined by
R = {(x, y) ∈ Z × Z : 5|(x − y) }
a. Find all elements in R of the form
(•) (3, −). Assume (3, x) ∈ R, then 5|(3−x). By the definition
of ”divisible” we have 3 − x = 5a for some a ∈ Z. This
implies x = 3 − 5a and therefore
x ∈ {3−5a : a ∈ Z} = {· · ·−17, −12, −7, −2, 3, 8, 13, 18, 23, 28, . . . }
..
.
(3, −17) ∈ R,
(3. − 12) ∈ R,
(3, −7) ∈ R,
(3, 3) ∈ R,
i.e.
(3, 8) ∈ R,
(3, 13) ∈ R,
(3, 18) ∈ R,
(3, 23) ∈ R,
..
.
(•) (−, 3). When we make use of analogical arguments, we
have (x, 3) ∈ R if x − 3 = 5a, or in other words x = 3 + 5a
for a ∈ Z, thus
x ∈ {3+5a : a ∈ Z} = {· · ·−17, −12, −7, −2, 3, 8, 13, 18, 23, 28, . . . }
Notice: This is the same set as above
..
.
(−17, 3) ∈ R,
(−12, 3) ∈ R,
(−7, , 3) ∈ R,
(3, 3) ∈ R,
i.e.
(8, 3) ∈ R,
(13, 3) ∈ R,
(18, 3) ∈ R,
(23, 3) ∈ R,
..
.
(•) (1, −). Similarly, (1, x) ∈ R if 1 − x = 5a, i.e. x = 1 − 5a
for a ∈ Z. Therefore
x ∈ {1−5a : a ∈ Z} = {. . . , −19, −14, −9, −4, 1, 6, 11, 16, 21, 26, . . . }
4
..
.
(1, −19) ∈ R,
(1, −14) ∈ R,
(1, −9) ∈ R,
(1, −4) ∈ R,
(1, 1) ∈ R,
i.e.
(1, 6) ∈ R,
(1, 11) ∈ R,
(1, 16) ∈ R,
(1, 21) ∈ R,
(1, 26) ∈ R,
..
.
(•) (0, −). Similarly, (0, x) ∈ R if 0 − x = 5a, i.e. x = −5a for
a ∈ Z. Therefore
x ∈ {−5a : a ∈ Z} = {. . . , −15, −10, −5, 0, 5, 10, 15, 20, . . . }
..
.
(0, −15) ∈ R,
(0, −10) ∈ R,
(0, −5) ∈ R,
(0, 0) ∈ R,
i.e.
(0, 5) ∈ R,
(0, 10) ∈ R,
(0, 15) ∈ R,
(0, 20) ∈ R,
..
.
(•) (24, −). Similarly, (24, x) ∈ R if 24 − x = 5a, i.e. x =
24 − 5a for a ∈ Z. Therefore
x ∈ {24−5a : a ∈ Z} = {· · ·−16, −11, −6, −1, 4, 9, 14, 19, 24, . . . }
5
..
.
(24, −16) ∈ R,
(24, −) ∈ R,
(24, −11) ∈ R,
(24, −6) ∈ R,
i.e. (24, −1) ∈ R,
(24, 4) ∈ R,
(24, 9) ∈ R,
(24, 14) ∈ R,
(24, 24) ∈ R,
..
.
(•) (−99, −). Similarly, (−99, x) ∈ R if −99 − x = 5a, i.e.
x = −99 − 5a for a ∈ Z. Therefore
x ∈ {−99−5a : a ∈ Z} = {· · ·−19, −14, −9, −4, 1, 6, 11, 16, 21, . . . }
..
.
(−99, −19) ∈ R,
(−99, −14) ∈ R,
(−99, −9) ∈ R,
(−99, −4) ∈ R,
i.e.
(−99, 1) ∈ R,
(−99, 11) ∈ R,
(−99, 16) ∈ R,
(−99, 21) ∈ R,
..
.
b. The relation R = {(x, y) ∈ Z × Z : 5|(x − y) } is reflexive.
(We have to show that for any integer x ∈ Z we have (x, x) ∈ R)
Proof. Let x ∈ Z. Since 5|0 and x − x = 0, we have 5|(x − x).
Therefore (x, x) ∈ R. Since x is arbitrarily selected, we have
shown (x, x) ∈ R for all x ∈ Z.
c. The relation R = {(x, y) ∈ Z × Z : 5|(x − y) } is symmetric.
(We have to show that if (m, n) ∈ R, then (n, m) ∈ R)
Proof. Assume (m, n) ∈ R, then 5|(m − n). By the definition
of ”divisible” there exists an integer a such that m − n = 5a.
Since n − m = −m + n = −(m − n) = −5a = 5(−a), we obtain
that 5 divides n − m. Therefore (n, m) ∈ R.
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d. The relation R = {(x, y) ∈ Z × Z : 5|(x − y) } is transitive.
(We have to show that if (m, n) and (n, r) are in R, then (m, r) ∈ R)
Proof. Assume (m, n), (n, r) ∈ R, then there exist a, b ∈ Z,
such that m − n = 5a and n − r = 5b. This can be rewritten as
m = n + 5a and r = n − 5b.
Since
m − r = (n + 5a) − (n − 5b) = 5a + 5b = 5(a + b),
we obtain that m − r is divisible by 5, i.e. 5|(m − r). Therefore
(m, r) ∈ R.
Using similar arguments prove that the relations
(I) R = {(x, y) ∈ Z × Z : 3|(x − y)},
(II) R = {(x, y) ∈ Z × Z : 7|(x − y)}
are equivalence relations on Z.
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