Discrete Mathematics with ApplicationsMATH236
Dr. Hung P. Tong-Viet
School of Mathematics, Statistics and Computer ScienceUniversity of KwaZulu-Natal
Pietermaritzburg Campus
Semester 1, 2013
Tong-Viet (UKZN) MATH236 Semester 1, 2013 1 / 21
Table of contents
1 Review of setsSpecial setsDeMorgan’s Law
2 Partitions
3 RelationDefinitions and examplesProperties of Relations
Tong-Viet (UKZN) MATH236 Semester 1, 2013 2 / 21
Review of sets Special sets
Some special sets
A numbers of sets occur frequently enough in mathematics that theyare given special names or symbols
The positive integers or natural numbers is the set:
N := {1, 2, 3, · · · }
The integers:
Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }
The real numbers: R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 21
Review of sets Special sets
Some special sets
A numbers of sets occur frequently enough in mathematics that theyare given special names or symbols
The positive integers or natural numbers is the set:
N := {1, 2, 3, · · · }
The integers:
Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }
The real numbers: R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 21
Review of sets Special sets
Some special sets
A numbers of sets occur frequently enough in mathematics that theyare given special names or symbols
The positive integers or natural numbers is the set:
N := {1, 2, 3, · · · }
The integers:
Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }
The real numbers: R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 21
Review of sets Special sets
Some special sets
A numbers of sets occur frequently enough in mathematics that theyare given special names or symbols
The positive integers or natural numbers is the set:
N := {1, 2, 3, · · · }
The integers:
Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }
The real numbers: R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N√
2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N
√2 ∈ R−Q
π ∈ R−QN ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N√
2 ∈ R−Q
π ∈ R−QN ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N√
2 ∈ R−Qπ ∈ R−Q
N ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N√
2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets Special sets
Some special sets, (cont.)
The rational numbers:
Q = {a
b: a, b ∈ Z, b 6= 0} ⊆ R
The irrational numbers: R−Q
Example
0,−6 ∈ Z− N√
2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}
S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}
S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}
S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}
We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
DeMorgan’s Law
Theorem (Demorgan’s Law)
If S1,S2, · · · , Sn are sets, then
S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn
Example
S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2
Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and
2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B
1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n
4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law
To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X
Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn
We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn
2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn
3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required
Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}
We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}
S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}
S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Review of sets DeMorgan’s Law
Proof of DeMorgan’s Law (cont.)
The proof that B ⊆ A is left as an exercise
We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.
Example
Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}
S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}
Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3
Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint
But {S1,S2,S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint
But {S1,S2,S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}
Then {S1,S2,S3} is pairwise disjoint
But {S1,S2,S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint
But {S1,S2,S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint
But {S1,S2, S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Disjoint sets
Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint
That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅
Example
S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint
But {S1,S2, S4} is not pairwise disjoint
Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 21
Partitions
Partitions
Definition (Partitions)
A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .
That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:
1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅
3⋃k
i=1 Si = S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 21
Partitions
Partitions
Definition (Partitions)
A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .
That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:
1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅
2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅
3⋃k
i=1 Si = S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 21
Partitions
Partitions
Definition (Partitions)
A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .
That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:
1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅
3⋃k
i=1 Si = S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 21
Partitions
Partitions
Definition (Partitions)
A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .
That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:
1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅
3⋃k
i=1 Si = S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 21
Partitions
Partitions
Definition (Partitions)
A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .
That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:
1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅
3⋃k
i=1 Si = S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 21
Partitions
Partitions (cont.)
If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S
Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition
Example
Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6},S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2,S5} are partitions of S
{S1,S2, S3,S4,S5} and {S2,S3,S5} are NOT partitions of S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 21
Partitions
Partitions (cont.)
If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S
Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition
Example
Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}
{S1,S2, S3,S4} and {S1,S2,S5} are partitions of S
{S1,S2, S3,S4,S5} and {S2,S3,S5} are NOT partitions of S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 21
Partitions
Partitions (cont.)
If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S
Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition
Example
Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2, S5} are partitions of S
{S1,S2, S3,S4,S5} and {S2,S3,S5} are NOT partitions of S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 21
Partitions
Partitions (cont.)
If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S
Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition
Example
Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2, S5} are partitions of S
{S1,S2, S3,S4,S5} and {S2, S3,S5} are NOT partitions of S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 21
Partitions
Partitions (cont.)
If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S
Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition
Example
Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2, S5} are partitions of S
{S1,S2, S3,S4,S5} and {S2, S3,S5} are NOT partitions of S
Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1,S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }
S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1,S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }
S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1,S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1, S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1, S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Partitions
Partitions (cont.)
Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,
S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }
Then {S0,S1, S2} is a partition of Z
Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 21
Relation Definitions and examples
Relation
Let S and T be nonempty sets
A relation R from S to T is a subset of the cartesian product S × T
That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T
If (s, t) ∈ R, we say that s is related to t under R and we write sRt
If (s, t) 6∈ R, then we write s��Rt
Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 21
Relation Definitions and examples
Relation
Let S and T be nonempty sets
A relation R from S to T is a subset of the cartesian product S × T
That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T
If (s, t) ∈ R, we say that s is related to t under R and we write sRt
If (s, t) 6∈ R, then we write s��Rt
Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 21
Relation Definitions and examples
Relation
Let S and T be nonempty sets
A relation R from S to T is a subset of the cartesian product S × T
That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T
If (s, t) ∈ R, we say that s is related to t under R and we write sRt
If (s, t) 6∈ R, then we write s��Rt
Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 21
Relation Definitions and examples
Relation
Let S and T be nonempty sets
A relation R from S to T is a subset of the cartesian product S × T
That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T
If (s, t) ∈ R, we say that s is related to t under R and we write sRt
If (s, t) 6∈ R, then we write s��Rt
Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 21
Relation Definitions and examples
Relation
Let S and T be nonempty sets
A relation R from S to T is a subset of the cartesian product S × T
That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T
If (s, t) ∈ R, we say that s is related to t under R and we write sRt
If (s, t) 6∈ R, then we write s��Rt
Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 21
Relation Definitions and examples
Relation (cont.)
The domain of R is domR = {s ∈ S : sRt for some t ∈ T}
The range of R is ranR = {t ∈ T : sRt for some s ∈ S}
If S = T , then we say that R is a relation on S
A relation in which each element of the domain is related to exactlyone element of the range is a function
Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 21
Relation Definitions and examples
Relation (cont.)
The domain of R is domR = {s ∈ S : sRt for some t ∈ T}
The range of R is ranR = {t ∈ T : sRt for some s ∈ S}
If S = T , then we say that R is a relation on S
A relation in which each element of the domain is related to exactlyone element of the range is a function
Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 21
Relation Definitions and examples
Relation (cont.)
The domain of R is domR = {s ∈ S : sRt for some t ∈ T}
The range of R is ranR = {t ∈ T : sRt for some s ∈ S}
If S = T , then we say that R is a relation on S
A relation in which each element of the domain is related to exactlyone element of the range is a function
Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 21
Relation Definitions and examples
Relation (cont.)
The domain of R is domR = {s ∈ S : sRt for some t ∈ T}
The range of R is ranR = {t ∈ T : sRt for some s ∈ S}
If S = T , then we say that R is a relation on S
A relation in which each element of the domain is related to exactlyone element of the range is a function
Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 21
Relation Definitions and examples
Relations: Examples
Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}
Then R ⊆ S × T and thus it is a relation from S to T
The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function
domR = {1, 2, 3} and ranR = {a, b, c}
We see that 1 is not related to b under R, that is, 1��Rb
Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 21
Relation Definitions and examples
Relations: Examples
Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}
Then R ⊆ S × T and thus it is a relation from S to T
The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function
domR = {1, 2, 3} and ranR = {a, b, c}
We see that 1 is not related to b under R, that is, 1��Rb
Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 21
Relation Definitions and examples
Relations: Examples
Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}
Then R ⊆ S × T and thus it is a relation from S to T
The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function
domR = {1, 2, 3} and ranR = {a, b, c}
We see that 1 is not related to b under R, that is, 1��Rb
Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 21
Relation Definitions and examples
Relations: Examples
Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}
Then R ⊆ S × T and thus it is a relation from S to T
The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function
domR = {1, 2, 3} and ranR = {a, b, c}
We see that 1 is not related to b under R, that is, 1��Rb
Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 21
Relation Definitions and examples
Relations: Examples
Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}
Then R ⊆ S × T and thus it is a relation from S to T
The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function
domR = {1, 2, 3} and ranR = {a, b, c}
We see that 1 is not related to b under R, that is, 1��Rb
Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.
Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.
We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations: Examples
Example
R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5
The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’
Similarly, ‘≥’ is a relation on R (and on Z,Q).
Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆
Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 21
Relation Definitions and examples
Relations (cont.)
Let a and b be integers
We say that a divides b and write a | b if b = at for some t ∈ Z
If a | b, then we say a is a divisor of b; and b is a multiple of a
If a | b, then we say that a is divisible by b and written a...b.
We write a - b if a is not a divisor of b
2 | 6 but 5 - 6
Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 21
Relation Definitions and examples
Relations (cont.)
Let a and b be integers
We say that a divides b and write a | b if b = at for some t ∈ Z
If a | b, then we say a is a divisor of b; and b is a multiple of a
If a | b, then we say that a is divisible by b and written a...b.
We write a - b if a is not a divisor of b
2 | 6 but 5 - 6
Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 21
Relation Definitions and examples
Relations (cont.)
Let a and b be integers
We say that a divides b and write a | b if b = at for some t ∈ Z
If a | b, then we say a is a divisor of b; and b is a multiple of a
If a | b, then we say that a is divisible by b and written a...b.
We write a - b if a is not a divisor of b
2 | 6 but 5 - 6
Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 21
Relation Definitions and examples
Relations (cont.)
Let a and b be integers
We say that a divides b and write a | b if b = at for some t ∈ Z
If a | b, then we say a is a divisor of b; and b is a multiple of a
If a | b, then we say that a is divisible by b and written a...b.
We write a - b if a is not a divisor of b
2 | 6 but 5 - 6
Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 21
Relation Definitions and examples
Relations (cont.)
Let a and b be integers
We say that a divides b and write a | b if b = at for some t ∈ Z
If a | b, then we say a is a divisor of b; and b is a multiple of a
If a | b, then we say that a is divisible by b and written a...b.
We write a - b if a is not a divisor of b
2 | 6 but 5 - 6
Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Definitions and examples
Relations (cont.)
Example
Let R be a relation on Z defined by xRy if and only if 2 | (x − y)
We have 2R4, 3R5 and (−1)R(−3)
But 1��R8 since 2 - −7 = 1− 8
Example
Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)
Is 2R2?
Is 1R3?
Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of relations
Let R be a relation on a set S
R is reflexive if xRx for all x ∈ S
R is symmetric if for all x , y ∈ S , xRy implies yRx
R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz
R is irreflexive if there is no x ∈ S with xRx
R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff y = x2
(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive
(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive
R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R
R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)
R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.
Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff x < y
R is NOT reflexive since no number is less than itself
R is irreflexive since (x , x) ∈∈ R for all x ∈ R
R is NOT symmetric since 2 < 4 but 4 6< 2
R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff x < y
R is NOT reflexive since no number is less than itself
R is irreflexive since (x , x) ∈∈ R for all x ∈ R
R is NOT symmetric since 2 < 4 but 4 6< 2
R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff x < y
R is NOT reflexive since no number is less than itself
R is irreflexive since (x , x) ∈∈ R for all x ∈ R
R is NOT symmetric since 2 < 4 but 4 6< 2
R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff x < y
R is NOT reflexive since no number is less than itself
R is irreflexive since (x , x) ∈∈ R for all x ∈ R
R is NOT symmetric since 2 < 4 but 4 6< 2
R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on R defined by xRy iff x < y
R is NOT reflexive since no number is less than itself
R is irreflexive since (x , x) ∈∈ R for all x ∈ R
R is NOT symmetric since 2 < 4 but 4 6< 2
R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R
Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on Z defined by xRy iff 2 | (x − y)
R is reflexive since 2 | x − x = 0 for all x ∈ Z
R is symmetric since if (x , y) ∈ R, then 2 | x − y so2 | −(x − y) = y − x hence yRx
R is transitive since if 2 | x − y and 2 | y − z , then2 | (x − y) + (y − z) = x − z so xRz
Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on Z defined by xRy iff 2 | (x − y)
R is reflexive since 2 | x − x = 0 for all x ∈ Z
R is symmetric since if (x , y) ∈ R, then 2 | x − y so2 | −(x − y) = y − x hence yRx
R is transitive since if 2 | x − y and 2 | y − z , then2 | (x − y) + (y − z) = x − z so xRz
Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on Z defined by xRy iff 2 | (x − y)
R is reflexive since 2 | x − x = 0 for all x ∈ Z
R is symmetric since if (x , y) ∈ R, then 2 | x − y so2 | −(x − y) = y − x hence yRx
R is transitive since if 2 | x − y and 2 | y − z , then2 | (x − y) + (y − z) = x − z so xRz
Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 21
Relation Properties of Relations
Properties of Relations (cont.)
Consider the relation R on Z defined by xRy iff 2 | (x − y)
R is reflexive since 2 | x − x = 0 for all x ∈ Z
R is symmetric since if (x , y) ∈ R, then 2 | x − y so2 | −(x − y) = y − x hence yRx
R is transitive since if 2 | x − y and 2 | y − z , then2 | (x − y) + (y − z) = x − z so xRz
Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 21