A directed graph, also called a digraph, is a graph in which the edges have a direction. This is usually indicated with an arrow on the edge; more formally, if \(v\) and \(w\) are vertices, an edge is an unordered pair \({v,w}\), while a directed edge, called an arc, is an ordered pair (v,w) or (w,v).
A relation \(R\) from a set A to a set B is a subset of the cartesian product A x B.
Thus R is a set of where the first element comes from A and the second element comes from B i.e. \((a, b)\)
If \((a, b) \in R\) we say that \(a\) is related to \(b\) and write \(aRb\).
If \((a, b) \notin R\), we say that \(a\) is not related to \(b\) and write \(aRb\). CHECK
If \(R\) is a relation from a set \(A\) to itself then we say that ``\(R\) is a relation on \(A\)".
Let R be an equivalence relation defined on a set S and let x 2 S. Then the equivalence class of x is the subset of S containing all elements of S which are related to x.
We denote this by [x]. Thus \[[x] = {y 2 S : yRx}\].
Solution:
We say that a relation R on a set S is anti-symmetric if for all x, y 2 S such that xRy and yRx, we have \(x = y\).
Thus R is anti-symmetric if and only if the relationship digraph of R has no directed cycles of length two.
< a href = =“https://www.youtube.com/watch?v=/4vSr6q5Nxqc”> Digraphs and Relations
< a href = =“https://www.youtube.com/watch?v=/dHzu897SdDA”> Cartesian Products
< a href = =“https://www.youtube.com/watch?v=/4vSr6q5Nxqc”> Digraphs and Relations -Example
< a href = =“https://www.youtube.com/watch?v=/eZIEc1O0W1w”> Relations Example 2
< a href = =“https://www.youtube.com/watch?v=/dHzu897SdDA”> Cartesian
< a href = =“https://www.youtube.com/watch?v=/4vSr6q5Nxqc”> Digraphs and Relations -Example
< a href = =“https://www.youtube.com/watch?v=/rG8mhlqnkr0”> Relations -Example
< a href = =“https://www.youtube.com/watch?v=/_QTKlTJYKRA”> Transitive Relations
< a href = =“https://www.youtube.com/watch?v=/A08P9ZJsLOs”> Relations -Example
Transitive relations are binary relations defined on a set such that if the first element is related to the second element, and the second element is related to the third element of the set, then the first element must be related to the third element. For example, if for three elements a, b, c in set A, if a = b and b = c, then a = c.
A partial order relation is a homogeneous relation that is transitive and antisymmetric. There are two common sub-definitions for a partial order relation, for reflexive and irreflexive partial order relations, also called “non-strict” and “strict” respectively.
Given a flock of chickens, between any two chickens one of them is dominant. A relation, R, is defined between chicken x and chicken y as xRy if x is dominant over y. This gives what is known as a pecking order to the flock. Home Farm has 5 chickens: Amy, Beth, Carol, Daisy and Eve, with the following relations:
Suppose \(A = \{1,2,3,4\}\). Consider the following relation in A
\[ \{ (1,1),(2,2),(2,3),(3,2),(4,2),(4,4)\} \]
Draw the direct graph of \(A\). Based on the Digraph of \(A\) discuss whether or not a relation that could be depicted by the digraph could be described as the following, justifying your answer.
Determine which of the following relations $ x R y$ are reflexive, transitive, symmetric, or antisymmetric on the following - there may be more than one characteristic.
Given a flock of chickens, between any two chickens one of them is dominant. A relation, R, is defined between chicken x and chicken y as xRy if x is dominant over y. This gives what is known as a pecking order to the flock. Home Farm has 5 chickens: Amy, Beth, Carol, Daisy and Eve, with the following relations:
For \(a, b \in Z\) define a R b to mean that a divides b.
% Solution: % % % a. Since 0 does not divide 0, therefore R is not reflexive. % b. 2 divides 4 so 2 R 4. But 4 does not divide 2, so 4 R 2. Thus, R is not symmetric. % c. To see that R is transitive, let a, b, c be integers. Suppose that a R b and b R c. Thus, % a divides b and b divides c so there exist integers k and l such that b = ak and c = bl. This % gives c = bl = (ak)l = a(kl). Therefore, a divides c so a R c %
Given the set \(S =\{g,e,r,b,i,l\}\).