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9.1 Relations and Their Properties
Binary Relation
Definition: Let A, B be any sets. A binary relation R from A to B, written R : A × B, is a subset of the set A × B.
Complementary Relation
Definition: Let R be the binary relation from A to B. Then the complement of R can be defined by R = {(a, b)|(a, b) 6 ∈ R} = (A × B) − R
Inverse Relation
Definition: Let R be the binary relation from A to B. Then the inverse of R can be defined by R−^1 = {(b, a)|(a, b) ∈ R}
Relations on a Set
Definition: A relation on a set A is a relation from A to A. In other words, a relation on a set A is a subset of A × A.
Digraph
Definition: A directed graph, or digraph, consists of a set V of vertices (or nodes) together with a set E of ordered pairs of elements of V called edges (or arcs). The vertex a is called the initial vertex of the edge (a, b), and the vertex b is called the terminal vertex of this edge.
Properties
Reflexive: A relation R on a set A is called reflexive if (a, a) ∈ R for every element a ∈ A.
a
Every vertex has a self-loop.
Symmetric: A relation R on a set A is called symmetric if (b, a) ∈ R whenever (a, b) ∈ R, for all a, b ∈ A.
a b
If there is an edge from one vertex to another, there is an edge in the opposite direction.
Antisymmetric: A relation R on a set A such that for all a, b ∈ A, if (a, b) ∈ R and (b, a) ∈ R, then a = b is called antisymmetric.
a b
There is at most one edge between distinct vertices.
Some notes on Symmetric and Antisymmetric:
- A relation can be both symmetric and antisymmetric.
- A relation can be neither symmetric nor antisymmetric.
Transitive: A relation R on a set A is called transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A.
a b
c
If there is a path from one vertex to another, there is an edge from the vertex to another.
Combining Relations
Since relations from A to B are subsets of A × B, two relations from A to B can be combined in any way two sets can be combined. Such as union, intersection, and set difference.
Composition
Definition: Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of ordered pairs (a, c), where a ∈ A, c ∈ C, and for which there exists an element b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. We denote the composite of R and S by S ◦ R.
Powers of a Relation
Let R be a relation on the set A. The powers Rn, n = 1, 2 , 3 , ..., are defined recursively by R^1 = R and Rn+1^ = Rn^ ◦ R.
9.1 pg. 581 # 3
For each of these relations on the set { 1 , 2 , 3 , 4 }, decide whether it is reflexive, whether it is sym- metric, whether it is antisymmetric, and whether it is transitive.
a {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)} Not reflexive because we do not have (1, 1), (3, 3), and (4, 4). Not symmetric because while we we have (3, 4), we do not have (4, 3). Not antisymmetric because we have both (2, 3) and (3, 2). Transitive because if we have (a, b) in this relation, then a will be either 2 or 3. Then (2, c)
b xy ≥ 1. Not reflexive because we can’t have (0, 0). Is symmetric because we have xy = yx. Not antisymmetric because we have xy = yx. Is transitive because if we have (a, b) ∈ R and that (b, c) ∈ R, it follows that (a, c) ∈ R. Note that in order for the relation to be true, a, b, and c will have to be all positive or all negative.
c x = y + 1 or x = y − 1. Not reflexive because we can’t have (1, 1) Is symmetric because we have x = y + 1 and y = x − 1. They are equivalent equations. Not antisymmetric because of the same reason above. Not transitive because if we have (1, 2 ) and (2, 1) in the relation, (1, 1) is not in relation.
g x = y^2. Not reflexive because (2, 2) does not satisfy. Not symmetric because although we can have (9, 3), we can’t have (3, 9). Is antisymmetric because each integer will map to another integer but not in reverse (besides 0 and 1). Not transitive because if we have (16, 4) and (4, 2), it’s not the case that 16 = 2^2.
h x ≥ y^2. Not reflexive because we can’t have (2, 2). Not symmetric because if we have (9, 3), we can’t have (3, 9). Is antisymmetric, because each integer will map to another integer but not in reverse (besides 0 and 1). Is transitive because if x ≥ y^2 and y ≥ z^2 , then x ≥ z^2.
9.1 pg. 582 # 27
Let R be the relation R = {(a, b)|a | b} on the set of positive integers. Find
a R−^1 R−^1 = {(b, a)|a | b} = {(a, b)|b | a}
b R R = {(a, b)|a - |b}.
