2.4 KiB
Lecture Topic: Relation examples
Relation Example 1
R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}
e.g (22,7) \in R because 22-7 = 5(3)
e.g (7,22) \in R because 7-22 = -15 = 5(-3)
e.g (22,9) \notin R because 22-9 = 13, which is not a multiple of 5
Reflexive? Yes
For every integer a, a-a = 0 = 5(0)
Therefor (a,a) \in R for every integer a
Symmetric? Yes Let a, b be any integer, such that (a,b) is in the relation (can we prove that (b,a) is in relation as well?)
Since (a,b) is in the relation, a-b = 5n, so, b-a = -5(n) = 5(-n)
Note: (b-a) = -(a-b)
Transitive? Yes
Let a,b,c be any integers such that (a,b) \in R and (b,c) \in R
Can we prove that (a,c) \in R ?
Since (a,b) \in R, we know a-b=5n for some n \in \mathbb{Z}
Since (b,c) \in R, we know b-c=5p for some n \in \mathbb{Z}
Now, (a-c) = (a-b) + (b-c) = 5n + 5p = 5(np)
Therefore, (a-c) \in R
Note: This is an integer because it is the sum of integers
Relation Example 2
R = \{(i,j) | i \in \mathbb{R}, j \in \mathbb{R}, i + 2 > j\}
Examples:
(5,2) \in Rbecause(5+2) > 2(3,4) \in Rbecause(3+2) > 4(0,7) \notin Rbecause(0+2) \ngtr 7
Reflexive? Yes Symmetric? No Transitive? No
Proof: We need to find real numbers where a,b,c where (a,b) \in R and (b,c) \in R but (a,c) \notin R
For example, a = 1, b = 2, c = 3
(1,2) \in R because 1+2 > 2
(2,3) \in R because 2+2 > 3
But (1,3) \notin R because (1+2) \ngtr 3
\therefore R is not transitive
Equivalence relations
Equivalence relations have all three of these properties Any equivalence relation R on a set A induces a partition of A
- Splits A into equivalence classes
- Each class contains elements that are related to themselves and to each other, but to nothing outside the class
So, for the above relation:
R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}
This forms an equivalence relation
So for the set of all integers, this forms 5 equivalence classes:
..., 0, 5, 10, ......, 1, 6, 11, ......, 2, 7, 12, ......, 3, 8, 13, ......, 4, 9, 14, ...
Example
A = \{-3, -2, -1, 0, 1, 2, 3\}
R = \{(i,j) | i \in \mathbb{A}, j \in \mathbb{A}, i^2 = j^2\}
So the equivalence classes induced by this relation:
- -3, 3
- -2, 2
- -1, 1
- 0