\end{array}} \right];}\], ${S = \left\{ {\left( {1,0} \right),\left( {1,1} \right),\left( {1,2} \right),\left( {2,2} \right)} \right\},}\;\; \Rightarrow {{M_S} = \left[ {\begin{array}{*{20}{c}} The converse relation $$S^T$$ is represented by the digraph with reversed edge directions. Reflexive and symmetric Relations on a set with n elements : 2n(n-1)/2. In that, there is no pair of distinct elements of A, each of which gets related by R to the other. Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity. 1&0&1\\ Here, x and y are nothing but the elements of set A. When there’s no element of set X is related or mapped to any element of X, then the relation R in A is an empty relation, and also called the void relation, i.e R= ∅. And as the relation is empty in both cases the antecedent is false hence the empty relation is symmetric and transitive. \end{array}} \right]. Hence, $$R \backslash S$$ does not contain the diagonal elements $$\left( {a,a} \right),$$ i.e. A (non-strict) partial order is a homogeneous binary relation ≤ over a set P satisfying particular axioms which are discussed below. 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It is mandatory to procure user consent prior to running these cookies on your website. 1&1&0 generate link and share the link here. 1&0&0\\ a. The relations $$R$$ and $$S$$ are represented in matrix form as follows: \[{R = \left\{ {\left( {a,a} \right),\left( {b,a} \right),\left( {b,d} \right),}\right.}\kern0pt{\left. {\left( {c,a} \right),\left( {c,d} \right),}\right.}\kern0pt{\left. i.e there is $$\{a,c\}\right arrow\{b}\}$$ and also $$\{b\}\right arrow\{a,c}\}$$.-The empty set is related to all elements including itself; every element is related to the empty set. 1&0&1&0 In the example: {(1,1), (2,2)} the statement "x <> y AND (x,y in R)" is always false, so the relation is antisymmetric. A set P of subsets of X, is a partition of X if 1. }$, The symmetric difference of two binary relations $$R$$ and $$S$$ is the binary relation defined as, ${R \,\triangle\, S = \left( {R \cup S} \right)\backslash \left( {R \cap S} \right),\;\;\text{or}\;\;}\kern0pt{R \,\triangle\, S = \left( {R\backslash S} \right) \cup \left( {S\backslash R} \right). Reflexive and symmetric Relations means (a,a) is included in R and (a,b)(b,a) pairs can be included or not. Domain and Range: 1&0&1 0&0&0&0\\ Irreflexive Relations on a set with n elements : 2n(n-1). Please anybody answer. 9. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION Elementary Mathematics Formal Sciences Mathematics 0&0&1 We also use third-party cookies that help us analyze and understand how you use this website. In Asymmetric Relations, element a can not be in relation with itself. This operation is called Hadamard product and it is different from the regular matrix multiplication. -This relation is symmetric, so every arrow has a matching cousin. 1&1&1\\ 1&0&0&0\\ Some specific relations. Hint: Start with small sets and check properties. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. B. Let's take an example to understand :— Question: Let R be a relation on a set A. New questions in Math. A compact way to define antisymmetry is: if $$x\,R\,y$$ and $$y\,R\,x$$, then we must have $$x=y$$. b. We can prove this by means of a counterexample. 1&0&1\\ Now for a Irreflexive relation, (a,a) must not be present in these ordered pairs means total n pairs of (a,a) is not present in R, So number of ordered pairs will be n2-n pairs. Asymmetry is not the same thing as "not symmetric ": the less-than-or-equal relation is an example of a relation that is neither symmetric nor asymmetric. In these notes, the rank of Mwill be denoted by 2n. However this contradicts to the fact that both differences of relations are irreflexive. Is It Possible For A Relation On An Empty Set Be Both Symmetric And Irreflexive? Important Points: If it is possible, give an example. Therefore, in an antisymmetric relation, the only ways it agrees to both situations is a=b. The relation is irreflexive and antisymmetric. 1&1&1\\ 1&0&0&1\\ Relation or Binary relation R from set A to B is a subset of AxB which can be defined as 1. Now a can be chosen in n ways and same for b. 0&0&1 Writing code in comment? where the product operation is performed as element-wise multiplication. Hence, $$R \cup S$$ is not antisymmetric. A relation $$R$$ on a set $$A$$ is an antisymmetric relation provided that for all $$x, y \in A$$, if $$x\ R\ y$$ and $$y\ R\ x$$, then $$x = y$$. \end{array}} \right]. 1&1&0&0 Experience. Is it possible for a relation on an empty set be both symmetric and irreflexive? 0&1&0\\ The question is whether these properties will persist in the combined relation? Therefore, when (x,y) is in relation to R, then (y, x) is not. 1&0&0&0\\ It is clearly irreflexive, hence not reflexive. Hence, $$R \cup S$$ is not antisymmetric. }$, Sometimes the converse relation is also called the inverse relation and denoted by $$R^{-1}.$$, A relation $$R$$ between sets $$A$$ and $$B$$ is called an empty relation if $$\require{AMSsymbols}{R = \varnothing. 9. (f) Let \(A = \{1, 2, 3\}$$. When we apply the algebra operations considered above we get a combined relation. A null set phie is subset of A * B. R = phie is empty relation. Since binary relations defined on a pair of sets $$A$$ and $$B$$ are subsets of the Cartesian product $$A \times B,$$ we can perform all the usual set operations on them. For Irreflexive relation, no (a,a) holds for every element a in R. It is also opposite of reflexive relation. If is an equivalence relation, describe the equivalence classes of . 2006, S. C. Sharma, Metric Space, Discovery Publishing House, page 73, (i) The identity relation on a set A is an antisymmetric relation. Similarly, we conclude that the difference of relations $$S \backslash R$$ is also irreflexive. For example, ${M = \left[ {\begin{array}{*{20}{c}} Antisymmetric Relation If (a,b), and (b,a) are in set Z, then a = b. Solution: The relation R is not antisymmetric as 4 ≠ 5 but (4, 5) and (5, 4) both belong to R. 5. Please use ide.geeksforgeeks.org, If we write it out it becomes: Dividing both sides by b gives that 1 = nm. There’s no possibility of finding a relation … So set of ordered pairs contains n2 pairs. So for (a,a), total number of ordered pairs = n and total number of relation = 2n. Empty Relation. What do you think is the relationship between the man and the boy? (f) Let $$A = \{1, 2, 3\}$$. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. \end{array}} \right].}$. Typically, relations can follow any rules. Proof: Similar to the argument for antisymmetric relations, note that there exists 3(n2 n)=2 asymmetric binary relations, as none of the diagonal elements are part of any asymmetric bi- naryrelations. The intersection of the relations $$R \cap S$$ is defined by, ${R \cap S }={ \left\{ {\left( {a,b} \right) \mid aRb \text{ and } aSb} \right\},}$. This website uses cookies to improve your experience. 0&0&0&1\\ }\], Converting back to roster form, we obtain, $R \cap S = \left\{ {\left( {b,a} \right),\left( {c,d} \right),\left( {d,a} \right)} \right\}.$. Consider the set $$A = \left\{ {0,1} \right\}$$ and two antisymmetric relations on it: ${R = \left\{ {\left( {1,2} \right),\left( {2,2} \right)} \right\},\;\;}\kern0pt{S = \left\{ {\left( {1,1} \right),\left( {2,1} \right)} \right\}. A relation is asymmetric if and only if it is both anti-symmetric and irreflexive. A relation is said to be asymmetric if it is both antisymmetric and irreflexive or else it is not. 6. 1&0&0&1\\ Four combinations are possible with a relation on a set of size two. The empty relation is symmetric and transitive. \end{array}} \right]. Discrete Mathematics Questions and Answers – Relations. 8. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Let $$R$$ and $$S$$ be two relations over the sets $$A$$ and $$B,$$ respectively. First we convert the relations $$R$$ and $$S$$ from roster to matrix form: \[{R = \left\{ {\left( {0,2} \right),\left( {1,0} \right),\left( {1,2} \right),\left( {2,0} \right)} \right\},}\;\; \Rightarrow {{M_R} = \left[ {\begin{array}{*{20}{c}} What do you think is the relationship between the man and the boy? it is irreflexive. Attention reader! Inverse of relation ... is antisymmetric relation. Empty RelationIf Relation has no elements,it is called empty relationWe write R = ∅Universal RelationIf relation has all the elements,it is a universal relationLet us take an exampleLet A = Set of all students in a girls school.We define relation R on set A asR = {(a, b): a and b are brothers}R’ = }$, Compose the union of the relations $$R$$ and $$S:$$, ${R \cup S }={ \left\{ {\left( {1,2} \right),\left( {2,2} \right)} \right\} }\cup{ \left\{ {\left( {1,1} \right),\left( {2,1} \right)} \right\} }={ \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right)} \right\}.}$. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. Then, ${R \,\triangle\, S }={ \left\{ {\left( {b,2} \right),\left( {c,3} \right)} \right\} }\cup{ \left\{ {\left( {b,1} \right),\left( {c,1} \right)} \right\} }={ \left\{ {\left( {b,1} \right),\left( {c,1} \right),\left( {b,2} \right),\left( {c,3} \right)} \right\}. 1&0&1&0 0&0&0\\ acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Mathematics | Some theorems on Nested Quantifiers, Mathematics | Set Operations (Set theory), Inclusion-Exclusion and its various Applications, Mathematics | Power Set and its Properties, Mathematics | Classes (Injective, surjective, Bijective) of Functions, Mathematics | Total number of possible functions, Discrete Maths | Generating Functions-Introduction and Prerequisites, Mathematics | Generating Functions – Set 2, Mathematics | Sequence, Series and Summations, Mathematics | Independent Sets, Covering and Matching, Mathematics | Rings, Integral domains and Fields, Mathematics | PnC and Binomial Coefficients, Number of triangles in a plane if no more than two points are collinear, Finding nth term of any Polynomial Sequence, Discrete Mathematics | Types of Recurrence Relations – Set 2, Mathematics | Graph Theory Basics – Set 1, Mathematics | Graph Theory Basics – Set 2, Mathematics | Euler and Hamiltonian Paths, Betweenness Centrality (Centrality Measure), Mathematics | Walks, Trails, Paths, Cycles and Circuits in Graph, Graph measurements: length, distance, diameter, eccentricity, radius, center, Relationship between number of nodes and height of binary tree, Mathematics | L U Decomposition of a System of Linear Equations, Bayes’s Theorem for Conditional Probability, Mathematics | Probability Distributions Set 1 (Uniform Distribution), Mathematics | Probability Distributions Set 2 (Exponential Distribution), Mathematics | Probability Distributions Set 3 (Normal Distribution), Mathematics | Probability Distributions Set 4 (Binomial Distribution), Mathematics | Probability Distributions Set 5 (Poisson Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange’s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Depth of the deepest odd level node in Binary Tree, Runge-Kutta 2nd order method to solve Differential equations, Difference between Spline, B-Spline and Bezier Curves, Regular Expressions, Regular Grammar and Regular Languages, Write Interview 1&0&0 The difference of the relations $$R \backslash S$$ consists of the elements that belong to $$R$$ but do not belong to $$S$$. And Then it is same as Anti-Symmetric Relations.(i.e. When a ≤ b, we say that a is related to b. Examples. 2. the empty relation is symmetric and transitive for every set A. But opting out of some of these cookies may affect your browsing experience. 0&1&0&0\\ So total number of symmetric relation will be 2n(n+1)/2. You also have the option to opt-out of these cookies. Now for a reflexive relation, (a,a) must be present in these ordered pairs. For example, if there are 100 mangoes in the fruit basket. If it is possible, give an example. For each of these relations on the set \{1,2,3,4\}, decide whether it is reflexive, whether it is symmetric, and whether it is antisymmetric, and whether it is transitive. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. 1&0&0&0\\ A relation can be antisymmetric and symmetric at the same time. Consider the relation ‘is divisible by,’ it’s a relation for ordered pairs in the set of integers. If It Is Not Possible, Explain Why. 0&0&1\\ Let R be any relation from A to B. {\left( {d,a} \right),\left( {d,c} \right)} \right\},}\;\; \Rightarrow {{M_R} = \left[ {\begin{array}{*{20}{c}} 4. The empty relation … 0&0&1&1\\ The empty relation between sets X and Y, or on E, is the empty set ... An order (or partial order) is a relation that is antisymmetric and transitive. 1&0&0&0\\ The empty relation is the only relation that is (vacuously) both symmetric and asymmetric. If a relation has a certain property, prove this is so; otherwise, provide a counterexample to show that it does not. The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). Empty Relation. A strict total order, also called strict semiconnex order, strict linear order, strict simple order, or strict chain, is a relation that … A transitive relation is asymmetric if it is irreflexive or else it is not. If it is not possible, explain why. there is no aRa ∀ a∈A relation.) https://tutors.com/math-tutors/geometry-help/antisymmetric-relation \end{array}} \right] }+{ \left[ {\begin{array}{*{20}{c}} Prove that 1. if A is non-empty, the empty relation is not reflexive on A. if there are two sets A and B and Relation from A to B is R(a,b), then domain is defined as the set { a | (a,b) € R for some b in B} and Range is defined as the set {b | (a,b) € R for some a in A}. An inverse of a relation is denoted by R^-1 which is the same set of pairs just written in different or reverse order. }$, To find the intersection $$R \cap S,$$ we multiply the corresponding elements of the matrices $$M_R$$ and $$M_S$$. 4. Furthermore, if A contains only one element, the proposition "x <> y" is always false, and the relation is also always antisymmetric. 4. 2. Or similarly, if R(x, y) and R(y, x), then x = y. A relation becomes an antisymmetric relation for a binary relation R on a set A. Definition: A relation R is antisymmetric if ... One combination is possible with a relation on an empty set. Their intersection $$R \cap S$$ will be the relation “is a friend and work colleague of“. An n-ary relation R between sets X 1, ... , and X n is a subset of the n-ary product X 1 ×...× X n, in which case R is a set of n-tuples. For a relation … Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. So, we have, ${{M_{R \cap S}} = {M_R} * {M_S} }={ \left[ {\begin{array}{*{20}{c}} Click or tap a problem to see the solution. (That means a is in relation with itself for any a). (selecting a pair is same as selecting the two numbers from n without repetition) As we have to find number of ordered pairs where a ≠ b. it is like opposite of symmetric relation means total number of ordered pairs = (n2) – symmetric ordered pairs(n(n+1)/2) = n(n-1)/2. 1&0&0&1\\ Hence, if an element a is related to element b, and element b is also related to element a, then a and b should be a similar element. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Rules of Antisymmetric Relation. 5. 0&0&1\\ Is It Possible For A Relation On An Empty Set Be Both Symmetric And Antisymmetric? Number of Asymmetric Relations on a set with n elements : 3n(n-1)/2. you have three choice for pairs (a,b) (b,a)). (In Symmetric relation for pair (a,b)(b,a) (considered as a pair). If R is a non-empty relation in A then [; R \cap R {-1} = I_A \Leftrightarrow R \text{ is antisymmetric } ;] Fair enough. The other combinations need a relation on a set of size three. For example, the union of the relations “is less than” and “is equal to” on the set of integers will be the relation “is less than or equal to“. 1&0&0 These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. Relations and their representations. }\), The universal relation between sets $$A$$ and $$B,$$ denoted by $$U,$$ is the Cartesian product of the sets: $$U = A \times B.$$, A relation $$R$$ defined on a set $$A$$ is called the identity relation (denoted by $$I$$) if $$I = \left\{ {\left( {a,a} \right) \mid \forall a \in A} \right\}.$$. \end{array}} \right] }={ \left[ {\begin{array}{*{20}{c}} In antisymmetric relation, it’s like a thing in one set has a relation with a different thing in another set. This section focuses on "Relations" in Discrete Mathematics. So total number of anti-symmetric relation is 2n.3n(n-1)/2. Asymmetric Relation: A relation R on a set A is called an Asymmetric Relation if for every (a, b) ∈ R implies that (b, a) does not belong to R. 6. If the relations $$R$$ and $$S$$ are defined by matrices $${M_R} = \left[ {{a_{ij}}} \right]$$ and $${M_S} = \left[ {{b_{ij}}} \right],$$ the matrix of their intersection $$R \cap S$$ is given by, \[{M_{R \cap S}} = {M_R} * {M_S} = \left[ {{a_{ij}} * {b_{ij}}} \right],$. For anti-symmetric relation, if (a,b) and (b,a) is present in relation R, then a = b. Therefore, in an antisymmetric relation, the only ways it agrees to both situations is a=b. The inverse of R denoted by R^-1 is the relation from B to A defined by: R^-1 = { (y, x) : yEB, xEA, (x, y) E R} 5. aRb ↔ (a,b) € R ↔ R(a,b). So total number of reflexive relations is equal to 2n(n-1). 0&0&1\\ 0&0&1&1\\ If it is not possible, explain why. 3. \end{array}} \right] }*{ \left[ {\begin{array}{*{20}{c}} A relation has ordered pairs (a,b). This is only possible if either matrix of $$R \backslash S$$ or matrix of $$S \backslash R$$ (or both of them) have $$1$$ on the main diagonal. So from total n2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. }\], Suppose that $$R$$ is a binary relation between two sets $$A$$ and $$B.$$ The complement of $$R$$ over $$A$$ and $$B$$ is the binary relation defined as, $\bar R = \left\{ {\left( {a,b} \right) \mid \text{not } aRb} \right\},$, For example, let $$A = \left\{ {1,2} \right\},$$ $$B = \left\{ {1,2,3} \right\}.$$ If a relation $$R$$ between sets $$A$$ and $$B$$ is given by, $R = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {2,2} \right),\left( {2,3} \right)} \right\},$, then the complement of $$R$$ has the form, $\bar R = \left\{ {\left( {1,1} \right),\left( {2,1} \right)} \right\}.$. Is it possible for a relation on an empty set be both symmetric and antisymmetric? A relation has ordered pairs (a,b). So there are three possibilities and total number of ordered pairs for this condition is n(n-1)/2. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above, Related Articles: {\left( {c,c} \right),\left( {c,d} \right),}\right.}\kern0pt{\left. In these notes, the rank of Mwill be denoted by 2n. This website uses cookies to improve your experience while you navigate through the website. We get the universal relation $$R \cup S = U,$$ which is always symmetric on an non-empty set. \end{array}} \right],\;\;}\kern0pt{{M^T} = \left[ {\begin{array}{*{20}{c}} Limitations and opposites of asymmetric relations are also asymmetric relations. \end{array}} \right];}\], ${S = \left\{ {\left( {a,b} \right),\left( {b,a} \right),}\right.}\kern0pt{\left. 1&0&0&1\\ 1&0&0&0\\ These cookies do not store any personal information. Therefore there are 3n(n-1)/2 Asymmetric Relations possible. To get the converse relation $$R^T,$$ we reverse the edge directions. The empty relation between sets X and Y, or on E, is the empty set ∅. Inverse of relation . 7. By definition, the symmetric difference of $$R$$ and $$S$$ is given by, \[R \,\triangle\, S = \left( {R \backslash S} \right) \cup \left( {S \backslash R} \right).$. Relations may also be of other arities. A relation has ordered pairs (a,b). A Binary relation R on a single set A is defined as a subset of AxA. If a relation $$R$$ is defined by a matrix $$M,$$ then the converse relation $$R^T$$ will be represented by the transpose matrix $$M^T$$ (formed by interchanging the rows and columns). 0&0&0\\ {\left( {1,1} \right),\left( {1,2} \right),}\right.}\kern0pt{\left. Antisymmetric Relation If (a,b), and (b,a) are in set Z, then a = b. Hence, if an element a is related to element b, and element b is also related to element a, then a and b should be a similar element. By adding the matrices $$M_R$$ and $$M_S$$ we find the matrix of the union of the binary relations: ${{M_{R \cup S}} = {M_R} + {M_S} }={ \left[ {\begin{array}{*{20}{c}} The original relations may have certain properties such as reflexivity, symmetry, or transitivity. By using our site, you Irreflective relation. A null set phie is subset of A * B. R = phie is empty relation. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. (e) Carefully explain what it means to say that a relation on a set $$A$$ is not antisymmetric. Now for a symmetric relation, if (a,b) is present in R, then (b,a) must be present in R. This lesson will talk about a certain type of relation called an antisymmetric relation. We'll assume you're ok with this, but you can opt-out if you wish. Empty Relation. Let $$R$$ be a binary relation on sets $$A$$ and $$B.$$ The converse relation or transpose of $$R$$ over $$A$$ and $$B$$ is obtained by switching the order of the elements: \[{R^T} = \left\{ {\left( {b,a} \right) \mid aRb} \right\},$, So, if $$R = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right)} \right\},$$ then the converse of $$R$$ is, ${R^T} = \left\{ {\left( {2,1} \right),\left( {3,1} \right),\left( {4,1} \right)} \right\}.$. Number of different relation from a set with n elements to a set with m elements is 2mn. whether it is included in relation or not) So total number of Reflexive and symmetric Relations is 2n(n-1)/2 . If It Is Not Possible, Explain Why. 0&1&1\\ The table below shows which binary properties hold in each of the basic operations. 1&0&1\\ A relation has ordered pairs (a,b). (e) Carefully explain what it means to say that a relation on a set $$A$$ is not antisymmetric. Number of Anti-Symmetric Relations on a set with n elements: 2n 3n(n-1)/2. A relation $$R$$ on a set $$A$$ is an antisymmetric relation provided that for all $$x, y \in A$$, if $$x\ R\ y$$ and $$y\ R\ x$$, then $$x = y$$. For two distinct set, A and B with cardinalities m and n, the maximum cardinality of the relation R from A to B is mn. }\], Then the relation differences $$R \backslash S$$ and $$S \backslash R$$ are given by, ${R\backslash S = \left\{ {\left( {b,2} \right),\left( {c,3} \right)} \right\},\;\;}\kern0pt{S\backslash R = \left\{ {\left( {b,1} \right),\left( {c,1} \right)} \right\}. Recommended Pages Number of Reflexive Relations on a set with n elements : 2n(n-1). For example, the inverse of less than is also asymmetric. We get the universal relation $$R \cup S = U,$$ which is always symmetric on an non-empty set. Formal definition. The difference of two relations is defined as follows: \[{R \backslash S }={ \left\{ {\left( {a,b} \right) \mid aRb \text{ and not } aSb} \right\},}$, ${S \backslash R }={ \left\{ {\left( {a,b} \right) \mid aSb \text{ and not } aRb} \right\},}$, Suppose $$A = \left\{ {a,b,c,d} \right\}$$ and $$B = \left\{ {1,2,3} \right\}.$$ The relations $$R$$ and $$S$$ have the form, \[{R = \left\{ {\left( {a,1} \right),\left( {b,2} \right),\left( {c,3} \right),\left( {d,1} \right)} \right\},\;\;}\kern0pt{S = \left\{ {\left( {a,1} \right),\left( {b,1} \right),\left( {c,1} \right),\left( {d,1} \right)} \right\}. The divisibility relation on the natural numbers is an important example of an antisymmetric relation.